SOLUTION: Please help, I'm very frustrated and don't understand how to solve linear systems with substitution. I have a lot of homework due tomorrow and don't understand what I'm supposed to

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Question 192200This question is from textbook Algebra 1
: Please help, I'm very frustrated and don't understand how to solve linear systems with substitution. I have a lot of homework due tomorrow and don't understand what I'm supposed to do.
x-y=0
12x-5y=-21 This question is from textbook Algebra 1

Found 2 solutions by Alan3354, RAY100:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
x-y=0
12x-5y=-21
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There's more than one way to do these, but substitution is the best for this one.
From eqn 1: x = y
Sub for x into the 2nd eqn
12y-5y=-21
7y = -21
y = -3
x = -3
That's all.

Answer by RAY100(1637) (Show Source):
You can put this solution on YOUR website!
1) x-y=0
2) 12x-5y=-21
In substitution we use one eqn to establish a x to y relationship.
Then we substitute this relationship in the other eqn
for example from (1) we know x-y=0, therefore x=y
substituting into (2)
12 (y) -5y = -21
simplify
7y=-21
divide by 7
y=(-3)
substitute this in either (1) or (2) to find x
using (1) x-y=0
x-(-3) =0
y=(-3)
checking
(1) (-3)-(-3)=0 ok
(2) 12(-3) -5(-3) =-21
-36 +15 = -21 ok