# Thread: Rational equation with two solutions

1. ## Rational equation with two solutions

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

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

2. ## Re: Rational equation with two solutions

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?

3. ## Re: Rational equation with two solutions

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

typo

5. ## Re: Rational equation with two solutions

Originally Posted by skeeter
typo
$2x^2+3x-2=0$

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

x = -2
&
x = 1/2