# What is the least common multiple for \frac{x}{x-2}+\frac{x}{x+3}=\frac{1}{x^2+x-6} and how do you solve the equations?

May 28, 2018

See explanation

#### Explanation:

$\left(x - 2\right) \left(x + 3\right)$ by FOIL (First, Outside, Inside, Last) is ${x}^{2} + 3 x - 2 x - 6$
which simplifies to ${x}^{2} + x - 6$. This will be your least common multiple (LCM)

Therefore you can find a common denominator in the LCM...
$\frac{x}{x - 2} \left(\frac{x + 3}{x + 3}\right) + \frac{x}{x + 3} \left(\frac{x - 2}{x - 2}\right) = \frac{1}{{x}^{2} + x - 6}$

Simplify to get:
$\frac{x \left(x + 3\right) + x \left(x - 2\right)}{{x}^{2} + x - 6} = \frac{1}{{x}^{2} + x - 6}$
You see the denominators are the same, so take them out.

Now you have the following -
$x \left(x + 3\right) + x \left(x - 2\right) = 1$

Let's distribute; now we have
${x}^{2} + 3 x + {x}^{2} - 2 x = 1$
Adding like terms, $2 {x}^{2} + x = 1$

Make one side equal to 0 and solve quadratic.
$2 {x}^{2} + x - 1 = 0$

Based on Symbolab, the answer is $x = - 1$ or $x = \frac{1}{2}$.