# How do you solve 5/x - 2 = 2/(x+3)?

Oct 20, 2015

${x}_{1 , 2} = \frac{- 3 \pm \sqrt{129}}{4}$

#### Explanation:

$\frac{5}{x} - 2 = \frac{2}{x + 3}$

Right from the start, you know that $x$ cannot take a value that would make the two denominators equal to zero. This means that you need

$x \ne 0 \text{ }$ and $\text{ } x + 3 \ne 0 \implies x \ne - 3$

With this in mind, use the common denominator $x \cdot \left(x + 3\right)$ to get rid of the denominators. Multiply the first fraction by $1 = \frac{x + 3}{x + 3}$, multiply $2$ by $1 = \frac{x \left(x + 3\right)}{x \left(x + 3\right)}$, and the second fraction by $1 = \frac{x}{x}$ to get

$\frac{5}{x} \cdot \frac{x + 3}{x + 3} - 2 \cdot \frac{x \left(x + 3\right)}{x \left(x + 3\right)} = \frac{2}{x + 3} \cdot \frac{x}{x}$

This is equivalent to

$\frac{5 \cdot \left(x + 3\right)}{x \left(x + 3\right)} - \frac{2 x \left(x + 3\right)}{x \left(x + 3\right)} = \frac{2 x}{x \left(x + 3\right)}$

You can thus say that

$5 \left(x + 3\right) - 2 x \left(x + 3\right) = 2 x$

Expand the parantheses to get

$5 x + 15 - 2 {x}^{2} - 6 x = 2 x$

Rearrange by getting all the terms on one side of the equation

$2 {x}^{2} + 3 x - 15 = 0$

Use the quadratic formula to find the two solutions

${x}_{1 , 2} = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \cdot 2 \cdot \left(- 15\right)}}{2 \cdot 2}$

${x}_{1 , 2} = \frac{- 3 \pm \sqrt{129}}{4}$

The two solutions to the original equation will thus be

${x}_{1} = \frac{- 3 - \sqrt{129}}{4} \text{ }$ and $\text{ } {x}_{2} = \frac{- 3 + \sqrt{129}}{4}$

Do a quick check to make sure that the calculations are correct - I'll do them for $x = \frac{- 3 - \sqrt{129}}{4}$

$\frac{5}{\frac{- 3 - \sqrt{129}}{4}} - 2 = \frac{2}{\frac{- 3 - \sqrt{129}}{4} + 3}$

$\frac{20}{- 3 - \sqrt{129}} = \frac{8}{9 - \sqrt{129}} + 2$

$- 1.39296944 \ldots = - 3.39296944 \ldots + 2 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$