# Rational equation with two solutions

• March 15th 2012, 12:56 PM
vaironxxrd
Rational equation with two solutions
Problem:
$\frac{5x}{x-1}-3=\frac{2x+5}{x^2-1}$

Solution
$x=2$
$x=-\frac{1}{2}$
• March 15th 2012, 01:11 PM
skeeter
Re: Rational equation with two solutions
Quote:

Originally Posted by vaironxxrd
Problem:
$\frac{5x}{x-1}-3=\frac{2x+5}{x^2-1}$

Solution
$x=2$
$x=-\frac{1}{2}$

$\frac{5x}{x-1}-3=\frac{2x+5}{x^2-1}$

common denominator ...

$\frac{5x(x+1)}{(x-1)(x+1)}-\frac{3(x^2-1)}{x^2-1}=\frac{2x+5}{x^2-1}$

numerators form the equation ...

$5x(x+1) - 3(x^2-1) = 2x+5 \, \, ; \, \, x \ne \pm 1$

take it from here?
• March 15th 2012, 01:25 PM
vaironxxrd
Re: Rational equation with two solutions
Quote:

Originally Posted by skeeter
$\frac{5x}{x-1}-3=\frac{2x+5}{x^2-1}$

common denominator ...

$\frac{5x(x+1)}{(x-1)(x+1)}-\frac{3(x^2-1)}{x^2-1}=\frac{2x+5}{x^2-1}$

numerators form the equation ...

$5x(x+1) - 3(x^2-1) = 2x-5 \, \, ; \, \, x \ne \pm 1$

take it from here?

I had a very similar equation, though I have 2x+5 at the end.
I cannot understand why was 2x+5 switched to 2x-5
• March 15th 2012, 01:28 PM
skeeter
Re: Rational equation with two solutions
typo
• March 15th 2012, 01:35 PM
vaironxxrd
Re: Rational equation with two solutions
Quote:

Originally Posted by skeeter
typo

$2x^2+3x-2=0$

$(x+2)(2x-1)$

x = -2
&
x = 1/2