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CLASS COPY. PLEASE DO NOT WRITE. USE A SEPARATE SHEET OF PAPER.
DOK 2] Solving Linear Systems by Multiplying First
Solve each system of equations. Explain your reasoning about multiplying.
3 x  4 y  2
1. 
6 x  4 y  3
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 y  x  17
3. 
2 y  3 x  11
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3 x  y  15
5. 
2 x  3 y  23
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2 x  2y  14
2. 
 x  4 y  13
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 x  6y  1
4. 
2 x  3 y  32
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5 x  2y  48
6. 
2 x  3 y  23
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Solve each system of equations. Check your answer by graphing.
4 x  3 y  9
7. 
5 x  y  8
3 x  3 y   1
8. 
12 x  2 y  16
Solve.
9. Ten bagels and four muffins cost $13. Five bagels and eight muffins
cost $14. What are the prices of a bagel and a muffin?
________________________________________________________________________________________
10. John can service a television and a cable box in one hour. It took him
four hours yesterday to service two televisions and ten cable boxes.
How many minutes does John need to service a cable box?
________________________________________________________________________________________
CLASS COPY. PLEASE DO NOT WRITE. USE A SEPARATE SHEET OF PAPER.
ANSWER KEY
 1 1
1.  , 
3 4
2. (3, 4)
3. (9, 8)
4. (13, 2)
5. (2, 9)
6. (10, 1)
7. (3, 7)
5 
8.  , 2 
3 
9. Bagel: $0.80; muffin: $1.25
10. 15 minutes
CLASS COPY. PLEASE DO NOT WRITE. USE A SEPARATE SHEET OF PAPER.
Practice and Problem Solving: Modified
1. 2nd; 3 or 3
2. 1st; 2 or 2
3. 1st; 4 or 4
4. 2nd; 4 or 4
5. (2, 3)
6. (10, 4)
7. (4, 1)
8. (2, 0)
9. One hot chocolate costs $2.
10. 9 two-point shots and 3 three-point shots
Reading Strategies
1. Multiply the first equation by 3 and the second equation by 5 to get a common coefficient of
15.
ì 4(9x - 10y = 7) ì36x - 40y = 28
2. í
Þí
î5(5x + 8y = 31)
î25x + 40y = 155
3. (1, 3)
4. (10, 10)
Success for English Learners
1. The y-variable is eliminated.
2. Multiply the equation by 3 so that the
y-variable is eliminated.
MODULE 11 Challenge
1.
x  y  z  12
(3 x  y  z  32)
4 x  2 y  44
4 x  2y  44
(6 x  2y  56)
3 x  y  z  32
(3 x  y  z  24)
6 x  2 y  56
4(6)  2 y  44
y  10
 2 x  12
x6
6  10  z  12
Solution: (6, 10, 4)
z  4
The solution checks in all three equations.
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