The whole question is as follows

Solve each system by substitution. If the system is inconsistent or has dependent equations, say so

24. -3x+y=-5

x+2y=0

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-3x + y = 5

x + 2y = 0 => x = -2y

substitue in first equation

-3(-2y) + y = 5

6y + y = 5

y = 5/7

substitute

-3x + 5/7 = 5

-3x = 5 - 5/7

-3x = 30/7

x = -30/21 = -10/7

check

substitute values in 1st equation

-3(-10/7) + 5/7 = 5

30/7 + 5/7 = 5

35/7 =5 correct

substitue in 2nd equation

-10/7 + 2(5/7) = 0

-10/7 + 10/7 = 0 correct

Hi Jason;

FIRST EQUATION...-3x+y=5

SECOND EQUATION...x+2y=0

Let's take the second equation and isolate x...

x=-2y

FIRST EQUATION...

-3x+y=5

[-3(-2y)]+y=5

6y+y=5

7y=5

FIRST EQUATION...

-3x+y=5

-3x+(5/7)=5

Let's eliminate annoying fractions and multiply both sides by 7...

7[-3x+(5/7)]=5(7)

-21x+5=35

-21x=30

SECOND EQUATION, to verify results of first equation...

x+2y=0

(-10/7)+[(2)(5/7)]=0

(-10/7)+(10/7)=0

0=0

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