Question 23686: solve the system by substitution
6(x+2)-y=31
5x-2(y-3)=23
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! 6(x+2)-y=31 --eqn1
5x-2(y-3)=23 --eqn2
eqn1 is re-arranged as follows:
6(x+2)-y=31
6(x+2)=31+y
6(x+2)-31=y
so, y = 6(x+2)-31. --eqn3
5x-2(y-3)=23
5x-2y+6 = 23
5x-2y = 17
Now sub eqn3 into our simplified version of eqn2 as follows:
5x-2y = 17 becomes
5x-2(6(x+2)-31) = 17
5x-2(6x+12)-31) = 17
5x-2(6x-19) = 17
5x-12x+38 = 17
-7x+38 = 17
-7x = -21
x = -21/-7
--> x = 3
So, now find y:
y = 6(x+2)-31
y = 6(3+2)-31
y = 6(5)-31
y = 30-31
--> y = -1
Now, check in the other equation that your solution works there too:
5x-2(y-3)=23
5(3)-2(-1-3)=23
15-2(-1-3)=23
15-2(-4)=23
15+8=23
23=23 --> correct
Now you KNOW that your answers are correct.
jon.
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