Algebra 1 - Solving Systems By Substitution

  • December 2019
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Algebra 1

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Solving Systems by Substitution

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Solve each system by substitution. 1) y = 5 x + 24 −x + 4 y = 1

2) −2 x + 8 y = 24 y = 3 x − 19

3) y = −5 x + 20 −8 x + 7 y = 11

4) 5 x + 2 y = 7 y = 6 x + 12

5) −2 x + 2 y = −6 y = 6 x + 12

6) y = x + 6 −2 x − 7 y = −24

7) −8 x − 8 y = −8 y = 5x + 7

8) 5 x + 2 y = 15 y = 3x + 2

9) y = 5 x − 17 4x − 3 y = 7

10) y = 3 x + 1 −5 x + 2 y = 2

11) − x + y = 0 2 x − 4 y = −16

12) −4 x − 7 y = 16 −2 x + y = −10

13) x − 4 y = 11 −6 x − 8 y = −2

14) x − 3 y = 18 −6 x − 5 y = −16

15) x + 7 y = −4 −x − 8 y = 5

16) x + y = 2 −6 x − 4 y = −12

17) 7 x + y = 14 2x − y = 4

18) 6 x − 6 y = −6 2x + y = 7

19) −6 x + 6 y = −18 2 x + y = −12

20) 2 x − 5 y = −14 7 x + y = −12

21) 7 x + 6 y = −20 8 x + y = 24

22) 4 x − 7 y = −16 x − 6 y = −4

23) 4 x + 4 y = −8 2 x + y = −4

24) x − 4 y = −5 − x − 6 y = −5

25) −2 x + 8 y = −14 x− y=4

26) − x + y = 5 −7 x + 2 y = 0

27) −5 x + 5 y = 5 x − 8y = 6

28) 8 x + y = 0 −8 x + 5 y = 0

29) 3 x + y = 7 − x − 3 y = −13

30) 6 x + 3 y = 6 x + 8y = 1

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