# How do you solve 2/(x(x+2))+3/x=4/(x-2)?

Nov 26, 2017

We can solve this by multiplying everything by the least common multiple $x \left(x + 2\right) \left(x - 2\right)$

#### Explanation:

Note: We have to reject the solutions $x = 0 , x = 2 , x = - 2$, as they would make one or more of the fractions undetermined.

The factors not in the denominators will not be canceled and end up in the numerators:
$\to \frac{2 \cdot \cancel{x} \left(\cancel{x + 2}\right) \left(x - 2\right)}{\cancel{x} \left(\cancel{x + 2}\right)} + \frac{3 \cdot \cancel{x} \left(x + 2\right) \left(x - 2\right)}{\cancel{x}} = \frac{4 \cdot x \left(x + 2\right) \left(\cancel{x - 2}\right)}{\cancel{x - 2}}$

$\to 2 \left(x - 2\right) + 3 \left(x + 2\right) \left(x - 2\right) = 4 x \left(x + 2\right)$

Work out the parentheses and put everything to one side:
$\to 2 x - 4 + 3 {x}^{2} - 12 = 4 {x}^{2} + 8 x$

$\to - {x}^{2} - 6 x - 16 = 0 \to {x}^{2} + 6 x + 16 = 0$

The Discriminant $D = {b}^{2} - 4 a c = {6}^{2} - 4 \cdot 1 \cdot 16 = - 28 < 0$

Since $D < 0$, there are no real solutions.