# basic algebra

• Dec 26th 2012, 07:27 AM
NormalKid
basic algebra
x-p/x+p = p+q/p-q

thnx

• Dec 26th 2012, 07:39 AM
Plato
Re: basic algebra
Quote:

Originally Posted by NormalKid
[SIZE=6]x-p/x+p = p+q/p-q

First of all, please use grouping symbols.

Is the problem $\displaystyle \frac{x-p}{x+p}=\frac{p+q}{p-q}~?$

The way you have posted it reads:
$\displaystyle x-\frac{x}{p}+p=p+\frac{q}{p}-q$.

Which is it?
• Dec 26th 2012, 07:59 AM
NormalKid
Re: basic algebra
the first one
how do u do thaat?
• Dec 26th 2012, 08:09 AM
Soroban
Re: 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 AM
NormalKid
Re: basic algebra
Quote:

Originally Posted by Soroban
Hello, NormalKid!

What is the question?
I'll take a guess . . .

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

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

good guessing Soroban sir.
so my guess was correct !
THANK YOU EVERYBODY. SP.THANKS TO YOU SIR. :)