Algebra 1 Linear Worksheet Answer Key – Free PDF

Contents

This algebra 1 linear worksheet answer key solves all 22 problems of the linear functions: slope-intercept, point-slope and standard form worksheet step by step.

Every solution below shows the setup, the arithmetic and a check. Compare your slopes, equations and graphs with these, and look for the step where a sign or a fraction went astray if an answer differs.

Algebra 1Common Core: F-IF.C.7a · F-LE.A.1a · F-LE.A.2 · A-CED.A.2

Algebra 1 Linear Worksheet Answer Key: problems 1–5

Problem 1: Slope from two points

$m=\frac{-6-9}{6-(-4)}=\frac{-15}{10}=-\frac{3}{2}$. The slope is $-\frac{3}{2}$.

Problem 2: A negative slope

$m=\frac{-7-5}{1-(-3)}=\frac{-12}{4}=-3$. The slope is $-3$. The line falls from left to right.

Problem 3: A fractional slope

$m=\frac{1-(-1)}{2-(-4)}=\frac{2}{6}=\frac{1}{3}$. The slope is $\frac{1}{3}$.

Problem 4: Horizontal and vertical lines

$m=\frac{5-5}{9-2}=\frac{0}{7}=0$. The slope is 0 (horizontal line, $y=5$).
$m=\frac{6-(-2)}{3-3}=\frac{8}{0}$, and division by zero is not defined. The slope is undefined (vertical line, $x=3$).

Problem 5: Read slope and intercept

The slope is $-\frac{3}{4}$ and the y-intercept is $5$, the point $(0,5)$.
$y=-\frac{3}{4}(8)+5=-6+5=-1$. So $y=-1$.

Algebra 1 Linear Worksheet Answer Key: problems 6–10

Problem 6: Equation from a graph

The line crosses the y-axis at $(0,2)$, so $b=2$. From $(0,2)$ to $(2,5)$, the rise is 3 and the run is 2, so $m=\frac{3}{2}$.

Check with $(-2,-1)$: $\frac{3}{2}(-2)+2=-1$. The equation is $y=\frac{3}{2}x+2$.

Problem 7: Slope and a point

Substitute into $y=mx+b$: $3=\frac{4}{5}(10)+b$, so $3=8+b$ and $b=-5$.

The equation is $y=\frac{4}{5}x-5$.

Problem 8: Point-slope form

Point-slope form: $y+1=3(x-2)$.

Distribute: $y+1=3x-6$. Subtract 1: $y=3x-7$.

Problem 9: Line through two points

Slope: $m=\frac{11-3}{5-1}=2$. Use point-slope with $(1,3)$: $y-3=2(x-1)$.

Distribute and add 3: $y=2x-2+3$, so $y=2x+1$. Check $(5,11)$: $2(5)+1=11$.

Problem 10: Standard to slope-intercept

Subtract $3x$: $4y=-3x+12$. Divide by 4: $y=-\frac{3}{4}x+3$.

The slope is $-\frac{3}{4}$ and the y-intercept is 3.

Algebra 1 Linear Worksheet Answer Key: problems 11–15

Problem 11: Slope-intercept to standard

Multiply every term by 5: $5y=-2x+20$. Add $2x$ to both sides: $2x+5y=20$.

Check $(0,4)$: $2(0)+5(4)=20$.

Problem 12: Intercepts and graph

Let $y=0$: $4x=24$, so $x=6$. Let $x=0$: $-6y=24$, so $y=-4$.

The intercepts are $(6,0)$ and $(0,-4)$. Draw the line through them.

The line has slope $\frac{2}{3}$.

Problem 13: Graph from slope-intercept

Plot the y-intercept $(0,1)$. The slope $-\frac{2}{3}$ means down 2, right 3: plot $(3,-1)$. Going the other way (up 2, left 3) gives $(-3,3)$.

Three points on the line.

Problem 14: Graph from point-slope

The equation is in point-slope form with $(x_{1},y_{1})=(-3,2)$ and $m=\frac{1}{2}$. Plot $(-3,2)$, then go up 1 and right 2 to $(-1,3)$, and again to $(1,4)$.

Three points on the line.

Problem 15: Parallel line

Parallel lines have equal slopes, so $m=4$. Point-slope form: $y+2=4(x-3)$.

Distribute: $y+2=4x-12$. Subtract 2: $y=4x-14$.

Algebra 1 Linear Worksheet Answer Key: problems 16–20

Problem 16: Perpendicular line

The slope of the given line is 2. The negative reciprocal is $-\frac{1}{2}$. Point-slope form: $y-1=-\frac{1}{2}(x-4)$.

Distribute: $y-1=-\frac{1}{2}x+2$. Add 1: $y=-\frac{1}{2}x+3$.

The two lines meet at a right angle at $(4,1)$.

Problem 17: Gym membership

The rate is $18 per month and the starting amount is $25: $C=18m+25$.
$C=18(9)+25=162+25=187$. The cost is $187.
$18m+25=277$, so $18m=252$ and $m=14$. It takes 14 months.

Problem 18: Data plan table

When $x$ increases by 2, $y$ increases by 11 each time (31, 42, 53, 64). The rate of change is constant at $\frac{11}{2}=5.5$, so the relationship is linear.
$m=5.5$. Use $(2,31)$: $31=5.5(2)+b$, so $b=20$. The equation is $y=5.5x+20$.
$y=5.5(15)+20=82.5+20=102.5$. It costs $102.50.

Problem 19: Road trip

$\frac{145-55}{3-1}=\frac{90}{2}=45$. The speed is 45 miles per hour.
$d=45t+b$. Using $(1,55)$: $55=45+b$, so $b=10$. The equation is $d=45t+10$ (Jamal starts 10 miles from home).
$d=45(5)+10=235$. The distance is 235 miles.

Problem 20: Temperature conversion

$m=\frac{212-32}{100-0}=\frac{180}{100}=\frac{9}{5}$. The y-intercept is 32 (the point $(0,32)$), so $F=\frac{9}{5}C+32$.
$F=\frac{9}{5}(25)+32=45+32=77$. That is $77^{\circ}\text{F}$.
$77=\frac{9}{5}C+32$, so $45=\frac{9}{5}C$ and $C=45\cdot\frac{5}{9}=25$. That is $25^{\circ}\text{C}$.

Algebra 1 Linear Worksheet Answer Key: problems 21–22

Problem 21: Notebooks and binders

If $y=0$: $3x=60$, so $x=20$: the intercept $(20,0)$ means 20 notebooks and no binders. If $x=0$: $5y=60$, so $y=12$: the intercept $(0,12)$ means 12 binders and no notebooks.
$3(5)+5y=60$, so $5y=45$ and $y=9$. She can buy 9 binders.

Problem 22: Two savings plans

Priya: $s=8w+20$. Ethan: $s=5w+50$.
Set them equal: $8w+20=5w+50$, so $3w=30$ and $w=10$. Then $s=8(10)+20=100$. After 10 weeks each has $100.

The graphs cross at $(10,100)$.

How to use this algebra 1 linear worksheet answer key

Read this algebra 1 linear worksheet answer key after trying each problem yourself, and redo any problem you missed.

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Linear Functions: Slope-Intercept, Point-Slope and Standard Form: Algebra 1 Math Worksheet Answer Key – PDF Download