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132
5.3
Solving Systems of Linear Equations by Elimination
For use with Exploration 5.3
Name _________________________________________________________ Date _________
Essential Question How can you use elimination to solve a system of
linear equations?
Work with a partner. You purchase a drink and a sandwich for $4.50. Your friend
purchases a drink and five sandwiches for $16.50. You want to determine the price of a
drink and the price of a sandwich.
a. Let x represent the price (in dollars) of one drink. Let y represent the price (in
dollars) of one sandwich. Write a system of equations for the situation. Use the
following verbal model.
Number Price Number of Price per Total
=
of drinks per drink sandwiches sandwich price
•+ •
Label one of the equations Equation 1 and the other equation Equation 2.
b. Subtract Equation 1 from Equation 2. Explain how you can use the result to
solve the system of equations. Then find and interpret the solution.
Work with a partner. Solve each system of linear equations using two methods.
Method 1 Subtract. Subtract Equation 2 from Equation 1.Then use the result to solve
the system.
Method 2 Add. Add the two equations. Then use the result to solve the system.
Is the solution the same using both methods? Which method do you prefer?
a.
36
30
xy
xy
−=
+=
b.
26
22
xy
xy
+=
−=
c.
27
25
xy
xy
− =−
+=
1 EXPLORATION: Writing and Solving a System of Equations
2 EXPLORATION: Using Elimination to Solve Systems
152
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133
5.3
Solvin
g
S
y
stems of Linear Equations b
y
Elimination (continued)
Name _________________________________________________________ Date __________
Work with a partner.
2 7 Equation 1
5 17 Equation 2
xy
xy
+=
+=
a. Can you eliminate a variable by adding or subtracting the equations as they
are? If not, what do you need to do to one or both equations so that you can?
b. Solve the system individually. Then exchange solutions with your partner and
compare and check the solutions.
Communicate Your Answer
4. How can you use elimination to solve a system of linear equations?
5. When can you add or subtract the equations in a system to solve the system?
When do you have to multiply first? Justify your answers with examples.
6. In Exploration 3, why can you multiply an equation in the system by a constant
and not change the solution of the system? Explain your reasoning.
3 EXPLORATION: Using Elimination to Solve a System
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5.3
Notetaking with Vocabulary
For use after Lesson 5.3
Name _________________________________________________________ Date _________
In your own words, write the meaning of each vocabulary term.
coefficient
Core Concepts
Solving a System of Linear Equations by Elimination
Step 1 Multiply, if necessary, one or both equations by a constant so at least one pair of
like terms has the same or opposite coefficients.
Step 2 Add or subtract the equations to eliminate one of the variables.
Step 3 Solve the resulting equation.
Step 4 Substitute the value from Step 3 into one of the original equations and solve for
the other variable.
Notes:
153
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Student Journal
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134
5.3
For use after Lesson 5.3
Name _________________________________________________________ Date _________
In your own words, write the meaning of each vocabulary term.
coefficient
Core Concepts
Solving a System of Linear Equations by Elimination
Step 1 Multiply, if necessary, one or both equations by a constant so at least one pair of
like terms has the same or opposite coefficients.
Step 2 Add or subtract the equations to eliminate one of the variables.
Step 3 Solve the resulting equation.
Step 4 Substitute the value from Step 3 into one of the original equations and solve for
the other variable.
Notes:
153
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Student Journal
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134
5.3
For use after Lesson 5.3
Name _________________________________________________________ Date _________
In your own words, write the meaning of each vocabulary term.
coefficient
Core Concepts
Solving a System of Linear Equations by Elimination
Step 1 Multiply, if necessary, one or both equations by a constant so at least one pair of
like terms has the same or opposite coefficients.
Step 2 Add or subtract the equations to eliminate one of the variables.
Step 3 Solve the resulting equation.
Step 4 Substitute the value from Step 3 into one of the original equations and solve for
the other variable.
Notes:
Worked-Out Examples
Example #1
Solve the system of linear equations by elimination. Check your solution.
Copyright © Big Ideas Learning, LLC Integrated Mathematics I 227
All rights reserved. Worked-Out Solutions
Chapter 5
4. Step 1 Step 2
5x + 2y = 235,000 −10x − 4y = −470,000
2x + 3y = 160,000
Multiply by 5.
10x + 15y = 800,000
0 + 11y = 330,000
Step 3 11y = 330,000
11y
—
11
=
330,000
—
11
y = 30,000
Step 4
2x + 3y = 160,000
2x + 3(30,000) = 160,000
2x + 90,000 = 160,000
−90,000 −90,000
2x = 70,000
2x
—
2
=
70,000
—
2
x = 35,000
The solution (35,000, 30,000) is the same. So, a large van
costs $35,000, and a small van costs $30,000.
5.3 Exercises (pp. 233– 234)
Vocabulary and Core Concept Check
1. Sample answer: Write a system in which at least one pair of
like terms has opposite coef cients.
2x − 3y = 2
−5x + 3y = −14
2. Sample answer: First, multiply each side of Equation 1 by3
so that the coef cients of the y-terms are 9 and −9. Then
add the equations to eliminate y. Solve the resulting equation
for x. Then substitute the value for x into one of the original
equations, and solve for y.
Monitoring Progress and Modeling with Mathematics
3. Step 2
x + 2y = 13
−x + y = 5
0 + 3y = 18
Step 3 3y = 18
3y
—
3
=
18
—
3
y = 6
Step 4
x + 2y = 13
x + 2(6) = 13
x + 12 = 13
−12 −12
x = 1
Check x + 2y = 13 −x + y = 5
1 + 2(6) =
?
13 −1 + 6 =
?
5
1 + 12 =
?
13 5 = 5 ✓
13
The solution is (1, 6).
4. Step 2
9x + y = 2
−4x − y = −17
5x + 0 = −15
Step 3 5x = −15
5x
—
5
=
−15
—
5
x = −3
Step 4
9x + y = 12
9(−3) + y = 12
−27 + y = 2
+27 +27
y = 29
Check 9x + y = 2 −4x − y = −17
9(−3) + 29 =
?
2 −4(−3) − 29 =
?
−17
−27 + 29 =
?
2 12 − 29 =
?
−17
2 = 2 ✓ −17 = −17 ✓
The solution is (−3, 29).
5.
Step 2
5x + 6y = 50
x − 6y = −26
6x + 0 = 24
Step 3 6x = 24
6x
—
6
=
24
—
6
x = 4
Step 4
x − 6y = −26
4 − 6y = −26
−4 −4
−6y = −30
−6y
—
−6
=
−30
—
−6
y = 5
Check 5x + 6y = 50 x − 6y = −26
5(4) + 6(5) =
?
50 4 − 6(5) =
?
−26
20 + 30 =
?
50 4 −30 =
?
−26
50 = 50 ✓ −26 = −26 ✓
The solution is (4, 5).
Multiply by 2.
Copyright © Big Ideas Learning, LLC Integrated Mathematics I 227
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Chapter 5
4. Step 1 Step 2
5x + 2y = 235,000 −10x − 4y = −470,000
2x + 3y = 160,000
Multiply by 5.
10x + 15y = 800,000
0 + 11y = 330,000
Step 3 11y = 330,000
11y
—
11
=
330,000
—
11
y = 30,000
Step 4
2x + 3y = 160,000
2x + 3(30,000) = 160,000
2x + 90,000 = 160,000
−90,000 −90,000
2x = 70,000
2x
—
2
=
70,000
—
2
x = 35,000
The solution (35,000, 30,000) is the same. So, a large van
costs $35,000, and a small van costs $30,000.
5.3 Exercises (pp. 233– 234)
Vocabulary and Core Concept Check
1. Sample answer: Write a system in which at least one pair of
like terms has opposite coef cients.
2x − 3y = 2
−5x + 3y = −14
2. Sample answer: First, multiply each side of Equation 1 by3
so that the coef cients of the y-terms are 9 and −9. Then
add the equations to eliminate y. Solve the resulting equation
for x. Then substitute the value for x into one of the original
equations, and solve for y.
Monitoring Progress and Modeling with Mathematics
3. Step 2
x + 2y = 13
−x + y = 5
0 + 3y = 18
Step 3 3y = 18
3y
—
3
=
18
—
3
y = 6
Step 4
x + 2y = 13
x + 2(6) = 13
x + 12 = 13
−12 −12
x = 1
Check x + 2y = 13 −x + y = 5
1 + 2(6) =
?
13 −1 + 6 =
?
5
1 + 12 =
?
13 5 = 5 ✓
13 = 13 ✓
The solution is (1, 6).
4. Step 2
9x + y = 2
−4x − y = −17
5x + 0 = −15
Step 3 5x = −15
5x
—
5
=
−15
—
5
x = −3
Step 4
9x + y = 12
9(−3) + y = 12
−27 + y = 2
+27 +27
Check 9x + y = 2 −4x − y = −17
9(−3) + 29 =
?
2 −4(−3) − 29 =
?
−17
−27 + 29 =
?
2 12 − 29 =
?
−17
2 = 2 ✓ −17 = −17 ✓
The solution is (−3, 29).
5. Step 2
5x + 6y = 50
x − 6y = −26
6x + 0 = 24
Step 3 6x = 24
6x
—
6
=
24
—
6
x = 4
Step 4
x − 6y = −26
4 − 6y = −26
−4 −4
−6y = −30
−6y
—
−6
=
−30
—
−6
y = 5
Check
5x
+
6y
=
50 x
−
6y
=
−
26
5(4) + 6(5) =
?
50 4 − 6(5) =
?
−26
20 + 30 =
?
50 4 −30 =
?
−26
50 = 50 ✓ −26 = −26 ✓
The solution is (4, 5).
Multiply by 2.
Practice
5x + 6y = 50
x − 6y = −26
154
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5.3
Notetakin
g
with Vocabular
y
(continued)
Name _________________________________________________________ Date __________
Extra Practice
In Exercises 1–18, solve the system of linear equations by elimination. Check your
solution.
1.
3 17
28
xy
xy
+=
−+ =
2.
25
5 16
xy
xy
−=
+=
3.
2 3 10
22
xy
xy
+=
− − =−
4.
43 6
33
xy
xy
+=
−− =
5.
5 2 28
53 8
xy
xy
+ =−
−+ =
6.
25 8
3 5 13
xy
xy
−=
+ =−
7.
2 12
3 18
xy
x
y
+=
−=
8.
4 3 14
2 64
xy
yx
+=
=+
9.
4 24
4 14
x
y
y
x
− =− +
− =−
Practice A
154
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135
5.3
Notetakin
g
with Vocabular
y
(continued)
Name _________________________________________________________ Date __________
Extra Practice
In Exercises 1–18, solve the system of linear equations by elimination. Check your
solution.
1.
3 17
28
xy
xy
+=
−+ =
2.
25
5 16
xy
xy
−=
+=
3.
2 3 10
22
xy
xy
+=
− − =−
4.
43 6
33
xy
xy
+=
−− =
5.
5 2 28
53 8
xy
xy
+ =−
−+ =
6.
25 8
3 5 13
xy
xy
−=
+ =−
7.
2 12
3 18
xy
x
y
+=
−=
8.
4 3 14
2 64
xy
yx
+=
=+
9.
4 24
4 14
x
y
y
x
− =− +
− =−
Practice A
154
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135
5.3
Name _________________________________________________________ Date __________
Extra Practice
In Exercises 1–18, solve the system of linear equations by elimination. Check your
solution.
1.
3 17
28
xy
xy
+=
−+ =
2.
25
5 16
xy
xy
−=
+=
3.
2 3 10
22
xy
xy
+=
− − =−
4.
43 6
33
xy
xy
+=
−− =
5.
5 2 28
53 8
xy
xy
+ =−
−+ =
6.
25 8
3 5 13
xy
xy
−=
+ =−
7.
2 12
3 18
xy
x
y
+=
−=
8.
4 3 14
2 64
xy
yx
+=
=+
9.
4 24
4 14
x
y
y
x
− =− +
− =−
Practice A
154
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135
5.3
Notetakin
g
with Vocabular
y
(continued)
Name _________________________________________________________ Date __________
Extra Practice
In Exercises 1–18, solve the system of linear equations by elimination. Check your
solution.
1.
3 17
28
xy
xy
+=
−+ =
2.
25
5 16
xy
xy
−=
+=
3.
2 3 10
22
xy
xy
+=
− − =−
4.
43 6
33
xy
xy
+=
−− =
5.
5 2 28
53 8
xy
xy
+ =−
−+ =
6.
25 8
3 5 13
xy
xy
−=
+ =−
7.
2 12
3 18
xy
x
y
+=
−=
8.
4 3 14
2 64
xy
yx
+=
=+
9.
4 24
4 14
x
y
y
x
− =− +
− =−
Practice A
154
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135
5.3
Notetakin
g
with Vocabular
y
(continued)
Name _________________________________________________________ Date __________
Extra Practice
In Exercises 1–18, solve the system of linear equations by elimination. Check your
solution.
1.
3 17
28
xy
xy
+=
−+ =
2.
25
5 16
xy
xy
−=
+=
3.
2 3 10
22
xy
xy
+=
− − =−
4.
43 6
33
xy
xy
+=
−− =
5.
5 2 28
53 8
xy
xy
+ =−
−+ =
6.
25 8
3 5 13
xy
xy
−=
+ =−
7.
2 12
3 18
xy
x
y
+=
−=
8.
4 3 14
2 64
xy
yx
+=
=+
9.
4 24
4 14
x
y
y
x
− =− +
− =−
Practice A
154
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All rights reserved. Student Journal
135
5.3
Notetakin
g
with Vocabular
y
(continued)
Name _________________________________________________________ Date __________
Extra Practice
In Exercises 1–18, solve the system of linear equations by elimination. Check your
solution.
1.
3 17
28
xy
xy
+=
−+ =
2.
25
5 16
xy
xy
−=
+=
3.
2 3 10
22
xy
xy
+=
− − =−
4.
43 6
33
xy
xy
+=
−− =
5.
5 2 28
53 8
xy
xy
+ =−
−+ =
6.
25 8
3 5 13
xy
xy
−=
+ =−
7.
2 12
3 18
xy
x
y
+=
−=
8.
4 3 14
2 64
xy
yx
+=
=+
9.
4 24
4 14
x
y
y
x
− =− +
− =−
Practice A
230 Integrated Mathematics I Copyright © Big Ideas Learning, LLC
Worked-Out Solutions All rights reserved.
Chapter 5
14.
Step 1 Step 2
10x − 9y = 46 10x − 9y = 46
−2x + 3y = 10
Multiply by 5.
−10x + 15y = 50
0 + 6y = 96
Step 3 6y = 96
6y
—
6
=
96
—
6
y = 16
Step 4
−2x + 3y = 10
−2x + 3(16) = 10
−2x + 48 = 10
−48 −48
−2x = −38
−2x
—
−2
=
−38
—
−2
x = 19
Check 10x − 9y = 46 −2x + 3y = 10
10(19) − 9(16) =
?
46 −2(19) + 3(16) =
?
10
190 − 144 =
?
46 −38 + 48 =
?
10
46 = 46 ✓ 10 = 10 ✓
The solution is (19, 16).
15. Step 1 Step 2
4x − 3y = 8
Multiply by 2.
8x −6y = 16
5x − 2y = −11
Multiply by 3.
−15x + 6y = 33
−7x + 0 = 49
Step 3
−
7
x =
49
−7x
—
−7
=
49
—
−7
x = −7
Step 4
4x − 3y = 8
4(−7) − 3y = 8
−28 − 3y = 8
+28 +28
−3y = 36
−3y
—
−3
=
36
—
−3
y = −12
Check 4x − 3y = 8 5x − 2y = −11
4(−7) − 3(−12) =
?
8 5(−7) − 2(−12) =
?
−11
−28 + 36 =
?
8 −35 + 24 =
?
−11
8 = 8 ✓ −11 = −11 ✓
The solution is (−7, −12).
16. Step 1
Step 2
−2x − 5y = 9
Multiply by 3.
−6x − 15y = 27
3x + 11y = 4
Multiply by 2.
6x + 22y = 8
0 + 7y = 35
Step 3
7y
—
7
=
35
—
7
y = 5
Step 4
−2x − 5y = 9
−2x − 5(5) = 9
−2x − 25 = 9
+25 +25
−2x = 34
−2x
—
−2
=
34
—
−2
x
=
−17
Check: −2x −5y = 9
3x + 11y = 4
−2(−17) − 5(5) =
?
9 3(
−17) + 11(5) =
?
4
34 − 25 =
?
9
−51 + 55 =
?
4
9 = 9 ✓
4 = 4 ✓
The solution is (−17, 5).
17. Step 1
Step 2
9x + 2y = 39
Multiply by 4.
36x + 8y = 156
6x + 13y = −9
Multiply by
6.
−36x − 78y = 54
0 − 70y = 210
Step 3 −70y = 210
−70y
—
−70
=
210
—
−70
y = −3
Step 4
9x + 2y = 39
9x + 2(−3) = 39
9
x −
6
=
39
+6 +6
9x = 45
9x
—
9
=
45
—
9
x = 5
Check 9x + 2y = 39 6x + 13y = −9
9(5) + 2(−3) =
?
39 6(5) + 13(−3) =
?
−9
45 − 6 =
?
39 30 − 39 =
?
−9
39 = 39 ✓ −9 = −9 ✓
The solution is (5, −3).
Example #2
Solve the system of linear equations by substitution. Check your solution.
230 Integrated Mathematics I Copyright © Big Ideas Learning, LLC
Worked-Out Solutions All rights reserved.
Chapter 5
14. Step 1 Step 2
10x − 9y = 46 10x − 9y = 46
−2x + 3y = 10
Multiply by 5.
−10x + 15y = 50
0 + 6y = 96
Step 3 6y = 96
6y
—
6
=
96
—
6
y = 16
Step 4
−2x + 3y = 10
−2x + 3(16) = 10
−2x + 48 = 10
−48 −48
−2x = −38
−2x
—
−2
=
−38
—
−2
x = 19
Check 10x − 9y = 46
−
10(19) − 9(16) =
?
46
−2(19) + 3(16) = 10
190 − 144 =
?
46
−38 + 48 =
?
10
46 = 46 ✓
10 = 10 ✓
The solution is (19, 16).
15. Step 1 Step 2
4x − 3y = 8
Multiply by 2.
8x −6y = 16
5x − 2y = −11
Multiply by 3.
−15
x + 6y = 33
−
7x + 0 = 49
Step 3
−7x = 49
−7x
—
−7
=
49
—
−7
x = −7
Step 4
4x − 3y = 8
4(−7) − 3y = 8
−28 − 3y = 8
+28 +28
−3y = 36
−3y
—
−
3
=
36
—
−
3
y = −12
Check 4x − 3y = 8 5
x − 2y = −11
4(−7) − 3(−12) =
?
8 5(−7) − 2(−12) =
?
−11
−28 + 36 =
?
8 −35 + 24 =
?
−11
8 = 8 ✓ −11 = −11 ✓
The solution is (−7, −12).
16. Step 1 Step 2
−2x − 5y = 9
Multiply by 3.
−6x − 15y = 27
3x + 11y = 4
Multiply by 2.
6x + 22y = 8
0 + 7y = 35
Step 3
7y
—
7
=
35
—
7
y = 5
Step 4
−2x − 5y = 9
−2x − 5(5) = 9
−2x − 25 = 9
+25 +25
−2x = 34
−2x
—
−2
=
34
—
−2
x = −17
Check: −2x −5y = 9 3x + 11y = 4
−2(−17) − 5(5) =
?
9 3(−17) + 11(5) =
?
4
34 − 25 =
?
9 −51 + 55 =
?
4
9 = 9 ✓ 4 = 4 ✓
The solution is (−17, 5).
17. Step 1 Step 2
9x + 2y = 39
Multiply by 4.
36x + 8y = 156
6x + 13y = −9
Multiply by 6.
−36x − 78y = 54
0 − 70y = 210
Step 3 −70y = 210
−70y
—
−70
=
210
—
−70
y = −3
Step 4
9x + 2y = 39
9x + 2(−3) = 39
9x − 6 = 39
+6 +6
9x = 45
9x
—
9
=
45
—
9
x = 5
Check 9x + 2y = 39 6x + 13y = −9
9(5) + 2(−3) =
?
39 6(5) + 13(−3) =
?
−9
45 − 6 =
?
39 30 − 39 =
?
−9
39 = 39 ✓ −9 = −9 ✓
The solution is (5, −3).
230 Integrated Mathematics I Copyright © Big Ideas Learning, LLC
Worked-Out Solutions All rights reserved.
Chapter 5
14. Step 1 Step 2
10x − 9y = 46 10x − 9y = 46
−2x + 3y = 10
Multiply by 5.
−10x + 15y = 50
0 + 6y = 96
6 =
6y
—
6
=
96
—
6
y = 16
Step 4
−2x + 3y = 10
−2x + 3(16) = 10
−2x + 48 = 10
−48 −48
−2x = −38
−2x
—
−2
=
−38
—
−2
x = 19
Check 10x − 9y = 46 −2x + 3y = 10
10(19) − 9(16) =
?
46 −2(19) + 3(16) =
?
10
190 − 144 =
?
46 −38 + 48 =
?
10
46 = 46 ✓
10
= 10 ✓
The solution is (19, 16).
15. Step 1 Step 2
4x − 3y = 8
Multiply by 2.
8x −6y = 16
5x − 2y = −11
Multiply by 3.
−15x + 6y = 33
−7x + 0 = 49
Step 3 −7x = 49
−7x
—
−7
=
49
—
−7
x = −7
Step 4
4x − 3y = 8
4(−7) − 3y = 8
−28 − 3y = 8
+28 +28
−3y = 36
−3y
—
−
3
=
36
—
−
3
y = −12
Check 4x − 3y = 8 5x − 2y = −11
4(−7) − 3(−12) =
?
8 5(−7) − 2(−12) =
?
−11
−28 + 36 =
?
8 −35 + 24 =
?
−11
8 = 8 ✓ −11 = −11 ✓
The solution is (−7, −12).
16. Step 1 Step 2
−2x − 5y = 9
Multiply by 3.
−6x − 15y = 27
3x + 11y = 4
Multiply by 2.
6x + 22y = 8
0 + 7y = 35
Step 3
7y
—
7
=
35
—
7
y = 5
Step 4
−2x − 5y = 9
−2x − 5(5) = 9
−2x − 25 = 9
+25 +25
−2x = 34
−2x
—
−2
=
34
—
−2
x = −17
Check: −2x −5y =
9 3x + 11y = 4
−2(−17) − 5(5)
=
?
9 3(− + 11(5) =
?
4
34 − 25
=
?
9 −51 + 55 =
?
4
9
=
9 ✓ 4 = 4 ✓
The solution is (−17, 5).
17. Step 1
Step 2
9x + 2
y
= 39
Multiply by 4.
36x + 8y = 156
6x + 13y = −9
Multiply by
6.
−36x − 78y = 54
0 − 70y = 210
Step 3 −70y = 210
−70y
—
−70
=
210
—
−70
y = −3
Step 4
9x + 2y = 39
9x + 2(−3) = 39
9x − 6 = 39
+6 +6
9x = 45
9x
—
9
=
45
—
9
x
=
5
Check 9x + 2y =
39 6x + 13y = −9
9(5) + 2(−3) =
?
39 6(5) + 13(−3) =
?
−9
45 − 6 =
?
39 30 − 39 =
?
−9
39 = 39 ✓ −9 = −9 ✓
The solution is (5, −3).
Practice (continued)
154
Copyright © Big Ideas Learning, LLC
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Copyright © Big Ideas Learning, LLC Integrated Mathematics I
All rights reserved. Student Journal
135
5.3
Notetakin
g
with Vocabular
y
(continued)
Name _________________________________________________________ Date __________
Extra Practice
In Exercises 1–18, solve the system of linear equations by elimination. Check your
solution.
1.
3 17
28
xy
xy
+=
−+ =
2.
25
5 16
xy
xy
−=
+=
3.
2 3 10
22
xy
xy
+=
− − =−
4.
43 6
33
xy
xy
+=
−− =
5.
5 2 28
53 8
xy
xy
+ =−
−+ =
6.
25 8
3 5 13
xy
xy
−=
+ =−
7.
2 12
3 18
xy
x
y
+=
−=
8.
4 3 14
2 64
xy
yx
+=
=+
9.
4 24
4 14
x
y
y
x
− =− +
− =−
Practice A
154
Copyright © Big Ideas Learning, LLC
All rights reserved.
Copyright © Big Ideas Learning, LLC Integrated Mathematics I
All rights reserved. Student Journal
135
5.3
Notetakin
g
with Vocabular
y
(continued)
Name _________________________________________________________ Date __________
Extra Practice
In Exercises 1–18, solve the system of linear equations by elimination. Check your
solution.
1.
3 17
28
xy
xy
+=
−+ =
2.
25
5 16
xy
xy
−=
+=
3.
2 3 10
22
xy
xy
+=
− − =−
4.
43 6
33
xy
xy
+=
−− =
5.
5 2 28
53 8
xy
xy
+ =−
−+ =
6.
25 8
3 5 13
xy
xy
−=
+ =−
7.
2 12
3 18
xy
x
y
+=
−=
8.
4 3 14
2 64
xy
yx
+=
=+
9.
4 24
4 14
x
y
y
x
− =− +
− =−
Practice A
154
Copyright © Big Ideas Learning, LLC
All rights reserved.
Copyright © Big Ideas Learning, LLC Integrated Mathematics I
All rights reserved. Student Journal
135
5.3
Notetakin
g
with Vocabular
y
(continued)
Name _________________________________________________________ Date __________
Extra Practice
In Exercises 1–18, solve the system of linear equations by elimination. Check your
solution.
1.
3 17
28
xy
xy
+=
−+ =
2.
25
5 16
xy
xy
−=
+=
3.
2 3 10
22
xy
xy
+=
− − =−
4.
43 6
33
xy
xy
+=
−− =
5.
5 2 28
53 8
xy
xy
+ =−
−+ =
6.
25 8
3 5 13
xy
xy
−=
+ =−
7.
2 12
3 18
xy
x
y
+=
−=
8.
4 3 14
2 64
xy
yx
+=
=+
9.
4 24
4 14
x
y
y
x
− =− +
− =−
Practice A
230 Integrated Mathematics I Copyright © Big Ideas Learning, LLC
Worked-Out Solutions All rights reserved.
Chapter 5
14.
Step 1 Step 2
10x − 9y = 46 10x − 9y = 46
−2x + 3y = 10
Multiply by 5.
−10x + 15y = 50
0 + 6y = 96
Step 3 6y = 96
6y
—
6
=
96
—
6
y = 16
Step 4
−2x + 3y = 10
−2x + 3(16) = 10
−2x + 48 = 10
−48 −48
−2x = −38
−2x
—
−2
=
−38
—
−2
x = 19
Check 10x − 9y = 46 −2x + 3y = 10
10(19) − 9(16) =
?
46 −2(19) + 3(16) =
?
10
190 − 144 =
?
46 −38 + 48 =
?
10
46 = 46 ✓ 10 = 10 ✓
The solution is (19, 16).
15. Step 1 Step 2
4x − 3y = 8
Multiply by 2.
8x −6y = 16
5x − 2y = −11
Multiply by 3.
−15x + 6y = 33
−7x + 0 = 49
Step 3
−
7
x =
49
−7x
—
−7
=
49
—
−7
x = −7
Step 4
4x − 3y = 8
4(−7) − 3y = 8
−28 − 3y = 8
+28 +28
−3y = 36
−3y
—
−3
=
36
—
−3
y = −12
Check 4x − 3y = 8 5x − 2y = −11
4(−7) − 3(−12) =
?
8 5(−7) − 2(−12) =
?
−11
−28 + 36 =
?
8 −35 + 24 =
?
−11
8 = 8 ✓ −11 = −11 ✓
The solution is (−7, −12).
16. Step 1
Step 2
−2x − 5y = 9
Multiply by 3.
−6x − 15y = 27
3x + 11y = 4
Multiply by 2.
6x + 22y = 8
0 + 7y = 35
Step 3
7y
—
7
=
35
—
7
y = 5
Step 4
−2x − 5y = 9
−2x − 5(5) = 9
−2x − 25 = 9
+25 +25
−2x = 34
−2x
—
−2
=
34
—
−2
x
=
−17
Check: −2x −5y = 9
3x + 11y = 4
−2(−17) − 5(5) =
?
9 3(
−17) + 11(5) =
?
4
34 − 25 =
?
9
−51 + 55 =
?
4
9 = 9 ✓
4 = 4 ✓
The solution is (−17, 5).
17. Step 1
Step 2
9x + 2y = 39
Multiply by 4.
36x + 8y = 156
6x + 13y = −9
Multiply by
6.
−36x − 78y = 54
0 − 70y = 210
Step 3 −70y = 210
−70y
—
−70
=
210
—
−70
y = −3
Step 4
9x + 2y = 39
9x + 2(−3) = 39
9
x −
6
=
39
+6 +6
9x = 45
9x
—
9
=
45
—
9
x = 5
Check 9x + 2y = 39 6x + 13y = −9
9(5) + 2(−3) =
?
39 6(5) + 13(−3) =
?
−9
45 − 6 =
?
39 30 − 39 =
?
−9
39 = 39 ✓ −9 = −9 ✓
The solution is (5, −3).
Example #2
Solve the system of linear equations by elimination. Check your solution.
230 Integrated Mathematics I Copyright © Big Ideas Learning, LLC
Worked-Out Solutions All rights reserved.
Chapter 5
14. Step 1 Step 2
10x − 9y = 46 10x − 9y = 46
−2x + 3y = 10
Multiply by 5.
−10x + 15y = 50
0 + 6y = 96
Step 3 6y = 96
6y
—
6
=
96
—
6
y = 16
Step 4
−2x + 3y = 10
−2x + 3(16) = 10
−2x + 48 = 10
−48 −48
−2x = −38
−2x
—
−2
=
−38
—
−2
x = 19
Check 10x − 9y = 46
10(19) − 9(16) =
?
46
−2(19) + 3(16) = 10
190 − 144 =
?
46
−38 + 48 =
?
10
46 = 46 ✓
10 = 10 ✓
The solution is (19, 16).
15. Step 1 Step 2
4x − 3y = 8
Multiply by 2.
8x −6y = 16
5x − 2y = −11
Multiply by 3.
−15
x + 6y = 33
−
7x + 0 = 49
Step 3
−7x = 49
−7x
—
−7
=
49
—
−7
x = −7
Step 4
4x − 3y = 8
4(−7) − 3y = 8
−28 − 3y = 8
+28 +28
−3y = 36
−3y
—
−
3
=
36
—
−
3
y = −12
Check 4x − 3y = 8 5
x − 2y = −11
4(−7) − 3(−12) =
?
8 5(−7) − 2(−12) =
?
−11
−28 + 36 =
?
8 −35 + 24 =
?
−11
8 = 8 ✓ −11 = −11 ✓
The solution is (−7, −12).
16. Step 1 Step 2
−2x − 5y = 9
Multiply by 3.
−6x − 15y = 27
3x + 11y = 4
Multiply by 2.
6x + 22y = 8
0 + 7y = 35
Step 3
7y
—
7
=
35
—
7
y = 5
Step 4
−2x − 5y = 9
−2x − 5(5) = 9
−2x − 25 = 9
+25 +25
−2x = 34
−2x
—
−2
=
34
—
−2
x = −17
Check: −2x −5y = 9 3x + 11y = 4
−2(−17) − 5(5) =
?
9 3(−17) + 11(5) =
?
4
34 − 25 =
?
9 −51 + 55 =
?
4
9 = 9 ✓ 4 = 4 ✓
The solution is (−17, 5).
17. Step 1 Step 2
9x + 2y = 39
Multiply by 4.
36x + 8y = 156
6x + 13y = −9
Multiply by 6.
−36x − 78y = 54
0 − 70y = 210
Step 3 −70y = 210
−70y
—
−70
=
210
—
−70
y = −3
Step 4
9x + 2y = 39
9x + 2(−3) = 39
9x − 6 = 39
+6 +6
9x = 45
9x
—
9
=
45
—
9
x = 5
Check 9x + 2y = 39 6x + 13y = −9
9(5) + 2(−3) =
?
39 6(5) + 13(−3) =
?
−9
45 − 6 =
?
39 30 − 39 =
?
−9
39 = 39 ✓ −9 = −9 ✓
The solution is (5, −3).
230 Integrated Mathematics I Copyright © Big Ideas Learning, LLC
Worked-Out Solutions All rights reserved.
Chapter 5
14. Step 1 Step 2
10x − 9y = 46 10x − 9y = 46
−2x + 3y = 10
Multiply by 5.
−10x + 15y = 50
0 + 6y = 96
Step 3 6y = 96
6y
—
6
=
96
—
6
y = 16
Step 4
−2x + 3y = 10
−2x + 3(16) = 10
−2x + 48 = 10
−48 −48
−2x = −38
−2x
—
−2
=
−38
—
−2
x = 19
Check 10x − 9y = 46 −2x + 3y = 10
10(19) − 9(16) =
?
46 −2(19) + 3(16) =
?
10
190 − 144 =
?
46 −38 + 48 =
?
10
46 = 46 ✓
10
= 10 ✓
The solution is (19, 16).
15. Step 1 Step 2
4x − 3y = 8
Multiply by 2.
8x −6y = 16
5x − 2y = −11
Multiply by 3.
−15x + 6y = 33
−7x + 0 = 49
Step 3 −7x = 49
−7x
—
−7
=
49
—
−7
x = −7
Step 4
4x − 3y = 8
4(−7) − 3y = 8
−28 − 3y = 8
+28 +28
−3y = 36
−3y
—
−
3
=
36
—
−
3
y = −12
Check 4x − 3y = 8 5x − 2y = −11
4(−7) − 3(−12) =
?
8 5(−7) − 2(−12) =
?
−11
−28 + 36 =
?
8 −35 + 24 =
?
−11
8 = 8 ✓ −11 = −11 ✓
The solution is (−7, −12).
16. Step 1 Step 2
−2x − 5y = 9
Multiply by 3.
−6x − 15y = 27
3x + 11y = 4
Multiply by 2.
6x + 22y = 8
0 + 7y = 35
Step 3
7y
—
7
=
35
—
7
y = 5
Step 4
−2x − 5y = 9
−2x − 5(5) = 9
−2x − 25 = 9
+25 +25
−2x = 34
−2x
—
−2
=
34
—
−2
x = −17
Check: −2x −5y =
9 3x + 11y = 4
−2(−17) − 5(5)
=
?
9 3( 17) + 11(5) =
?
4
34 − 25
=
?
9 −51 + 55 =
?
4
9
=
9 ✓ 4 = 4 ✓
The solution is (−17, 5).
17. Step 1
Step 2
9x + 2
y
= 39
Multiply by 4.
36x + 8y = 156
6x + 13y = −9
Multiply by
6.
−36x − 78y = 54
0 − 70y = 210
Step 3 −70y = 210
−70y
—
−70
=
210
—
−70
y = −3
Step 4
9x + 2y = 39
9x + 2(−3) = 39
9x − 6 = 39
+6 +6
9x = 45
9x
—
9
=
45
—
9
x
=
5
Check 9x + 2y =
39 6x + 13y = −9
9(5) + 2(−3) =
?
39 6(5) + 13(−3) =
?
−9
45 − 6 =
?
39 30 − 39 =
?
−9
39 = 39 ✓ −9 = −9 ✓
The solution is (5, −3).
10 9 = 46 x y−
−2x + 3y = 10
155
Copyright © Big Ideas Learning, LLC
All rights reserved.
Integrated Mathematics I Copyright © Big Ideas Learning, LLC
Student Journal
All rights reserved.
136
5.3
Name _________________________________________________________ Date _________
10.
2 20
2 19
xy
xy
+=
+=
11.
32 2
43 4
xy
xy
− =−
− =−
12. 9 4 11
3 10 2
xy
xy
+=
− =−
13.
4 3 21
5 2 21
xy
xy
+=
+=
14.
35 7
43 2
xy
xy
− − =−
− − =−
15.
8 4 12
7 3 10
xy
xy
+=
+=
16.
43 7
25 7
xy
xy
+ =−
−− =
17.
83 9
5 4 12
xy
xy
− =−
+=
18.
35 2
22 1
xy
xy
− + =−
−=
19. The sum of two numbers is 22. The difference is 6. What are the two numbers?
Practice (continued)
156
Copyright © Big Ideas Learning, LLC
All rights reserved.
Copyright © Big Ideas Learning, LLC Integrated Mathematics I
All rights reserved. Resources by Chapter
159
Practice B
5.3 5.3
Name _________________________________________________________ Date __________
In Exercises 1–6, solve the system of linear equations by elimination. Check your
solution.
1.
2 10
5 11
xy
xy
+=
−=
2.
3 2 14
4 2 16
xy
xy
−+ =
− =−
3.
27
13 5
xy
yx
+=
−=
4.
10 11 3
5 5 10
x
y
yx
− =−
− =−
5.
2 43
2 62
yx
x
y
−=
−=
6.
83 5
34
xy
yx
+ =−
=+
In Exercises 7–12, solve the system of linear equations by elimination. Check your
solution.
7.
3 4 19
6 9 21
xy
xy
−=
+=
8.
45 3
3 2 38
xy
xy
+=
−+ =
9.
8 2 22
5 3 35
xy
xy
+=
−=
10.
47 1
6 3 15
xy
xy
+=
−=
11.
21 11 9
14 8 4
xy
xy
− =−
−+=
12.
36 6
2 9 24
xy
xy
+=
− − =−
13. Describe and correct the error in solving for one of the variables in the linear
system
4 5 10 and 2 4 9.xy xy+ =− − =
In Exercises 14–16, solve the system of linear equations using any method.
Explain why you chose the method.
14.
1
3
3
5
xy
xy
−=
=+
15.
5
2
2
35 2
xy
xy
+=
−=
16.
45 3
14 2 9
xy
xy
− =−
+=
17. You and your friend are making 30 liters of sodium water. You have liters
of 10% sodium and your friend has liters of 22% sodium. How many of your
liters and how many of your friend's liters should you mix to make 30 liters
of 15% sodium?
Step 1
Step 2
Step 3
Practice B
