Students can use the Spectrum Math Grade 8 Answer Key Chapter 3 Posttest as a quick guide to resolve any of their doubts.
Spectrum Math Grade 8 Chapter 3 Posttest Answers Key
Find the rate of change, or slope, for the situation.
Question 1.
A sales person earns commission based on total sales.
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.
\(\frac{Change in sale amount}{Change in commission}\) = \(\frac{250-100}{12.50-5}\) = \(\frac{150}{7.5}\) =20
\(\frac{Change in sale amount}{Change in commission}\)= \(\frac{500-250}{25-12.5}\) = \(\frac{250}{12.5}\) =20
The rate of change for this situation is constant.
Find the value of the variable in each equation.
Question 2.
a. 12 + n = 37 ____
Answer: n = 25
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:
12 + n = 37
Subracting 12 on both sides
12 + n – 12 = 37 – 12
n = 25
b. \(\frac{a}{3}\) = 15 ____
Answer: a = 45
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.
\(\frac{a}{3}\) = 15
First, undo the division by multiplying.
\(\frac{a}{3}\) x 3 = 15 x 3
a = 45
c. 3x + 2 = 56 _____
Answer: x =18
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.
3x + 2 = 56
First, undo the subtraction by adding.
3x + 2 – 2 = 56 – 2
3x = 54
Then, undo the multiplication by dividing.
3x ÷ 3 = 54 ÷ 3
x = 18
Question 3.
a. p – 17 = 26 ____
Answer: p = 43
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:
p – 17 = 26
Adding 17 on both sides
p – 17 + 17 = 26 + 17
p = 43
b. 9 – \(\frac{m}{4}\) = 4 ____
Answer: m = 20
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.
9 – \(\frac{m}{4}\) = 4
First, undo the addition by subtracting.
9 – \(\frac{m}{4}\) – 9 = 4 – 9
– \(\frac{m}{4}\) = -5
Then, undo the division by multiplying.
– \(\frac{m}{4}\) x 4 = -5 x 4
-m = -20
m = 20
c. q + 27 = 35 ____
Answer: q = 8
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:
q + 27 = 35
Subracting 27 on both sides
q + 27 – 27 = 35 – 27
q = 8
Question 4.
a. 7b – 5 = 44 _____
Answer: b = 7
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.
7b – 5 = 44
First, undo the subtraction by adding.
7b – 5 + 5 = 44 + 5
7b = 49
Then, undo the multiplication by dividing.
7b ÷ 7 = 49 ÷ 7
b = 7
b. n × 4 = 52 _____
Answer: n = 13
n × 4 = 52
undo the division by multiplying.
n × 4 ÷ 4 = 52 ÷ 4
n = 13
c. \(\frac{e}{8}\) + 4 = 11 ____
Answer: e = 56
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.
\(\frac{e}{8}\) + 4 = 11
First, undo the addition by subtracting.
\(\frac{e}{8}\) + 4 – 4 = 11 – 4
\(\frac{e}{8}\) = 7
Then, undo the division by multiplying.
\(\frac{e}{8}\) x 8 = 7 x 8
e = 56
Use the slope-intercept form of equations to draw lines on the grids below.
Question 5.
a.
y = –\(\frac{1}{3}\)x + 7
Answer:
From the given slope-intercept equation, find out the slope and y-intercept
Slope = –\(\frac{1}{3}\)
Y-intercept = (0,7)
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 (3, 6)
b. y = 3x + 1
Answer:
From the given slope-intercept equation, find out the slope and y-intercept
Slope = 3
Y-intercept = (0,1)
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 (2,7)
Complete the table. Then, graph the equation.
Question 6.
a. y = -x + 3
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 – 2y = 14
-10x + 7y = -25
x = ____, y = ____
Answer: x = -1; y =-5
-4x – 2y = 14
-10x + 7y = -25
Use inverse operations to isolate one variable on one side of the equation.
-4x – 2y = 14
-4x – 2y + 4x= 14 + 4x
– 2y = 14 + 4x
-y = 7 + 2x
y = -7 – 2x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
-10x + 7y = -25
-10x + 7(-7 – 2x) = -25
-10x -49 -14x= -25
-10x -14x= -25 + 49
-24x = 24
24x = -24
x = -1
Substitute the value of the variable in one of the equations and solve.
y = -7 – 2x
y = -7 – 2(-1)
y = -7 + 2
y = -5
Therefore, x = -1; y =-5
b.
y = \(\frac{3}{2}\)x – 2
y = x + 2
x = ____, y = ____
Answer: x =8; y =10
y = \(\frac{3}{2}\)x – 2
y = x + 2
Substitute the equation in place of the appropriate variable so there is only one variable in the new equation.
x + 2 = \(\frac{3}{2}\)x – 2
x – \(\frac{3}{2}\)x = – 2 -2
\(\frac{2 – 3}{2}\)x = – 4
\(\frac{-1}{2}\)x = – 4
\(\frac{1}{2}\)x = 4
x = 8
Substitute the value of the variable in one of the equations and solve.
y = x + 2
y = 8 + 2
y = 10
Therefore, x =8; y =10
Question 8.
a. 5x + y = -14
y = 1 – \(\frac{1}{2}\)x
x = ____, y = ____
Answer: x = \(\frac{-10}{3}\); y = \(\frac{-2}{3}\)
5x + y = -14
y = 1 – \(\frac{1}{2}\)x
Use inverse operations to isolate one variable on one side of the equation.
5x + y = -14
5x + y -5x= -14 -5x
y = -14 -5x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
y = 1 – \(\frac{1}{2}\)x
-14 -5x = 1 – \(\frac{1}{2}\)x
-5x + \(\frac{1}{2}\)x= 1 +14
\(\frac{-10 + 1}{2}\)x = 15
\(\frac{-9}{2}\)x = 15
-9x = 30
9x = -30
3x = -10
x = \(\frac{-10}{3}\)
Substitute the value of the variable in one of the equations and solve.
y = 1 – \(\frac{1}{2}\)x
y = 1 – \(\frac{1}{2}\){\(\frac{-10}{3}\)}
y = 1 – \(\frac{-10}{6}\)
y = \(\frac{6-10}{6}\)
y = \(\frac{-4}{6}\)
y = \(\frac{-2}{3}\)
Therefore, x = \(\frac{-10}{3}\); y = \(\frac{-2}{3}\)
b. -20y – 7x = -14
10y – 2x = -4
x = ____, y = ____
Answer: x = 2, y =0
-20y – 7x = -14
10y – 2x = -4
Use inverse operations to isolate one variable on one side of the equation.
-20y – 7x = -14
-20y – 7x + 7x= -14 + 7x
-20y = -14 + 7x
20y = 14 – 7x
y = \(\frac{14}{20}\) – \(\frac{7}{20}\)x
y = \(\frac{7}{10}\) – \(\frac{7}{20}\)x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
10y – 2x = -4
10{\(\frac{7}{10}\) – \(\frac{7}{20}\)x} – 2x = -4
7 – \(\frac{7}{2}\)x – 2x = -4
– \(\frac{7}{2}\)x – 2x = -4 -7
\(\frac{7}{2}\)x + 2x = 4 +7
\(\frac{7 + 4}{2}\)x= 11
\(\frac{11}{2}\)x= 11
11x= 22
x = \(\frac{22}{11}\)
x = 2
Substitute the value of the variable in one of the equations and solve.
y = \(\frac{7}{10}\) – \(\frac{7}{20}\)x
y = \(\frac{7}{10}\) – \(\frac{7}{20}\)2
y = \(\frac{7}{10}\) – \(\frac{7}{10}\)
y = 0
Therefore, x = 2, y =0
Use slope-intercept form to graph each system of equations and solve the system.
Question 9.
a.
-2x + 5y = 15
y = -x – 4
x: ____;
y: _____
Answer: x:-5;
y: 1
-2x + 5y = 15
y = -x – 4
Step 1: Isolate y in both equations by using inverse operations to create slope-intercept form.
y = 3 + \(\frac{2}{5}\)x
y = -x – 4
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.
(-5,1)
x:-5;
y: 1
b.
-x + 5y = 30
22x + y = 6
x: ____;
y: _____
Answer: x: 0;
y: 6
-x + 5y = 30
22x + y = 6
Step 1: Isolate y in both equations by using inverse operations to create slope-intercept form.
y = 6 + \(\frac{1}{5}\)x
y = 6 – 22x
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.
(0,6)
x: 0;
y: 6
Set up a system of equations to solve the word problem.
Question 10.
Kim scored 33 points at her basketball game with a combination of 2-point shots and 3-point shots. If she made a total of 15 baskets, how many of each kind of shot did Kim make? Use t to represent 2-point shots and x to represent 3-point shots.
Equation 1: _______
Equation 2: ______
t = ____; x = ____
Answer: Equation 1: t + x = 15
Equation 2: 2t + 3x = 33
t = 12; x = 3
Kim scored 33 points at her basketball game with a combination of 2-point shots and 3-point shots. She made a total of 15 baskets
Use t to represent 2-point shots and x to represent 3-point shots.
Equation 1: t + x = 15
Equation 2: 2t + 3x = 33
Use inverse operations to isolate one variable on one side of the equation.
t + x = 15
t + x -x= 15 -x
t = 15 -x
Substitute the new equation in place of the appropriate variable so there is only one variable in the new equation.
2t + 3x = 33
2(15 -x) + 3x = 33
30 – 2x + 3x = 33
-2x + 3x = 33 – 30
x = 3
Substitute the value of the variable in one of the equations and solve.
t = 15 -x
t = 15 – 3
t = 12
Therefore, x = 3, t = 12