1. ## Simultaneous equations?

Hm... I've been asked to find x and y in terms of a, b, c, d, e, f. It's supposed to be 'simple', but I keep getting the wrong answer...I've tried equating the equations and multiplying out, but there're just too many variables, please help .

$\displaystyle y=a+\left( \frac{b+c-2d}{e+f-2d} \right)(x-d)$

$\displaystyle y=e+\left( \frac{a+c-2e}{d+f-2e} \right)(x-e)$

The solution should be:

$\displaystyle x=\frac{d+e+f}{3}$

$\displaystyle y=\frac{a+b+c}{3}$

Ah well, thanks in advance to whoever can be bothered to solve it .

2. Originally Posted by SuumEorum
Hm... I've been asked to find x and y in terms of a, b, c, d, e, f. It's supposed to be 'simple', but I keep getting the wrong answer...I've tried equating the equations and multiplying out, but there're just too many variables, please help .

$\displaystyle y=a+\left( \frac{b+c-2d}{e+f-2d} \right)(x-d)$

$\displaystyle y=e+\left( \frac{a+c-2e}{d+f-2e} \right)(x-e)$

The solution should be:

$\displaystyle x=\frac{d+e+f}{3}$

$\displaystyle y=\frac{a+b+c}{3}$
Just looking at the given solution, it seems that you have written the equations wrongly. I think they should be

$\displaystyle y=a+\left( \frac{b+c-2{\color{red}a}}{e+f-2d} \right)(x-d)$

$\displaystyle y={\color{red}b}+\left( \frac{a+c-2{\color{red}b}}{d+f-2e} \right)(x-e)$

3. Aah I don't believe it, I've been solving the wrong equations, no wonder they've been impossible. Thanks for spotting that for me .