# Thread: Hard simultaneous equation.

1. ## Hard simultaneous equation.

For the simultaneous equations ax+by=p and bx-ay=q, show that x=(ap+bq)/(a^2+b^2) and y=(bp-aq)/(a^2+b^2)
I just can't figure out which method to use and how to do it. Because I've never done one without numbers before.
Many thanks in advance.

2. Do it the same way as you would with numbers...

$\displaystyle a\,x + b\,y = p$
$\displaystyle b\,x - a\,y = q$.

Multiply equation 1 by $\displaystyle b$ and multiply equation 2 by $\displaystyle a$ to get

$\displaystyle ab\,x + b^2y = bp$
$\displaystyle ab\,x - a^2y = aq$.

Now subtract equation 2 from equation 1 to eliminate the $\displaystyle x$ term.

3. Thanks for that. I ended up getting y=(aq-bp)/(a^2+b^2).
But now how do I get the x value, I know I have to sub in the y but which equation do I sub it into?

4. Either one, since the solution satisfies both equations at the same time.

BTW you should have gotten $\displaystyle y = \frac{bp-aq}{a^2 + b^2}$.

5. Ok, thanks so much.
I was wondering why mine wasn't working.
I got it all fixed up now.