How do you solve the system of equations 2a-4b=62a4b=6 and a+3b=-7a+3b=7?

1 Answer
Mar 5, 2018

a=-1a=1, b=-2b=2

Explanation:

Multiplying the second equation by 2, we have 2a+6b=-142a+6b=14. Subtracting this equation into the first equation that was given in the problem => 10b=-2010b=20 => b=-2b=2.
We now know b=-2b=2, solving for aa should be easy. Plug b=-2b=2 into any one of the equation, and then solve for aa => a+3(-2)=-7a+3(2)=7 => a-6=-7a6=7 => a=-1a=1