Student copies include complete teaching and a separate answer section. Tutor copies place worked solutions beside each question.
Take the time you need. Follow every step, practise, and return to anything you cannot yet explain.
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In this level
I can read a function, work out an output, find a simple input and check what the answer means.
Learning objectives:
I can identify input, rule and output in a function.
I can substitute a positive or negative input and calculate in order.
I can find an input for a stated linear output using equal operations.
I can interpret and check a function model in its stated context.
A rule connects inputs to outputs. A reliable solution starts by deciding which quantity the question gives you.
Prerequisite review
Signed numbers, squares and order of operations — Use this bridge whenever a readiness answer or its reasoning is uncertain. Complete the worked model, fill the gaps, then try the fresh check.
factor, term, square, inverse operation: Factors multiply; terms add. A square has two equal factors. An inverse operation undoes another operation.
Watch the prerequisite 1
Work out 2 × (−3) + 5 and (−3)².
Step 1
Before: 2 × (−3)
Action: Add two groups of negative three.
After: (−3) + (−3) = −6.
Why: Two steps of three to the left of zero end at negative six.
Track each operation
Step 2
Before: −6 + 5
Action: Move five units right from negative six.
After: −5, −4, −3, −2, −1.
Why: Adding a positive number moves right. The result is negative one.
Step 3
Before: (−3)²
Action: Write the two equal factors.
After: (−3) × (−3) = 9.
Why: A square multiplies the whole bracketed number by itself. Two negative factors give a positive product.
Step 4
Before: −3²
Action: Compare without the brackets.
After: −(3 × 3) = −9.
Why: An exponent acts before an unbracketed leading minus. These expressions are different.
Step 5
Before: Multiply negative two by a decreasing integer.
Action: Keep the same change between neighbouring products.
Keep an equation balanced — Use this bridge whenever a readiness answer or its reasoning is uncertain. Complete the worked model, fill the gaps, then try the fresh check.
equation, equals, nonzero: An equation says two amounts are equal. Apply the same valid operation to both sides. Nonzero means not zero; division by zero is undefined.
Watch the prerequisite 1
Solve 3x − 10 = 8.
Step 1
Before: 3x − 10 = 8
Action: Add ten to both sides.
After: 3x − 10 + 10 = 8 + 10; 3x = 18.
Why: The opposite terms −10 and +10 sum to zero.
Apply +10 to both sides
Step 2
Before: 3x = 18
Action: Divide both sides by three.
After: (3 × x)/3 = 18/3; 1 × x = 6; x = 6.
Why: The matching nonzero factors give 3/3 = 1, not zero.
Factors reduce to one
Step 3
Before: x = 6
Action: Substitute into the original equation.
After: 3 × 6 − 10 = 18 − 10 = 8.
Why: The original sides agree.
x = 6.
Complete the missing step 1
Complete: 2x − 6 = 4. Add __ to both sides; 2x = __; x = __.
Identify what is given.
Follow the matching bridge model.
Perform the displayed operation, then check the original expression.
2x − 6 = 4.
Add 6 on both sides:
2x − 6 + 6 = 4 + 6.
2x + 0 = 10; 2x = 10.
Divide both sides by 2:
(2 × x)/(2) = 10/(2).
(2/2) × x = 5; 1 × x = 5.
x = 5.
Check: 2 × (5) − 6 = 10 − 6 = 4.
Try a fresh question 1
Solve 4x + 1 = 13.
Identify the operation first.
Use brackets and equal operations where needed.
Check each arithmetic line separately.
4x + 1 = 13.
Subtract 1 on both sides:
4x + 1 − 1 = 13 − 1.
4x + 0 = 12; 4x = 12.
Divide both sides by 4:
(4 × x)/(4) = 12/(4).
(4/4) × x = 3; 1 × x = 3.
x = 3.
Check: 4 × (3) + 1 = 12 + 1 = 13.
A half as an input — Use this bridge before substituting a fractional input.
numerator and denominator: In 1/2, the denominator 2 divides a whole into two equal parts. The numerator 1 counts one of those parts.
Watch the prerequisite 1
Find six halves, then subtract one.
Step 1
Before: 6 × (1/2)
Action: Write the six equal parts.
After: 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2.
Why: A half is one of two equal parts of a whole.
Pair the halves
Step 2
Before: 6/2
Action: Group the halves in pairs.
After: 6/2 = 3.
Why: Two halves make each whole; six halves make three wholes.
Step 3
Before: 3 − 1
Action: Subtract one whole.
After: 3 − 1 = 2.
Why: Check by adding the removed whole: 2 + 1 = 3.
6 × (1/2) − 1 = 6/2 − 1 = 3 − 1 = 2.
Complete the missing step 1
Complete: 8 × (1/2) = 8/2 = __ wholes.
Each whole needs two halves.
Group eight parts in pairs.
Count the pairs: (1, 2), (3, 4), (5, 6), (7, 8).
Eight halves form four pairs, so 8/2 = 4.
Try a fresh question 1
Find 10 × (1/2) − 2.
Calculate the halves before subtracting.
Divide ten by two.
Five wholes minus two wholes leaves the answer.
Ten halves form five pairs: 10/2 = 5. Subtract two: 5 − 2 = 3. Check 3 + 2 = 5.
Readiness check
Watch — I do
Functions, relations and function notation
Read the function notation
A function assigns exactly one output to each permitted input. Its name is f; the bracket holds the input. Different inputs may share one output. One input cannot have two outputs under the same rule.
f(3) means the output when x = 3.
Input → rule → output
When the input is known
Write the rule. Replace every x by the bracketed input. Calculate powers, then multiplication or division, then addition or subtraction. Write the result using function notation.
f(x) = x² + 2x; f(−2) = (−2)² + 2(−2)
A negative input is one number
When the output is known
Set the rule equal to the output. Undo the added term on both sides, then undo multiplication on both sides. Show the cancellation and check by substituting your input.
f(x) = 3x − 10; f(x) = 8 ⇒ 3x − 10 = 8
Apply +10 to both sides
Choose what to do
If the question gives f(4), the input is four: substitute. If it gives f(x)=4, the output is four: solve the equation for x. A table lookup is enough only when the needed input is listed.
Known input → substitute; known output → solve.
Input or output?
Read a model in context
Define the variable, its unit and permitted values. A rule for whole tickets has whole-number inputs. A temperature rule may accept negative values. An algebraic answer can be impossible in the situation.
Cost C(n) = 4n + 7 dollars; n is a nonnegative whole number.
Input → rule → output
Check rather than memorise a shortcut
f(x) is not f times x. f(−3) is not necessarily −f(3). A pair of outputs can be equal for different inputs, such as (−2)² and 2². Division cancels common factors of a whole expression, not one term of a sum.
(x + 6)/3 = x/3 + 2, not x + 2.
Divide every term
Worked examples
I do
For f(x) = 2x + 1, find f(3).
Step 1
Before: f(x) = 2x + 1; find f(3).
Action: Read the input inside the brackets.
After: The input is 3.
Why: f names the rule. f(x) is its output; it is not f multiplied by x.
Step 2
Before: 2x + 1
Action: Replace every x with the input in brackets.
After: f(3) = 2(3) + 1
Why: Brackets keep a negative input or a whole expression together.
Keep the input together
Step 3
Before: 2(3) + 1
Action: Carry out one operation.
After: 6 + 1
Why: Multiply two by three first.
Step 4
Before: 6 + 1
Action: Carry out one operation.
After: 7
Why: Add the remaining one.
Step 5
Before: f(3) = 7.
Action: Check against the rule.
After: Start with three; doubling gives six; adding one gives seven.
Why: Recalculate in the original rule, including each sign.
f(3) = 7. Start with three; doubling gives six; adding one gives seven.
I do
For f(x) = x² + 2x, find f(−2).
Step 1
Before: f(x) = x² + 2x; find f(−2).
Action: Read the input inside the brackets.
After: The input is −2.
Why: f names the rule. f(x) is its output; it is not f multiplied by x.
Step 2
Before: x² + 2x
Action: Replace every x with the input in brackets.
After: f(−2) = (−2)² + 2(−2)
Why: Brackets keep a negative input or a whole expression together.
Keep the input together
Step 3
Before: (−2)² + 2(−2)
Action: Carry out one operation.
After: 4 + 2(−2)
Why: Square the bracketed input: (−2)×(−2)=4.
Step 4
Before: 4 + 2(−2)
Action: Carry out one operation.
After: 4 + (−4)
Why: Multiply 2×(−2)=−4.
Step 5
Before: 4 + (−4)
Action: Carry out one operation.
Why: Opposite terms sum to zero.
Step 6
Before: f(−2) = 0.
Action: Check against the rule.
After: The two terms are four and negative four, which balance to zero.
Why: Recalculate in the original rule, including each sign.
f(−2) = 0. The two terms are four and negative four, which balance to zero.
No exact $17 order is allowed. The equation gives 2.5 tickets, which violates the input condition.
I do
Does the table (1, 4), (2, 5), (3, 5) describe a function? Each pair is (input, output).
Step 1
Before: Pairs (1, 4), (2, 5), (3, 5)
Action: Group rows by their input.
After: Input 1→4; input 2→5; input 3→5.
Why: Check outputs for the same input, not inputs for the same output.
One output for each input
Step 2
Before: Each input appears with one output.
Action: Apply the definition.
After: This is a function on inputs {1, 2, 3}.
Why: Inputs 2 and 3 sharing output 5 is allowed.
Yes, each listed input has exactly one output.
I do
Does the table (1, 4), (1, 6), (2, 5) describe a function?
Step 1
Before: Pairs (1, 4), (1, 6), (2, 5)
Action: Group the repeated input.
After: Input 1→4 and input 1→6.
Why: The same input must not demand different outputs.
One input, two different outputs
Step 2
Before: One input has two outputs.
Action: Apply the definition.
After: This relation is not a function of the input.
Why: The row for input 2 cannot repair the conflict at input 1.
No: input 1 is assigned both 4 and 6.
I do
Only f(0)=1 and f(1)=2 are known. Must f(2) equal three?
Step 1
Before: Two input-output pairs, no rule.
Action: Separate the facts from the proposed pattern.
After: Known: 0 → 1 and 1 → 2. Unknown: the rule and output for input 2.
Why: A table lookup cannot supply a missing row.
Step 2
Before: Try f(x)=x+1.
Action: Check both known inputs.
After: f(0)=0+1=1; f(1)=1+1=2. Then f(2)=2+1=3.
Why: This rule fits, but one fitting rule need not be the only rule.
One possible rule
Step 3
Before: Try g(x)=x²+1.
Action: Check the same inputs with another rule.
After: g(0)=0²+1=0+1=1; g(1)=1²+1=1+1=2. But g(2)=2²+1=4+1=5.
Why: Both rules fit the facts but give different missing outputs.
A second possible rule
Step 4
Before: Possible outputs include 3 and 5.
Action: State what the evidence supports.
After: The output is not determined. Ask for the rule or more conditions.
Why: Do not turn a plausible pattern into a claim of certainty.
No. The two given pairs alone do not determine f(2).
Together — We do
Worked examples
We do
For f(x)=2x+3, complete f(5).
Step 1
Before: f(x)=2x+3;input 5
Action: Replace each x with five.
After: f(5)=2(__)+3=__+3=__.
Why: Multiply first, then add the fixed term.
f(5)=2(5)+3=10+3=13. Check 5+5+3=13.
For f(x)=3x+2, complete f(4)=3(__)+2=__+2=__.
Input → rule → outputHint 1
The bracket gives input four.
Hint 2
Replace x in 3x + 2 by (4).
Hint 3
3 × 4 = 12. Finish 12 + 2, then check.
Show the complete working
Step 1
Before: f(4)
Action: Read the supplied input.
Operation: f(4) → x=4
After: x=4
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: f(x)=3x+2
Action: Replace each occurrence of the input variable.
Operation: f(x)=3x+2 → f(4)=3(4) + 2
After: f(4)=3(4) + 2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (3)×(4)
Action: Multiply the coefficient and the input.
Operation: (3)×(4) → 12
After: 12
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 4
Before: (12)+(2)
Action: Apply the constant term.
Operation: (12)+(2) → 14
After: 14
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 5
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(4)=14
After: f(4)=14
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 6
Before: Fill the original gaps
Action: Use input, product and final output in order.
Operation: Fill the original gaps → f(4)=3(4)+2=12+2=14; 4+4+4+2=14
After: f(4)=3(4)+2=12+2=14; 4+4+4+2=14
Why: The repeated-addition check agrees with the coefficient product.
Use input, product and final output in order.
1. Read the supplied input. f(4) → x=4 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. f(x)=3x+2 → f(4)=3(4) + 2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Multiply the coefficient and the input. (3)×(4) → 12 Complete multiplication before addition or subtraction.
4. Apply the constant term. (12)+(2) → 14 The constant is added once after the variable terms have been evaluated.
5. Use the original function name. Label the computed output → f(4)=14 The requested input has now passed through every operation in the rule.
6. Use input, product and final output in order. Fill the original gaps → f(4)=3(4)+2=12+2=14; 4+4+4+2=14 The repeated-addition check agrees with the coefficient product.
Answer: f(4)=14
The input is 4. f(4)=3(4)+2.
Multiply 3×4=12.
Add 2:14.
Check repeated addition 4+4+4+2=14.
Review your response against these criteria
Selects the correct method.
Shows intermediate calculations.
Checks notation and allowed inputs.
For g(x)=x²−1, complete g(−3)=(__)²−1=__−1=__.
Input → rule → outputHint 1
The input is negative three.
Hint 2
Write (−3)² − 1, keeping the brackets.
Hint 3
(−3) × (−3) = 9. Finish 9 − 1.
Show the complete working
Step 1
Before: g(−3)
Action: Read the supplied input.
Operation: g(−3) → x=−3
After: x=−3
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: g(x)=x²−1
Action: Replace each occurrence of the input variable.
Operation: g(x)=x²−1 → g(−3)=(−3)² + −1
After: g(−3)=(−3)² + −1
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (−3)²
Action: Write the square as a product.
Operation: (−3)² → (−3)×(−3)
After: (−3)×(−3)
Why: Both factors contain the whole input, including its sign.
Write the square as a product.
Step 4
Before: (−3)×(−3)
Action: Multiply the two identical inputs.
Operation: (−3)×(−3) → 9
After: 9
Why: Two negative factors give a positive product.
Multiply the two identical inputs.
Step 5
Before: (9)+(−1)
Action: Apply the constant term.
Operation: (9)+(−1) → 8
After: 8
Why: Adding a negative constant is the same as subtracting its positive magnitude.
Apply the constant term.
Step 6
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → g(−3)=8
After: g(−3)=8
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 7
Before: Fill the original gaps
Action: Keep the negative input in its bracket.
Operation: Fill the original gaps → g(−3)=(−3)²−1=9−1=8
After: g(−3)=(−3)²−1=9−1=8
Why: Without brackets, −3² means the negative of 3², a different expression.
Keep the negative input in its bracket.
1. Read the supplied input. g(−3) → x=−3 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. g(x)=x²−1 → g(−3)=(−3)² + −1 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Write the square as a product. (−3)² → (−3)×(−3) Both factors contain the whole input, including its sign.
4. Multiply the two identical inputs. (−3)×(−3) → 9 Two negative factors give a positive product.
5. Apply the constant term. (9)+(−1) → 8 Adding a negative constant is the same as subtracting its positive magnitude.
6. Use the original function name. Label the computed output → g(−3)=8 The requested input has now passed through every operation in the rule.
7. Keep the negative input in its bracket. Fill the original gaps → g(−3)=(−3)²−1=9−1=8 Without brackets, −3² means the negative of 3², a different expression.
Answer: g(−3)=8
Put −3 in brackets: (−3)²−1.
Multiply (−3)×(−3)=9.
Subtract 1:8.
Check 9−1=8; using −3² would be a different expression.
Review your response against these criteria
Selects the correct method.
Shows intermediate calculations.
Checks notation and allowed inputs.
For h(x)=2x −5, find x when h(x)=9. Fill the same added quantity under both sides.
Hint 1
Nine is the output; x is unknown.
Hint 2
Set 2x − 5 = 9 and add five on both sides.
Hint 3
2x = 14. Divide both sides by two and check.
Show the complete working
Step 1
Before: h(x)=9
Action: The output is known; the input is unknown.
Operation: h(x)=9 → 2x−5=9
After: 2x−5=9
Why: Set the rule equal to the known output. Do not substitute the output in place of the input.
The output is known; the input is unknown.
Step 2
Before: 2x − 5=9
Action: Apply the same inverse addition to both sides.
Operation: 2x − 5=9 → 2x − 5+(5)=9+(5)
After: 2x − 5+(5)=9+(5)
Why: Equality is preserved because both sides receive the same operation.
Apply + 5 to both sides
Step 3
Before: −5+(5)
Action: Cancel the additive inverse pair.
Operation: -5+(5) → -5+(5)=0; 2x+0=9+(5)
After: −5+(5)=0; 2x+0=9+(5)
Why: Opposite added terms become zero. The variable term remains.
Cancel the additive inverse pair.
Step 4
Before: 9+(5)
Action: Calculate the right-hand side.
Operation: 9+(5) → 14
After: 14
Why: Finish this arithmetic before dividing.
Calculate the right-hand side.
Step 5
Before: 2x=14
Action: Divide both sides by the nonzero coefficient.
Operation: 2x=14 → (2x)/(2)=(14)/(2)
After: (2x)/(2)=(14)/(2)
Why: The same nonzero divisor is applied to both sides.
Divide both sides; equal nonzero factors make one
Step 6
Before: (2/2)x=(14)/(2)
Action: Reduce the coefficient quotient to one.
Operation: (2/2)x=(14)/(2) → 1×x=(14)/(2)
After: 1×x=(14)/(2)
Why: 2/2=1, not zero; 2 is nonzero.
Reduce the coefficient quotient to one.
Step 7
Before: x=(14)/(2)
Action: Evaluate the quotient.
Operation: x=(14)/(2) → x=7
After: x=7
Why: Keep an exact fractional value if the division is not integral.
Evaluate the quotient.
Step 8
Before: x=7
Action: Substitute the candidate into the original rule.
Operation: x=7 → h(7)=(2)(7)+(-5)
After: h(7)=(2)(7)+(−5)
Why: Check the rule before applying any contextual restriction.
Substitute the candidate into the original rule.
Step 9
Before: (2)(7)
Action: Calculate the variable term.
Operation: (2)(7) → 14
After: 14
Why: The product recovers the right-hand side before the constant was removed.
Calculate the variable term.
Step 10
Before: 14+(−5)
Action: Restore the constant and compare with the target.
Operation: 14+(-5) → 9=9
After: 9=9
Why: The algebraic candidate produces the required output.
Restore the constant and compare with the target.
Step 11
Before: State the requested result
Action: Report the input with its meaning and unit.
Operation: State the requested result → x=7
After: x=7
Why: The final statement answers the original question after both the algebraic and domain checks.
Report the input with its meaning and unit.
1. The output is known; the input is unknown. h(x)=9 → 2x−5=9 Set the rule equal to the known output. Do not substitute the output in place of the input.
2. Apply the same inverse addition to both sides. 2x − 5=9 → 2x − 5+(5)=9+(5) Equality is preserved because both sides receive the same operation.
3. Cancel the additive inverse pair. -5+(5) → -5+(5)=0; 2x+0=9+(5) Opposite added terms become zero. The variable term remains.
4. Calculate the right-hand side. 9+(5) → 14 Finish this arithmetic before dividing.
5. Divide both sides by the nonzero coefficient. 2x=14 → (2x)/(2)=(14)/(2) The same nonzero divisor is applied to both sides.
6. Reduce the coefficient quotient to one. (2/2)x=(14)/(2) → 1×x=(14)/(2) 2/2=1, not zero; 2 is nonzero.
7. Evaluate the quotient. x=(14)/(2) → x=7 Keep an exact fractional value if the division is not integral.
8. Substitute the candidate into the original rule. x=7 → h(7)=(2)(7)+(-5) Check the rule before applying any contextual restriction.
9. Calculate the variable term. (2)(7) → 14 The product recovers the right-hand side before the constant was removed.
10. Restore the constant and compare with the target. 14+(-5) → 9=9 The algebraic candidate produces the required output.
11. Report the input with its meaning and unit. State the requested result → x=7 The final statement answers the original question after both the algebraic and domain checks.
Answer: x=7
2x − 5 = 9.
Add 5 on both sides:
2x − 5 + 5 = 9 + 5.
2x + 0 = 14; 2x = 14.
Divide both sides by 2:
(2 × x)/(2) = 14/(2).
(2/2) × x = 7; 1 × x = 7.
x = 7.
Check: 2 × (7) − 5 = 14 − 5 = 9.
Review your response against these criteria
Selects the correct method.
Shows intermediate calculations.
Checks notation and allowed inputs.
With help
Worked examples
You do
Find x if f(x)=3x −2 and f(x)=10.
Step 1
Before: 3x −2=10
Action: Choose the operation that removes negative two.
After: 3x=__ after adding the same amount to both sides.
Why: Then divide both sides by the coefficient.
3x − 2 = 10.
Add 2 on both sides:
3x − 2 + 2 = 10 + 2.
3x + 0 = 12; 3x = 12.
Divide both sides by 3:
(3 × x)/(3) = 12/(3).
(3/3) × x = 4; 1 × x = 4.
x = 4.
Check: 3 × (4) − 2 = 12 − 2 = 10.
For p(x)=−3x+2, find p(−2), using a hint only if needed.
Hint 1
Use the entire input negative two.
Hint 2
Substitute: −3(−2) + 2.
Hint 3
The two negative factors give six. Finish 6 + 2.
Show the complete working
Step 1
Before: p(−2)
Action: Read the supplied input.
Operation: p(−2) → x=−2
After: x=−2
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: p(x)=−3x+2
Action: Replace each occurrence of the input variable.
Operation: p(x)=−3x+2 → p(−2)=−3(−2) + 2
After: p(−2)=−3(−2) + 2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (−3)×(−2)
Action: Multiply the coefficient and the input.
Operation: (-3)×(−2) → 6
After: 6
Why: Two negative factors give a positive product.
Multiply the coefficient and the input.
Step 4
Before: (6)+(2)
Action: Apply the constant term.
Operation: (6)+(2) → 8
After: 8
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 5
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → p(−2)=8
After: p(−2)=8
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
1. Read the supplied input. p(−2) → x=−2 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. p(x)=−3x+2 → p(−2)=−3(−2) + 2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Multiply the coefficient and the input. (-3)×(−2) → 6 Two negative factors give a positive product.
4. Apply the constant term. (6)+(2) → 8 The constant is added once after the variable terms have been evaluated.
5. Use the original function name. Label the computed output → p(−2)=8 The requested input has now passed through every operation in the rule.
Answer: p(−2)=8
p(−2)=−3(−2)+2.
Two negative factors give 6.
Then 6+2=8.
Check the ordered pair (−2, 8) fits the rule.
Review your response against these criteria
Selects the correct method.
Shows intermediate calculations.
Checks notation and allowed inputs.
A table lists (0, 2), (1, 2), (1, 3). Is it a function? Explain the deciding input.
Hint 1
Check outputs for each input, not inputs for each output.
Hint 2
Group the two rows that begin with one.
Hint 3
Input one has outputs two and three. Compare that with exactly one output.
Action: Interpret each ordered pair as an association.
Operation: Input set [0, 1]; pairs [(0, 2), (1, 2), (1, 3)] → The first coordinate is the input x; the second is its associated output y.
After: The first coordinate is the input x; the second is its associated output y.
Why: A relation is an association between elements of the two sets.
One relation in three representations
Step 2
Before: Input x=0
Action: Collect its distinct associated outputs.
Operation: Input x=0 → Outputs [2]; count 1
After: Outputs [2]; count 1
Why: Repeated identical pairs do not create a second distinct output; different inputs may share an output.
Collect its distinct associated outputs.
Step 3
Before: Input x=1
Action: Collect its distinct associated outputs.
Operation: Input x=1 → Outputs [2, 3]; count 2
After: Outputs [2, 3]; count 2
Why: Repeated identical pairs do not create a second distinct output; different inputs may share an output.
Collect its distinct associated outputs.
Step 4
Before: Apply exactly one output to every stated input
Action: State the decision and its reason.
Operation: Apply exactly one output to every stated input → Not a function on the stated input set: input 1 has 2 distinct outputs
After: Not a function on the stated input set: input 1 has 2 distinct outputs
Why: A single missing or multiple-output input breaks the function condition on the specified set.
State the decision and its reason.
1. Interpret each ordered pair as an association. Input set [0, 1]; pairs [(0, 2), (1, 2), (1, 3)] → The first coordinate is the input x; the second is its associated output y. A relation is an association between elements of the two sets.
2. Collect its distinct associated outputs. Input x=0 → Outputs [2]; count 1 Repeated identical pairs do not create a second distinct output; different inputs may share an output.
3. Collect its distinct associated outputs. Input x=1 → Outputs [2, 3]; count 2 Repeated identical pairs do not create a second distinct output; different inputs may share an output.
4. State the decision and its reason. Apply exactly one output to every stated input → Not a function on the stated input set: input 1 has 2 distinct outputs A single missing or multiple-output input breaks the function condition on the specified set.
Answer: Not a function on the stated input set: input 1 has 2 distinct outputs
Input 0 has output 2.
Input 1 has both 2 and 3.
Because the same input has two different outputs, the table is not a function.
Review your response against these criteria
Selects the correct method.
Shows intermediate calculations.
Checks notation and allowed inputs.
For C(n)=5n+2 dollars and whole n≥0, could the exact cost be $14?
Hint 1
The cost is given and n must be whole.
Hint 2
Set 5n + 2 = 14; subtract two on both sides.
Hint 3
Divide 5n = 12 by five. Decide whether that input is permitted.
Show the complete working
Step 1
Before: C(n)=14
Action: The output is known; the input is unknown.
Operation: C(n)=14 → 5n+2=14
After: 5n+2=14
Why: Set the rule equal to the known output. Do not substitute the output in place of the input.
The output is known; the input is unknown.
Step 2
Before: 5n + 2=14
Action: Apply the same inverse addition to both sides.
Operation: 5n + 2=14 → 5n + 2+(-2)=14+(-2)
After: 5n + 2+(−2)=14+(−2)
Why: Equality is preserved because both sides receive the same operation.
Apply − 2 to both sides
Step 3
Before: 2+(−2)
Action: Cancel the additive inverse pair.
Operation: 2+(-2) → 2+(-2)=0; 5n+0=14+(-2)
After: 2+(−2)=0; 5n+0=14+(−2)
Why: Opposite added terms become zero. The variable term remains.
Cancel the additive inverse pair.
Step 4
Before: 14+(−2)
Action: Calculate the right-hand side.
Operation: 14+(-2) → 12
After: 12
Why: Finish this arithmetic before dividing.
Calculate the right-hand side.
Step 5
Before: 5n=12
Action: Divide both sides by the nonzero coefficient.
Operation: 5n=12 → (5n)/(5)=(12)/(5)
After: (5n)/(5)=(12)/(5)
Why: The same nonzero divisor is applied to both sides.
Divide both sides; equal nonzero factors make one
Step 6
Before: (5/5)n=(12)/(5)
Action: Reduce the coefficient quotient to one.
Operation: (5/5)n=(12)/(5) → 1×n=(12)/(5)
After: 1×n=(12)/(5)
Why: 5/5=1, not zero; 5 is nonzero.
Reduce the coefficient quotient to one.
Step 7
Before: n=(12)/(5)
Action: Evaluate the quotient.
Operation: n=(12)/(5) → n=12/5
After: n=12/5
Why: Keep an exact fractional value if the division is not integral.
Evaluate the quotient.
Step 8
Before: n=12/5
Action: Substitute the candidate into the original rule.
Operation: n=12/5 → C(12/5)=(5)(12/5)+(2)
After: C(12/5)=(5)(12/5)+(2)
Why: Check the rule before applying any contextual restriction.
Substitute the candidate into the original rule.
Step 9
Before: (5)(12/5)
Action: Calculate the variable term.
Operation: (5)(12/5) → 12
After: 12
Why: The product recovers the right-hand side before the constant was removed.
Calculate the variable term.
Step 10
Before: 12+(2)
Action: Restore the constant and compare with the target.
Operation: 12+(2) → 14=14
After: 14=14
Why: The algebraic candidate produces the required output.
Restore the constant and compare with the target.
Step 11
Before: Candidate n=12/5
Action: Check the whole-number, nonnegative input condition.
Operation: Candidate n=12/5 → Not permitted: this input is not a nonnegative whole number.
After: Not permitted: this input is not a nonnegative whole number.
Why: A fractional number of whole items cannot be ordered.
Check the whole-number, nonnegative input condition.
Step 12
Before: C(2)=(5)(2)+(2)
Action: Multiply the whole-item count by the per-item amount.
Operation: C(2)=(5)(2)+(2) → 10+(2)
After: 10+(2)
Why: This is a permitted neighbouring count.
Multiply the whole-item count by the per-item amount.
Step 13
Before: 10+(2)
Action: Add the fixed amount.
Operation: 10+(2) → C(2)=12 dollars
After: C(2)=12 dollars
Why: Compare this attainable cost with the required exact cost.
Add the fixed amount.
Step 14
Before: C(3)=(5)(3)+(2)
Action: Multiply the whole-item count by the per-item amount.
Operation: C(3)=(5)(3)+(2) → 15+(2)
After: 15+(2)
Why: This is a permitted neighbouring count.
Multiply the whole-item count by the per-item amount.
Step 15
Before: 15+(2)
Action: Add the fixed amount.
Operation: 15+(2) → C(3)=17 dollars
After: C(3)=17 dollars
Why: Compare this attainable cost with the required exact cost.
Add the fixed amount.
Step 16
Before: Both neighbouring whole counts miss the target
Action: State the contextual conclusion.
Operation: Both neighbouring whole counts miss the target → No exact 14-dollar order is available.
After: No exact 14-dollar order is available.
Why: Every additional item changes the cost by the positive fixed per-item amount; no whole count lies between the neighbours.
State the contextual conclusion.
Step 17
Before: State the requested result
Action: Report the input with its meaning and unit.
Operation: State the requested result → No exact 14-dollar order is available.
After: No exact 14-dollar order is available.
Why: The final statement answers the original question after both the algebraic and domain checks.
Report the input with its meaning and unit.
1. The output is known; the input is unknown. C(n)=14 → 5n+2=14 Set the rule equal to the known output. Do not substitute the output in place of the input.
2. Apply the same inverse addition to both sides. 5n + 2=14 → 5n + 2+(-2)=14+(-2) Equality is preserved because both sides receive the same operation.
3. Cancel the additive inverse pair. 2+(-2) → 2+(-2)=0; 5n+0=14+(-2) Opposite added terms become zero. The variable term remains.
4. Calculate the right-hand side. 14+(-2) → 12 Finish this arithmetic before dividing.
5. Divide both sides by the nonzero coefficient. 5n=12 → (5n)/(5)=(12)/(5) The same nonzero divisor is applied to both sides.
6. Reduce the coefficient quotient to one. (5/5)n=(12)/(5) → 1×n=(12)/(5) 5/5=1, not zero; 5 is nonzero.
7. Evaluate the quotient. n=(12)/(5) → n=12/5 Keep an exact fractional value if the division is not integral.
8. Substitute the candidate into the original rule. n=12/5 → C(12/5)=(5)(12/5)+(2) Check the rule before applying any contextual restriction.
9. Calculate the variable term. (5)(12/5) → 12 The product recovers the right-hand side before the constant was removed.
10. Restore the constant and compare with the target. 12+(2) → 14=14 The algebraic candidate produces the required output.
11. Check the whole-number, nonnegative input condition. Candidate n=12/5 → Not permitted: this input is not a nonnegative whole number. A fractional number of whole items cannot be ordered.
12. Multiply the whole-item count by the per-item amount. C(2)=(5)(2)+(2) → 10+(2) This is a permitted neighbouring count.
13. Add the fixed amount. 10+(2) → C(2)=12 dollars Compare this attainable cost with the required exact cost.
14. Multiply the whole-item count by the per-item amount. C(3)=(5)(3)+(2) → 15+(2) This is a permitted neighbouring count.
15. Add the fixed amount. 15+(2) → C(3)=17 dollars Compare this attainable cost with the required exact cost.
16. State the contextual conclusion. Both neighbouring whole counts miss the target → No exact 14-dollar order is available. Every additional item changes the cost by the positive fixed per-item amount; no whole count lies between the neighbours.
17. Report the input with its meaning and unit. State the requested result → No exact 14-dollar order is available. The final statement answers the original question after both the algebraic and domain checks.
Answer: No exact 14-dollar order is available.
5n + 2 = 14.
Subtract 2 on both sides:
5n + 2 − 2 = 14 − 2.
5n + 0 = 12; 5n = 12.
Divide both sides by 5:
(5 × n)/(5) = 12/(5).
(5/5) × n = 2.4; 1 × n = 2.4.
n = 2.4.
Check: 5 × (2.4) + 2 = 12 + 2 = 14.
The equation gives 2.4, but tickets must be whole.
C(2) = 5(2) + 2 = 10 + 2 = 12 dollars.
C(3) = 5(3) + 2 = 15 + 2 = 17 dollars.
No allowed input gives exactly 14 dollars.
Review your response against these criteria
Selects the correct method.
Shows intermediate calculations.
Checks notation and allowed inputs.
On my own
f(x)=4x −3. Find f(2).
Input → rule → outputHint 1
The input is two.
Hint 2
Replace x with (2): 4(2) − 3.
Hint 3
4 × 2 = 8. Subtract three and check.
Show the complete working
Step 1
Before: f(2)
Action: Read the supplied input.
Operation: f(2) → x=2
After: x=2
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: f(x)=4x−3
Action: Replace each occurrence of the input variable.
Operation: f(x)=4x−3 → f(2)=4(2) + −3
After: f(2)=4(2) + −3
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (4)×(2)
Action: Multiply the coefficient and the input.
Operation: (4)×(2) → 8
After: 8
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 4
Before: (8)+(−3)
Action: Apply the constant term.
Operation: (8)+(−3) → 5
After: 5
Why: Adding a negative constant is the same as subtracting its positive magnitude.
Apply the constant term.
Step 5
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(2)=5
After: f(2)=5
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
1. Read the supplied input. f(2) → x=2 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. f(x)=4x−3 → f(2)=4(2) + −3 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Multiply the coefficient and the input. (4)×(2) → 8 Complete multiplication before addition or subtraction.
4. Apply the constant term. (8)+(−3) → 5 Adding a negative constant is the same as subtracting its positive magnitude.
5. Use the original function name. Label the computed output → f(2)=5 The requested input has now passed through every operation in the rule.
Answer: f(2)=5
f(2)=4(2)−3=8−3=5.
Check two groups of four minus three leaves five.
Review your response against these criteria
Uses the relevant method.
Shows the working.
Explains a valid check.
g(x)=x²+x. Find g(−3).
Input → rule → outputHint 1
Use negative three in both occurrences of x.
Hint 2
Write (−3)² + (−3).
Hint 3
The square is nine. Finish 9 + (−3).
Show the complete working
Step 1
Before: g(−3)
Action: Read the supplied input.
Operation: g(−3) → x=−3
After: x=−3
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: g(x)=x²+x
Action: Replace each occurrence of the input variable.
Operation: g(x)=x²+x → g(−3)=(−3)² + (−3)
After: g(−3)=(−3)² + (−3)
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (−3)²
Action: Write the square as a product.
Operation: (−3)² → (−3)×(−3)
After: (−3)×(−3)
Why: Both factors contain the whole input, including its sign.
Write the square as a product.
Step 4
Before: (−3)×(−3)
Action: Multiply the two identical inputs.
Operation: (−3)×(−3) → 9
After: 9
Why: Two negative factors give a positive product.
Multiply the two identical inputs.
Step 5
Before: (1)×(−3)
Action: Multiply the coefficient and the input.
Operation: (1)×(−3) → −3
After: −3
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 6
Before: (9)+(−3)
Action: Add the two evaluated variable terms.
Operation: (9)+(−3) → 6
After: 6
Why: The squared input and the unsquared input have different roles; evaluate both before adding.
Add the two evaluated variable terms.
Step 7
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → g(−3)=6
After: g(−3)=6
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
1. Read the supplied input. g(−3) → x=−3 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. g(x)=x²+x → g(−3)=(−3)² + (−3) Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Write the square as a product. (−3)² → (−3)×(−3) Both factors contain the whole input, including its sign.
4. Multiply the two identical inputs. (−3)×(−3) → 9 Two negative factors give a positive product.
5. Multiply the coefficient and the input. (1)×(−3) → −3 Complete multiplication before addition or subtraction.
6. Add the two evaluated variable terms. (9)+(−3) → 6 The squared input and the unsquared input have different roles; evaluate both before adding.
7. Use the original function name. Label the computed output → g(−3)=6 The requested input has now passed through every operation in the rule.
Answer: g(−3)=6
g(−3)=(−3)²+(−3)=9−3=6.
Squaring the whole negative input is essential.
Review your response against these criteria
Uses the relevant method.
Shows the working.
Explains a valid check.
h(x)=5x −7. Find the input for output 18.
Hint 1
Eighteen is an output.
Hint 2
Set 5x − 7 = 18; add seven to both sides.
Hint 3
5x = 25. Divide both sides by five and substitute to check.
Show the complete working
Step 1
Before: h(x)=18
Action: The output is known; the input is unknown.
Operation: h(x)=18 → 5x−7=18
After: 5x−7=18
Why: Set the rule equal to the known output. Do not substitute the output in place of the input.
The output is known; the input is unknown.
Step 2
Before: 5x − 7=18
Action: Apply the same inverse addition to both sides.
Operation: 5x − 7=18 → 5x − 7+(7)=18+(7)
After: 5x − 7+(7)=18+(7)
Why: Equality is preserved because both sides receive the same operation.
Apply + 7 to both sides
Step 3
Before: −7+(7)
Action: Cancel the additive inverse pair.
Operation: -7+(7) → -7+(7)=0; 5x+0=18+(7)
After: −7+(7)=0; 5x+0=18+(7)
Why: Opposite added terms become zero. The variable term remains.
Cancel the additive inverse pair.
Step 4
Before: 18+(7)
Action: Calculate the right-hand side.
Operation: 18+(7) → 25
After: 25
Why: Finish this arithmetic before dividing.
Calculate the right-hand side.
Step 5
Before: 5x=25
Action: Divide both sides by the nonzero coefficient.
Operation: 5x=25 → (5x)/(5)=(25)/(5)
After: (5x)/(5)=(25)/(5)
Why: The same nonzero divisor is applied to both sides.
Divide both sides; equal nonzero factors make one
Step 6
Before: (5/5)x=(25)/(5)
Action: Reduce the coefficient quotient to one.
Operation: (5/5)x=(25)/(5) → 1×x=(25)/(5)
After: 1×x=(25)/(5)
Why: 5/5=1, not zero; 5 is nonzero.
Reduce the coefficient quotient to one.
Step 7
Before: x=(25)/(5)
Action: Evaluate the quotient.
Operation: x=(25)/(5) → x=5
After: x=5
Why: Keep an exact fractional value if the division is not integral.
Evaluate the quotient.
Step 8
Before: x=5
Action: Substitute the candidate into the original rule.
Operation: x=5 → h(5)=(5)(5)+(-7)
After: h(5)=(5)(5)+(−7)
Why: Check the rule before applying any contextual restriction.
Substitute the candidate into the original rule.
Step 9
Before: (5)(5)
Action: Calculate the variable term.
Operation: (5)(5) → 25
After: 25
Why: The product recovers the right-hand side before the constant was removed.
Calculate the variable term.
Step 10
Before: 25+(−7)
Action: Restore the constant and compare with the target.
Operation: 25+(-7) → 18=18
After: 18=18
Why: The algebraic candidate produces the required output.
Restore the constant and compare with the target.
Step 11
Before: State the requested result
Action: Report the input with its meaning and unit.
Operation: State the requested result → x=5
After: x=5
Why: The final statement answers the original question after both the algebraic and domain checks.
Report the input with its meaning and unit.
1. The output is known; the input is unknown. h(x)=18 → 5x−7=18 Set the rule equal to the known output. Do not substitute the output in place of the input.
2. Apply the same inverse addition to both sides. 5x − 7=18 → 5x − 7+(7)=18+(7) Equality is preserved because both sides receive the same operation.
3. Cancel the additive inverse pair. -7+(7) → -7+(7)=0; 5x+0=18+(7) Opposite added terms become zero. The variable term remains.
4. Calculate the right-hand side. 18+(7) → 25 Finish this arithmetic before dividing.
5. Divide both sides by the nonzero coefficient. 5x=25 → (5x)/(5)=(25)/(5) The same nonzero divisor is applied to both sides.
6. Reduce the coefficient quotient to one. (5/5)x=(25)/(5) → 1×x=(25)/(5) 5/5=1, not zero; 5 is nonzero.
7. Evaluate the quotient. x=(25)/(5) → x=5 Keep an exact fractional value if the division is not integral.
8. Substitute the candidate into the original rule. x=5 → h(5)=(5)(5)+(-7) Check the rule before applying any contextual restriction.
9. Calculate the variable term. (5)(5) → 25 The product recovers the right-hand side before the constant was removed.
10. Restore the constant and compare with the target. 25+(-7) → 18=18 The algebraic candidate produces the required output.
11. Report the input with its meaning and unit. State the requested result → x=5 The final statement answers the original question after both the algebraic and domain checks.
Answer: x=5
5x − 7 = 18.
Add 7 on both sides:
5x − 7 + 7 = 18 + 7.
5x + 0 = 25; 5x = 25.
Divide both sides by 5:
(5 × x)/(5) = 25/(5).
(5/5) × x = 5; 1 × x = 5.
x = 5.
Check: 5 × (5) − 7 = 25 − 7 = 18.
Review your response against these criteria
Uses the relevant method.
Shows the working.
Explains a valid check.
C(n)=6n+4 dollars, with whole n≥0. An order costs 28 dollars. How many tickets?
Hint 1
Find the whole-number ticket input.
Hint 2
Set 6n + 4 = 28; subtract four on both sides.
Hint 3
6n = 24. Divide both sides by six, then state the ticket count.
Show the complete working
Step 1
Before: C(n)=28
Action: The output is known; the input is unknown.
Operation: C(n)=28 → 6n+4=28
After: 6n+4=28
Why: Set the rule equal to the known output. Do not substitute the output in place of the input.
The output is known; the input is unknown.
Step 2
Before: 6n + 4=28
Action: Apply the same inverse addition to both sides.
Operation: 6n + 4=28 → 6n + 4+(-4)=28+(-4)
After: 6n + 4+(−4)=28+(−4)
Why: Equality is preserved because both sides receive the same operation.
Apply − 4 to both sides
Step 3
Before: 4+(−4)
Action: Cancel the additive inverse pair.
Operation: 4+(-4) → 4+(-4)=0; 6n+0=28+(-4)
After: 4+(−4)=0; 6n+0=28+(−4)
Why: Opposite added terms become zero. The variable term remains.
Cancel the additive inverse pair.
Step 4
Before: 28+(−4)
Action: Calculate the right-hand side.
Operation: 28+(-4) → 24
After: 24
Why: Finish this arithmetic before dividing.
Calculate the right-hand side.
Step 5
Before: 6n=24
Action: Divide both sides by the nonzero coefficient.
Operation: 6n=24 → (6n)/(6)=(24)/(6)
After: (6n)/(6)=(24)/(6)
Why: The same nonzero divisor is applied to both sides.
Divide both sides; equal nonzero factors make one
Step 6
Before: (6/6)n=(24)/(6)
Action: Reduce the coefficient quotient to one.
Operation: (6/6)n=(24)/(6) → 1×n=(24)/(6)
After: 1×n=(24)/(6)
Why: 6/6=1, not zero; 6 is nonzero.
Reduce the coefficient quotient to one.
Step 7
Before: n=(24)/(6)
Action: Evaluate the quotient.
Operation: n=(24)/(6) → n=4
After: n=4
Why: Keep an exact fractional value if the division is not integral.
Evaluate the quotient.
Step 8
Before: n=4
Action: Substitute the candidate into the original rule.
Operation: n=4 → C(4)=(6)(4)+(4)
After: C(4)=(6)(4)+(4)
Why: Check the rule before applying any contextual restriction.
Substitute the candidate into the original rule.
Step 9
Before: (6)(4)
Action: Calculate the variable term.
Operation: (6)(4) → 24
After: 24
Why: The product recovers the right-hand side before the constant was removed.
Calculate the variable term.
Step 10
Before: 24+(4)
Action: Restore the constant and compare with the target.
Operation: 24+(4) → 28=28
After: 28=28
Why: The algebraic candidate produces the required output.
Restore the constant and compare with the target.
Step 11
Before: Candidate n=4
Action: Check the whole-number, nonnegative input condition.
Operation: Candidate n=4 → Permitted: a nonnegative whole number.
After: Permitted: a nonnegative whole number.
Why: A fractional number of whole items cannot be ordered.
Check the whole-number, nonnegative input condition.
Step 12
Before: State the requested result
Action: Report the input with its meaning and unit.
Operation: State the requested result → n=4 tickets
After: n=4 tickets
Why: The final statement answers the original question after both the algebraic and domain checks.
Report the input with its meaning and unit.
1. The output is known; the input is unknown. C(n)=28 → 6n+4=28 Set the rule equal to the known output. Do not substitute the output in place of the input.
2. Apply the same inverse addition to both sides. 6n + 4=28 → 6n + 4+(-4)=28+(-4) Equality is preserved because both sides receive the same operation.
3. Cancel the additive inverse pair. 4+(-4) → 4+(-4)=0; 6n+0=28+(-4) Opposite added terms become zero. The variable term remains.
4. Calculate the right-hand side. 28+(-4) → 24 Finish this arithmetic before dividing.
5. Divide both sides by the nonzero coefficient. 6n=24 → (6n)/(6)=(24)/(6) The same nonzero divisor is applied to both sides.
6. Reduce the coefficient quotient to one. (6/6)n=(24)/(6) → 1×n=(24)/(6) 6/6=1, not zero; 6 is nonzero.
7. Evaluate the quotient. n=(24)/(6) → n=4 Keep an exact fractional value if the division is not integral.
8. Substitute the candidate into the original rule. n=4 → C(4)=(6)(4)+(4) Check the rule before applying any contextual restriction.
9. Calculate the variable term. (6)(4) → 24 The product recovers the right-hand side before the constant was removed.
10. Restore the constant and compare with the target. 24+(4) → 28=28 The algebraic candidate produces the required output.
11. Check the whole-number, nonnegative input condition. Candidate n=4 → Permitted: a nonnegative whole number. A fractional number of whole items cannot be ordered.
12. Report the input with its meaning and unit. State the requested result → n=4 tickets The final statement answers the original question after both the algebraic and domain checks.
Answer: n=4 tickets
6n + 4 = 28.
Subtract 4 on both sides:
6n + 4 − 4 = 28 − 4.
6n + 0 = 24; 6n = 24.
Divide both sides by 6:
(6 × n)/(6) = 24/(6).
(6/6) × n = 4; 1 × n = 4.
n = 4.
Check: 6 × (4) + 4 = 24 + 4 = 28.
Four is a nonnegative whole number, so four tickets are allowed and cost exactly 28 dollars.
Review your response against these criteria
Uses the relevant method.
Shows the working.
Explains a valid check.
Does (−1, 1), (0, 0), (1, 1) describe a function? Explain.
Yes: every input has one output.
No: two inputs share an output.
No: negative inputs are never allowed.
Hint 1
A shared output is allowed.
Hint 2
Group rows by their first number.
Hint 3
Each of −1, 0 and 1 has just one output. Apply the definition.
Action: Interpret each ordered pair as an association.
Operation: Input set [-1, 0, 1]; pairs [(-1, 1), (0, 0), (1, 1)] → The first coordinate is the input x; the second is its associated output y.
After: The first coordinate is the input x; the second is its associated output y.
Why: A relation is an association between elements of the two sets.
One relation in three representations
Step 2
Before: Input x=−1
Action: Collect its distinct associated outputs.
Operation: Input x=-1 → Outputs [1]; count 1
After: Outputs [1]; count 1
Why: Repeated identical pairs do not create a second distinct output; different inputs may share an output.
Collect its distinct associated outputs.
Step 3
Before: Input x=0
Action: Collect its distinct associated outputs.
Operation: Input x=0 → Outputs [0]; count 1
After: Outputs [0]; count 1
Why: Repeated identical pairs do not create a second distinct output; different inputs may share an output.
Collect its distinct associated outputs.
Step 4
Before: Input x=1
Action: Collect its distinct associated outputs.
Operation: Input x=1 → Outputs [1]; count 1
After: Outputs [1]; count 1
Why: Repeated identical pairs do not create a second distinct output; different inputs may share an output.
Collect its distinct associated outputs.
Step 5
Before: Apply exactly one output to every stated input
Action: State the decision and its reason.
Operation: Apply exactly one output to every stated input → Function on the stated input set; f(-1)=1, f(0)=0, f(1)=1
After: Function on the stated input set; f(−1)=1, f(0)=0, f(1)=1
Why: A single missing or multiple-output input breaks the function condition on the specified set.
State the decision and its reason.
1. Interpret each ordered pair as an association. Input set [-1, 0, 1]; pairs [(-1, 1), (0, 0), (1, 1)] → The first coordinate is the input x; the second is its associated output y. A relation is an association between elements of the two sets.
2. Collect its distinct associated outputs. Input x=-1 → Outputs [1]; count 1 Repeated identical pairs do not create a second distinct output; different inputs may share an output.
3. Collect its distinct associated outputs. Input x=0 → Outputs [0]; count 1 Repeated identical pairs do not create a second distinct output; different inputs may share an output.
4. Collect its distinct associated outputs. Input x=1 → Outputs [1]; count 1 Repeated identical pairs do not create a second distinct output; different inputs may share an output.
5. State the decision and its reason. Apply exactly one output to every stated input → Function on the stated input set; f(-1)=1, f(0)=0, f(1)=1 A single missing or multiple-output input breaks the function condition on the specified set.
Answer: Function on the stated input set; f(-1)=1, f(0)=0, f(1)=1
Every listed input has one output. −1 and 1 share output 1, which is allowed.
It is a function on the three listed inputs.
Check your reasoning:
Uses the relevant method.
Shows the working.
Explains a valid check.
A learner claims (x+8)/4=x+2. Test x=4 and repair the expression.
Hint 1
Test both expressions at the same input.
Hint 2
At x = 4, the left is 12/4 and the claimed right is 4 + 2.
Hint 3
They differ. Divide each numerator term by four: x/4 + 8/4. Simplify the constant.
Why: One failed input disproves a claim stated for every input.
Test the same input x=4 on both sides.
Step 2
Before: (4+8)/4
Action: Add inside the numerator first.
Operation: (4+8)/4 → 12/4
After: 12/4
Why: The fraction bar groups the entire sum.
Add inside the numerator first.
Step 3
Before: 12/4
Action: Divide the whole numerator by four.
Operation: 12/4 → 3
After: 3
Why: Division applies after the numerator is computed.
Divide the whole numerator by four.
Step 4
Before: 4+2
Action: Evaluate the claimed right side.
Operation: 4+2 → 6
After: 6
Why: This is a separate calculation with the same input.
Evaluate the claimed right side.
Step 5
Before: Left 3, right 6
Action: Compare the two results.
Operation: Left 3, right 6 → 3≠6; the claimed identity fails
After: 3≠6; the claimed identity fails
Why: The first term was not divided by four in the claimed expression.
Compare the two results.
Step 6
Before: (x+8)/4
Action: Distribute division across both added terms.
Operation: (x+8)/4 → x/4+8/4
After: x/4+8/4
Why: For a nonzero common denominator, every numerator term is divided by it.
Distribute division across both added terms.
Step 7
Before: 8/4
Action: Reduce only the numerical quotient.
Operation: 8/4 → 2; therefore (x+8)/4=x/4+2
After: 2; therefore (x+8)/4=x/4+2
Why: x/4 remains x/4; it does not become x.
Reduce only the numerical quotient.
Step 8
Before: x/4+2 with x=4
Action: Check the repaired expression.
Operation: x/4+2 with x=4 → 4/4+2
After: 4/4+2
Why: Use the same deciding input as before.
Check the repaired expression.
Step 9
Before: 4/4+2
Action: Complete division before addition.
Operation: 4/4+2 → 1+2=3
After: 1+2=3
Why: The repaired form now agrees with the original left side.
Complete division before addition.
1. Test the same input x=4 on both sides. (x+8)/4 = x+2? → Left: (4+8)/4; right:4+2 One failed input disproves a claim stated for every input.
2. Add inside the numerator first. (4+8)/4 → 12/4 The fraction bar groups the entire sum.
3. Divide the whole numerator by four. 12/4 → 3 Division applies after the numerator is computed.
4. Evaluate the claimed right side. 4+2 → 6 This is a separate calculation with the same input.
5. Compare the two results. Left 3, right 6 → 3≠6; the claimed identity fails The first term was not divided by four in the claimed expression.
6. Distribute division across both added terms. (x+8)/4 → x/4+8/4 For a nonzero common denominator, every numerator term is divided by it.
7. Reduce only the numerical quotient. 8/4 → 2; therefore (x+8)/4=x/4+2 x/4 remains x/4; it does not become x.
8. Check the repaired expression. x/4+2 with x=4 → 4/4+2 Use the same deciding input as before.
9. Complete division before addition. 4/4+2 → 1+2=3 The repaired form now agrees with the original left side.
Answer: The claim is false. The correct identity is (x+8)/4=x/4+2.
Left:(4+8)/4=12/4=3.
Claimed right:4+2=6.
They differ.
Divide both terms:(x+8)/4=x/4+8/4=x/4+2.
At x = 4 this gives 1+2=3.
Review your response against these criteria
Uses the relevant method.
Shows the working.
Explains a valid check.
Come back
Return on another day. Try these fresh questions before revealing a hint or answer. Explain what changed in your approach.
f(x)=2x+5. Find f(0).
Hint 1
Zero is the input.
Hint 2
Replace x with zero in 2x + 5.
Hint 3
2 × 0 = 0. Keep the constant and finish the sum.
Show the complete working
Step 1
Before: f(0)
Action: Read the supplied input.
Operation: f(0) → x=0
After: x=0
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: f(x)=2x+5
Action: Replace each occurrence of the input variable.
Operation: f(x)=2x+5 → f(0)=2(0) + 5
After: f(0)=2(0) + 5
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (2)×(0)
Action: Multiply the coefficient and the input.
Operation: (2)×(0) → 0
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 4
Before: (0)+(5)
Action: Apply the constant term.
Operation: (0)+(5) → 5
After: 5
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 5
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(0)=5
After: f(0)=5
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
1. Read the supplied input. f(0) → x=0 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. f(x)=2x+5 → f(0)=2(0) + 5 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Multiply the coefficient and the input. (2)×(0) → 0 Complete multiplication before addition or subtraction.
4. Apply the constant term. (0)+(5) → 5 The constant is added once after the variable terms have been evaluated.
5. Use the original function name. Label the computed output → f(0)=5 The requested input has now passed through every operation in the rule.
Answer: f(0)=5
f(0)=2(0)+5=0+5=5.
Zero input removes the variable term, not the constant.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
f(x)=2x+5. Find f(−4).
Hint 1
The input is negative four.
Hint 2
Write 2(−4) + 5.
Hint 3
Two groups of negative four give −8. Add five, then check.
Show the complete working
Step 1
Before: f(−4)
Action: Read the supplied input.
Operation: f(−4) → x=−4
After: x=−4
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: f(x)=2x+5
Action: Replace each occurrence of the input variable.
Operation: f(x)=2x+5 → f(−4)=2(−4) + 5
After: f(−4)=2(−4) + 5
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (2)×(−4)
Action: Multiply the coefficient and the input.
Operation: (2)×(−4) → −8
After: −8
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 4
Before: (−8)+(5)
Action: Apply the constant term.
Operation: (−8)+(5) → −3
After: −3
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 5
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(−4)=−3
After: f(−4)=−3
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
1. Read the supplied input. f(−4) → x=−4 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. f(x)=2x+5 → f(−4)=2(−4) + 5 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Multiply the coefficient and the input. (2)×(−4) → −8 Complete multiplication before addition or subtraction.
4. Apply the constant term. (−8)+(5) → −3 The constant is added once after the variable terms have been evaluated.
5. Use the original function name. Label the computed output → f(−4)=−3 The requested input has now passed through every operation in the rule.
Answer: f(−4)=−3
f(−4)=2(−4)+5=−8+5=−3.
Check: from −8 move five units right to −3.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
g(x)=x²+1. Find g(−5).
Hint 1
The entire negative input is squared.
Hint 2
Write (−5)² + 1.
Hint 3
(−5)(−5) = 25. Add the constant.
Show the complete working
Step 1
Before: g(−5)
Action: Read the supplied input.
Operation: g(−5) → x=−5
After: x=−5
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: g(x)=x²+1
Action: Replace each occurrence of the input variable.
Operation: g(x)=x²+1 → g(−5)=(−5)² + 1
After: g(−5)=(−5)² + 1
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (−5)²
Action: Write the square as a product.
Operation: (−5)² → (−5)×(−5)
After: (−5)×(−5)
Why: Both factors contain the whole input, including its sign.
Write the square as a product.
Step 4
Before: (−5)×(−5)
Action: Multiply the two identical inputs.
Operation: (−5)×(−5) → 25
After: 25
Why: Two negative factors give a positive product.
Multiply the two identical inputs.
Step 5
Before: (25)+(1)
Action: Apply the constant term.
Operation: (25)+(1) → 26
After: 26
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 6
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → g(−5)=26
After: g(−5)=26
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
1. Read the supplied input. g(−5) → x=−5 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. g(x)=x²+1 → g(−5)=(−5)² + 1 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Write the square as a product. (−5)² → (−5)×(−5) Both factors contain the whole input, including its sign.
4. Multiply the two identical inputs. (−5)×(−5) → 25 Two negative factors give a positive product.
5. Apply the constant term. (25)+(1) → 26 The constant is added once after the variable terms have been evaluated.
6. Use the original function name. Label the computed output → g(−5)=26 The requested input has now passed through every operation in the rule.
Answer: g(−5)=26
g(−5)=(−5)²+1=(−5)(−5)+1=25+1=26.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
g(x)=x²−3x. Find g(2).
Hint 1
Use two in both places.
Hint 2
Write 2² − 3(2).
Hint 3
The square is four and the product is six. Finish 4 − 6.
Show the complete working
Step 1
Before: g(2)
Action: Read the input.
Operation: g(2) → x=2
After: x=2
Why: Both appearances of x receive the same whole input.
Read the input.
Step 2
Before: g(x)=x²−3x
Action: Substitute both input occurrences.
Operation: g(x)=x²−3x → g(2)=(2)²−3(2)
After: g(2)=(2)²−3(2)
Why: Keep the subtraction in the original rule visible.
Substitute both input occurrences.
Step 3
Before: (2)²
Action: Write the square as repeated multiplication.
Operation: (2)² → (2)(2)
After: (2)(2)
Why: The sign is inside both factors.
Write the square as repeated multiplication.
Step 4
Before: (2)(2)
Action: Evaluate the square.
Operation: (2)(2) → 4
After: 4
Why: Two positive factors produce a positive product.
Evaluate the square.
Step 5
Before: 3(2)
Action: Evaluate the separate linear product.
Operation: 3(2) → 6
After: 6
Why: The original minus sign still stands before this whole product.
Evaluate the separate linear product.
Step 6
Before: 4−(6)
Action: Apply the original subtraction.
Operation: 4−(6) → 4−6
After: 4−6
Why: The subtraction produces a negative result because six is greater than four.
Apply the original subtraction.
Step 7
Before: 4−6
Action: Calculate the final output.
Operation: 4−6 → -2
After: −2
Why: Only now are the two evaluated parts combined.
Calculate the final output.
Step 8
Before: Label the result
Action: Write the original function statement.
Operation: Label the result → g(2)=-2
After: g(2)=−2
Why: This output corresponds to the stated input.
Write the original function statement.
1. Read the input. g(2) → x=2 Both appearances of x receive the same whole input.
2. Substitute both input occurrences. g(x)=x²−3x → g(2)=(2)²−3(2) Keep the subtraction in the original rule visible.
3. Write the square as repeated multiplication. (2)² → (2)(2) The sign is inside both factors.
4. Evaluate the square. (2)(2) → 4 Two positive factors produce a positive product.
5. Evaluate the separate linear product. 3(2) → 6 The original minus sign still stands before this whole product.
6. Apply the original subtraction. 4−(6) → 4−6 The subtraction produces a negative result because six is greater than four.
7. Calculate the final output. 4−6 → -2 Only now are the two evaluated parts combined.
8. Write the original function statement. Label the result → g(2)=-2 This output corresponds to the stated input.
Answer: g(2)=-2
g(2)=2²−3(2)=4−6=−2.
Evaluate the square and product before subtraction.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
g(x)=x²−3x. Find g(−2).
Hint 1
Keep negative two together in brackets.
Hint 2
Write (−2)² − 3(−2).
Hint 3
The square is four and the product is −6. Subtracting −6 adds six.
Show the complete working
Step 1
Before: g(−2)
Action: Read the input.
Operation: g(-2) → x=-2
After: x=−2
Why: Both appearances of x receive the same whole input.
Read the input.
Step 2
Before: g(x)=x²−3x
Action: Substitute both input occurrences.
Operation: g(x)=x²−3x → g(-2)=(-2)²−3(-2)
After: g(−2)=(−2)²−3(−2)
Why: Keep the subtraction in the original rule visible.
Substitute both input occurrences.
Step 3
Before: (−2)²
Action: Write the square as repeated multiplication.
Operation: (-2)² → (-2)(-2)
After: (−2)(−2)
Why: The sign is inside both factors.
Write the square as repeated multiplication.
Step 4
Before: (−2)(−2)
Action: Evaluate the square.
Operation: (-2)(-2) → 4
After: 4
Why: Two negative factors produce a positive product.
Evaluate the square.
Step 5
Before: 3(−2)
Action: Evaluate the separate linear product.
Operation: 3(-2) → -6
After: −6
Why: The original minus sign still stands before this whole product.
Evaluate the separate linear product.
Step 6
Before: 4−(−6)
Action: Apply the original subtraction.
Operation: 4−(-6) → 4+6
After: 4+6
Why: Subtracting a negative is adding its opposite.
Apply the original subtraction.
Step 7
Before: 4+6
Action: Calculate the final output.
Operation: 4+6 → 10
After: 10
Why: Only now are the two evaluated parts combined.
Calculate the final output.
Step 8
Before: Label the result
Action: Write the original function statement.
Operation: Label the result → g(-2)=10
After: g(−2)=10
Why: This output corresponds to the stated input.
Write the original function statement.
1. Read the input. g(-2) → x=-2 Both appearances of x receive the same whole input.
2. Substitute both input occurrences. g(x)=x²−3x → g(-2)=(-2)²−3(-2) Keep the subtraction in the original rule visible.
3. Write the square as repeated multiplication. (-2)² → (-2)(-2) The sign is inside both factors.
4. Evaluate the square. (-2)(-2) → 4 Two negative factors produce a positive product.
5. Evaluate the separate linear product. 3(-2) → -6 The original minus sign still stands before this whole product.
6. Apply the original subtraction. 4−(-6) → 4+6 Subtracting a negative is adding its opposite.
7. Calculate the final output. 4+6 → 10 Only now are the two evaluated parts combined.
8. Write the original function statement. Label the result → g(-2)=10 This output corresponds to the stated input.
Answer: g(-2)=10
g(−2)=(−2)²−3(−2)=4−(−6)=4+6=10.
Subtracting a negative adds its opposite.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
p(x)=6x −1. Find p(1/2).
Hint 1
One half is the input.
Hint 2
Replace x by 1/2 in 6x − 1.
Hint 3
Six halves make three wholes. Subtract one.
Show the complete working
Step 1
Before: p(1/2)
Action: Read the supplied input.
Operation: p(1/2) → x=1/2
After: x=1/2
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: p(x)=6x−1
Action: Replace each occurrence of the input variable.
Operation: p(x)=6x−1 → p(1/2)=6(1/2) + −1
After: p(1/2)=6(1/2) + −1
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (6)×(1/2)
Action: Write the integer as a fraction and multiply numerators.
Operation: (6)×(1/2) → (6×1)/2=6/2
After: (6×1)/2=6/2
Why: The denominator counts the equal fractional parts.
Write the integer as a fraction and multiply numerators.
Step 4
Before: 6/2
Action: Divide the numerator by the denominator.
Operation: 6/2 → 3
After: 3
Why: This finishes the variable term before the constant is applied.
Divide the numerator by the denominator.
Step 5
Before: (3)+(−1)
Action: Apply the constant term.
Operation: (3)+(−1) → 2
After: 2
Why: Adding a negative constant is the same as subtracting its positive magnitude.
Apply the constant term.
Step 6
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → p(1/2)=2
After: p(1/2)=2
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
1. Read the supplied input. p(1/2) → x=1/2 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. p(x)=6x−1 → p(1/2)=6(1/2) + −1 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Write the integer as a fraction and multiply numerators. (6)×(1/2) → (6×1)/2=6/2 The denominator counts the equal fractional parts.
4. Divide the numerator by the denominator. 6/2 → 3 This finishes the variable term before the constant is applied.
5. Apply the constant term. (3)+(−1) → 2 Adding a negative constant is the same as subtracting its positive magnitude.
6. Use the original function name. Label the computed output → p(1/2)=2 The requested input has now passed through every operation in the rule.
Answer: p(1/2)=2
p(1/2)=6(1/2)−1=(6/2)−1=3−1=2.
Six halves make three.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
f(x)=3x+1. Decide whether f(2) means “find the input” or “find the output”, then solve.
Hint 1
The number inside f(2) is the input.
Hint 2
Substitute two into 3x + 1.
Hint 3
3 × 2 = 6. Add one and name the output.
Show the complete working
Step 1
Before: f(2)
Action: Decide which quantity is already given.
Operation: f(2) → Input 2 is known; find its output
After: Input 2 is known; find its output
Why: The bracket supplies the input. In contrast, f(x)=2 would specify an output.
Decide which quantity is already given.
Step 2
Before: f(2)
Action: Read the supplied input.
Operation: f(2) → x=2
After: x=2
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 3
Before: f(x)=3x+1
Action: Replace each occurrence of the input variable.
Operation: f(x)=3x+1 → f(2)=3(2) + 1
After: f(2)=3(2) + 1
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 4
Before: (3)×(2)
Action: Multiply the coefficient and the input.
Operation: (3)×(2) → 6
After: 6
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 5
Before: (6)+(1)
Action: Apply the constant term.
Operation: (6)+(1) → 7
After: 7
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 6
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(2)=7
After: f(2)=7
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
1. Decide which quantity is already given. f(2) → Input 2 is known; find its output The bracket supplies the input. In contrast, f(x)=2 would specify an output.
2. Read the supplied input. f(2) → x=2 The number inside the brackets is the input; the required result is the output.
3. Replace each occurrence of the input variable. f(x)=3x+1 → f(2)=3(2) + 1 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
4. Multiply the coefficient and the input. (3)×(2) → 6 Complete multiplication before addition or subtraction.
5. Apply the constant term. (6)+(1) → 7 The constant is added once after the variable terms have been evaluated.
6. Use the original function name. Label the computed output → f(2)=7 The requested input has now passed through every operation in the rule.
Answer: f(2)=7
The bracket gives input 2.
Find output:f(2)=3(2)+1=6+1=7.
It does not ask for 3x+1=2.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
For the same f(x)=3x+1, find x if f(x)=7.
Hint 1
Seven is the output, not the input.
Hint 2
Set 3x + 1 = 7 and subtract one on both sides.
Hint 3
3x = 6. Divide both sides by three and check.
Show the complete working
Step 1
Before: f(x)=7
Action: The output is known; the input is unknown.
Operation: f(x)=7 → 3x+1=7
After: 3x+1=7
Why: Set the rule equal to the known output. Do not substitute the output in place of the input.
The output is known; the input is unknown.
Step 2
Before: 3x + 1=7
Action: Apply the same inverse addition to both sides.
Operation: 3x + 1=7 → 3x + 1+(-1)=7+(-1)
After: 3x + 1+(−1)=7+(−1)
Why: Equality is preserved because both sides receive the same operation.
Apply − 1 to both sides
Step 3
Before: 1+(−1)
Action: Cancel the additive inverse pair.
Operation: 1+(-1) → 1+(-1)=0; 3x+0=7+(-1)
After: 1+(−1)=0; 3x+0=7+(−1)
Why: Opposite added terms become zero. The variable term remains.
Cancel the additive inverse pair.
Step 4
Before: 7+(−1)
Action: Calculate the right-hand side.
Operation: 7+(-1) → 6
After: 6
Why: Finish this arithmetic before dividing.
Calculate the right-hand side.
Step 5
Before: 3x=6
Action: Divide both sides by the nonzero coefficient.
Operation: 3x=6 → (3x)/(3)=(6)/(3)
After: (3x)/(3)=(6)/(3)
Why: The same nonzero divisor is applied to both sides.
Divide both sides; equal nonzero factors make one
Step 6
Before: (3/3)x=(6)/(3)
Action: Reduce the coefficient quotient to one.
Operation: (3/3)x=(6)/(3) → 1×x=(6)/(3)
After: 1×x=(6)/(3)
Why: 3/3=1, not zero; 3 is nonzero.
Reduce the coefficient quotient to one.
Step 7
Before: x=(6)/(3)
Action: Evaluate the quotient.
Operation: x=(6)/(3) → x=2
After: x=2
Why: Keep an exact fractional value if the division is not integral.
Evaluate the quotient.
Step 8
Before: x=2
Action: Substitute the candidate into the original rule.
Operation: x=2 → f(2)=(3)(2)+(1)
After: f(2)=(3)(2)+(1)
Why: Check the rule before applying any contextual restriction.
Substitute the candidate into the original rule.
Step 9
Before: (3)(2)
Action: Calculate the variable term.
Operation: (3)(2) → 6
After: 6
Why: The product recovers the right-hand side before the constant was removed.
Calculate the variable term.
Step 10
Before: 6+(1)
Action: Restore the constant and compare with the target.
Operation: 6+(1) → 7=7
After: 7=7
Why: The algebraic candidate produces the required output.
Restore the constant and compare with the target.
Step 11
Before: State the requested result
Action: Report the input with its meaning and unit.
Operation: State the requested result → x=2
After: x=2
Why: The final statement answers the original question after both the algebraic and domain checks.
Report the input with its meaning and unit.
1. The output is known; the input is unknown. f(x)=7 → 3x+1=7 Set the rule equal to the known output. Do not substitute the output in place of the input.
2. Apply the same inverse addition to both sides. 3x + 1=7 → 3x + 1+(-1)=7+(-1) Equality is preserved because both sides receive the same operation.
3. Cancel the additive inverse pair. 1+(-1) → 1+(-1)=0; 3x+0=7+(-1) Opposite added terms become zero. The variable term remains.
4. Calculate the right-hand side. 7+(-1) → 6 Finish this arithmetic before dividing.
5. Divide both sides by the nonzero coefficient. 3x=6 → (3x)/(3)=(6)/(3) The same nonzero divisor is applied to both sides.
6. Reduce the coefficient quotient to one. (3/3)x=(6)/(3) → 1×x=(6)/(3) 3/3=1, not zero; 3 is nonzero.
7. Evaluate the quotient. x=(6)/(3) → x=2 Keep an exact fractional value if the division is not integral.
8. Substitute the candidate into the original rule. x=2 → f(2)=(3)(2)+(1) Check the rule before applying any contextual restriction.
9. Calculate the variable term. (3)(2) → 6 The product recovers the right-hand side before the constant was removed.
10. Restore the constant and compare with the target. 6+(1) → 7=7 The algebraic candidate produces the required output.
11. Report the input with its meaning and unit. State the requested result → x=2 The final statement answers the original question after both the algebraic and domain checks.
Answer: x=2
3x + 1 = 7.
Subtract 1 on both sides:
3x + 1 − 1 = 7 − 1.
3x + 0 = 6; 3x = 6.
Divide both sides by 3:
(3 × x)/(3) = 6/(3).
(3/3) × x = 2; 1 × x = 2.
x = 2.
Check: 3 × (2) + 1 = 6 + 1 = 7.
The unknown input is two; the specified output was seven.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
h(x)=4x −9. Find x when h(x)=11.
Hint 1
Eleven is the given output.
Hint 2
Set 4x − 9 = 11 and add nine on both sides.
Hint 3
4x = 20. Divide both sides by four and check.
Show the complete working
Step 1
Before: h(x)=11
Action: The output is known; the input is unknown.
Operation: h(x)=11 → 4x−9=11
After: 4x−9=11
Why: Set the rule equal to the known output. Do not substitute the output in place of the input.
The output is known; the input is unknown.
Step 2
Before: 4x − 9=11
Action: Apply the same inverse addition to both sides.
Operation: 4x − 9=11 → 4x − 9+(9)=11+(9)
After: 4x − 9+(9)=11+(9)
Why: Equality is preserved because both sides receive the same operation.
Apply + 9 to both sides
Step 3
Before: −9+(9)
Action: Cancel the additive inverse pair.
Operation: -9+(9) → -9+(9)=0; 4x+0=11+(9)
After: −9+(9)=0; 4x+0=11+(9)
Why: Opposite added terms become zero. The variable term remains.
Cancel the additive inverse pair.
Step 4
Before: 11+(9)
Action: Calculate the right-hand side.
Operation: 11+(9) → 20
After: 20
Why: Finish this arithmetic before dividing.
Calculate the right-hand side.
Step 5
Before: 4x=20
Action: Divide both sides by the nonzero coefficient.
Operation: 4x=20 → (4x)/(4)=(20)/(4)
After: (4x)/(4)=(20)/(4)
Why: The same nonzero divisor is applied to both sides.
Divide both sides; equal nonzero factors make one
Step 6
Before: (4/4)x=(20)/(4)
Action: Reduce the coefficient quotient to one.
Operation: (4/4)x=(20)/(4) → 1×x=(20)/(4)
After: 1×x=(20)/(4)
Why: 4/4=1, not zero; 4 is nonzero.
Reduce the coefficient quotient to one.
Step 7
Before: x=(20)/(4)
Action: Evaluate the quotient.
Operation: x=(20)/(4) → x=5
After: x=5
Why: Keep an exact fractional value if the division is not integral.
Evaluate the quotient.
Step 8
Before: x=5
Action: Substitute the candidate into the original rule.
Operation: x=5 → h(5)=(4)(5)+(-9)
After: h(5)=(4)(5)+(−9)
Why: Check the rule before applying any contextual restriction.
Substitute the candidate into the original rule.
Step 9
Before: (4)(5)
Action: Calculate the variable term.
Operation: (4)(5) → 20
After: 20
Why: The product recovers the right-hand side before the constant was removed.
Calculate the variable term.
Step 10
Before: 20+(−9)
Action: Restore the constant and compare with the target.
Operation: 20+(-9) → 11=11
After: 11=11
Why: The algebraic candidate produces the required output.
Restore the constant and compare with the target.
Step 11
Before: State the requested result
Action: Report the input with its meaning and unit.
Operation: State the requested result → x=5
After: x=5
Why: The final statement answers the original question after both the algebraic and domain checks.
Report the input with its meaning and unit.
1. The output is known; the input is unknown. h(x)=11 → 4x−9=11 Set the rule equal to the known output. Do not substitute the output in place of the input.
2. Apply the same inverse addition to both sides. 4x − 9=11 → 4x − 9+(9)=11+(9) Equality is preserved because both sides receive the same operation.
3. Cancel the additive inverse pair. -9+(9) → -9+(9)=0; 4x+0=11+(9) Opposite added terms become zero. The variable term remains.
4. Calculate the right-hand side. 11+(9) → 20 Finish this arithmetic before dividing.
5. Divide both sides by the nonzero coefficient. 4x=20 → (4x)/(4)=(20)/(4) The same nonzero divisor is applied to both sides.
6. Reduce the coefficient quotient to one. (4/4)x=(20)/(4) → 1×x=(20)/(4) 4/4=1, not zero; 4 is nonzero.
7. Evaluate the quotient. x=(20)/(4) → x=5 Keep an exact fractional value if the division is not integral.
8. Substitute the candidate into the original rule. x=5 → h(5)=(4)(5)+(-9) Check the rule before applying any contextual restriction.
9. Calculate the variable term. (4)(5) → 20 The product recovers the right-hand side before the constant was removed.
10. Restore the constant and compare with the target. 20+(-9) → 11=11 The algebraic candidate produces the required output.
11. Report the input with its meaning and unit. State the requested result → x=5 The final statement answers the original question after both the algebraic and domain checks.
Answer: x=5
4x − 9 = 11.
Add 9 on both sides:
4x − 9 + 9 = 11 + 9.
4x + 0 = 20; 4x = 20.
Divide both sides by 4:
(4 × x)/(4) = 20/(4).
(4/4) × x = 5; 1 × x = 5.
x = 5.
Check: 4 × (5) − 9 = 20 − 9 = 11.
The original output is recovered.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
h(x)=−2x+3. Find x when h(x)=9.
Hint 1
Find the input producing output nine.
Hint 2
Set −2x + 3 = 9; subtract three on both sides.
Hint 3
−2x = 6. Divide both sides by −2; use the sign bridge if needed.
Show the complete working
Step 1
Before: h(x)=9
Action: The output is known; the input is unknown.
Operation: h(x)=9 → -2x+3=9
After: −2x+3=9
Why: Set the rule equal to the known output. Do not substitute the output in place of the input.
The output is known; the input is unknown.
Step 2
Before: −2x + 3=9
Action: Apply the same inverse addition to both sides.
Operation: -2x + 3=9 → -2x + 3+(-3)=9+(-3)
After: −2x + 3+(−3)=9+(−3)
Why: Equality is preserved because both sides receive the same operation.
Apply − 3 to both sides
Step 3
Before: 3+(−3)
Action: Cancel the additive inverse pair.
Operation: 3+(-3) → 3+(-3)=0; -2x+0=9+(-3)
After: 3+(−3)=0; −2x+0=9+(−3)
Why: Opposite added terms become zero. The variable term remains.
Cancel the additive inverse pair.
Step 4
Before: 9+(−3)
Action: Calculate the right-hand side.
Operation: 9+(-3) → 6
After: 6
Why: Finish this arithmetic before dividing.
Calculate the right-hand side.
Step 5
Before: −2x=6
Action: Divide both sides by the nonzero coefficient.
Operation: -2x=6 → (-2x)/(-2)=(6)/(-2)
After: (−2x)/(−2)=(6)/(−2)
Why: The same nonzero divisor is applied to both sides.
Divide both sides; equal nonzero factors make one
Step 6
Before: (−2/−2)x=(6)/(−2)
Action: Reduce the coefficient quotient to one.
Operation: (-2/-2)x=(6)/(-2) → 1×x=(6)/(-2)
After: 1×x=(6)/(−2)
Why: −2/−2=1, not zero; −2 is nonzero.
Reduce the coefficient quotient to one.
Step 7
Before: x=(6)/(−2)
Action: Evaluate the quotient.
Operation: x=(6)/(-2) → x=−3
After: x=−3
Why: Keep an exact fractional value if the division is not integral.
Evaluate the quotient.
Step 8
Before: x=−3
Action: Substitute the candidate into the original rule.
Operation: x=−3 → h(−3)=(-2)(−3)+(3)
After: h(−3)=(−2)(−3)+(3)
Why: Check the rule before applying any contextual restriction.
Substitute the candidate into the original rule.
Step 9
Before: (−2)(−3)
Action: Calculate the variable term.
Operation: (-2)(−3) → 6
After: 6
Why: The product recovers the right-hand side before the constant was removed.
Calculate the variable term.
Step 10
Before: 6+(3)
Action: Restore the constant and compare with the target.
Operation: 6+(3) → 9=9
After: 9=9
Why: The algebraic candidate produces the required output.
Restore the constant and compare with the target.
Step 11
Before: State the requested result
Action: Report the input with its meaning and unit.
Operation: State the requested result → x=−3
After: x=−3
Why: The final statement answers the original question after both the algebraic and domain checks.
Report the input with its meaning and unit.
1. The output is known; the input is unknown. h(x)=9 → -2x+3=9 Set the rule equal to the known output. Do not substitute the output in place of the input.
2. Apply the same inverse addition to both sides. -2x + 3=9 → -2x + 3+(-3)=9+(-3) Equality is preserved because both sides receive the same operation.
3. Cancel the additive inverse pair. 3+(-3) → 3+(-3)=0; -2x+0=9+(-3) Opposite added terms become zero. The variable term remains.
4. Calculate the right-hand side. 9+(-3) → 6 Finish this arithmetic before dividing.
5. Divide both sides by the nonzero coefficient. -2x=6 → (-2x)/(-2)=(6)/(-2) The same nonzero divisor is applied to both sides.
6. Reduce the coefficient quotient to one. (-2/-2)x=(6)/(-2) → 1×x=(6)/(-2) -2/-2=1, not zero; -2 is nonzero.
7. Evaluate the quotient. x=(6)/(-2) → x=−3 Keep an exact fractional value if the division is not integral.
8. Substitute the candidate into the original rule. x=−3 → h(−3)=(-2)(−3)+(3) Check the rule before applying any contextual restriction.
9. Calculate the variable term. (-2)(−3) → 6 The product recovers the right-hand side before the constant was removed.
10. Restore the constant and compare with the target. 6+(3) → 9=9 The algebraic candidate produces the required output.
11. Report the input with its meaning and unit. State the requested result → x=−3 The final statement answers the original question after both the algebraic and domain checks.
Answer: x=−3
−2x + 3 = 9.
Subtract 3 on both sides:
−2x + 3 − 3 = 9 − 3.
−2x + 0 = 6; −2x = 6.
Divide both sides by −2:
(−2 × x)/(−2) = 6/(−2).
(−2/−2) × x = −3; 1 × x = −3.
x = −3.
Check: −2 × (−3) + 3 = 6 + 3 = 9.
Dividing a positive number by a negative number gives a negative input.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
A learner writes f(−2)=−f(2) for every function. Test f(x)=x²+1.
Hint 1
Compare the two claimed values separately.
Hint 2
Compute f(−2) and f(2) using x² + 1.
Hint 3
Both outputs are five, but −f(2) is negative five. Decide whether the claim survives.
Show the complete working
Step 1
Before: f(−2)
Action: Read the supplied input.
Operation: f(−2) → x=−2
After: x=−2
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: f(x)=x²+1
Action: Replace each occurrence of the input variable.
Operation: f(x)=x²+1 → f(−2)=(−2)² + 1
After: f(−2)=(−2)² + 1
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (−2)²
Action: Write the square as a product.
Operation: (−2)² → (−2)×(−2)
After: (−2)×(−2)
Why: Both factors contain the whole input, including its sign.
Write the square as a product.
Step 4
Before: (−2)×(−2)
Action: Multiply the two identical inputs.
Operation: (−2)×(−2) → 4
After: 4
Why: Two negative factors give a positive product.
Multiply the two identical inputs.
Step 5
Before: (4)+(1)
Action: Apply the constant term.
Operation: (4)+(1) → 5
After: 5
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 6
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(−2)=5
After: f(−2)=5
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 7
Before: f(2)
Action: Read the supplied input.
Operation: f(2) → x=2
After: x=2
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 8
Before: f(x)=x²+1
Action: Replace each occurrence of the input variable.
Operation: f(x)=x²+1 → f(2)=(2)² + 1
After: f(2)=(2)² + 1
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 9
Before: (2)²
Action: Write the square as a product.
Operation: (2)² → (2)×(2)
After: (2)×(2)
Why: Both factors contain the whole input, including its sign.
Write the square as a product.
Step 10
Before: (2)×(2)
Action: Multiply the two identical inputs.
Operation: (2)×(2) → 4
After: 4
Why: A square is nonnegative.
Multiply the two identical inputs.
Step 11
Before: (4)+(1)
Action: Apply the constant term.
Operation: (4)+(1) → 5
After: 5
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 12
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(2)=5
After: f(2)=5
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 13
Before: −f(2)
Action: Apply the minus outside the already calculated output.
Operation: −f(2) → −(5)=−5
After: −(5)=−5
Why: An outside minus reverses the output sign; it does not change the input.
Apply the minus outside the already calculated output.
Step 14
Before: f(−2)=5; −f(2)=−5
Action: Compare the two instructions.
Operation: f(−2)=5; −f(2)=−5 → 5≠−5
After: 5≠−5
Why: This single function and input are a counterexample to a claim about every function.
Compare the two instructions.
1. Read the supplied input. f(−2) → x=−2 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. f(x)=x²+1 → f(−2)=(−2)² + 1 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Write the square as a product. (−2)² → (−2)×(−2) Both factors contain the whole input, including its sign.
4. Multiply the two identical inputs. (−2)×(−2) → 4 Two negative factors give a positive product.
5. Apply the constant term. (4)+(1) → 5 The constant is added once after the variable terms have been evaluated.
6. Use the original function name. Label the computed output → f(−2)=5 The requested input has now passed through every operation in the rule.
7. Read the supplied input. f(2) → x=2 The number inside the brackets is the input; the required result is the output.
8. Replace each occurrence of the input variable. f(x)=x²+1 → f(2)=(2)² + 1 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
9. Write the square as a product. (2)² → (2)×(2) Both factors contain the whole input, including its sign.
10. Multiply the two identical inputs. (2)×(2) → 4 A square is nonnegative.
11. Apply the constant term. (4)+(1) → 5 The constant is added once after the variable terms have been evaluated.
12. Use the original function name. Label the computed output → f(2)=5 The requested input has now passed through every operation in the rule.
13. Apply the minus outside the already calculated output. −f(2) → −(5)=−5 An outside minus reverses the output sign; it does not change the input.
14. Compare the two instructions. f(−2)=5; −f(2)=−5 → 5≠−5 This single function and input are a counterexample to a claim about every function.
Answer: The universal claim fails: f(−2)=5 but −f(2)=−5.
f(−2)=(−2)²+1=4+1=5. f(2)=2²+1=5, so −f(2)=−5.
Five is not −5; the claim fails for this rule.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
Does the table (2, 4),(3, 4),(4, 4) represent a function on these inputs?
Hint 1
Check each input separately.
Hint 2
Group rows for inputs two, three and four.
Hint 3
Each has output four only. Decide whether sharing an output violates the definition.
Action: Interpret each ordered pair as an association.
Operation: Input set [2, 3, 4]; pairs [(2, 4), (3, 4), (4, 4)] → The first coordinate is the input x; the second is its associated output y.
After: The first coordinate is the input x; the second is its associated output y.
Why: A relation is an association between elements of the two sets.
One relation in three representations
Step 2
Before: Input x=2
Action: Collect its distinct associated outputs.
Operation: Input x=2 → Outputs [4]; count 1
After: Outputs [4]; count 1
Why: Repeated identical pairs do not create a second distinct output; different inputs may share an output.
Collect its distinct associated outputs.
Step 3
Before: Input x=3
Action: Collect its distinct associated outputs.
Operation: Input x=3 → Outputs [4]; count 1
After: Outputs [4]; count 1
Why: Repeated identical pairs do not create a second distinct output; different inputs may share an output.
Collect its distinct associated outputs.
Step 4
Before: Input x=4
Action: Collect its distinct associated outputs.
Operation: Input x=4 → Outputs [4]; count 1
After: Outputs [4]; count 1
Why: Repeated identical pairs do not create a second distinct output; different inputs may share an output.
Collect its distinct associated outputs.
Step 5
Before: Apply exactly one output to every stated input
Action: State the decision and its reason.
Operation: Apply exactly one output to every stated input → Function on the stated input set; f(2)=4, f(3)=4, f(4)=4
After: Function on the stated input set; f(2)=4, f(3)=4, f(4)=4
Why: A single missing or multiple-output input breaks the function condition on the specified set.
State the decision and its reason.
1. Interpret each ordered pair as an association. Input set [2, 3, 4]; pairs [(2, 4), (3, 4), (4, 4)] → The first coordinate is the input x; the second is its associated output y. A relation is an association between elements of the two sets.
2. Collect its distinct associated outputs. Input x=2 → Outputs [4]; count 1 Repeated identical pairs do not create a second distinct output; different inputs may share an output.
3. Collect its distinct associated outputs. Input x=3 → Outputs [4]; count 1 Repeated identical pairs do not create a second distinct output; different inputs may share an output.
4. Collect its distinct associated outputs. Input x=4 → Outputs [4]; count 1 Repeated identical pairs do not create a second distinct output; different inputs may share an output.
5. State the decision and its reason. Apply exactly one output to every stated input → Function on the stated input set; f(2)=4, f(3)=4, f(4)=4 A single missing or multiple-output input breaks the function condition on the specified set.
Answer: Function on the stated input set; f(2)=4, f(3)=4, f(4)=4
Input 2→4, input 3→4, input 4→4.
Each input has exactly one output, so yes.
Equal outputs do not break the definition.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
Does (2, 4),(2, 7),(3, 4) represent a function? Name the exact conflict.
Hint 1
Look for one input with different outputs.
Hint 2
Compare the two rows that start with two.
Hint 3
Those outputs are four and seven. Apply exactly one output per input.
Action: Interpret each ordered pair as an association.
Operation: Input set [2, 3]; pairs [(2, 4), (2, 7), (3, 4)] → The first coordinate is the input x; the second is its associated output y.
After: The first coordinate is the input x; the second is its associated output y.
Why: A relation is an association between elements of the two sets.
One relation in three representations
Step 2
Before: Input x=2
Action: Collect its distinct associated outputs.
Operation: Input x=2 → Outputs [4, 7]; count 2
After: Outputs [4, 7]; count 2
Why: Repeated identical pairs do not create a second distinct output; different inputs may share an output.
Collect its distinct associated outputs.
Step 3
Before: Input x=3
Action: Collect its distinct associated outputs.
Operation: Input x=3 → Outputs [4]; count 1
After: Outputs [4]; count 1
Why: Repeated identical pairs do not create a second distinct output; different inputs may share an output.
Collect its distinct associated outputs.
Step 4
Before: Apply exactly one output to every stated input
Action: State the decision and its reason.
Operation: Apply exactly one output to every stated input → Not a function on the stated input set: input 2 has 2 distinct outputs
After: Not a function on the stated input set: input 2 has 2 distinct outputs
Why: A single missing or multiple-output input breaks the function condition on the specified set.
State the decision and its reason.
1. Interpret each ordered pair as an association. Input set [2, 3]; pairs [(2, 4), (2, 7), (3, 4)] → The first coordinate is the input x; the second is its associated output y. A relation is an association between elements of the two sets.
2. Collect its distinct associated outputs. Input x=2 → Outputs [4, 7]; count 2 Repeated identical pairs do not create a second distinct output; different inputs may share an output.
3. Collect its distinct associated outputs. Input x=3 → Outputs [4]; count 1 Repeated identical pairs do not create a second distinct output; different inputs may share an output.
4. State the decision and its reason. Apply exactly one output to every stated input → Not a function on the stated input set: input 2 has 2 distinct outputs A single missing or multiple-output input breaks the function condition on the specified set.
Answer: Not a function on the stated input set: input 2 has 2 distinct outputs
Input 2 has outputs 4 and 7.
One input cannot have two different outputs, so this is not a function of the stated input.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
A printer model is C(n)=2n+6 dollars for n whole posters. Find the cost of 5 posters.
Hint 1
Five posters is the input; dollars are the output.
Hint 2
Write C(5) = 2(5) + 6.
Hint 3
Five two-dollar posters cost ten dollars. Add the fixed fee once.
Show the complete working
Step 1
Before: C(5)
Action: Read the supplied input.
Operation: C(5) → n=5
After: n=5
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: C(n)=2n+6
Action: Replace each occurrence of the input variable.
Operation: C(n)=2n+6 → C(5)=2(5) + 6
After: C(5)=2(5) + 6
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (2)×(5)
Action: Multiply the coefficient and the input.
Operation: (2)×(5) → 10
After: 10
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 4
Before: (10)+(6)
Action: Apply the constant term.
Operation: (10)+(6) → 16
After: 16
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 5
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → C(5)=16
After: C(5)=16
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 6
Before: C(5)=16
Action: Interpret the output with its unit.
Operation: C(5)=16 → Five posters cost 16 dollars
After: Five posters cost 16 dollars
Why: The input is a whole poster count and the output is the order cost.
Interpret the output with its unit.
1. Read the supplied input. C(5) → n=5 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. C(n)=2n+6 → C(5)=2(5) + 6 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Multiply the coefficient and the input. (2)×(5) → 10 Complete multiplication before addition or subtraction.
4. Apply the constant term. (10)+(6) → 16 The constant is added once after the variable terms have been evaluated.
5. Use the original function name. Label the computed output → C(5)=16 The requested input has now passed through every operation in the rule.
6. Interpret the output with its unit. C(5)=16 → Five posters cost 16 dollars The input is a whole poster count and the output is the order cost.
Answer: 16 dollars
C(5)=2(5)+6=10+6=16 dollars.
Five two-dollar posters plus one six-dollar setup fee gives 16.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
Under C(n)=2n+6, how many posters give an exact 20-dollar order?
Hint 1
Twenty dollars is the output.
Hint 2
Set 2n + 6 = 20 and subtract six on both sides.
Hint 3
2n = 14. Divide both sides by two; check the input is whole.
Show the complete working
Step 1
Before: C(n)=20
Action: The output is known; the input is unknown.
Operation: C(n)=20 → 2n+6=20
After: 2n+6=20
Why: Set the rule equal to the known output. Do not substitute the output in place of the input.
The output is known; the input is unknown.
Step 2
Before: 2n + 6=20
Action: Apply the same inverse addition to both sides.
Operation: 2n + 6=20 → 2n + 6+(-6)=20+(-6)
After: 2n + 6+(−6)=20+(−6)
Why: Equality is preserved because both sides receive the same operation.
Apply − 6 to both sides
Step 3
Before: 6+(−6)
Action: Cancel the additive inverse pair.
Operation: 6+(-6) → 6+(-6)=0; 2n+0=20+(-6)
After: 6+(−6)=0; 2n+0=20+(−6)
Why: Opposite added terms become zero. The variable term remains.
Cancel the additive inverse pair.
Step 4
Before: 20+(−6)
Action: Calculate the right-hand side.
Operation: 20+(-6) → 14
After: 14
Why: Finish this arithmetic before dividing.
Calculate the right-hand side.
Step 5
Before: 2n=14
Action: Divide both sides by the nonzero coefficient.
Operation: 2n=14 → (2n)/(2)=(14)/(2)
After: (2n)/(2)=(14)/(2)
Why: The same nonzero divisor is applied to both sides.
Divide both sides; equal nonzero factors make one
Step 6
Before: (2/2)n=(14)/(2)
Action: Reduce the coefficient quotient to one.
Operation: (2/2)n=(14)/(2) → 1×n=(14)/(2)
After: 1×n=(14)/(2)
Why: 2/2=1, not zero; 2 is nonzero.
Reduce the coefficient quotient to one.
Step 7
Before: n=(14)/(2)
Action: Evaluate the quotient.
Operation: n=(14)/(2) → n=7
After: n=7
Why: Keep an exact fractional value if the division is not integral.
Evaluate the quotient.
Step 8
Before: n=7
Action: Substitute the candidate into the original rule.
Operation: n=7 → C(7)=(2)(7)+(6)
After: C(7)=(2)(7)+(6)
Why: Check the rule before applying any contextual restriction.
Substitute the candidate into the original rule.
Step 9
Before: (2)(7)
Action: Calculate the variable term.
Operation: (2)(7) → 14
After: 14
Why: The product recovers the right-hand side before the constant was removed.
Calculate the variable term.
Step 10
Before: 14+(6)
Action: Restore the constant and compare with the target.
Operation: 14+(6) → 20=20
After: 20=20
Why: The algebraic candidate produces the required output.
Restore the constant and compare with the target.
Step 11
Before: Candidate n=7
Action: Check the whole-number, nonnegative input condition.
Operation: Candidate n=7 → Permitted: a nonnegative whole number.
After: Permitted: a nonnegative whole number.
Why: A fractional number of whole items cannot be ordered.
Check the whole-number, nonnegative input condition.
Step 12
Before: State the requested result
Action: Report the input with its meaning and unit.
Operation: State the requested result → n=7 posters
After: n=7 posters
Why: The final statement answers the original question after both the algebraic and domain checks.
Report the input with its meaning and unit.
1. The output is known; the input is unknown. C(n)=20 → 2n+6=20 Set the rule equal to the known output. Do not substitute the output in place of the input.
2. Apply the same inverse addition to both sides. 2n + 6=20 → 2n + 6+(-6)=20+(-6) Equality is preserved because both sides receive the same operation.
3. Cancel the additive inverse pair. 6+(-6) → 6+(-6)=0; 2n+0=20+(-6) Opposite added terms become zero. The variable term remains.
4. Calculate the right-hand side. 20+(-6) → 14 Finish this arithmetic before dividing.
5. Divide both sides by the nonzero coefficient. 2n=14 → (2n)/(2)=(14)/(2) The same nonzero divisor is applied to both sides.
6. Reduce the coefficient quotient to one. (2/2)n=(14)/(2) → 1×n=(14)/(2) 2/2=1, not zero; 2 is nonzero.
7. Evaluate the quotient. n=(14)/(2) → n=7 Keep an exact fractional value if the division is not integral.
8. Substitute the candidate into the original rule. n=7 → C(7)=(2)(7)+(6) Check the rule before applying any contextual restriction.
9. Calculate the variable term. (2)(7) → 14 The product recovers the right-hand side before the constant was removed.
10. Restore the constant and compare with the target. 14+(6) → 20=20 The algebraic candidate produces the required output.
11. Check the whole-number, nonnegative input condition. Candidate n=7 → Permitted: a nonnegative whole number. A fractional number of whole items cannot be ordered.
12. Report the input with its meaning and unit. State the requested result → n=7 posters The final statement answers the original question after both the algebraic and domain checks.
Answer: n=7 posters
2n + 6 = 20.
Subtract 6 on both sides:
2n + 6 − 6 = 20 − 6.
2n + 0 = 14; 2n = 14.
Divide both sides by 2:
(2 × n)/(2) = 14/(2).
(2/2) × n = 7; 1 × n = 7.
n = 7.
Check: 2 × (7) + 6 = 14 + 6 = 20.
Seven whole posters are allowed.
The order costs exactly 20 dollars.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
Under the same printer model, can an exact 15-dollar order be bought?
Hint 1
An algebraic input must also be a whole poster count.
Hint 2
Set 2n + 6 = 15 and subtract six on both sides.
Hint 3
2n = 9. Divide both sides by two, then decide whether that exact input is allowed.
Show the complete working
Step 1
Before: C(n)=15
Action: The output is known; the input is unknown.
Operation: C(n)=15 → 2n+6=15
After: 2n+6=15
Why: Set the rule equal to the known output. Do not substitute the output in place of the input.
The output is known; the input is unknown.
Step 2
Before: 2n + 6=15
Action: Apply the same inverse addition to both sides.
Operation: 2n + 6=15 → 2n + 6+(-6)=15+(-6)
After: 2n + 6+(−6)=15+(−6)
Why: Equality is preserved because both sides receive the same operation.
Apply − 6 to both sides
Step 3
Before: 6+(−6)
Action: Cancel the additive inverse pair.
Operation: 6+(-6) → 6+(-6)=0; 2n+0=15+(-6)
After: 6+(−6)=0; 2n+0=15+(−6)
Why: Opposite added terms become zero. The variable term remains.
Cancel the additive inverse pair.
Step 4
Before: 15+(−6)
Action: Calculate the right-hand side.
Operation: 15+(-6) → 9
After: 9
Why: Finish this arithmetic before dividing.
Calculate the right-hand side.
Step 5
Before: 2n=9
Action: Divide both sides by the nonzero coefficient.
Operation: 2n=9 → (2n)/(2)=(9)/(2)
After: (2n)/(2)=(9)/(2)
Why: The same nonzero divisor is applied to both sides.
Divide both sides; equal nonzero factors make one
Step 6
Before: (2/2)n=(9)/(2)
Action: Reduce the coefficient quotient to one.
Operation: (2/2)n=(9)/(2) → 1×n=(9)/(2)
After: 1×n=(9)/(2)
Why: 2/2=1, not zero; 2 is nonzero.
Reduce the coefficient quotient to one.
Step 7
Before: n=(9)/(2)
Action: Evaluate the quotient.
Operation: n=(9)/(2) → n=9/2
After: n=9/2
Why: Keep an exact fractional value if the division is not integral.
Evaluate the quotient.
Step 8
Before: n=9/2
Action: Substitute the candidate into the original rule.
Operation: n=9/2 → C(9/2)=(2)(9/2)+(6)
After: C(9/2)=(2)(9/2)+(6)
Why: Check the rule before applying any contextual restriction.
Substitute the candidate into the original rule.
Step 9
Before: (2)(9/2)
Action: Calculate the variable term.
Operation: (2)(9/2) → 9
After: 9
Why: The product recovers the right-hand side before the constant was removed.
Calculate the variable term.
Step 10
Before: 9+(6)
Action: Restore the constant and compare with the target.
Operation: 9+(6) → 15=15
After: 15=15
Why: The algebraic candidate produces the required output.
Restore the constant and compare with the target.
Step 11
Before: Candidate n=9/2
Action: Check the whole-number, nonnegative input condition.
Operation: Candidate n=9/2 → Not permitted: this input is not a nonnegative whole number.
After: Not permitted: this input is not a nonnegative whole number.
Why: A fractional number of whole items cannot be ordered.
Check the whole-number, nonnegative input condition.
Step 12
Before: C(4)=(2)(4)+(6)
Action: Multiply the whole-item count by the per-item amount.
Operation: C(4)=(2)(4)+(6) → 8+(6)
After: 8+(6)
Why: This is a permitted neighbouring count.
Multiply the whole-item count by the per-item amount.
Step 13
Before: 8+(6)
Action: Add the fixed amount.
Operation: 8+(6) → C(4)=14 dollars
After: C(4)=14 dollars
Why: Compare this attainable cost with the required exact cost.
Add the fixed amount.
Step 14
Before: C(5)=(2)(5)+(6)
Action: Multiply the whole-item count by the per-item amount.
Operation: C(5)=(2)(5)+(6) → 10+(6)
After: 10+(6)
Why: This is a permitted neighbouring count.
Multiply the whole-item count by the per-item amount.
Step 15
Before: 10+(6)
Action: Add the fixed amount.
Operation: 10+(6) → C(5)=16 dollars
After: C(5)=16 dollars
Why: Compare this attainable cost with the required exact cost.
Add the fixed amount.
Step 16
Before: Both neighbouring whole counts miss the target
Action: State the contextual conclusion.
Operation: Both neighbouring whole counts miss the target → No exact 15-dollar order is available.
After: No exact 15-dollar order is available.
Why: Every additional item changes the cost by the positive fixed per-item amount; no whole count lies between the neighbours.
State the contextual conclusion.
Step 17
Before: State the requested result
Action: Report the input with its meaning and unit.
Operation: State the requested result → No exact 15-dollar order is available.
After: No exact 15-dollar order is available.
Why: The final statement answers the original question after both the algebraic and domain checks.
Report the input with its meaning and unit.
1. The output is known; the input is unknown. C(n)=15 → 2n+6=15 Set the rule equal to the known output. Do not substitute the output in place of the input.
2. Apply the same inverse addition to both sides. 2n + 6=15 → 2n + 6+(-6)=15+(-6) Equality is preserved because both sides receive the same operation.
3. Cancel the additive inverse pair. 6+(-6) → 6+(-6)=0; 2n+0=15+(-6) Opposite added terms become zero. The variable term remains.
4. Calculate the right-hand side. 15+(-6) → 9 Finish this arithmetic before dividing.
5. Divide both sides by the nonzero coefficient. 2n=9 → (2n)/(2)=(9)/(2) The same nonzero divisor is applied to both sides.
6. Reduce the coefficient quotient to one. (2/2)n=(9)/(2) → 1×n=(9)/(2) 2/2=1, not zero; 2 is nonzero.
7. Evaluate the quotient. n=(9)/(2) → n=9/2 Keep an exact fractional value if the division is not integral.
8. Substitute the candidate into the original rule. n=9/2 → C(9/2)=(2)(9/2)+(6) Check the rule before applying any contextual restriction.
9. Calculate the variable term. (2)(9/2) → 9 The product recovers the right-hand side before the constant was removed.
10. Restore the constant and compare with the target. 9+(6) → 15=15 The algebraic candidate produces the required output.
11. Check the whole-number, nonnegative input condition. Candidate n=9/2 → Not permitted: this input is not a nonnegative whole number. A fractional number of whole items cannot be ordered.
12. Multiply the whole-item count by the per-item amount. C(4)=(2)(4)+(6) → 8+(6) This is a permitted neighbouring count.
13. Add the fixed amount. 8+(6) → C(4)=14 dollars Compare this attainable cost with the required exact cost.
14. Multiply the whole-item count by the per-item amount. C(5)=(2)(5)+(6) → 10+(6) This is a permitted neighbouring count.
15. Add the fixed amount. 10+(6) → C(5)=16 dollars Compare this attainable cost with the required exact cost.
16. State the contextual conclusion. Both neighbouring whole counts miss the target → No exact 15-dollar order is available. Every additional item changes the cost by the positive fixed per-item amount; no whole count lies between the neighbours.
17. Report the input with its meaning and unit. State the requested result → No exact 15-dollar order is available. The final statement answers the original question after both the algebraic and domain checks.
Answer: No exact 15-dollar order is available.
2n + 6 = 15.
Subtract 6 on both sides:
2n + 6 − 6 = 15 − 6.
2n + 0 = 9; 2n = 9.
Divide both sides by 2:
(2 × n)/(2) = 9/(2).
(2/2) × n = 4.5; 1 × n = 4.5.
n = 4.5.
Check: 2 × (4.5) + 6 = 9 + 6 = 15.
The equation gives 4.5, but whole posters are required.
C(4) = 2(4) + 6 = 8 + 6 = 14 dollars.
C(5) = 2(5) + 6 = 10 + 6 = 16 dollars.
No whole input gives exactly 15 dollars.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
A simplified sensor model is V(t)=3t −2. t is seconds, 0≤t≤4; V is volts. Find V(4) and state whether t=6 is permitted.
Hint 1
The rule is restricted to times from zero to four seconds.
Hint 2
Substitute four: 3(4) − 2.
Hint 3
Finish 12 − 2, attach volts, then compare six seconds with the domain.
Show the complete working
Step 1
Before: V(4)
Action: Read the supplied input.
Operation: V(4) → t=4
After: t=4
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 2
Before: V(t)=3t−2
Action: Replace each occurrence of the input variable.
Operation: V(t)=3t−2 → V(4)=3(4) + −2
After: V(4)=3(4) + −2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 3
Before: (3)×(4)
Action: Multiply the coefficient and the input.
Operation: (3)×(4) → 12
After: 12
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 4
Before: (12)+(−2)
Action: Apply the constant term.
Operation: (12)+(−2) → 10
After: 10
Why: Adding a negative constant is the same as subtracting its positive magnitude.
Apply the constant term.
Step 5
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → V(4)=10
After: V(4)=10
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 6
Before: t=4
Action: Check the input before reporting the model output.
Operation: t=4 → 0≤4≤4; the endpoint is included
After: 0≤4≤4; the endpoint is included
Why: The ≤ symbols include both stated endpoints.
Check the input before reporting the model output.
Step 7
Before: V(4)=10
Action: Label the model output.
Operation: V(4)=10 → 10 volts at 4 seconds
After: 10 volts at 4 seconds
Why: The function output is a voltage, not another time.
Label the model output.
Step 8
Before: Proposed input t=6
Action: Compare it with the permitted upper bound.
Operation: Proposed input t=6 → 6>4; t=6 is outside 0≤t≤4
After: 6>4; t=6 is outside 0≤t≤4
Why: The supplied model is authorised only on its stated interval.
Compare it with the permitted upper bound.
Step 9
Before: The time lies outside the stated domain
Action: State the restriction.
Operation: The time lies outside the stated domain → t=6 is not permitted by this model
After: t=6 is not permitted by this model
Why: Do not use an extrapolated number as a valid output under the given model.
State the restriction.
1. Read the supplied input. V(4) → t=4 The number inside the brackets is the input; the required result is the output.
2. Replace each occurrence of the input variable. V(t)=3t−2 → V(4)=3(4) + −2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
3. Multiply the coefficient and the input. (3)×(4) → 12 Complete multiplication before addition or subtraction.
4. Apply the constant term. (12)+(−2) → 10 Adding a negative constant is the same as subtracting its positive magnitude.
5. Use the original function name. Label the computed output → V(4)=10 The requested input has now passed through every operation in the rule.
6. Check the input before reporting the model output. t=4 → 0≤4≤4; the endpoint is included The ≤ symbols include both stated endpoints.
7. Label the model output. V(4)=10 → 10 volts at 4 seconds The function output is a voltage, not another time.
8. Compare it with the permitted upper bound. Proposed input t=6 → 6>4; t=6 is outside 0≤t≤4 The supplied model is authorised only on its stated interval.
9. State the restriction. The time lies outside the stated domain → t=6 is not permitted by this model Do not use an extrapolated number as a valid output under the given model.
Answer: V(4)=10 volts; t=6 is not permitted.
V(4)=3(4)−2=12−2=10 volts.
Four lies in the stated interval. t=6 is outside it; the supplied model does not authorise a prediction there.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
The same sensor reports 7 volts. Find the time and check the permitted interval.
Hint 1
Seven volts is an output, so solve for time.
Hint 2
Set 3t − 2 = 7 and add two to both sides.
Hint 3
3t = 9. Divide both sides by three and check the time lies in the domain.
Show the complete working
Step 1
Before: V(t)=7
Action: The output is known; the input is unknown.
Operation: V(t)=7 → 3t−2=7
After: 3t−2=7
Why: Set the rule equal to the known output. Do not substitute the output in place of the input.
The output is known; the input is unknown.
Step 2
Before: 3t − 2=7
Action: Apply the same inverse addition to both sides.
Operation: 3t − 2=7 → 3t − 2+(2)=7+(2)
After: 3t − 2+(2)=7+(2)
Why: Equality is preserved because both sides receive the same operation.
Apply + 2 to both sides
Step 3
Before: −2+(2)
Action: Cancel the additive inverse pair.
Operation: -2+(2) → -2+(2)=0; 3t+0=7+(2)
After: −2+(2)=0; 3t+0=7+(2)
Why: Opposite added terms become zero. The variable term remains.
Cancel the additive inverse pair.
Step 4
Before: 7+(2)
Action: Calculate the right-hand side.
Operation: 7+(2) → 9
After: 9
Why: Finish this arithmetic before dividing.
Calculate the right-hand side.
Step 5
Before: 3t=9
Action: Divide both sides by the nonzero coefficient.
Operation: 3t=9 → (3t)/(3)=(9)/(3)
After: (3t)/(3)=(9)/(3)
Why: The same nonzero divisor is applied to both sides.
Divide both sides; equal nonzero factors make one
Step 6
Before: (3/3)t=(9)/(3)
Action: Reduce the coefficient quotient to one.
Operation: (3/3)t=(9)/(3) → 1×t=(9)/(3)
After: 1×t=(9)/(3)
Why: 3/3=1, not zero; 3 is nonzero.
Reduce the coefficient quotient to one.
Step 7
Before: t=(9)/(3)
Action: Evaluate the quotient.
Operation: t=(9)/(3) → t=3
After: t=3
Why: Keep an exact fractional value if the division is not integral.
Evaluate the quotient.
Step 8
Before: t=3
Action: Substitute the candidate into the original rule.
Operation: t=3 → V(3)=(3)(3)+(-2)
After: V(3)=(3)(3)+(−2)
Why: Check the rule before applying any contextual restriction.
Substitute the candidate into the original rule.
Step 9
Before: (3)(3)
Action: Calculate the variable term.
Operation: (3)(3) → 9
After: 9
Why: The product recovers the right-hand side before the constant was removed.
Calculate the variable term.
Step 10
Before: 9+(−2)
Action: Restore the constant and compare with the target.
Operation: 9+(-2) → 7=7
After: 7=7
Why: The algebraic candidate produces the required output.
Restore the constant and compare with the target.
Step 11
Before: t=3
Action: Check the model’s stated input interval.
Operation: t=3 → 0≤3≤4
After: 0≤3≤4
Why: The time satisfies both bounds and is therefore permitted by this model.
Check the model’s stated input interval.
Step 12
Before: State the requested result
Action: Report the input with its meaning and unit.
Operation: State the requested result → t=3 seconds
After: t=3 seconds
Why: The final statement answers the original question after both the algebraic and domain checks.
Report the input with its meaning and unit.
1. The output is known; the input is unknown. V(t)=7 → 3t−2=7 Set the rule equal to the known output. Do not substitute the output in place of the input.
2. Apply the same inverse addition to both sides. 3t − 2=7 → 3t − 2+(2)=7+(2) Equality is preserved because both sides receive the same operation.
3. Cancel the additive inverse pair. -2+(2) → -2+(2)=0; 3t+0=7+(2) Opposite added terms become zero. The variable term remains.
4. Calculate the right-hand side. 7+(2) → 9 Finish this arithmetic before dividing.
5. Divide both sides by the nonzero coefficient. 3t=9 → (3t)/(3)=(9)/(3) The same nonzero divisor is applied to both sides.
6. Reduce the coefficient quotient to one. (3/3)t=(9)/(3) → 1×t=(9)/(3) 3/3=1, not zero; 3 is nonzero.
7. Evaluate the quotient. t=(9)/(3) → t=3 Keep an exact fractional value if the division is not integral.
8. Substitute the candidate into the original rule. t=3 → V(3)=(3)(3)+(-2) Check the rule before applying any contextual restriction.
9. Calculate the variable term. (3)(3) → 9 The product recovers the right-hand side before the constant was removed.
10. Restore the constant and compare with the target. 9+(-2) → 7=7 The algebraic candidate produces the required output.
11. Check the model’s stated input interval. t=3 → 0≤3≤4 The time satisfies both bounds and is therefore permitted by this model.
12. Report the input with its meaning and unit. State the requested result → t=3 seconds The final statement answers the original question after both the algebraic and domain checks.
Answer: t=3 seconds
3t − 2 = 7.
Add 2 on both sides:
3t − 2 + 2 = 7 + 2.
3t + 0 = 9; 3t = 9.
Divide both sides by 3:
(3 × t)/(3) = 9/(3).
(3/3) × t = 3; 1 × t = 3.
t = 3.
Check: 3 × (3) − 2 = 9 − 2 = 7.
The time is 3 seconds.
Check the domain: 0 ≤ 3 ≤ 4.
The voltage is 7 volts.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
A table shows only f(0) = 2 and f(1) = 3. Can you determine f(2) without a rule? Explain with two possible rules.
Hint 1
Only two input-output pairs are given.
Hint 2
Compare the candidate rules x + 2 and x² + 2 at inputs zero and one.
Hint 3
Both fit. Their values at two are 2 + 2 and 2² + 2. Decide whether the missing output is determined.
Show the complete working
Step 1
Before: Candidate rule f(x)=x+2
Action: Test all supplied rows before using the new input.
Operation: Candidate rule f(x)=x+2 → Check inputs 0 and 1, then calculate input 2
After: Check inputs 0 and 1, then calculate input 2
Why: A candidate must match both given values; otherwise it is irrelevant.
Test all supplied rows before using the new input.
Step 2
Before: f(0)
Action: Read the supplied input.
Operation: f(0) → x=0
After: x=0
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 3
Before: f(x)=x+2
Action: Replace each occurrence of the input variable.
Operation: f(x)=x+2 → f(0)=(0) + 2
After: f(0)=(0) + 2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 4
Before: (1)×(0)
Action: Multiply the coefficient and the input.
Operation: (1)×(0) → 0
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 5
Before: (0)+(2)
Action: Apply the constant term.
Operation: (0)+(2) → 2
After: 2
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 6
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(0)=2
After: f(0)=2
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 7
Before: f(1)
Action: Read the supplied input.
Operation: f(1) → x=1
After: x=1
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 8
Before: f(x)=x+2
Action: Replace each occurrence of the input variable.
Operation: f(x)=x+2 → f(1)=(1) + 2
After: f(1)=(1) + 2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 9
Before: (1)×(1)
Action: Multiply the coefficient and the input.
Operation: (1)×(1) → 1
After: 1
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 10
Before: (1)+(2)
Action: Apply the constant term.
Operation: (1)+(2) → 3
After: 3
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 11
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(1)=3
After: f(1)=3
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 12
Before: f(2)
Action: Read the supplied input.
Operation: f(2) → x=2
After: x=2
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 13
Before: f(x)=x+2
Action: Replace each occurrence of the input variable.
Operation: f(x)=x+2 → f(2)=(2) + 2
After: f(2)=(2) + 2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 14
Before: (1)×(2)
Action: Multiply the coefficient and the input.
Operation: (1)×(2) → 2
After: 2
Why: Complete multiplication before addition or subtraction.
Multiply the coefficient and the input.
Step 15
Before: (2)+(2)
Action: Apply the constant term.
Operation: (2)+(2) → 4
After: 4
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 16
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(2)=4
After: f(2)=4
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 17
Before: Candidate rule g(x)=x²+2
Action: Test all supplied rows before using the new input.
Operation: Candidate rule g(x)=x²+2 → Check inputs 0 and 1, then calculate input 2
After: Check inputs 0 and 1, then calculate input 2
Why: A candidate must match both given values; otherwise it is irrelevant.
Test all supplied rows before using the new input.
Step 18
Before: g(0)
Action: Read the supplied input.
Operation: g(0) → x=0
After: x=0
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 19
Before: g(x)=x²+2
Action: Replace each occurrence of the input variable.
Operation: g(x)=x²+2 → g(0)=(0)² + 2
After: g(0)=(0)² + 2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 20
Before: (0)²
Action: Write the square as a product.
Operation: (0)² → (0)×(0)
After: (0)×(0)
Why: Both factors contain the whole input, including its sign.
Write the square as a product.
Step 21
Before: (0)×(0)
Action: Multiply the two identical inputs.
Operation: (0)×(0) → 0
Why: A square is nonnegative.
Multiply the two identical inputs.
Step 22
Before: (0)+(2)
Action: Apply the constant term.
Operation: (0)+(2) → 2
After: 2
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 23
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → g(0)=2
After: g(0)=2
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 24
Before: g(1)
Action: Read the supplied input.
Operation: g(1) → x=1
After: x=1
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 25
Before: g(x)=x²+2
Action: Replace each occurrence of the input variable.
Operation: g(x)=x²+2 → g(1)=(1)² + 2
After: g(1)=(1)² + 2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 26
Before: (1)²
Action: Write the square as a product.
Operation: (1)² → (1)×(1)
After: (1)×(1)
Why: Both factors contain the whole input, including its sign.
Write the square as a product.
Step 27
Before: (1)×(1)
Action: Multiply the two identical inputs.
Operation: (1)×(1) → 1
After: 1
Why: A square is nonnegative.
Multiply the two identical inputs.
Step 28
Before: (1)+(2)
Action: Apply the constant term.
Operation: (1)+(2) → 3
After: 3
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 29
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → g(1)=3
After: g(1)=3
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 30
Before: g(2)
Action: Read the supplied input.
Operation: g(2) → x=2
After: x=2
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 31
Before: g(x)=x²+2
Action: Replace each occurrence of the input variable.
Operation: g(x)=x²+2 → g(2)=(2)² + 2
After: g(2)=(2)² + 2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 32
Before: (2)²
Action: Write the square as a product.
Operation: (2)² → (2)×(2)
After: (2)×(2)
Why: Both factors contain the whole input, including its sign.
Write the square as a product.
Step 33
Before: (2)×(2)
Action: Multiply the two identical inputs.
Operation: (2)×(2) → 4
After: 4
Why: A square is nonnegative.
Multiply the two identical inputs.
Step 34
Before: (4)+(2)
Action: Apply the constant term.
Operation: (4)+(2) → 6
After: 6
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 35
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → g(2)=6
After: g(2)=6
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 36
Before: Both rules match output 2 at input 0 and output 3 at input 1
Action: Compare their predictions at input 2.
Operation: Both rules match output 2 at input 0 and output 3 at input 1 → First rule gives 4; second rule gives 6
After: First rule gives 4; second rule gives 6
Why: The two specified rows do not distinguish these rules.
Compare their predictions at input 2.
Step 37
Before: Different valid rules give different next outputs
Action: State what the table determines.
Operation: Different valid rules give different next outputs → f(2) cannot be determined uniquely without more information
After: f(2) cannot be determined uniquely without more information
Why: The two possible rules are examples of compatible choices, not two simultaneous definitions of the same known function.
State what the table determines.
1. Test all supplied rows before using the new input. Candidate rule f(x)=x+2 → Check inputs 0 and 1, then calculate input 2 A candidate must match both given values; otherwise it is irrelevant.
2. Read the supplied input. f(0) → x=0 The number inside the brackets is the input; the required result is the output.
3. Replace each occurrence of the input variable. f(x)=x+2 → f(0)=(0) + 2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
4. Multiply the coefficient and the input. (1)×(0) → 0 Complete multiplication before addition or subtraction.
5. Apply the constant term. (0)+(2) → 2 The constant is added once after the variable terms have been evaluated.
6. Use the original function name. Label the computed output → f(0)=2 The requested input has now passed through every operation in the rule.
7. Read the supplied input. f(1) → x=1 The number inside the brackets is the input; the required result is the output.
8. Replace each occurrence of the input variable. f(x)=x+2 → f(1)=(1) + 2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
9. Multiply the coefficient and the input. (1)×(1) → 1 Complete multiplication before addition or subtraction.
10. Apply the constant term. (1)+(2) → 3 The constant is added once after the variable terms have been evaluated.
11. Use the original function name. Label the computed output → f(1)=3 The requested input has now passed through every operation in the rule.
12. Read the supplied input. f(2) → x=2 The number inside the brackets is the input; the required result is the output.
13. Replace each occurrence of the input variable. f(x)=x+2 → f(2)=(2) + 2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
14. Multiply the coefficient and the input. (1)×(2) → 2 Complete multiplication before addition or subtraction.
15. Apply the constant term. (2)+(2) → 4 The constant is added once after the variable terms have been evaluated.
16. Use the original function name. Label the computed output → f(2)=4 The requested input has now passed through every operation in the rule.
17. Test all supplied rows before using the new input. Candidate rule g(x)=x²+2 → Check inputs 0 and 1, then calculate input 2 A candidate must match both given values; otherwise it is irrelevant.
18. Read the supplied input. g(0) → x=0 The number inside the brackets is the input; the required result is the output.
19. Replace each occurrence of the input variable. g(x)=x²+2 → g(0)=(0)² + 2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
20. Write the square as a product. (0)² → (0)×(0) Both factors contain the whole input, including its sign.
21. Multiply the two identical inputs. (0)×(0) → 0 A square is nonnegative.
22. Apply the constant term. (0)+(2) → 2 The constant is added once after the variable terms have been evaluated.
23. Use the original function name. Label the computed output → g(0)=2 The requested input has now passed through every operation in the rule.
24. Read the supplied input. g(1) → x=1 The number inside the brackets is the input; the required result is the output.
25. Replace each occurrence of the input variable. g(x)=x²+2 → g(1)=(1)² + 2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
26. Write the square as a product. (1)² → (1)×(1) Both factors contain the whole input, including its sign.
27. Multiply the two identical inputs. (1)×(1) → 1 A square is nonnegative.
28. Apply the constant term. (1)+(2) → 3 The constant is added once after the variable terms have been evaluated.
29. Use the original function name. Label the computed output → g(1)=3 The requested input has now passed through every operation in the rule.
30. Read the supplied input. g(2) → x=2 The number inside the brackets is the input; the required result is the output.
31. Replace each occurrence of the input variable. g(x)=x²+2 → g(2)=(2)² + 2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
32. Write the square as a product. (2)² → (2)×(2) Both factors contain the whole input, including its sign.
33. Multiply the two identical inputs. (2)×(2) → 4 A square is nonnegative.
34. Apply the constant term. (4)+(2) → 6 The constant is added once after the variable terms have been evaluated.
35. Use the original function name. Label the computed output → g(2)=6 The requested input has now passed through every operation in the rule.
36. Compare their predictions at input 2. Both rules match output 2 at input 0 and output 3 at input 1 → First rule gives 4; second rule gives 6 The two specified rows do not distinguish these rules.
37. State what the table determines. Different valid rules give different next outputs → f(2) cannot be determined uniquely without more information The two possible rules are examples of compatible choices, not two simultaneous definitions of the same known function.
Answer: No unique f(2): x+2 gives 4 whereas x²+2 gives 6; both match the two supplied rows.
No.
Test f(x) = x + 2: f(0) = 0 + 2 = 2 and f(1) = 1 + 2 = 3, then f(2) = 2 + 2 = 4.
Both fit the known rows but disagree at input two.
Ask for the rule or more conditions.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
A machine squares its input, then adds two. Write its rule and compare the outputs for −3 and 3. Explain whether equal outputs break the function rule.
Hint 1
The order is square first, then add two.
Hint 2
Write the rule x² + 2; substitute −3 and 3 separately.
Hint 3
Both squares are nine. Add two to each; explain the one-output rule for each input.
Show the complete working
Step 1
Before: Square the input, then add two
Action: Write the rule in the stated order.
Operation: Square the input, then add two → f(x)=x²+2
After: f(x)=x²+2
Why: The entire input is squared before the constant is added.
Write the rule in the stated order.
Step 2
Before: f(−3)
Action: Read the supplied input.
Operation: f(−3) → x=−3
After: x=−3
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 3
Before: f(x)=x²+2
Action: Replace each occurrence of the input variable.
Operation: f(x)=x²+2 → f(−3)=(−3)² + 2
After: f(−3)=(−3)² + 2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 4
Before: (−3)²
Action: Write the square as a product.
Operation: (−3)² → (−3)×(−3)
After: (−3)×(−3)
Why: Both factors contain the whole input, including its sign.
Write the square as a product.
Step 5
Before: (−3)×(−3)
Action: Multiply the two identical inputs.
Operation: (−3)×(−3) → 9
After: 9
Why: Two negative factors give a positive product.
Multiply the two identical inputs.
Step 6
Before: (9)+(2)
Action: Apply the constant term.
Operation: (9)+(2) → 11
After: 11
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 7
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(−3)=11
After: f(−3)=11
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 8
Before: f(3)
Action: Read the supplied input.
Operation: f(3) → x=3
After: x=3
Why: The number inside the brackets is the input; the required result is the output.
Read the supplied input.
Step 9
Before: f(x)=x²+2
Action: Replace each occurrence of the input variable.
Operation: f(x)=x²+2 → f(3)=(3)² + 2
After: f(3)=(3)² + 2
Why: Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
Replace each occurrence of the input variable.
Step 10
Before: (3)²
Action: Write the square as a product.
Operation: (3)² → (3)×(3)
After: (3)×(3)
Why: Both factors contain the whole input, including its sign.
Write the square as a product.
Step 11
Before: (3)×(3)
Action: Multiply the two identical inputs.
Operation: (3)×(3) → 9
After: 9
Why: A square is nonnegative.
Multiply the two identical inputs.
Step 12
Before: (9)+(2)
Action: Apply the constant term.
Operation: (9)+(2) → 11
After: 11
Why: The constant is added once after the variable terms have been evaluated.
Apply the constant term.
Step 13
Before: Label the computed output
Action: Use the original function name.
Operation: Label the computed output → f(3)=11
After: f(3)=11
Why: The requested input has now passed through every operation in the rule.
Use the original function name.
Step 14
Before: f(−3)=11 and f(3)=11
Action: Compare the two outputs.
Operation: f(−3)=11 and f(3)=11 → Different inputs −3 and 3 share output 11
After: Different inputs −3 and 3 share output 11
Why: Squaring removes the difference between these opposite input signs.
Compare the two outputs.
Step 15
Before: Check the function condition input by input
Action: Count outputs for each input.
Operation: Check the function condition input by input → Each input still has exactly one output
After: Each input still has exactly one output
Why: The function condition prohibits two outputs for one input, not two inputs sharing one output.
Count outputs for each input.
1. Write the rule in the stated order. Square the input, then add two → f(x)=x²+2 The entire input is squared before the constant is added.
2. Read the supplied input. f(−3) → x=−3 The number inside the brackets is the input; the required result is the output.
3. Replace each occurrence of the input variable. f(x)=x²+2 → f(−3)=(−3)² + 2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
4. Write the square as a product. (−3)² → (−3)×(−3) Both factors contain the whole input, including its sign.
5. Multiply the two identical inputs. (−3)×(−3) → 9 Two negative factors give a positive product.
6. Apply the constant term. (9)+(2) → 11 The constant is added once after the variable terms have been evaluated.
7. Use the original function name. Label the computed output → f(−3)=11 The requested input has now passed through every operation in the rule.
8. Read the supplied input. f(3) → x=3 The number inside the brackets is the input; the required result is the output.
9. Replace each occurrence of the input variable. f(x)=x²+2 → f(3)=(3)² + 2 Keep a negative or fractional input inside a complete bracket; the constant remains unchanged.
10. Write the square as a product. (3)² → (3)×(3) Both factors contain the whole input, including its sign.
11. Multiply the two identical inputs. (3)×(3) → 9 A square is nonnegative.
12. Apply the constant term. (9)+(2) → 11 The constant is added once after the variable terms have been evaluated.
13. Use the original function name. Label the computed output → f(3)=11 The requested input has now passed through every operation in the rule.
14. Compare the two outputs. f(−3)=11 and f(3)=11 → Different inputs −3 and 3 share output 11 Squaring removes the difference between these opposite input signs.
15. Count outputs for each input. Check the function condition input by input → Each input still has exactly one output The function condition prohibits two outputs for one input, not two inputs sharing one output.
Answer: f(x)=x²+2; f(−3)=f(3)=11. Shared outputs do not violate the function rule.
Each input has one output, so equal outputs are allowed.
Review your response against these criteria
Uses the method appropriate to the question.
Shows all necessary arithmetic or reasoning.
Checks the original rule, units and permitted inputs.
Extension
Construct two different functions that both send 0 to 1 and 1 to 2 but send 2 to different outputs. Explain why two rows do not determine a rule.
Optional extension task. It does not change lesson access.
For f(x)=5x −4, find f(−2).
Compare your explanation
f(−2)=5(−2)−4=−10−4=−14. The bracket contains the input; multiply before subtracting.
A table lists (4, 7), (5, 7), (6, 8). Is it a function?
Compare your explanation
Yes. Each input has one output. Inputs four and five sharing output seven is allowed.
Whole notebooks cost C(n)=3n+5 dollars. Is an exact $16 order possible?
Compare your explanation
3n + 5 = 16. Subtract five on both sides: 3n + 5 − 5 = 16 − 5; 3n + 0 = 11; 3n = 11. Divide both sides by three: (3 × n)/3 = 11/3; (3/3) × n = 11/3; 1 × n = 11/3; n = 11/3. This is not whole, so no. C(3) = 3(3) + 5 = 9 + 5 = 14 dollars; C(4) = 3(4) + 5 = 12 + 5 = 17 dollars.
Only f(0)=3 and f(1)=4 are known. Explain why f(2) need not be five.
Compare your explanation
The rule f(x)=x+3 gives 0+3=3 and 1+3=4, then 2+3=5. The rule g(x)=x²+3 gives 0²+3=3 and 1²+3=4, then 2²+3=7. Both fit the known data, so the missing output is undetermined.
Viewing examples, using help and independent correct working are different kinds of evidence. Completing these pages does not automatically mark this skill as mastered.