Thread: Simultaneous Equations

$\displaystyle \cosh x + \cosh y = 4$
$\displaystyle \sinh x + \sinh y = 2$

I didn't get very far after writing it out in exponential form.
Anyone willing to give me a pointer?

$\displaystyle \cosh x + \cosh y = 4$
$\displaystyle \sinh x + \sinh y = 2$

We know that $\displaystyle \cosh x + \sinh x = e^x$ and $\displaystyle \cosh x - \sinh x = e^{-x}$.

Then,
$\displaystyle \cosh x + \cosh y = 4$
$\displaystyle \sinh x + \sinh y = 2$

Add:
$\displaystyle \cosh x + \cosh y + \sinh x + \sinh y = 6$
$\displaystyle e^x + e^y = 6$

Subtract:
$\displaystyle \cosh x + \cosh y - \sinh x - \sinh y = 2$
$\displaystyle e^{-x} + e^{-y} = 2$

So we have,
$\displaystyle e^x + e^y = 6$

$\displaystyle e^{-x} + e^{-y} = 2$

$\displaystyle \frac{1}{e^x} + \frac{1}{e^y} = 2$

$\displaystyle \frac{e^x + e^y}{e^x e^y} = 2$

$\displaystyle \frac{6}{e^x e^y} = 2$

$\displaystyle e^x e^y = 3$

$\displaystyle e^y = \frac{3}{e^x}$

$\displaystyle e^x + e^y = 6$

$\displaystyle e^x + \frac{3}{e^x} = 6$

Let $\displaystyle a = e^x$

$\displaystyle a + \frac{3}{a} = 6$

$\displaystyle a^2 + 3 = 6a$

$\displaystyle a^2 - 6a + 3 = 0$

$\displaystyle a = \frac{6 \mp \sqrt{24}}{2} = 3 \mp \sqrt{6}$

$\displaystyle x = \ln (3 \mp \sqrt{6})$

I think you can find $\displaystyle y$s for the $\displaystyle x$s we found