SOLUTION: solve each system of linear equations by elimination 3a-2b+4=0 2a-5b-1=0

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Question 902851: solve each system of linear equations by elimination
3a-2b+4=0
2a-5b-1=0
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Multiply the first equation by 2.
6a-4b%2B8=0
Multiply the second equation by (-3).
-6a%2B15b%2B3=0
Add them together to eliminate a.
6a-4b%2B8-6a%2B15b%2B3=0
11b%2B11=0
11b=-11
b=-1
Then use either equation to solve for b,
6a-4%28-1%29%2B8=0
6a%2B4%2B8=0
6a%2B12=0
6a=-12
a=-2