Math Problem, Zelaron Come Together

It matters not whether you win or lose; what matters is whether I win or lose.

Your method ends up 2x + y = 5. Which is what you started with.



Also, your first line finishes the equation:

x + y = 3 ; 2x + y = 5 | x = 2 ; y = 1

If we're given x = 2 and y = 1, then we're finished right there - problem solved.

It matters not whether you win or lose; what matters is whether I win or lose.

Um... you did take Algebra in high school, right?

There's nothing mysterious or groundbreaking about substitution and manipulation.

It matters not whether you win or lose; what matters is whether I win or lose.

Graphing works for that. But beyond that, you're starting out with

Ax + By = C
Dx + Ey = F




The proper way is to do this:

y = -Ax/B + C/B
y = -Dx/E + F/E

-Ax/B + C/B = -Dx/E + F/E
-(-Ax/B + C/B) = -(-Dx/E + F/E)
Ax/B - C/B = Dx/E - F/E
Ax/B - Dx/E = C/B - F/E

Then you'd solve it from there. Actually we can go farther:

BE(Ax/B - Dx/E) =BE(C/B - F/E)
EAx - BDx = EC - BF

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Then you could just throw x into either of the starter statements. Doing that with your question:

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1x + 1y = 3
2x + 1y = 5

1*1x - 1*2x = 1*3 - 1*5
1x-2x = 3 - 5
-x = -2
x = 2

And you can find y easily here:

2 + 1y = 3
y = 1

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There is no way to get either x or y without getting rid of one somehow. You can't just throw the number away, it must be set as something equal. That's subbing.