How do you solve the rational equation 1/(x-1)+3/(x+1)=2?

Dec 18, 2015

$x = 0 , x = 2$

Explanation:

Step 1 : Identify the restricted value.

This is done by set the denominator equal to zero like this

$x - 1 = 0 \iff x = 1$
$x + 1 = 0 \iff x = - 2$

The idea of restricted value, is to narrow down what value our variable can't be (aka domain)

Step 2: Multiply the equation by $\textcolor{red}{L C D}$

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

color(red)((x-1)(x+1))(1/(x-1)) +color(red)( (x-1)(x+1))(3/(x+1)) = 2color(red)((x-1)(x+1)

color(red)(cancel(x-1)(x+1))(1/cancel(x-1)) +color(red)( (x-1)cancel(x+1))(3/cancel(x+1)) = 2color(red)((x-1)(x+1)

$\left(x + 1\right) + 3 \left(x - 1\right) = 2 \left(x - 1\right) \left(x + 1\right)$

Step 3: Multiply and combine like terms

$x + 1 + 3 x - 3 = 2 \left({x}^{2} - x + x - 1\right)$

$4 x - 2 = 2 \left({x}^{2} - 1\right)$

$4 x - 2 = 2 {x}^{2} - 2$

$0 = 2 {x}^{2} - 4 x$

Step 4: Solve the quadratic equation

$2 {x}^{2} - 4 x = 0$
$2 x \left(x - 2\right) = 0$

$2 x = 0 \implies \textcolor{b l u e}{x = 0}$

$x - 2 = 0 \implies \textcolor{b l u e}{x = 2}$

Step 5 Check your solution ..

Check to see if the answer from Step 4 is the same a restricted value.

If it's not, the the solution is $x = 0 , x = 2$