Algebra 1 Writing Linear Equations and Functions Test Solutions and Scoring (PDF)

Algebra 1 math test solutions: Writing Linear Equations and Functions – Maths PDF

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This writing linear test answers for Algebra 1 covers writing linear equations and functions (Common Core F-LE.A.2, F-BF.A.1a, A-CED.A.2), with a free PDF to print.

Writing linear test answers (Algebra 1) – first page of the printable PDF

These are worked solutions for the writing linear equations test with a scoring guide for every question. Award points for correct method as well as for the final answer. The test totals 20 points, so add the question scores to find the final grade.

Algebra 1 · Common Core State Standards: F-LE.A.2, F-BF.A.1a, A-CED.A.2

Writing linear test answers: worked solutions

Question 1 – Slope-intercept form [2 pts]

a) $y = 5x – 1$.

b) $y = -\frac{1}{3}x + 4$.

Scoring: 1 point for each equation.

Question 2 – Point-slope [2 pts]

Point-slope: $y – 6 = 4(x + 1)$.

Distribute: $y – 6 = 4x + 4$, so $y = 4x + 10$.

$y – 6 = 4(x + 1)$; $y = 4x + 10$

Scoring: 1 point for each form.

Question 3 – Two points [2 pts]

Slope: $\frac{5 – (-3)}{6 – 2} = 2$. Substitute $(2, -3)$: $-3 = 2(2) + b$, so $b = -7$.

$y = 2x – 7$

Scoring: 1 point for the slope; 1 point for the equation.

Question 4 – Equation from a graph [2 pts]

Slope $\frac{3}{2}$, intercept -1.

The y-intercept is $-1$. From $(0, -1)$ to $(2, 2)$: rise 3, run 2.

$y = \frac{3}{2}x – 1$

Scoring: 1 point for the slope; 1 point for the intercept and equation.

Question 5 – Equation from a table [2 pts]

When $x$ increases by 2, $y$ increases by 8, so $m = 4$.

Substitute $(1, 5)$: $5 = 4(1) + b$, so $b = 1$.

$y = 4x + 1$

Scoring: 1 point for the slope; 1 point for the equation.

Question 6 – Parallel line [2 pts]

The slope is $-3$. Substitute $(2, 5)$: $5 = -3(2) + b$, so $b = 11$.

$y = -3x + 11$

Scoring: 1 point for the slope; 1 point for the equation.

Question 7 – Perpendicular line [3 pts]

Dashed line: perpendicular through (6, 2).

a) The negative reciprocal of 3 is $-\frac{1}{3}$.

b) $2 = -\frac{1}{3}(6) + b$, so $2 = -2 + b$ and $b = 4$.

$y = -\frac{1}{3}x + 4$

Scoring: part a, 1 point; part b, 1 point for substituting and 1 point for the equation.

Question 8 – Taxi fare model [3 pts]

a) $\frac{28 – 13}{10 – 4} = \frac{15}{6} = 2.5$ dollars per mile.

b) $13 = 2.5(4) + b$, so $b = 3$: $C = 2.5m + 3$.

c) $2.5(15) + 3 = 40.5$.

$2.50 per mile; $C = 2.5m + 3$; $40.50

Scoring: 1 point per part.

Question 9 – Standard form [2 pts]

Multiply by 5: $5y = -3x + 20$. Add $3x$ to both sides.

$3x + 5y = 20$

Scoring: 1 point for clearing the fraction; 1 point for the standard form.

Total score

Total

20 points

Key facts behind this writing linear test answers

Key facts

Slope from two points: m = (y₂ – y₁)/(x₂ – x₁).
Find b by substituting a point into y = mx + b.
Point-slope form: y – y₁ = m(x – x₁).
Parallel lines have equal slopes; perpendicular slopes are negative reciprocals.
In a model, the rate is the slope and the starting amount is the y-intercept.

How to use this writing linear test answers

Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.

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Algebra 1 Writing Linear Equations and Functions Test Solutions and Scoring (PDF)

Writing linear test answers: worked solutions

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How to use this writing linear test answers

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