# Some quick help on solving rational equations

• Oct 16th 2011, 03:56 PM
Dante
Some quick help on solving rational equations
How can I solve for X with a question with a form like this :

(2/x+1)+5 = 1/x

and also

x - 5/x = 4

I seem to struggle when they single something out with a denominator of 1 for example the +5 in (2/x+1)+5 and x in x - 5/x

First time posting btw :)
• Oct 16th 2011, 04:02 PM
skeeter
Re: Some quick help on solving rational equations
Quote:

Originally Posted by Dante
How can I solve for X with a question with a form like this :

(2/x+1)+5 = 1/x

and also

x - 5/x = 4

I seem to struggle when they single something out with a denominator of 1 for example the +5 in (2/x+1)+5 and x in x - 5/x

First time posting btw :)

$\displaystyle \frac{2}{x+1} + 5 = \frac{1}{x}$

common denominator ... common denominator ... common denominator

$\displaystyle \frac{2x}{x(x+1)} + \frac{5x(x+1)}{x(x+1)} = \frac{x+1}{x(x+1)}$

the numerators form the equation ...

$\displaystyle 2x + 5x(x+1) = x+1$

solve the resulting quadratic equation keeping in mind that x cannot equal 0 or -1
• Oct 16th 2011, 04:15 PM
Dante
Re: Some quick help on solving rational equations
Quote:

Originally Posted by skeeter
$\displaystyle \frac{2}{x+1} + 5 = \frac{1}{x}$

common denominator ... common denominator ... common denominator

$\displaystyle \frac{2x}{x(x+1)} + \frac{5x(x+1)}{x(x+1)} = \frac{x+1}{x(x+1)}$

the numerators form the equation ...

$\displaystyle 2x + 5x(x+1) = x+1$

solve the resulting quadratic equation keeping in mind that x cannot equal 0 or -1

Oh thank you! Your reply made it a lot clearer, just one question though ... How did the right side turn into x+1 shouldn't it cancel out and turn into x ?

Edit : Oh never mind, I see what happened now. Thanks !