x-p/x+p = p+q/p-q

sorry about the...inaccuracy.please solve this I tried this like 10 times already and i always get 2(xq+p^{2}) or something and I know its wrong. please help me

thnx

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- Dec 26th 2012, 07:27 AMNormalKidbasic algebra
x-p/x+p = p+q/p-q

sorry about the...inaccuracy.please solve this I tried this like 10 times already and i always get 2(xq+p^{2}) or something and I know its wrong. please help me

thnx

- Dec 26th 2012, 07:39 AMPlatoRe: basic algebra
- Dec 26th 2012, 07:59 AMNormalKidRe: basic algebra
the first one

how do u do thaat? - Dec 26th 2012, 08:09 AMSorobanRe: basic algebra
Hello, NormalKid!

What is the question?

I'll take a guess . . .

Quote:

$\displaystyle \text{Solve for }x\!:\;\frac{x-p}{x+p} \:=\:\frac{p+q}{p-q}$

Cross-multiply: .$\displaystyle (x-p)(p-q) \:=\:(x+p)(p+q)$

.n. . . . . . . . $\displaystyle {\color{red}\rlap{//}}px - qx - p^2 + {\color{green}\rlap{//}}pq \:=\:{\color{red}\rlap{//}}px + qx + p^2 + {\color{green}\rlap{//}}pq$

n . . . . . . . . . . . . . . . . . . $\displaystyle 2qx \:=\:-2p^2$

. . . . . . . . . . . . . . . . . . . . . $\displaystyle x \:=\:-\frac{p^2}{q}$

- Dec 26th 2012, 08:14 AMNormalKidRe: basic algebra