basic algebra

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December 26th 2012, 07:27 AM
NormalKid

basic 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+p²) or something and I know its wrong. please help me
thnx
December 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 $\frac{x-p}{x+p}=\frac{p+q}{p-q}~?$

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

Which is it?
December 26th 2012, 07:59 AM
NormalKid

Re: basic algebra

the first one
how do u do thaat?
December 26th 2012, 08:09 AM
Soroban

Re: basic algebra

Hello, NormalKid!

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

Quote:

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

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

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

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

. . . . . . . . . . . . . . . . . . . . . $x \:=\:-\frac{p^2}{q}$
December 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: . $(x-p)(p-q) \:=\:(x+p)(p+q)$

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

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

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

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