Question #5eee2

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Answer 1 · 2014-08-18T03:13:51

I will assume that you are treating $a$ and $b$ as constants, and are trying to solve for $x$ and $y$ .

We will solve the first equation for $x$ :

$x = yab$

Now, substitution into the second equation gives:

$a^2 by + by = a^2 + b^2$

Factor out $y$ :

$y(a^2 b + b) = a^2 + b^2$

And lastly, divide:

$y = (a^2 + b^2)/(a^2 b + b)$

So there's $y$ . To solve for $x$ , we will revisit the first equation, this time solving for $y$ :

$y = x/(ab)$

Substitute into the second equation:

$ax + x/a = a^2 + b^2$

Factor $x$ :

$x(a + 1/a) = a^2 + b^2$

And lastly, divide to isolate $x$ :

$x = (a^2 + b^2)/(a + 1/a)$

If we would like, we can rewrite $a + 1/a$ as $(a^2 + 1)/a$ :

$x = (a^2 + b^2)/((a^2 + 1)/a)$

And then simplify a bit, just to make the equation a little prettier:

$x = (a^3 + b^2 a)/(a^2 + 1)$