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Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations

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Go Math Grade 8 Chapter 8 Solving Systems of Linear Equations Answer Key

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Lesson 1: Solving Systems of Linear Equations by Graphing

Lesson 2: Solving Systems by Substitution

Lesson 3: Solving Systems by Elimination

Lesson 4: Solving Systems by Elimination with Multiplication

Lesson 5: Solving Solving Special Systems

Model Quiz

Review

Guided Practice – Solving Systems of Linear Equations by Graphing – Page No. 232

Solve each system by graphing.

Question 1.
\(\left\{\begin{array}{l}y=3 x-4 \\y=x+2\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 1
Type below:
______________

Answer:
grade 8 chapter 8 image 1

Explanation:
y = 3x – 4
y = x + 2
The solution of thr linear system of equations is the intersection point of the two equations.
(3, 5) is the solution of the system of equations.
If x = 3, y = 3(3) – 4 = 9 – 4 = 5; y = 3 + 2 = 5
5 = 5; True

Question 2.
\(\left\{\begin{array}{l}x-3 y=2 \\-3x+9y=-6\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 2
Type below:
______________

Answer:
grade 8 chapter 8 image 2
Infinitely many solutions

Explanation:
x – 3y = 2
-3x + 9y = -6
x – 3y – x = -x + 2
-3y = -x + 2
y = 1/3 . x – 2/3
-3x + 9y + 3x = 3x – 6
9y = 3x – 6
y = 3/9 . x – 6/9
y = 1/3 . x – 2/3
The solution of the linear system of equations is the intersection of the two equations.
Infinitely many solutions

Lesson 8.1 Solve Systems of Equations Algebraically Answer Key Question 3.
Mrs. Morales wrote a test with 15 questions covering spelling and vocabulary. Spelling questions (x) are worth 5 points and vocabulary questions (y) are worth 10 points. The maximum number of points possible on the test is 100.
a. Write an equation in slope-intercept form to represent the number of questions on the test.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 3
Type below:
______________

Answer:
y = -x + 15

Explanation:
Mrs. Morales wrote a test with 15 questions covering spelling and vocabulary. Spelling questions (x) are worth 5 points and vocabulary questions (y) are worth 10 points.
x + y = 15
x + y – x = -x + 15
y = -x + 15

Question 3.
b. Write an equation in slope-intercept form to represent the total number of points on the test.
Type below:
______________

Answer:
y = -1/2 . x + 10

Explanation:
The total number of points on the test is 100
5x + 10y = 100
5x + 10y – 5x = -5x + 100
10y = -5x + 100
y = -5/10 . x + 100/10
y = -1/2 . x + 10

Question 3.
c. Graph the solutions of both equations.
Type below:
______________

Answer:
grade 8 chapter 8 image 3

Question 3.
d. Use your graph to tell how many of each question type are on the test.
_________ spelling questions
_________ vocabulary questions

Answer:
10 spelling questions
5 vocabulary questions

ESSENTIAL QUESTION CHECK-IN

Question 4.
When you graph a system of linear equations, why does the intersection of the two lines represent the solution of the system?
Type below:
______________

Answer:
Solving a system of linear equations means finding the solutions that satisfy all the equations of that system. When we graph a system of linear equations, the intersection point lies on the line of each equation, which means that satisfies all the equations. Therefore, it is considered to be the solution to that system.

Solving Systems of Linear Equations by Graphing – Page No. 233

Question 5.
Vocabulary
A_________________ is a set of equations that have the same variables.
______________

Answer:
system of equations

Explanation:
A system of equations is a set of equations that have the same variables.

Lesson 8.1 Solving Systems of Linear Equations by Graphing Question 6.
Eight friends started a business. They will wear either a baseball cap or a shirt imprinted with their logo while working. They want to spend exactly $36 on the shirts and caps. The shirts cost $6 each and the caps cost $3 each.
a. Write a system of equations to describe the situation. Let x represent the number of shirts and let y represent the number of caps.
______________

Answer:
6x + 3y = 36

Explanation:
The sum of caps and shirts is 8. The total cost of caps and shirts is $36.
x + y = 8
6x + 3y = 36

Question 6.
b. Graph the system. What is the solution and what does it represent?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 4
Type below:
______________

Answer:
The solution is (4, 4)
grade 8 chapter 8 image 4

Explanation:
x + y – x = -x + 8
y = -x + 8
6x + 3y – 6x = -6x + 36
3y = -6x + 36
y = -6/2 . x + 36/3
y = -2x + 12
(4, 4). They should order 4 shirts and 4 caps.

Question 7.
Multistep The table shows the cost of bowling at two bowling alleys.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 5
a. Write a system of equations, with one equation describing the cost to bowl at Bowl-o-Rama and the other describing the cost to bowl at Bowling Pinz. For each equation, let x represent the number of games played and let y represent the total cost.
Type below:
______________

Answer:
y = 2.5x + 2
y = 2x + 4

Explanation:
Cost at Bowl-o-Rama => y = 2.5x + 2
Cost at Bowling Pinz => y = 2x + 4

Question 7.
b. Graph the system. What is the solution and what does it represent?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 6
Type below:
______________

Answer:
grade 8 chapter 8 image 5

Explanation:
The solution of the linear system of equations is the intersection of the two equations.
(4, 12)
When 4 games are played, the total cost is $12.

Solving Systems of Linear Equations by Graphing – Page No. 234

Question 8.
Multi-Step Jeremy runs 7 miles per week and increases his distance by 1 mile each week. Tony runs 3 miles per week and increases his distance by 2 miles each week. In how many weeks will Jeremy and Tony be running the same distance? What will that distance be?
Type below:
______________

Answer:
After 4 weeks Jeremy and Tony will be running the same distance and that distance would be 11 miles.

Explanation:
Multi-Step Jeremy runs 7 miles per week and increases his distance by 1 mile each week.
y = x + 7
Tony runs 3 miles per week and increases his distance by 2 miles each week.
y = 2x + 3
grade 8 chapter 8 image 6
The solution of the system of linear equation is (4, 11) which means that after 4 weeks Jeremy and Tony will be running the same distance and that distance would be 11 miles.

Lesson 1 Solve Systems of Equations Algebraically Answers Question 9.
Critical Thinking Write a real-world situation that could be represented by the system of equations shown below.
\(\left\{\begin{array}{l}y=4 x+10 \\y=3x+15\end{array}\right.\)
Type below:
______________

Answer:
The entry fee of the first gym is $10 and for every hour that you spend there, you pay an extra $4. If we denote with x the number of hours that somebody spends at the gym and with y the total cost is
y = 4x + 10
The entry fee of the second gym is $15 and for every hour that you spend there, you pay an extra $3. If we denote with x the number of hours that somebody spends at the gym and with y the total cost is
y = 3x + 15
y = 4x + 10
y = 3x + 15

FOCUS ON HIGHER ORDER THINKING

Question 10.
Multistep The table shows two options provided by a high-speed Internet provider.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson1: Solving Systems of Linear Equations by Graphing img 7
a. In how many months will the total cost of both options be the same? What will that cost be?
________ months
$ ________

Answer:
5 months
$ 200

Explanation:
Let y be the total cost after x month
y = 30x + 50
Let y be the total cost after x month
y = 40x
Substitute y = 40x in y = 30x + 50
40x = 30x + 50
40x – 30x = 50
10x = 50
x = 50/10
x = 5
The total cost of both options will be the same after 5 months. Total cost would be y = 40(5) = $200.

Question 10.
b. If you plan to cancel your Internet service after 9 months, which is the cheaper option? Explain.
______________

Answer:
When x = 9 months
y = 30(9) + 50 = $320
y = 40(9) = $360
$320 < $360
Option 1 is cheaper as the total cost is less for option 1

Lesson 1 Solve Systems of Equations by Graphing Question 11.
Draw Conclusions How many solutions does the system formed by x − y = 3 and ay − ax + 3a = 0 have for a nonzero number a? Explain.
Type below:
______________

Answer:
x – y = 3
ay – ax + 3a =0
ay – ax + 3a – 3a = 0 – 3a
ay – ax = – 3a
a(y – x) = -3a
y – x = -3
x – y = 3
Both equations are the same. The system of linear equations has infinitely many solutions.

Guided Practice – Solving Systems by Substitution – Page No. 240

Solve each system of linear equations by substitution.

Question 1.
\(\left\{\begin{array}{l}3x-2y=9 \\y=2x-7\end{array}\right.\)
x = ________
y = ________

Answer:
x = 5
y = 3

Explanation:
\(\left\{\begin{array}{l}3x-2y=9 \\y=2x-7\end{array}\right.\)
Substitute 2x – 7 in 3x – 2y = 9
3x – 2(2x – 7) = 9
3x – 4x + 14 = 9
-x + 14 = 9
-x + 14 – 14 = 9 – 14
-x = -5
x = -5/-1 = 5
y = 2(5) – 7 = 3
Solution is (5, 3)

Question 2.
\(\left\{\begin{array}{l}y=x-4 \\2x+y=5\end{array}\right.\)
x = ________
y = ________

Answer:
x = 3
y = -1

Explanation:
\(\left\{\begin{array}{l}y=x-4 \\2x+y=5\end{array}\right.\)
2x + x – 4 = 5
3x – 4 = 5
3x – 4 + 4 = 5 + 4
3x = 9
x = 9/3 = 3
y = 3 – 4 = -1
The solution is (3, -1)

8.2 Solving Systems by Substitution Answer Key Question 3.
\(\left\{\begin{array}{l}x+4y=6 \\y=-x+3\end{array}\right.\)
x = ________
y = ________

Answer:
x = 2
y = 1

Explanation:
\(\left\{\begin{array}{l}x+4y=6 \\y=-x+3\end{array}\right.\)
Substitute y = -x + 3 in x + 4y = 6
x + 4(-x + 3) = 6
x – 4x + 12 = 6
-3x + 12 = 6
-3x + 12 – 12 = 6 – 12
-3x = -6
x = -6/-3 = 2
y = -2 + 3 = 1
The solution is (2, 1)

Question 4.
\(\left\{\begin{array}{l}x+2y=6 \\x-y=3\end{array}\right.\)
x = ________
y = ________

Answer:
x = 4
y = 1

Explanation:
\(\left\{\begin{array}{l}x+2y=6 \\x-y=3\end{array}\right.\)
y = x – 3
Substitute y = x – 3 in x + 2y = 6
x + 2(x – 3) = 6
x + 2x – 6 = 6
3x = 12
x = 12/3
x = 4
4 – y = 3
-y = 3 – 4
-y = -1
y = 1
The solution is (4, 1)

Solve each system. Estimate the solution first.

Question 5.
\(\left\{\begin{array}{l}6x+y=4 \\x-4y=19\end{array}\right.\)
Estimate ______________
Solution ______________
Type below:
______________

Answer:
Estimate (2, -5)
Solution (1.4, -4.4)

Explanation:
\(\left\{\begin{array}{l}6x+y=4 \\x-4y=19\end{array}\right.\)
Let’s find the estimation by graphing the equations
Estimate: (2, -5)
grade 8 chapter 8 image 7
x = 4y + 19
6(4y + 19) + y = 4
24y + 114 + y = 4
25y + 114 = 4
25y = 4 – 114
25y = -110
y = -110/25
y = -4.4
x + 4(-4.4) = 19
x + 17.6 = 19
x = 19 – 17.6
x = 1.4
The solution is (1.4, -4.4)

8.2 Solving Systems by Substitution Question 6.
\(\left\{\begin{array}{l}x+2y=8 \\3x+2y=6\end{array}\right.\)
Estimate ______________
Solution ______________
Type below:
______________

Answer:
Estimate (-1, 5)
Solution (-1, 4.5)

Explanation:
\(\left\{\begin{array}{l}x+2y=8 \\3x+2y=6\end{array}\right.\)
Let’s find the estimation by graphing the equations
Estimate: (-1, 5)
grade 8 chapter 8 image 8
x = -2y + 8
Substitute the equation x = -2y + 8 in 3x + 2y = 6
3(-2y + 8) + 2y = 6
-6y + 24 + 2y = 6
-4y = 6 – 24
-4y = -18
y = -18/-4
y = 4.5
x + 2(4.5) = 8
x + 9 = 8
x = 8 – 9
x = -1
The solution is (-1, 4.5)

Question 7.
\(\left\{\begin{array}{l}3x+y=4 \\5x-y=22\end{array}\right.\)
Estimate ______________
Solution ______________
Type below:
______________

Answer:
Estimate (3, -6)
Solution (3.25, -5.75)

Explanation:
\(\left\{\begin{array}{l}3x+y=4 \\5x-y=22\end{array}\right.\)
Find the Estimation using graphing the equations.
Estimate: (3, -6)
grade 8 chapter 8 image 9
y = -3x + 4
Substitute y = -3x + 4 in 5x – y = 22
5x – (-3x + 4) = 22
5x + 3x -4 = 22
8x = 26
x = 26/8
x = 3.25
3(3.25) + y = 4
9.75 + y = 4
y = 4 – 9.75
y = -5.75
The solution is (3.25, -5.75)

Question 8.
\(\left\{\begin{array}{l}2x+7y=2 \\x+y=-1\end{array}\right.\)
Estimate ______________
Solution ______________
Type below:
______________

Answer:
Estimate (-2, 1)
Solution (-1.8, 0.8)

Explanation:
\(\left\{\begin{array}{l}2x+7y=2 \\x+y=-1\end{array}\right.\)
Find the Estimation using graphing the equations.
Estimate: (-2, 1)
grade 8 chapter 8 image 10
y = -x -1
Substitute y = -x – 1 in 2x + 7y = 2
2x + 7(-x – 1) = 2
2x – 7x -7 = 2
-5x = 2 + 7
-5x = 9
x = -9/5
x = -1.8
-1.8 + y = -1
y = -1 + 1.8
y = 0.8
The solution is (-1.8, 0.8)

Question 9.
Adult tickets to Space City amusement park cost x dollars. Children’s tickets cost y dollars. The Henson family bought 3 adult and 1 child tickets for $163. The Garcia family bought 2 adult and 3 child tickets for $174.
a. Write equations to represent the Hensons’ cost and the Garcias’ cost.
Hensons’ cost: ________________
Garcias’ cost:__________________
Type below:
______________

Answer:
Hensons’ cost: 3x + y = 163
Garcias’ cost: 2x + 3y = 174

Explanation:
Henson’s cost
3x + y = 163
Garcia’s cost
2x + 3y = 174

Question 9.
b. Solve the system.
adult ticket price: $ _________
Garcias’ cost: $ _________

Answer:
adult ticket price: $ 45
Garcias’ cost: $ 28

Explanation:
y = -3x + 163
Substitute y = -3x + 163 in 2x + 3y = 174
2x + 3(-3x + 163) = 174
2x -9x + 489 = 174
-7x = -315
x = -315/-7 = 45
3(45) + y = 163
135 + y = 163
y = 163 – 135
y = 28
Adult ticket price: $ 45
Garcias’ cost: $ 28

ESSENTIAL QUESTION CHECK-IN

Question 10.
How can you decide which variable to solve for first when you are solving a linear system by substitution?
Type below:
______________

Answer:
The variable with the unit coefficient should be solved first when solving a linear system by substitution.

8.2 Independent Practice – Solving Systems by Substitution – Page No. 241

Question 11.
Check for Reasonableness Zach solves the system
\(\left\{\begin{array}{l}x+y=-3 \\x-y=1\end{array}\right.\)
and finds the solution (1, -2). Use a graph to explain whether Zach’s solution is reasonable.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 2: Solving Systems by Substitution img 8
Type below:
______________

Answer:
grade 8 chapter 8 image 11

Explanation:
\(\left\{\begin{array}{l}x+y=-3 \\x-y=1\end{array}\right.\)
The x coordinate of the solution is negative, hence Zach’s solution is not reasonable.

Represent Real-World Problems Angelo bought apples and bananas at the fruit stand. He bought 20 pieces of fruit and spent $11.50. Apples cost $0.50 and bananas cost $0.75 each.
a. Write a system of equations to model the problem. (Hint: One equation will represent the number of pieces of fruit. A second equation will represent the money spent on the fruit.)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 2: Solving Systems by Substitution img 9
Type below:
______________

Answer:
x + y = 20
0.5x + 0.75y = 11.5

Explanation:
x + y = 20
0.5x + 0.75y = 11.5
where c is the number of Apples and y is the number of Bananas.

Question 12.
b. Solve the system algebraically. Tell how many apples and bananas Angelo bought.
________ apples
________ bananas

Answer:
14 apples
6 bananas

Explanation:
y = -x + 20
Substitute y = -x + 20 in 0.5x + 0.75y = 11.5
0.5x + 0.75(-x + 20) = 11.5
0.5x – 0.75x + 15 = 11.5
-0.25x + 15 = 11.5
-0.25x = 11.5 – 15
-0.25x = -3.5
x = -3.5/-0.25
x = 14
14 + y = 20
y = 6
Angelo bought 14 apples and 6 bananas.

Solving Systems by Substitution Lesson 8.2 Answer Key Question 13.
Represent Real-World Problems A jar contains n nickels and d dimes. There is a total of 200 coins in the jar. The value of the coins is $14.00. How many nickels and how many dimes are in the jar?
________ nickels
________ dimes

Answer:
120 nickels
80 dimes

Explanation:
A jar contains n nickels and d dimes. There is a total of 200 coins in the jar. The value of the coins is $14.00.
$14 = 1400 cents
n + d = 200
5n + 10d = 1400
d = -n + 200
5n + 10(-n + 200) = 1400
5n – 10n + 2000 = 1400
-5n = -600
n = -600/-5
n = 120
120 + d = 200
d = 200 – 120
d = 80
There are 120 nickles and 80 dimes in the jar.

Question 14.
Multistep The graph shows a triangle formed by the x-axis, the line 3x−2y=0, and the line x+2y=10. Follow these steps to find the area of the triangle.
a. Find the coordinates of point A by solving the system
\(\left\{\begin{array}{l}3x-2y=0 \\x-2y=10\end{array}\right.\)
Point A: ____________________
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 2: Solving Systems by Substitution img 10
Type below:
______________

Answer:
Point A: (2.5, 3.75)Coordinate of A is (2.5, 3.75)

Explanation:
\(\left\{\begin{array}{l}3x-2y=0 \\x-2y=10\end{array}\right.\)
x = -2y + 10
Substitute x = -2y + 10 in 3x – 2y = 0
3(-2y + 10) -2y = 0
-6y + 30 – 2y = 0
-8y = -30
y = -30/-8 = 3.75
x + 2(3.75) = 10
x + 7.5 = 10
x = 10 – 7.5
x = 2.5
Coordinate of A is (2.5, 3.75)

Question 14.
b. Use the coordinates of point A to find the height of the triangle.
height:__________________
height: \(\frac{â–¡}{â–¡}\) units

Answer:
height: 3.75
height: \(\frac{15}{4}\) units

Explanation:
The height of the triangle is the y coordinate of A
Height = 3.75

Question 14.
c. What is the length of the base of the triangle?
base:________________
base: ______ units

Answer:
base: 10 units

Explanation:
Length of the base = 10

Question 14.
d. What is the area of the triangle?
A = ______ \(\frac{â–¡}{â–¡}\) square units

Answer:
A = 18.75 square units
A = 18 \(\frac{3}{4}\) square units

Explanation:
Area of the triangle = 1/2 . Height . Base
Area = 1/2 . 3.75 . 10 = 18.75

Solving Systems by Substitution – Page No. 242

Question 15.
Jed is graphing the design for a kite on a coordinate grid. The four vertices of the kite are at A(−\(\frac{4}{3}\), \(\frac{2}{3}\)), B(\(\frac{14}{3}\), −\(\frac{4}{3}\)), C(\(\frac{14}{3}\), −\(\frac{16}{3}\)), and D(\(\frac{2}{3}\), −\(\frac{16}{3}\)). One kite strut will connect points A and C. The other will connect points B and D. Find the point where the struts cross.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 2: Solving Systems by Substitution img 11
Type below:
______________

Answer:
The struts cross as (8/3, 10/3)

Explanation:
1. From AC
Slope = (y2 – y1)/(x2 – x1) = [(-16/3)-(2/3)] ÷ [(14/3) – (-4/3)] = (-18/3) ÷ (18/3) = -1
y = mx + b
2/3 = -1(-4/3) + b
2/3 = 4/3 + b
1. From BD
Slope = (y2 – y1)/(x2 – x1) = [(-16/3)-(-4/3)] ÷ [(2/3) – (144/3)] = (-12/3) ÷ (-12/3) = 1
y = mx + b
-4/3 = 1(14/3) + b
-4/3 = 14/3 + b
-18/3 = b
-6 = b
y = mx + b
y = x -6
3. y = -x -2/3
y = x – 6
4. y = -x – 2/3
x – 6 = -x – 2/3
x = -x – 2/3 + 6
x = – x + 16/3
2x = 16/3
x = 16/6
x = 8/3
then y = x – 6
y = 8/3 – 18/3
y = -10/3
The struts cross as (8/3, 10/3)

FOCUS ON HIGHER ORDER THINKING

Question 16.
Analyze Relationships Consider the system
\(\left\{\begin{array}{l}6x-3y=15 \\x+3y=-8\end{array}\right.\)
Describe three different substitution methods that can be used to solve this system. Then solve the system.
Type below:
______________

Answer:
(1, -3) is the answer.

Explanation:
As there are three different substitution methods, we can write
Solve for y in the first equation, then substitute that value into the second equation.
Solve for x in the second equation, then substitute that value into the first equation.
Solve either equation for 3y, then substitute that value into the other equation.
From the Second method,
x + 3y = -8
x = -3y – 8
6x – 3y = 15
6 (-3y – 8) -3y = 15
-18y – 48 -3y = 15
-21y – 48 = 15
-21y = 63
y = -3
x + 3y = -8
x + 3(-3) = -8
x – 9 = -8
x = 1
(1, -3) is the answer.

Question 17.
Communicate Mathematical Ideas Explain the advantages, if any, that solving a system of linear equations by substitution has over solving the same system by graphing.
Type below:
______________

Answer:
The advantage of solving a system of linear equations by graphing is that it is relatively easy to do and requires very little algebra.

Question 18.
Persevere in Problem Solving Create a system of equations of the form
\(\left\{\begin{array}{l}Ax+By=C \\Dx+Ey=F\end{array}\right.\)
that has (7, −2) as its solution. Explain how you found the system.
Type below:
______________

Answer:
x + y = 5
x – y = 9
solves in :
x = (5+9)/2 = 7
y = 5-9)/2 = -2
A=1, B=2, C= 5
D=1, E= -1, F=9
x = 7
y = -2
IS a system (even if it is a trivial one) of equations so this answer would be acceptable.
The target for a system is to find it SOLUTION SET and not to conclude with x=a and y=b

Guided Practice – Solving Systems by Elimination – Page No. 248

Question 1.
Solve the system
\(\left\{\begin{array}{l}4x+3y=1 \\x-3y=-11\end{array}\right.\)
by adding.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 3: Solving Systems by Elimination img 12
Type below:
______________

Answer:
4x + 3y = 1
x – 3y = -11
Add the above two equations
4x + 3y = 1
+(x – 3y = -11)
Add to eliminate the variable y
5x + 0y = -10
Simplify and solve for x
5x = -10
Divide both sided by 5
x = -10/5 = -2
Substitute into one of the original equations and solve for y.
4(-2) + 3y = 1
-8 + 3y = 1
3y = 9
y = 9/3 = 3
So, (-2, 3) is the solution of the system.

Solve each system of equations by adding or subtracting.

Question 2.
\(\left\{\begin{array}{l}x+2y=-2 \\-3x+2y=-10\end{array}\right.\)
x = ________
y = ________

Answer:
x = 2
y = -2

Explanation:
\(\left\{\begin{array}{l}x+2y=-2 \\-3x+2y=-10\end{array}\right.\)
Subtract the equations
x + 2y = -2
-(-3x + 2y = -10)
y is eliminated as it has reversed coefficients. Solve for x
x + 2y + 3x – 2y = -2 + 10
4x = 8
x = 8/4 = 2
Substituting x in either of the equations to find y
2 + 2y = -2
2 + 2y -2 = -2 -2
2y = -4
y = -4/2 = -2
(2, -2) is the answer.

Lesson 3 Solve Systems of Equations Algebraically Answer Key Question 3.
\(\left\{\begin{array}{l}3x+y=23 \\3x-2y=8\end{array}\right.\)
(________ , ________)

Answer:
(6, 5)

Explanation:
\(\left\{\begin{array}{l}3x+y=23 \\3x-2y=8\end{array}\right.\)
Subtract the equations
3x + y = 23
-(3x – 2y = 8)
x is eliminated as it has reversed coefficients. Solve for y
3x + y – 3x + 2y = 23 – 8
3y = 15
y = 15/3 = 5
Substituting y in either of the equations to find x
3x + 5 = 23
3x + 5 – 5 = 23 – 5
3x = 18
x = 18/3 = 6
Solution is (6, 5)

Question 4.
\(\left\{\begin{array}{l}-4x-5y=7 \\3x+5y=-14\end{array}\right.\)
(________ , ________)

Answer:
(7, -7)

Explanation:
\(\left\{\begin{array}{l}-4x-5y=7 \\3x+5y=-14\end{array}\right.\)
Add the equations
-4x – 5y = 7
+(3x + 5y = -14)
y is eliminated as it has reversed coefficients. Solve for x
-4x -5y +3x + 5y = 7 -14
-x = -7
x = -7/-1 = 7
Substituting x in either of the equations to find y
3(7) + 5y = -14
21 + 5y -21 = -14 -21
5y = -35
y = -35/5 = -7
The answer is (7, -7)

Question 5.
\(\left\{\begin{array}{l}x-2y=-19 \\5x+2y=1\end{array}\right.\)
(________ , ________)

Answer:
(-3, 8)

Explanation:
\(\left\{\begin{array}{l}x-2y=-19 \\5x+2y=1\end{array}\right.\)
Add the equations
x – 2y = -19
+(5x + 2y = 1)
y is eliminated as it has reversed coefficients. Solve for x
x – 2y + 5x + 2y = -19 + 1
6x = -18
x = -18/6 = -3
Substituting x in either of the equations to find y
-3 -2y = -19
-3 -2y + 3 = -19 + 3
-2y = -16
y = -16/-2 = 8
The answer is (-3, 8)

Question 6.
\(\left\{\begin{array}{l}3x+4y=18 \\-2x+4y=8\end{array}\right.\)
(________ , ________)

Answer:
(2, 3)

Explanation:
\(\left\{\begin{array}{l}3x+4y=18 \\-2x+4y=8\end{array}\right.\)
Subtract the equations
3x + 4y = 18
-(-2x + 4y = 8)
y is eliminated as it has reversed coefficients. Solve for x
3x + 4y + 2x – 4y = 18 – 8
5x = 10
x = 10/5 = 2
Substituting x in either of the equations to find y
3(2) + 4y = 18
6 + 4y – 6 = 18 – 6
4y = 12
y = 12/4 =3
Solution is (2, 3)

Question 7.
\(\left\{\begin{array}{l}-5x+7y=11 \\-5x+3y=19\end{array}\right.\)
(________ , ________)

Answer:
(-5, -2)

Explanation:
\(\left\{\begin{array}{l}-5x+7y=11 \\-5x+3y=19\end{array}\right.\)
Subtract the equations
-5x + 7y = 11
-(-5x + 3y = 19)
x is eliminated as it has reversed coefficients. Solve for y
-5x + 7y + 5x – 3y = 11 – 19
4y = -8
y = -8/4 = -2
Substituting y in either of the equations to find x
-5x + 7(-2) = 11
-5x -14 + 14 = 11 + 14
-5x = 25
x = 25/-5 = -5
The solution is (-5, -2)

Question 8.
The Green River Freeway has a minimum and a maximum speed limit. Tony drove for 2 hours at the minimum speed limit and 3.5 hours at the maximum limit, a distance of 355 miles. Rae drove 2 hours at the minimum speed limit and 3 hours at the maximum limit, a distance of 320 miles. What are the two-speed limits?
a. Write equations to represent Tony’s distance and Rae’s distance.
Type below:
______________

Answer:
Tony’s distance: 2x + 3.5y = 355
Rae’s distance: 2x + 3y = 320
where x is the minimum speed and y is the maximum speed.

Question 8.
b. Solve the system.
minimum speed limit:______________
maximum speed limit______________
minimum speed limit: ________ mi/h
maximum speed limit: ________ mi/h

Answer:
minimum speed limit:55
maximum speed limit70
minimum speed limit: 55mi/h
Maximum speed limit: 70mi/h

Explanation:
Subtract the equations
2x + 3.5y = 355
-(2x + 3y = 320)
x is eliminated as it has reversed coefficients. Solve for y
2x + 3.5y – 2x – 3y = 355 – 320
0.5y = 35
y = 35/0.5 = 70
Substituting y in either of the equation to find x
2x + 3(70) = 320
2x + 210 – 210 = 320 – 210
2x = 110
x = 110/2 = 55
Minimum speed limit: 55 miles per hour
Maximum speed limit: 70 miles per hour

ESSENTIAL QUESTION CHECK-IN

Question 9.
Can you use addition or subtraction to solve any system? Explain.
________

Answer:
No. One of the variables should have the same coefficient in order to add or subtract the system.

8.3 Independent Practice – Solving Systems by Elimination – Page No. 249

Question 10.
Represent Real-World Problems Marta bought new fish for her home aquarium. She bought 3 guppies and 2 platies for a total of $13.95. Hank also bought guppies and platies for his aquarium. He bought 3 guppies and 4 platies for a total of $18.33. Find the price of a guppy and the price of a platy.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 3: Solving Systems by Elimination img 13
Guppy: $ ________
Platy: $ ________

Answer:
Guppy: $ 3.19
Platy: $ 2.19

Explanation:
3x + 2y = 13.95
3x + 4y = 18.33
where x is the unit price of guppy and y is the unit price of platy
Subtract the equations
3x + 2y = 13.95
-(3x + 4y = 18.33)
x is eliminated as it has reversed coefficients. Solve for y
3x + 2y – 3x – 4y = 13.95 – 18.33
-2y = -4.38
y = -4.38/-2 = 2.19
Substituting y in either of the equations to find x
3x + 2(2.19) = 13.95
3x + 4.38 – 4.38 = 13.95 – 4.38
3x = 9.57
x = 9.57/3 = 3.19
The price of a guppy is $3.19 and the price of platy is $2.19

Practice 8.3 Systems of Equations (Elimination) Answers Question 11.
Represent Real-World Problems The rule for the number of fish in a home aquarium is 1 gallon of water for each inch of fish length. Marta’s aquarium holds 13 gallons and Hank’s aquarium holds 17 gallons. Based on the number of fish they bought in Exercise 10, how long is a guppy and how long is a platy?
Length of a guppy = ________ inches
Length of a platy = ________ inches

Answer:
Length of a guppy = 3 inches
Length of a platy = 2 inches

Explanation:
3x + 2y = 13
3x + 4y = 17
where x is the length of guppy and y is the length of a platy
Subtract the equations
3x + 2y = 13
-(3x + 4y = 17)
x is eliminated as it has reversed coefficients. Solve for y
3x + 2y – 3x – 4y = 13 – 17
-2y = -4
y = -4/-2 = 2
Substituting y in either of the equation to find x
3x + 2(2) = 13
3x + 4 – 4 = 13 – 4
3x = 9
x = 9/3 = 3
The length of a guppy is 3 inches and price of a platy is 2 inches

Question 12.
Line m passes through the points (6, 1) and (2, -3). Line n passes through the points (2, 3) and (5, -6). Find the point of intersection of these lines.
Type below:
________________

Answer:
The intersection of these lines is (3.5, -1.5)

Explanation:
Find the slope of line m = (y2 – y1)/(x2 – x1) where (x2, y2) = (2, -3) and (x1, y1) = (6, 1)
Slope = (-3 -1)/(2 – 6) = -4/-4 = 1
Substitute the value of m and any of the given ordered pair (x, y) in the point-slope form of an equation: y – y1 = m(x – x1)
y – 1 = 1(x – 6)
y – 1 = x – 6
y = x – 6 + 1
x – y = 5
Find the slope of line n = (y2 – y1)/(x2 – x1) where (x2, y2) = (5, -6) and (x1, y1) = (2, 3)
Slope = (-6 -3)/(5 – 2) = -9/3 = -3
Substitute the value of m and any of the given ordered pair (x, y) in the point-slope form of an equation: y – y1 = m(x – x1)
y – 3 = -3(x – 2)
y – 3 = -3x + 6
y = -3x + 6 + 3
3x + y = 9
Add the equations
x – y = 5
+(3x + y = 9)
y is eliminated as it has reversed coefficients. Solve for x
x – y + 3x + y = 5 + 9
4x = 14
x = 14/4 = 3.5
Substituting x in either of the equations to find y
3.5 – y = 5
3.5 – y – 3.5 = 5 – 3.5
-y = 1.5
y = -1.5
The intersection of these lines is (3.5, -1.5)

Question 13.
Represent Real-World Problems Two cars got an oil change at the same auto shop. The shop charges customers for each quart of oil plus a flat fee for labor. The oil change for one car required 5 quarts of oil and cost $22.45. The oil change for the other car required 7 quarts of oil and cost $25.45. How much is the labor fee and how much is each quart of oil?
Labor fee: $ ________
Quart of oil: $ ________

Answer:
Labor fee: $ 14.95
Quart of oil: $ 1.5

Explanation:
5x + y = 22.45
7x + y = 25.45
where x is the unit cost of quarts of oil and y is the flat fee for labor
Subtract the equations
5x + y = 22.45
-(7x + y = 25.45)
y is eliminated as it has reversed coefficients. Solve for x
5x + y – 7x – y = 22.45 – 25.45
-2x = -3
x = -3/-2 = 1.5
Substituting x in either of the equations to find y
5(1.5) + y = 22.45
7.5 + y – 7.5 = 22.45 – 7.5
y = 14.95
Labor fee is $14.95 and the unit cost of a quart of oil is $1.5

Solving System of Equations by Elimination Worksheet Answers Question 14.
Represent Real-World Problems A sales manager noticed that the number of units sold for two T-shirt styles, style A and style B, was the same during June and July. In June, total sales were $2779 for the two styles, with A selling for $15.95 per shirt and B selling for $22.95 per shirt. In July, total sales for the two styles were $2385.10, with A selling at the same price and B selling at a discount of 22% off the June price. How many T-shirts of each style were sold in June and July combined?
________ T-shirts of style A and style B were sold in June and July.

Answer:
15.95x + 22.95y = 2779
15.95x + 17.9y = 2385.10
where x is the number of style A shirts and y is the number of style B shirts
In July, the price of style B shirt is 22% of the price of style B shirt in June, hence 0.78(22.95) = 17.90
Subtract the equations
15.95x + 22.95y = 2779
-(15.95x + 17.9y = 2385.10)
x is eliminated as it has reversed coefficients. Solve for y
15.95x + 22.95 – 15.95x – 17.9y = 2779 – 2385.10
5.05y = 393.9
y = 393.9/5.05 = 78
Substituting y in either of the equations to find x
15.95x +22.95(78) = 2779
15.95x + 1790.1 – 1790.1 = 2779 – 1790.1
15.95x = 988.9
x = 988.9/15.95 = 62
The number of style A T-shirt sold in June is 62.
Since the number of T-shirts sold in both numbers is the same, the total number = 2. 62 = 124.
The number of style B T-shirts sold in June is 78.
Since the number of T-shirts sold in both numbers is the same, the total number = 2. 78 = 156.

Question 15.
Represent Real-World Problems Adult tickets to a basketball game cost $5. Student tickets cost $1. A total of $2,874 was collected on the sale of 1,246 tickets. How many of each type of ticket were sold?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 3: Solving Systems by Elimination img 14img 14
________ adult tickets
________ student tickets

Answer:
407 adult tickets
839 student tickets

Explanation:
x + y = 1246
5x + y = 2874
where x is the number of adult tickets sold and y is the number of student tickets sold.
Subtract the equations
x + y = 1246
-(5x + y = 2874)
y is eliminated as it has reversed coefficients. Solve for x
x + y – 5x – y = 1246 – 2874
-4x = -1628
x = -1628/-4 = 407
Substituting x in either of the equations to find y
407 + y = 1246
407 + y – 407 = 1246 – 407
y = 839
The number of adult tickets sold is 407 and student tickets sold is 839.

FOCUS ON HIGHER ORDER THINKING – Solving Systems by Elimination – Page No. 250

Question 16.
Communicate Mathematical Ideas Is it possible to solve the system
\(\left\{\begin{array}{l}3x-2y=10 \\x+2y=6\end{array}\right.\)
by using substitution? If so, explain how. Which method, substitution or elimination, is more efficient? Why?
________

Answer:
The system can be solved by substitution as x in equation 2 can be isolated.
3x – 2y = 10
x + 2y = 6
Solve the equation for x in the equation.
x = -2y + 6
Substitute the expression for x in the other equation and solve.
3(-2y + 6) -2y = 10
-6y + 18 – 2y = 10
-8y + 18 = 10
-8y = -8
y = -8/-8 = 1
Substitute the values of y into one of the equations and solve for the other variable x.
x + 2(1) = 6
x = 4
The solution is (4, 1)
As the coefficient if variable y is opposite, it will be eliminated and solved for x in less number of steps.
Elimination would be more efficient.

Question 17.
Jenny used substitution to solve the system
\(\left\{\begin{array}{l}2x+y=8 \\x-y=1\end{array}\right.\). Her solution is shown below.
Step 1: y = -2x + 8               Solve the first equation for y.
Step 2: 2x + (-2x + 8) = 8     Substitute the value of y in an original equation.
Step 3: 2x – 2x + 8 = 8          Use the Distributive Property.
Step 4: 8 = 8                         Simplify.
a. Explain the Error Explain the error Jenny made. Describe how to correct it.
Type below:
______________

Answer:
2x + y = 8
x – y = 1
The rewritten equation should be substituted in the other original equation
Error is that Jenny solved for y in the first equation and substituted it in the original equation.
x – (-2x + 8) = 1
3x – 8 = 1
3x = 9
x = 9/3 = 3
x = 3

Question 17.
b. Communicate Mathematical Ideas Would adding the equations have been a better method for solving the system? If so, explain why.
________

Answer:
Yes

Explanation:
As the coefficient, if variable y is the opposite, it will be eliminated and solved for x in less number of steps.

Guided Practice – Solving Systems by Elimination with Multiplication – Page No. 256

Question 1.
Solve the system
\(\left\{\begin{array}{l}3x-y=8 \\-2x+4y=-12\end{array}\right.\)
by multiplying and adding.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 4: Solving Systems by Elimination with Multiplication img 15
Type below:
______________

Answer:
\(\left\{\begin{array}{l}3x-y=8 \\-2x+4y=-12\end{array}\right.\)
Multiply each term in the first equation by 4 to get opposite coefficients for the y-terms.
4(3x – y = 8)
12x – 4y = 32
Add the second equation to the new equation
12x – 4y = 32
+(-2x + 4y = -12)
Add to eliminate the variable y
10x = 20
Divide both sides by 10
x = 20/10 = 2
Substitute into one of the original equations and solve for y
y = 3(2) – 8 = -1
S0, (2, -2)is the solution of the system.

Solve each system of equations by multiplying first.

Question 2.
\(\left\{\begin{array}{l}x+4y=2 \\2x+5y=7\end{array}\right.\)
(________ , ________ )

Answer:
(6, -1)

Explanation:
x + 4y = 2
2x + 5y = 7
To eliminate x terms, multiply the 2nd equation by 2
2(x + 4y = 2)
2x + 8y = 4
Subtract the equations
2x + 8y = 4
-(2x + 5y = 7)
x is eliminated as it has reversed coefficients. Solve for y
2x + 8y – 2x – 5y = 4 – 7
3y = -3
y = -3/3 = -1
Substituting y in either of the equations to find x
x + 4(-1) = 2
x – 4 + 4 = 2 + 4
x = 6
Solution: (6, -1)

Question 3.
\(\left\{\begin{array}{l}3x+y=-1 \\2x+3y=18\end{array}\right.\)
(________ , ________ )

Answer:
(-3, 8)

Explanation:
\(\left\{\begin{array}{l}3x+y=-1 \\2x+3y=18\end{array}\right.\)
To eliminate y terms, multiply the 1st equation by 3
3(3x + y = -1)
9x + 3y = -3
Subtract the equations
9x + 3y = -3
-(2x + 3y = 18)
y is eliminated as it has reversed coefficients. Solve for x
9x + 3y – 2x – 3y = -3 -18
7x = -21
x = -21/7
x = -3
Substituting x in either of the equations to find y
3(-3) + y = -1
-9 + y + 9 = -1 + 9
y = 8
Solution: (-3, 8)

Question 4.
\(\left\{\begin{array}{l}2x+8y=21 \\6x-4y=14\end{array}\right.\)
Type below:
______________

Answer:
The solution is (3.5, 1.75)

Explanation:
\(\left\{\begin{array}{l}2x+8y=21 \\6x-4y=14\end{array}\right.\)
To eliminate y terms, multiply the 2nd equation by 2
2(6x – 4y = 14)
2x + 8y = 21
Add the equations
2x + 8y = 21
+(12x – 8y = 28)
y is eliminated it has reversed coefficients. Solve for x
2x + 8y + 12x – 8y = 21 + 28
14x = 49
x = 49/14 = 3.5
Substituting x in either of the equations to find y
6(3.5) – 4y = 14
21 – 4y – 21 = 14 – 21
-4y = -7
y = -7/-4 = 1.75
The solution is (3.5, 1.75)

Question 5.
\(\left\{\begin{array}{l}2x+y=3 \\-x+3y=-12\end{array}\right.\)
(________ , ________ )

Answer:

Explanation:
\(\left\{\begin{array}{l}2x+y=3 \\-x+3y=-12\end{array}\right.\)
To eliminate x terms, multiply the 2nd equation by 2
2(-x + 3y = -12)
-2x + 6y = -24
Add the equations
2x + y = 3
+(-2x + 6y = -24)
x is eliminated it has reversed coefficients. Solve for y
2x + y – 2x + 6y = 3 – 24
7y = -21
y = -21/7 = -3
Substituting y in either of the equations to find x
-x + 3(-3) = -12
-x -9 + 9 = -12 + 9
-x = -3
x = 3
The soultion is (3, -3)

Question 6.
\(\left\{\begin{array}{l}6x+5y=19 \\2x+3y=5\end{array}\right.\)
(________ , ________ )

Answer:
The solution is (4, -1)

Explanation:
\(\left\{\begin{array}{l}6x+5y=19 \\2x+3y=5\end{array}\right.\)
To eliminate x terms, multiply the 2nd equation by 3
3(2x + 3y = 5)
6x + 9y = 15
Subtract the equations
6x + 5y = 19
-(6x + 9y = 15)
x is eliminated it has reversed coefficients. Solve for y
6x + 5y – 6x – 9y = 19 – 15
-4y = 4
y = 4/-4 = -1
Substituting y in either of the equations to find x
2x + 3(-1) = 5
2x – 3 + 3 = 5 + 3
2x = 8
x = 8/2 = 4
The solution is (4, -1)

Question 7.
\(\left\{\begin{array}{l}2x+5y=16 \\-4x+3y=20\end{array}\right.\)
(________ , ________ )

Answer:
The solution is (-2, 4)

Explanation:
\(\left\{\begin{array}{l}2x+5y=16 \\-4x+3y=20\end{array}\right.\)
To eliminate x terms, multiply the 1st equation by 2
2(2x + 5y = 16)
4x + 10y = 32
Add the equations
4x + 10y = 32
+(-4x + 3y = 20)
x is eliminated it has reversed coefficients. Solve for y
10y + 3y = 32 + 20
13y = 52
y = 52/13 = 4
Substituting y in either of the equations to find x
2x + 5(4) = 16
2x + 20 – 20 = 16 – 20
2x = -4
x = -4/2 = -2
The solution is (-2, 4)

Question 8.
Bryce spent $5.26 on some apples priced at $0.64 each and some pears priced at $0.45 each. At another store he could have bought the same number of apples at $0.32 each and the same number of pears at $0.39 each, for a total cost of $3.62. How many apples and how many pears did Bryce buy?
a. Write equations to represent Bryce’s expenditures at each store
First store: _____________
Second store: _____________
Type below:
_____________

Answer:
First store: 0.64x + 0.45y = 5.26
Second store: 0.32x + 0.39y = 3.62

Explanation:
First store = 0.64x + 0.45y = 5.26
Second store = 0.32x + 0.39y = 3.62
where x is the number of apples and y is the number of pears.

Question 8.
b. Solve the system.
Number of apples: _______
Number of pears: _______

Answer:
Number of apples: 4
Number of pears: 6

Explanation:
First store = 0.64x + 0.45y = 5.26
Second store = 0.32x + 0.39y = 3.62
Multiply by 100
64x + 45y = 526
32x + 39y = 362
To eliminate x terms, multiply the 2nd equation by 2
2(32x + 39y = 362)
64x + 45y = 526
Subtract the equations
64x + 45y = 526
-(64x + 78y = 724)
x is eliminated it has reversed coefficients. Solve for y
64x + 45y – 64x – 78y = 526 – 724
-33y = -198
y = -198/-33 = 6
Substituting y in either of the equation to find x
32x + 39(6) = 362
32x + 234 – 234 = 362 – 234
32x = 128
x = 128/32 = 4
He bought 4 apples and 6 pears.

ESSENTIAL QUESTION CHECK-IN

Question 9.
When solving a system by multiplying and then adding or subtracting, how do you decide whether to add or subtract?
Type below:
_____________

Answer:
If the variable with the same coefficient but a reversed sign, we add and if they have the same sign, we subtract.

Solving Systems by Elimination with Multiplication – Page No. 257

Question 10.
Explain the Error Gwen used elimination with multiplication to solve the system
\(\left\{\begin{array}{l}2x+6y=3 \\x-3y=-1\end{array}\right.\)
Her work to find x is shown. Explain her error. Then solve the system.
2(x − 3y) = -1
2x − 6y = -1
+2x + 6y = 3
_____________
4x + 0y = 2
x = \(\frac{1}{2}\)
Type below:
____________

Answer:
2x + 6y = 3
x – 3y = -1
To eliminate x terms, multiply the 2nd equation by 2
2(x – 3y = -1)
2x – 6y = -2
Error is the Gnew did not multiply the entire expression with 2.
Add the equations
2x + 6y = 3
+(2x – 6y = -2)
y is eliminated it has reversed coefficients. Solve for x
2x + 6y + 2x – 6y = 3 – 2
4x = 1
x = 1/4
Substituting x in either of the equations to find y
x – 3y = -1
1/4 – 3y – 1/4 = -1 -1/4
-3y = -5/4
y = -5/4(-3) = 5/12

Question 11.
Represent Real-World Problems At Raging River Sports, polyester-fill sleeping bags sell for $79. Down-fill sleeping bags sell for $149. In one week the store sold 14 sleeping bags for $1,456.
a. Let x represent the number of polyester-fill bags sold and let y represent the number of down-fill bags sold. Write a system of equations you can solve to find the number of each type sold.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 4: Solving Systems by Elimination with Multiplication img 16
Type below:
____________

Answer:
x + y = 14
79x + 149y = 1456
where x is the polyester-fill bags and y is the number of down-fill bags

Question 11.
b. Explain how you can solve the system for y by multiplying and subtracting.
Type below:
____________

Answer:
x + y = 14
79x + 149y = 1456
Multiply the second equation by 79. Subtract the new equation from the first equation and solve the resulting equation for y.

Question 11.
c. Explain how you can solve the system for y using substitution.
Type below:
____________

Answer:
Solve the second equation for x. Substitute the expression for x , in the first equation and solve the resulting equation for y.

Question 11.
d. How many of each type of bag were sold?
_______ polyester-fill
_______ down-fill

Answer:
9 polyester-fill
5 down-fill

Explanation:
x + y = 14
79x + 149y = 1456
To eliminate x terms, multiply the 2nd equation by 2
79(x + y = 14)
79x + 149y = 1456
Subtract the equations
79x + 79y = 1106
-(79x + 149y = 1456)
x is eliminated it has reversed coefficients. Solve for y
79x + 79y – 79x – 149y = 1106 – 1456
-70y = -350
y = -350/-70 = 5
Substituting y in either of the equations to find x
x + 5 = 14
x = 14 – 5
x = 9
There were 9 polyester-fill bags and 5 down-fill bags sold.

Question 12.
Twice a number plus twice a second number is 310. The difference between the numbers is 55. Find the numbers by writing and solving a system of equations. Explain how you solved the system.
x = _______
y = _______

Answer:
x = 105
y = 50

Explanation:
2x + 2y = 310
x – y = 55
To eliminate y terms, multiply the 2nd equation by 2
2(x – y = 55)
2x – 2y = 110
Add the equations
2x + 2y = 310
+ (2x – 2y = 110)
y is eliminated it has reversed coefficients. Solve for x
2x + 2y + 2x – 2y = 310 + 110
4x = 420
x = 420/4 = 105
Substituting x in either of the equations to find y
105 – y = 55
y = 105 – 55
y = 50
The solution is (105, 50)

Solving Systems by Elimination with Multiplication – Page No. 258

Question 13.
Represent Real-World Problems A farm stand sells apple pies and jars of applesauce. The table shows the number of apples needed to make a pie and a jar of applesauce. Yesterday, the farm picked 169 Granny Smith apples and 95 Red Delicious apples. How many pies and jars of applesauce can the farm make if every apple is used?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 4: Solving Systems by Elimination with Multiplication img 17
_______ pies
_______ jars of applesauce

Answer:
21 pies
16 jars of applesauce

Explanation:
5x + 4y = 169
3x + 2y = 95
where x is the number of apples needed for pie and y is the number of apples for jar of applesauce
To eliminate y terms, multiply the 2nd equation by 2
2(3x + 2y = 95)
6x + 4y = 190
Subtract the equations
5x + 4y = 169
– (6x + 4y = 190)
y is eliminated it has reversed coefficients. Solve for x
5x + 4y – 6x – 4y = 169 – 190
-x = -21
x = -21/-1 = 21
Substituting x in either of the equation to find y
5(21) + 4y = 169
105 + 4y – 105 = 169 – 105
4y = 64
y = 64/4 = 16
The number of apples needed for pie is 21 and the number of apples for jar of applesauce is 16.

FOCUS ON HIGHER ORDER THINKING

Question 14.
Make a Conjecture Lena tried to solve a system of linear equations algebraically and in the process found the equation 5 = 9. Lena thought something was wrong, so she graphed the equations and found that they were parallel lines. Explain what Lena’s graph and equation could mean.
Type below:
____________

Answer:
Lena’s graph is a parallel line which means the graph does not intersect each other, hence they have no solutions. Equation 5 = 9 means variables are eliminated and this statement is not true. This linear system has no solution.

Question 15.
Consider the system
\(\left\{\begin{array}{l}2x+3y=6 \\3x+7y=-1\end{array}\right.\)
a. Communicate Mathematical Ideas Describe how to solve the system by multiplying the first equation by a constant and subtracting. Why would this method be less than ideal?
Type below:
____________

Answer:
Multiplying the first equation by a constant and subtracting
2x + 3y = 6
3x + 7y = -1
Multiply the first equation by 1.5 and subtract. This would be less than ideal because you would introduce decimals into the solution process.

Question 15.
b. Draw Conclusions Is it possible to solve the system by multiplying both equations by integer constants? If so, explain how.
Type below:
____________

Answer:
Yes

Explanation:
Multiply the first equation by 3 and the second equation by 2. Both x-term coefficients would be 6. Solve by eliminating the x-terms using subtraction.

Question 15.
c. Use your answer from part b to solve the system.
(_______ , _______)

Answer:
(9, -4)

Explanation:
2x + 3y = 6
3x + 7y = -1
Multiply the first equation by 3 and the second equation by 2.
3(2x + 3y = 6)
2(3x + 7y = -1)
Subtract the equations
6x + 9y = 18
-(6x + 14y = -2)
x is eliminated it has reversed coefficients. Solve for y
6x + 9y – 6x – 14y = 18 + 2
-5y = 20
y = 20/-5 = -4
Substituting y in either of the equation to find x
2x + 3(-4) = 6
2x = 18
x = 18/2 = 9
The solution is (9, -4)

Guided Practice – Solving Solving Special Systems – Page No. 262

Use the graph to solve each system of linear equations

Question 1.
A. \(\left\{\begin{array}{l}4x-2y=-6 \\2x-y=4\end{array}\right.\)
B. \(\left\{\begin{array}{l}4x-2y=-6 \\x+y=6\end{array}\right.\)
C. \(\left\{\begin{array}{l}2x-y=4 \\6x-3y=-12\end{array}\right.\)
STEP 1 Decide if the graphs of the equations in each system intersect, are parallel, or are the same line.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 5: Solving Solving Special Systems img 18
System A: The graphs __________
System B: The graphs __________
System C: The graphs __________

Answer:
System A: The graphs are parallel
System B: The graphs are intersecting
System C: The graphs are  the same line

Explanation:
System A: 4x – 2y = -6
2x – y = 4
System B: 4x – 2y = -6
x + y = 6
System C: 2x – y = 4
6x – 3y = 12

Question 1.
STEP 2 Decide how many points the graphs have in common.
a. Intersecting lines have _______________ point(s) in common.
b. Parallel lines have _______________ point(s) in common.
c. The same lines have ___________ point(s) in common.
a. __________
b. __________
c. __________

Answer:
a. Intersecting lines have one point(s) in common.
b. Parallel lines have no point(s) in common.
c. The same lines have infinitely many points (s) in common.

Explanation:
From the graphs,
Intersecting lines have one point(s) in common
Parallel lines have no point(s) in common
The same lines have infinitely many points (s) in common

Question 1.
STEP 3 Solve each system.
System A has __________ points in common, so it has __________ solution.
System B has __________ point in common. That point is the solution, __________.
System C has __________ points in common. ________ ordered pairs on the line will make both equations true.
Type below:
___________

Answer:
System A has no points in common, so it has no solution. System B has one point in common. That point is the solution, (1,5). System C has an infinite number of points in common. All ordered pairs on the line will make both equations true.

Explanation:
Number of solutions for each system
System A has no points in common, so it has no solution. System B has one point in common. That point is the solution, (1,5). System C has an infinite number of points in common. All ordered pairs on the line will make both equations true.

Solve each system. Tell how many solutions each system has.

Question 2.
\(\left\{\begin{array}{l}x-3y=4 \\-5x+15y=-20\end{array}\right.\)
___________

Answer:
infinitely many solutions

Explanation:
x – 3y = 4
-5x + 15y = -20
To eliminate y terms, multiply the 1st equation by 5
5(x – 3y = 4)
5x – 15y = 20
Add the equations
5x – 15y = 20
+(-5x + 15y = -20)
x and y is eliminated as it has reversed coefficients.
5x – 15y – 5x + 15y = 20 – 20
0 = 0
The statement is true, hence the solution has infinitely many solutions.

Question 3.
\(\left\{\begin{array}{l}6x+2y=-4 \\3x+y=4\end{array}\right.\)
___________

Answer:
no solution

Explanation:
6x + 2y = -4
3x + y = 4
To eliminate y terms, multiply the 2nd equation by 5
2(3x + y = 4)
6x + 2y = 8
Subtract the equations
6x + 2y = -4
-(6x + 2y = 8)
x and y is eliminated as it has reversed coefficients.
6x + 2y – 6x – 2y = -4 -8
0 = -12
The statement is false, hence the solution has no solution.

Question 4.
\(\left\{\begin{array}{l}6x-2y=-10 \\3x+4y=-25\end{array}\right.\)
___________

Answer:
one solution

Explanation:
6x – 2y = -10
3x + 4y = -25
To eliminate y terms, multiply the 1st equation by 2
2(6x – 2y = -10)
12x – 4y = -20
Add the equations
12x – 4y = -20
+(3x + 4y = -25)
y is eliminated as it has reversed coefficients. Solve for x.
12x – 4y + 3x + 4y = -20 – 25
15x = -45
x = -45/15 = -3
Substitute x in any one of the original equations and solve for y
3(-3) + 4y = -25
-9 + 4y + 9 = -25 + 9
4y = -16
y = -16/4
y = -4
There is one solution, (-3, -4)

ESSENTIAL QUESTION CHECK-IN

Question 5.
When you solve a system of equations algebraically, how can you tell whether the system has zero, one, or an infinite number of solutions?
Type below:
___________

Answer:
When x and y are eliminated and the statement is true, the system has infinitely many solutions.
When x and y are eliminated and the statement is false, the system has no solutions.
When the system has one solution by solving, the system has one solution.

8.5 Independent Practice – Solving Solving Special Systems – Page No. 263

Solve each system by graphing. Check your answer algebraically.

Question 6.
\(\left\{\begin{array}{l}-2x+6y=12 \\x-3y=3\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 5: Solving Solving Special Systems img 19
Solution: ______________
___________

Answer:
\(\left\{\begin{array}{l}-2x+6y=12 \\x-3y=3\end{array}\right.\)
Graph the equations on same coordinate plane
No solution as equations are parallel
grade 8 chapter 8 image 1
To eliminate y terms, multiply the 2nd equation by 2
2(x – 3y = 3)
2x – 6y = 6
Add the equations
-2x + 6y = 12
2x – 6y = 6
x and y is eliminated as it has reversed coefficients.
-2x + 6y + 2x – 6y = 12 + 6
0 = 18
The statement is false, hence the system has no solution.

Question 7.
\(\left\{\begin{array}{l}15x+5y=5 \\3x+y=1\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 5: Solving Solving Special Systems img 20
Solution: ______________
___________

Answer:
\(\left\{\begin{array}{l}15x+5y=5 \\3x+y=1\end{array}\right.\)
Graph the equations on same coordinate plane
grade 8 chapter 8 image 2
Infinitely many solutions as equations are overlapping
To eliminate y terms, multiply the 2nd equation by 5
5(3x + y = 1)
15x + 5y = 5
Subtarct the equations
15x + 5y = 5
-(15x + 5y = 5)
x and y is eliminated as it has reversed coefficients.
15x + 5y -15x – 5y = 5 – 5
0 = 0
The statement is true, hence the system has infinitely many solutions.

For Exs. 8–14, state the number of solutions for each system of linear equations

Question 8.
a system whose graphs have the same slope but different y-intercepts
___________

Answer:
No solutions

Explanation:
Equations are parallel
No solutions

Question 9.
a system whose graphs have the same y-intercepts but different slopes
___________

Answer:
One solution

Explanation:
Equations are intersecting
One solution

Question 10.
a system whose graphs have the same y-intercepts and the same slopes
___________

Answer:
Infinitely many solutions

Explanation:
Equations are overlapping
Infinitely many solutions

Question 11.
a system whose graphs have different y-intercepts and different slopes
___________

Answer:
One solution

Explanation:
Equations are intersecting
One solution

Question 12.
the system
\(\left\{\begin{array}{l}y=2 \\y=-3\end{array}\right.\)
___________

Answer:
No solutions

Explanation:
Equations are parallel
No solutions

Question 13.
the system
\(\left\{\begin{array}{l}y=2 \\y=-3\end{array}\right.\)
___________

Answer:
One solution

Explanation:
Equations are intersecting
One solution

Question 14.
the system whose graphs were drawn using these tables of values:
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 5: Solving Solving Special Systems img 21
___________

Answer:
No solutions

Explanation:
Equations are parallel The slope is the same for both equations but the y-intercept is different.
No solutions

Question 15.
Draw Conclusions The graph of a linear system appears in a textbook. You can see that the lines do not intersect on the graph, but also they do not appear to be parallel. Can you conclude that the system has no solution? Explain.
___________

Answer:

No; although the lines do not intersect on the graph, they intersect at a point that is not on the graph. To prove that a system has no solution, you must do so algebraically

Solving Solving Special Systems – Page No. 264

Question 16.
Represent Real-World Problems Two school groups go to a roller skating rink. One group pays $243 for 36 admissions and 21 skate rentals. The other group pays $81 for 12 admissions and 7 skate rentals. Let x represent the cost of admission and let y represent the cost of a skate rental. Is there enough information to find values for x and y? Explain.
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Lesson 5: Solving Solving Special Systems img 22
___________

Answer:
36x + 21y = 243
12x + 7y = 81
where x is the cost of admission and y is the cost of stake rentals.
Although the information can be used to develop a system of linear equations, where each equation has two variables when the system is solved, the number of solutions is infinite, Mee the values of x and y cannot be determined.
No

System of Linear Equations Answer Key Question 17.
Represent Real-World Problems Juan and Tory are practicing for a track meet. They start their practice runs at the same point, but Tory starts 1 minute after Juan. Both run at a speed of 704 feet per minute. Does Tory catch up to Juan? Explain.
___________

Answer:
No; Both Juan and Tory run at the same rate, so the lines representing the distances each has run are parallel. There is no solution to the system

FOCUS ON HIGHER ORDER THINKING

Question 18.
Justify Reasoning A linear system with no solution consists of the equation y = 4x − 3 and a second equation of the form y = mx + b. What can you say about the values of m and b? Explain your reasoning.
Type below:
___________

Answer:
y = 4x – 3
y = mx + b
Since the system has no solutions, the two equations are parallel. The value of the slope, m would be the same i.e. 4.
The value of y-intercept, b can be any number except -3 as b is different for parallel lines.

Question 19.
Justify Reasoning A linear system with infinitely many solutions consists of the equation 3x + 5 = 8 and a second equation of the form Ax + By = C. What can you say about the values of A, B, and C? Explain your reasoning.
Type below:
___________

Answer:
3x + 5 = 8
Ax + By = C
Since the system has infinitely many solutions, the values of A, B, and C must all be the same multiple of 3, 5, and 8, respectively. The two equations represent a single line, so the coefficients and constants of one equation must be a multiple of the other.

Question 20.
Draw Conclusions Both points (2, -2) and (4, -4) are solutions of a system of linear equations. What conclusions can you make about the equations and their graphs?
Type below:
___________

Answer:
If a system has more than one solution, the equations represent the same line and have infinitely many solutions.

Ready to Go On? – Model Quiz – Page No. 265

8.1 Solving Systems of Linear Equations by Graphing

Solve each system by graphing.

Question 1.
\(\left\{\begin{array}{l}y=x-1 \\y=2x-3\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Model Quiz img 23
(________ , ________)

Answer:
(2, 1)

Explanation:
y = x – 1
y = 2x – 3
Graph the equations on the same coordinate plane
grade 8 chapter 8 image 3
The solution of the system is the point of intersection
The solution is (2, 1)

Lesson 1 Solve Systems of Equations by Graphing Answers Question 2.
\(\left\{\begin{array}{l}x+2y=1 \\-x+y=2\end{array}\right.\)
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Model Quiz img 24
(________ , ________)

Answer:
(-1, 1)

Explanation:
x + 2y = 1
-x + y = 2
Graph the equations on same coordinate plane
grade 8 chapter 8 image 4
The solution of the system is the point of intersection
The solution is (-1, 1)

8.2 Solving Systems by Substitution

Solve each system of equations by substitution.

Question 3.
\(\left\{\begin{array}{l}y=2x \\x+y=-9\end{array}\right.\)
(________ , ________)

Answer:
(-3, -6)

Explanation:
y = 2x
x + y = -9
Substitute y from equation 1 in the other equation.
x + 2x = -9
3x = -9
x = -9/3
x = -3
Then, y = 2(-3) = -6
The Solution is (-3, -6)

Question 4.
\(\left\{\begin{array}{l}3x-2y=11 \\x+2y=9\end{array}\right.\)
(________ , ________)

Answer:
(5, 2)

Explanation:
3x – 2y = 11
x + 2y = 9
Solve for x in equation 2
x = – 2y + 9
Substitute x from equation 2 in the other equation
3(-2y + 9) – 2y = 11
-6y + 27 -2y = 11
-8y = -16
y = -16/-8 = 2
Substitute y in any of the equations to find x
x + 2(2) = 9
x + 4 – 4 = 9 – 4
x = 5
The solution is (5, 2)

8.3 Solving Systems by Elimination

Solve each system of equations by adding or subtracting.

Question 5.
\(\left\{\begin{array}{l}3x+y=9 \\2x+y=5\end{array}\right.\)
(________ , ________)

Answer:
(4, -3)

Explanation:
\(\left\{\begin{array}{l}3x+y=9 \\2x+y=5\end{array}\right.\)
Subtract the equations
3x + y = 9
-(2x + y = 5)
y is eliminated as it has reversed coefficients. Solve for x
3x + y – 2x – y = 9 – 5
x = 4
Substituting x in either of the equation to find y
2(4) + y = 5
8 + y – 8 = 5 – 8
y = -3
The solution is (4, -3)

Question 6.
\(\left\{\begin{array}{l}-x-2y=4 \\3x+2y=4\end{array}\right.\)
(________ , ________)

Answer:
(4, -4)

Explanation:
\(\left\{\begin{array}{l}-x-2y=4 \\3x+2y=4\end{array}\right.\)
Add the equations
-x – 2y = 4
+(3x + 2y = 4)
y is eliminated as it has reversed coefficients. Solve for x
-x – 2y + 3x + 2y = 4 + 4
2x = 8
x = 8/2 = 4
Substituting x in either of the equation to find y
3(4) + 2y = 4
12 + 2y – 12 = 4 – 12
2y = -8
y = -8/2 = -4
The solution is (4, -4)

8.4 Solving Systems by Elimination with Multiplication

Solve each system of equations by multiplying first.

Question 7.
\(\left\{\begin{array}{l}x+3y=-2 \\3x+4y=-1\end{array}\right.\)
(________ , ________)

Answer:
(1, -1)

Explanation:
\(\left\{\begin{array}{l}x+3y=-2 \\3x+4y=-1\end{array}\right.\)
Subtract the equations
3x + 9y = -6
-(3x + 4y = -1)
x is eliminated as it has reversed coefficients. Solve for y
3x + 9y – 3x – 4y = -6 + 1
5y = -5
y = -5/5
y = -1
Substituting y in either of the equation to find x
x + 3(-1) = -2
x – 3 = -2
x = -2 + 3
x = 1
The solution is (1, -1)

Solving Systems of Equations Algebraically Worksheet Question 8.
\(\left\{\begin{array}{l}2x+8y=22 \\3x-2y=5\end{array}\right.\)
(________ , ________)

Answer:
(3, 2)

Explanation:
\(\left\{\begin{array}{l}2x+8y=22 \\3x-2y=5\end{array}\right.\)
Multiply equation 2 by 4 so that y can be eliminated
4(3x – 2y = 5)
12x – 8y = 20
Add the equations
2x + 8y = 22
+(12x – 8y = 20)
y is eliminated as it has reversed coefficients. Solve for x
2x + 8y + 12x – 8y = 22 + 20
14x = 42
x = 42/14
x = 3
Substituting y in either of the equation to find x
2(3) + 8y = 22
6 + 8y = 22
8y = 22 – 6
8y = 16
y = 16/8
y = 2
The solution is (3, 2)

8.5 Solving Special Systems

Solve each system. Tell how many solutions each system has.

Question 9.
\(\left\{\begin{array}{l}-2x+8y=5 \\x-4y=-3\end{array}\right.\)
_____________

Answer:
no solution

Explanation:
\(\left\{\begin{array}{l}-2x+8y=5 \\x-4y=-3\end{array}\right.\)
Multiply equation 2 by 2 so that y can be eliminated
2(x – 4y = -3)
2x – 8y = -6
Add the equations
-2x + 8y = 5
+(2x – 8y = -6)
x and y is eliminated
-2x + 8y + 2x – 8y = 5 – 6
0 = -1
The statement is false. Hence, the system has no solution.

Question 10.
\(\left\{\begin{array}{l}6x+18y=-12 \\x+3y=-2\end{array}\right.\)
_____________

Answer:
infinitely many solutions

Explanation:
\(\left\{\begin{array}{l}6x+18y=-12 \\x+3y=-2\end{array}\right.\)
Multiply equation 2 by 6 so that x can be eliminated
6(x + 3y = -2)
6x + 18y = -12
Subtract the equations
6x + 18y = -12
-(6x + 18y = -12)
x and y is eliminated
6x + 18y -6x -18y = -12 + 12
0 = 0
The statement is true. Hence, the system has infinitely many solutions.

ESSENTIAL QUESTION

Question 11.
What are the possible solutions to a system of linear equations, and what do they represent graphically?
Type below:
___________

Answer:
A system of linear equations can have no solution, which is represented by parallel lines; one solution, which is represented by intersecting lines; and infinitely many solutions, which is represented by overlapping lines.

Selected Response – Mixed Review – Page No. 266

Question 1.
The graph of which equation is shown?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Mixed Review img 25
Options:
A. y = −2x + 2
B. y = −x + 2
C. y = 2x + 2
D. y = 2x + 1

Answer:
C. y = 2x + 2

Explanation:
Options A and B are eliminated as the slope of the graph is 2.
Option D is eliminated as the y-intercept from the graph should be 2.
Option C is the equation of the graph

Question 2.
Which best describes the solutions to the system
\(\left\{\begin{array}{l}x+y=-4 \\-2x-2y=0\end{array}\right.\)
Options:
A. one solution
B. no solution
C. infinitely many
D. (0, 0)

Answer:
B. no solution

Explanation:
\(\left\{\begin{array}{l}x+y=-4 \\-2x-2y=0\end{array}\right.\)
Multiply equation 1 by 2 so that x can be eliminated
2(x + y = -4)
2x + 2y = -8
Add the equations
2x + 2y = -8
-2x – 2y = 0
x and y is eliminated
2x + 2y – 2x -2y = -8 + 0
0 = -8
The statement is false. Hence, the system has no solution.

Question 3.
Which of the following represents 0.000056023 written in scientific notation?
Options:
A. 5.6023 × 105
B. 5.6023 × 104
C. 5.6023 × 10-4
D. 5.6023 × 10-5

Answer:
D. 5.6023 × 10-5

Explanation:
Move the decimal 5 points right to get the equation.
D. 5.6023 × 10-5

Question 4.
What is the solution to
\(\left\{\begin{array}{l}2x-y=1 \\4x+y=11\end{array}\right.\)
Options:
A. (2, 3)
B. (3, 2)
C. (-2, 3)
D. (3, -2)

Answer:
A. (2, 3)

Explanation:
\(\left\{\begin{array}{l}2x-y=1 \\4x+y=11\end{array}\right.\)
Add the equations
2x – y = 1
4x + y = 11
y is eliminated as it has reversed coefficients. Solve for x.
2x – y + 4x + y = 1 + 11
6x = 12
x = 12/6 = 2
Substituting x in either of the equations to find y
4(2) + y = 11
8 + y = 11
y = 11 – 8
y = 3
The solution is (2, 3)

Question 5.
Which expression can you substitute in the indicated equation to solve
\(\left\{\begin{array}{l}3x-y=5 \\x+2y=4\end{array}\right.\)
Options:
A. 2y – 4 for x in 3x – y = 5
B. 4 – x for y in 3x – y = 5
C. 3x – 5 for y in 3x – y = 5
D. 3x – 5 for y in x + 2y = 4

Answer:
D. 3x – 5 for y in x + 2y = 4

Explanation:
\(\left\{\begin{array}{l}3x-y=5 \\x+2y=4\end{array}\right.\)
Solve for y in equation 1
y = 3x – 5
Substitute in other equation x + 2y = 4

Question 6.
What is the solution to the system of linear equations shown on the graph?
Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations Mixed Review img 26
Options:
A. -1
B. -2
C. (-1, -2)
D. (-2, -1)

Answer:
C. (-1, -2)

Explanation:
The point of intersection is (-1, -2), which is the solution of the system

Question 7.
Which step could you use to start solving
\(\left\{\begin{array}{l}x-6y=8 \\2x-5y=3\end{array}\right.\)
Options:
A. Add 2x – 5y = 3 to x – 6y = 8.
B. Multiply x – 6y = 8 by 2 and add it to 2x – 5y = 3.
C. Multiply x – 6y = 8 by 2 and subtract it from 2x – 5y = 3.
D. Substitute x = 6y – 8 for x in 2x – 5y = 3.

Answer:
C. Multiply x – 6y = 8 by 2 and subtract it from 2x – 5y = 3.

Explanation:
x – 6y = 8
2x – 5y = 3
Multiply the 1st equation by 2 so that the coefficient of variable x is the same in both equations
Subtract the equations as x has the same sign.

Mini-Task

Question 8.
A hot-air balloon begins rising from the ground at 4 meters per second at the same time a parachutist’s chute opens at a height of 200 meters. The parachutist descends at 6 meters per second.
a. Define the variables and write a system that represents the situation.
Type below:
_____________

Answer:
y represents the distance from the ground and x represents the time in seconds
y = 4x
y = -6x + 200

Question 8.
b. Find the solution. What does it mean?
Type below:
_____________

Answer:
Substitute y from the equation 1 in the equation 2
4x = -6x + 200
4x + 6x = -6x + 200 + 6x
10x = 200
x = 200/10 = 20
Substitute x in any one of the equations and solve for x
y = 4(20) = 80
The solution is (20, 80)
The balloon and parachute meet after 20sec at 80m from the ground.

Conclusion:

Go Math Grade 8 Answer Key Chapter 8 Solving Systems of Linear Equations PDF for all the students who want to learn maths. See the Grade 8 Chapter 8 questions along with answers and explanations. Immediately start your practice now.

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