Simultaneous Equations

**bobak** · Feb 1st 2008, 06:55 AM

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

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

**wingless** · Feb 1st 2008, 10:35 AM

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

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

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

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

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

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

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

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

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

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

$e^x e^y = 3$

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

$e^x + e^y = 6$

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

Let $a = e^x$

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

$a^2 + 3 = 6a$

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

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

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

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

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