Students can use the Spectrum Math Grade 8 Answer Key Chapter 3 Pretest as a quick guide to resolve any of their doubts.
Spectrum Math Grade 8 Chapter 3 Pretest Answers Key
Check What You Know
Linear Equations
Find the rate of change, or slope, for the situation.
Question 1.
A cell-phone company charges per minute for each call.
The rate of change, or slope, for this situation is ____.
Answer: The rate of change, or slope, for this situation is constant.
The slope of a line on a coordinate grid can be found by determining the rate of change.
To find rate of change, divide the rate of change for the dependent variable by the rate of change for the independent variable.
\(\frac{Change in time}{Change in cost}\) = \(\frac{15-10}{1.50-1.00}\) = \(\frac{5}{0.5}\) =10
\(\frac{Change in time}{Change in cost}\) = \(\frac{25-20}{2.50-2.00}\) = \(\frac{5}{0.5}\) =10
The rate of change for this situation is constant.
Find the value of the variable in each equation.
Question 2.
a. a + 13 = 27 ____
Answer: a = 14
The Addition and Subtraction Properties of Equality state that when the same number is added to both sides of on equation, the two sides remain equal. When the same number is subtracted from both sides of an equation, the two sides remain equal. Use these properties to determine the value of variables:
a + 13 = 27
Subtracting 13 on both sides
a + 13 – 13 = 27 -13
a = 14
b. 2n – 2 = 10 ____
Answer: n = 6
Some problems with variables require more than one step to solve. Use the properties of equality to undo each step and find the value of the variable.
2n – 2 = 10
First, undo the subtraction by adding.
2n – 2 + 2 = 10 + 2
2n = 12
Then, undo the multiplication by dividing.
2n ÷ 2 = 12 ÷ 2
n = 6
c. \(\frac{x}{4}\) + 4 = 12 _____
Answer: x = 32
Some problems with variables require more than one step to solve. Use the properties of equality to undo each step and find the value of the variable.
\(\frac{x}{4}\) + 4 = 12
First, undo the addition by subtracting.
\(\frac{x}{4}\) + 4 – 4 = 12 – 4
\(\frac{x}{4}\) = 8
Then, undo the division by multiplying.
\(\frac{x}{4}\) x 4 = 8 x 4
x = 32
Question 3.
a. 18 – 2p = 10 _______
Answer: p = 4
Some problems with variables require more than one step to solve. Use the properties of equality to undo each step and find the value of the variable.
18 – 2p = 10
First, undo the addition by subtracting.
18 – 2p – 18 = 10 – 18
-2p = -8
Then, undo the multiplication by dividing.
-2p ÷ -2= -8 ÷ -2
p = 4
b. \(\frac{n}{24}\) = 3 ________
Answer: n = 72
Some problems with variables require more than one step to solve. Use the properties of equality to undo each step and find the value of the variable.
\(\frac{n}{24}\) = 3
First, undo the division by multiplying.
\(\frac{n}{24}\) x 24 = 3 x 24
n = 72
c. n – 33 = 19 ________
Answer: n = 52
The Addition and Subtraction Properties of Equality state that when the same number is added to both sides of on equation, the two sides remain equal. When the same number is subtracted from both sides of an equation, the two sides remain equal. Use these properties to determine the value of variables:
n – 33 = 19
Adding 33 on both sides
n – 33 + 33 = 19 + 33
n = 52
Question 4.
a. f + 22 = 45 _____
Answer: f = 23
The Addition and Subtraction Properties of Equality state that when the same number is added to both sides of on equation, the two sides remain equal. When the same number is subtracted from both sides of an equation, the two sides remain equal. Use these properties to determine the value of variables:
f + 22 = 45
Subtracting 22 on both sides
f + 22 – 22 = 45 – 22
f = 23
b. \(\frac{r}{16}\) + 3 = 6 _____
Answer: r = 48
Some problems with variables require more than one step to solve. Use the properties of equality to undo each step and find the value of the variable.
\(\frac{r}{16}\) + 3 = 6
First, undo the addition by subtracting.
\(\frac{r}{16}\) + 3 – 3 = 6 – 3
\(\frac{r}{16}\) = 3
Then, undo the division by multiplying.
\(\frac{r}{16}\) x 16 = 3 x 16
r = 48
c. s × 4 + 2 = 46 _____
Answer: s = 11
Some problems with variables require more than one step to solve. Use the properties of equality to undo each step and find the value of the variable.
s × 4 + 2 = 46
First, undo the subtraction by adding.
s × 4 + 2 – 2 = 46-2
s × 4 = 44
Then, undo the multiplication by dividing.
4s ÷ 4 = 44 ÷ 4
s = 11
Use the slope-intercept form of equations to draw lines on the grids below.
Question 5.
a.
Answer:
From the given slope-intercept equation, find out the slope and y-intercept
Slope = 3
Y-intercept = (0,2)
Any line can be graphed using two points. select the other point using slope and y-intercept in order to plot the graph.
The other point would be (1, 5)
b.
Answer:
From the given slope-intercept equation, find out the slope and y-intercept
Slope = -2
Y-intercept = (0,8)
Any line can be graphed using two points. select the other point using slope and y-intercept in order to plot the graph.
The other point would be (1, 6)
Complete the table. Then, graph the equation.
Question 6.
Answer:
A linear equation is an equation that creates a straight line when graphed on a coordinate plane. To graph a linear equation, create a function table with at least 3 ordered pairs. Then, plot these ordered pairs on a coordinate plane.
Draw a line through the points. In the table are some points for this linear function:
Solve each system of equations.
Question 7.
a. -4x – 15y = -17
-x + 5y = -13
x = ____, y = _____
Answer: x = 8; y = 1
-4x – 15y = -17
-x + 5y = -13
Use inverse operations to isolate one variable on one side of the equation.
-4x – 15y = -17
-4x – 15y + 4x= -17 + 4x
– 15y = -17 + 4x
15y = 17 – 4x
y = \(\frac{17}{15}\) – \(\frac{4}{15}\)x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
-x + 5y = -13
-x + 5{\(\frac{17}{15}\) – \(\frac{4}{15}\)x} = -13
-x + \(\frac{17}{3}\) – \(\frac{4}{3}\)x = -13
-x – \(\frac{4}{3}\)x = -13 – \(\frac{17}{3}\)
x + \(\frac{4}{3}\)x = 13 + \(\frac{17}{3}\)
\(\frac{3 + 4}{3}\)x = \(\frac{39 + 17}{3}\)
\(\frac{7}{3}\)x = \(\frac{56}{3}\)
7x = 56
x = \(\frac{56}{7}\)
x = 8
Substitute the value of the variable in one of the equations and solve.
y = \(\frac{17}{15}\) – \(\frac{4}{15}\)x
y = \(\frac{17}{15}\) – \(\frac{4}{15}\)8
y = \(\frac{17}{15}\) – \(\frac{32}{15}\)
y = \(\frac{17 – 32}{15}\)
y = \(\frac{15}{15}\)
y = 1
Therefore, x = 8; y = 1
b. -x – 7y = 14
-4x – 14y = 28
x = ___, y = ____
Answer: x = 0; y = -2
-x – 7y = 14
-4x – 14y = 28
Use inverse operations to isolate one variable on one side of the equation.
-x – 7y = 14
-x – 7y + x= 14 + x
– 7y = 14 + x
7y = -14 – x
y = \(\frac{-14}{7}\) – \(\frac{1}{7}\)x
y = -2 – \(\frac{1}{7}\)x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
-4x – 14y = 28
-4x – 14{-2 – \(\frac{1}{7}\)x} = 28
-4x +28 + \(\frac{14}{7}\)x = 28
-4x + 2x = 28-28
-2x = 0
x = 0
Substitute the value of the variable in one of the equations and solve.
y = -2 – \(\frac{1}{7}\)x
y = -2 – \(\frac{1}{7}\)0
y = -2
Therefore, x = 0; y = -2
Question 8.
a. y = -1\(\frac{1}{8}\) – \(\frac{7}{8}\)x
-4x + 9y = -22
x = ___, y = ____
Answer: x =1; y = -2
y = -1\(\frac{1}{8}\) – \(\frac{7}{8}\)x
y = \(\frac{-9}{8}\) – \(\frac{7}{8}\)x
-4x + 9y = -22
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
-4x + 9y = -22
-4x + 9{\(\frac{-9}{8}\) – \(\frac{7}{8}\)x}= -22
-4x – \(\frac{81}{8}\) – \(\frac{63}{8}\)x= -22
-4x – \(\frac{63}{8}\)x= -22 + \(\frac{81}{8}\)
\(\frac{-32 – 63}{8}\)x = \(\frac{-176 + 81}{8}\)
\(\frac{-95}{8}\)x = \(\frac{-95}{8}\)
x = 1
Substitute the value of the variable in one of the equations and solve.
y = \(\frac{-9}{8}\) – \(\frac{7}{8}\)x
y = \(\frac{-9}{8}\) – \(\frac{7}{8}\)1
y = \(\frac{-9-7}{8}\)
y = \(\frac{-16}{8}\)
y = -2
Therefore, x =1; y = -2
b. y = \(\frac{1}{3}\)x + 2
5x + 4y = -30
x = ____, y = ____
Answer: x =-16; y = \(\frac{-10}{3}\)
y = \(\frac{1}{3}\)x + 2
5x + 4y = -30
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
5x + 4y = -30
5x + 4{\(\frac{1}{3}\)x + 2} = -30
5x + \(\frac{4}{3}\)x + 8= -30
5x + \(\frac{4}{3}\)x = -30 -8
\(\frac{15 +4}{8}\)x = -38
\(\frac{19}{8}\)x = -38
19x = -304
x = -16
Substitute the value of the variable in one of the equations and solve.
y = \(\frac{1}{3}\)x + 2
y = \(\frac{1}{3}\)(-16) + 2
y = \(\frac{-16 + 6}{3}\)
y = \(\frac{-10}{3}\)
Therefore, x =-16; y = \(\frac{-10}{3}\)
Use slope-intercept form to graph each system of equations and solve the system.
Question 9.
a. y = 2x + 3
y = 3
x: ____;
y: _____
Answer: x: 0;
y: 3
y = 2x + 3
y = 3
Step 1: Isolate y in both equations by using inverse operations to create slope-intercept form.
y = 2x + 3
y = 3
Step 2: Graph the first, line in the system using slope intercept form as a guide.
Step 3: Graph the second line in the system using slope-intercept form as a guide.
Step 4: Find the point of intersection to solve the equation system.
(0,3)
x: 0;
y: 3
b. y = –\(\frac{1}{2}\)x + 4
y = x – 2
x: ____;
y: _____
Answer: x: 4;
y: 2
y = –\(\frac{1}{2}\)x + 4
y = x – 2
Step 1: Isolate y in both equations by using inverse operations to create slope-intercept form.
y = –\(\frac{1}{2}\)x + 4
y = x – 2
Step 2: Graph the first, line in the system using slope intercept form as a guide.
Step 3: Graph the second line in the system using slope-intercept form as a guide.
Step 4: Find the point of intersection to solve the equation system.
(4,2)
x: 4;
y: 2
Set up a system of equations to solve the word problem.
Question 10.
At Billy’s school, 80 students come to school by bicycle or by car. Together, the vehicles they arrive to school in have 270 wheels. How many of each are used? Use b to represent the number of bicycles and c to represent the number of cars.
Equation 1: _______
Equation 2: _______
b = _____; c = _____
Answer: Equation 1: b+ c = 80
Equation 2: 2b + 4c = 270
b = 25; c = 55
At Billy’s school, 80 students come to school by bicycle or by car. Together, the vehicles they arrive to school in have 270 wheels.
Use b to represent the number of bicycles and c to represent the number of cars.
bicycle have 2 wheels and car have 4 wheels
Equation 1: b+ c = 80
Equation 2: 2b + 4c = 270
Use inverse operations to isolate one variable on one side of the equation.
b+ c = 80
b+ c -c= 80 -c
b = 80 -c
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
2b + 4c = 270
2(80 -c) + 4c = 270
160 – 2c + 4c = 270
-2c + 4c = 270 -160
2c = 110
c = 55
Substitute the value of the variable in one of the equations and solve.
b = 80 -c
b = 80 – 55
b = 25
Therefore, b = 25; c = 55