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## okay so using the process of substitution,solve -3x+y=5 and x+2y=0, I know it seems simple, but every time i go to check my answer it comes up wrong,

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

-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
y=5/7
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
x=30/-21=-10/7
SECOND EQUATION, to verify results of first equation...
x+2y=0
(-10/7)+[(2)(5/7)]=0
(-10/7)+(10/7)=0
0=0