<span>Multiply one of the equations so that both equations share a common complementary coefficient.
In order to solve using the elimination method, you need to have a matching coefficient that will cancel out a variable when you add the equations together. For the 2 equations given, you have a huge number of choices. I'll just mention a few of them.
You can multiply the 1st equation by -2/5 to allow cancelling the a term.
You can multiply the 1st equation by 5/3 to allow cancelling the b term.
You can multiply the 2nd equation by -2.5 to allow cancelling the a term.
You can multiply the 2nd equation by 3/5 to allow cancelling the b term.
You can even multiply both equations. For instance, multiply the 1st equation by 5 and the second by 3. And in fact, let's do that.
5a + 3b = –9
2a – 5b = –16
5*(5a + 3b = -9) = 25a + 15b = -45
3*(2a - 5b = -16) = 6a - 15b = -48
Then add the equations
25a + 15b = -45
6a - 15b = -48
=
31a = -93
a = -3
And then plug in the discovered value of a into one of the original equations and solve for b.</span>
