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Old 05-14-2016, 03:34 AM   #2649933  /  #5
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Join Date: Apr 2010
Location: So. Cal. USA
Posts: 322

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 = 4-2b

Then substituting this for a into the other equation:

2(4-2b) +b =5 yields b=1

And so forth, same results...

ETA: Oh yeah, "Bite me Heinz" you so-called superior member! Heh.

Last edited by MikeB; 05-14-2016 at 03:38 AM.
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