This grade 8 linear worksheet answer key solves all 23 problems of the linear functions and modeling worksheet step by step.
Every problem from the linear functions worksheet is solved here step by step. Check the rate of change and the initial value first, then compare your equation with the one shown, and use the figures to see rise and run.
Grade 8 Linear Worksheet Answer Key: problems 1–5
Problem 1: Rate from a table
Pick two columns: \(m=\frac{7-4}{1-0}=3\). Check another pair: \(\frac{13-10}{3-2}=3\).
The rate of change is \(3\).
Problem 2: Read the equation
The equation has the form \(y=mx+b\).
(a) \(m=5\). (b) At \(x=0\), \(y=5(0)+8=8\).
Rate of change \(5\), initial value \(8\).
Problem 3: Slope between two points
\(m=\frac{12-3}{4-1}=\frac{9}{3}=3\).
The rate of change is \(3\).
Problem 4: A negative rate
\(m=\frac{1-9}{6-2}=\frac{-8}{4}=-2\).
The rate is negative, so the function is decreasing with rate of change \(-2\).
Problem 5: Linear or nonlinear?
Table A: the changes in \(y\) are \(2\), \(4\), \(8\). They are not equal, so A is nonlinear.
Table B: the changes are \(4\), \(4\), \(4\). They are equal, so B is linear with rate \(4\).
Grade 8 Linear Worksheet Answer Key: problems 6–10
Problem 6: Increasing or decreasing
Graph A falls from left to right in a straight line: decreasing and linear (rate \(-1\)).
Graph B rises and keeps getting steeper, so the rate of change is not constant: increasing and nonlinear.
Problem 7: Equation from a table
\(x\) increases by 2 and \(y\) decreases by 4 each time, so \(m=\frac{-4}{2}=-2\).
The column with \(x=0\) gives \(b=14\).
Equation: \(y=-2x+14\). Check: \(-2(6)+14=2\).
Problem 8: Equation from a graph
The line crosses the \(y\)-axis at \(-2\), so \(b=-2\).
From \((0,-2)\) to \((2,4)\): run 2, rise 6, so \(m=\frac{6}{2}=3\).
Equation: \(y=3x-2\).
Problem 9: Two points to equation
\(m=\frac{11-5}{3-1}=\frac{6}{2}=3\).
Use \((1,5)\): \(5=3(1)+b\), so \(b=2\).
Equation: \(y=3x+2\). Check: \(3(3)+2=11\).
Problem 10: A fractional rate
\(m=\frac{8-4}{9-3}=\frac{4}{6}=\frac{2}{3}\).
Use \((3,4)\): \(4=\frac{2}{3}(3)+b\), so \(4=2+b\) and \(b=2\).
Equation: \(y=\frac{2}{3}x+2\). Check: \(\frac{2}{3}(9)+2=8\).
Grade 8 Linear Worksheet Answer Key: problems 11–15
Problem 11: A point and a rate
Start with \(y=-2x+b\) and substitute \((4,10)\): \(10=-2(4)+b\).
\(10=-8+b\), so \(b=18\).
Equation: \(y=-2x+18\).
Problem 12: The burning candle
The initial value is 12 and the rate is \(-1.5\) inches per hour, so \(h=12-1.5t\).
At \(t=4\): \(h=12-1.5(4)=12-6=6\).
The candle is 6 inches tall after 4 hours.
Problem 13: Read a story graph
From 0 to 4 minutes the depth is increasing: \(\frac{8-0}{4-0}=2\) inches per minute.
From 4 to 7 minutes the depth is constant at 8 inches (rate 0).
From 7 to 9 minutes the depth is decreasing: \(\frac{0-8}{9-7}=-4\) inches per minute.
Rates: \(2\text{ in/min}\) and \(-4\text{ in/min}\).
Problem 14: Match the graph to the story
The trip has three parts: rising (moving away), level (resting), and falling to the start (going home).
Graph B never comes home and Graph C never rests, so they do not fit. In Graph A the falling part is steeper than the rising part, matching a faster walk home.
The answer is Graph A.
Problem 15: Sketch a trip
Three parts: from 0 to 10 minutes the graph rises to 0.5 mile; from 10 to 30 minutes it is level at 0.5 mile; from 30 to 40 minutes it falls back to 0.
The rates are \(\frac{0.5}{10}=0.05\) mile per minute, 0, and \(-0.05\) mile per minute.
Answer: rise, level, fall as sketched.
Grade 8 Linear Worksheet Answer Key: problems 16–20
Problem 16: Sketch a cooling curve
The graph starts high and falls steeply, then flattens out as it nears room temperature.
The steepness changes, so the rate of change is not constant: the function is nonlinear.
Problem 17: Taxi fare
(a) \(C=2m+3.5\).
(b) The rate is $2 per mile. The initial value is $3.50, the flat fee.
(c) \(C=2(9)+3.5=18+3.5=21.5\).
The ride costs $21.50.
Problem 18: Ethan's savings
(a) \(m=\frac{165-120}{6-3}=\frac{45}{3}=15\). Then \(120=15(3)+b\) gives \(b=75\). So \(y=15x+75\).
(b) He saves $15 per week and started with $75.
(c) \(15x+75=300\), so \(15x=225\) and \(x=15\).
He reaches $300 after 15 weeks.
Problem 19: A draining tank
(a) \(m=\frac{160-240}{7-3}=\frac{-80}{4}=-20\). Then \(240=-20(3)+b\) gives \(b=300\). So \(y=-20x+300\).
(b) The initial value is 300 gallons.
(c) \(-20x+300=0\), so \(x=15\). The tank is empty after 15 hours.
Problem 20: Kayak rental graph
(a) The line crosses the vertical axis at 8, so the initial value is $8. From \((0,8)\) to \((4,32)\): \(m=\frac{24}{4}=6\) dollars per hour.
(b) \(C=6h+8\).
(c) \(C=6(7)+8=42+8=50\). The cost is $50.
Grade 8 Linear Worksheet Answer Key: problems 21–23
Problem 21: Filling a pool
(a) For equal steps of 2 minutes, \(g\) changes by \(30\), \(30\), \(30\). The change is constant, so the function is linear with \(m=\frac{30}{2}=15\).
(b) \(50=15(2)+b\) gives \(b=20\). So \(g=15t+20\).
(c) \(g=15(25)+20=375+20=395\). About 395 gallons.
Problem 22: Two dog walkers
Noah: \(N\) changes by 7 each hour, so \(m=7\). From \(17=7(1)+b\), \(b=10\). So \(N=7h+10\).
For 10 hours: \(S=9(10)+12=102\) and \(N=7(10)+10=80\).
Sofia earns $22 more than Noah.
Problem 23: Area or perimeter?
Perimeter: \(P\) changes by 4 every time \(s\) increases by 1 (4, 8, 12, 16), so it is linear: \(P=4s\), rate 4 inches per inch, initial value 0.
Area: the changes are \(3\), \(5\), \(7\), which are not equal, so the area function is nonlinear.
How to use this grade 8 linear worksheet answer key
Read this grade 8 linear worksheet answer key after trying each problem yourself, and redo any problem you missed.
See every chapter on the Grade 8 math page.