# How do you solve 1/(x-1)+3/(x+1)=2?

Oct 13, 2015

$x = 0 \text{ }$ or $\text{ } x = 2$

#### Explanation:

You know that your ultimate goal here is to isolate $x$ on one side of the equation.

Start by getting rid of the denominators. Rewrite the equation as

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

The three denominators you're dealing with are

$\left(x - 1\right) \text{ }$, $\text{ "(x+1)" }$, and $\text{ } 1$

To find the common denominator of the three fractions, you need to find the least common multiple of these expressions. Notice that if you multiply the first one by $\left(x + 1\right)$, the second one by $\left(x - 1\right)$, and the third one by $\left(x - 1\right) \left(x + 1\right)$, you get

$\left(x - 1\right) \cdot \left(x + 1\right) \cdot 1 = \left(x - 1\right) \left(x + 1\right)$

This will be your common denominator.

The equation can thus be written as

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

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

This is equivalent to

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

Expand the parantheses and rearrange to get all the terms on one side of the equation

$x + 1 + 3 x - 3 = 2 {x}^{2} - 2$

$2 {x}^{2} - 4 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} = 0$

This is equivalent to

$2 x \left(x - 2\right) = 0$

The two solutions of this equaion will be

$2 x = 0 \implies x = \textcolor{g r e e n}{0}$

and

$x - 2 = 0 \implies x = \textcolor{g r e e n}{2}$

Do quick check to make sure the calculations are correct

$x = 0 \implies \frac{1}{0 - 1} + \frac{3}{0 + 1} = 2$

$- 1 + 3 = 2 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$

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

$1 + 1 = 2 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$