Question #5eee2

Aug 18, 2014

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 $x$:

$x = y a b$

Now, substitution into the second equation gives:

${a}^{2} b y + b y = {a}^{2} + {b}^{2}$

Factor out $y$:

$y \left({a}^{2} b + b\right) = {a}^{2} + {b}^{2}$

And lastly, divide:

$y = \frac{{a}^{2} + {b}^{2}}{{a}^{2} b + b}$

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

$y = \frac{x}{a b}$

Substitute into the second equation:

$a x + \frac{x}{a} = {a}^{2} + {b}^{2}$

Factor $x$:

$x \left(a + \frac{1}{a}\right) = {a}^{2} + {b}^{2}$

And lastly, divide to isolate $x$:

$x = \frac{{a}^{2} + {b}^{2}}{a + \frac{1}{a}}$

If we would like, we can rewrite $a + \frac{1}{a}$ as $\frac{{a}^{2} + 1}{a}$:

$x = \frac{{a}^{2} + {b}^{2}}{\frac{{a}^{2} + 1}{a}}$

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

$x = \frac{{a}^{3} + {b}^{2} a}{{a}^{2} + 1}$