This is a bit different from the solution path I was taught. I'm used to solving for one variable in terms of the other in one equation and substituting into the other equation.
So a + 2b = 4 resolves to a = 42b
Then substituting this for a into the other equation:
2(42b) +b =5 yields b=1
And so forth, same results...
ETA: Oh yeah, "Bite me Heinz" you socalled superior member! Heh.
Last edited by MikeB; 05142016 at 03:38 AM.
