How do you solve 3/(5-x)+2/(4-x) = 8/(x+2) ?

Sep 17, 2016

$x = 2 \text{ }$ or $\text{ } x = \frac{58}{13}$

Explanation:

The quickest way to find at least one solution is to "guess":

Note that $\frac{3}{5 - x}$ will divide evenly if $5 - x = 3$, i.e. when $x = \textcolor{b l u e}{2}$

Then we find:

$\frac{3}{5 - \textcolor{b l u e}{2}} + \frac{2}{4 - \textcolor{b l u e}{2}} = \frac{3}{3} + \frac{2}{2} = 1 + 1 = 2 = \frac{8}{4} = \frac{8}{\textcolor{b l u e}{2} + 2}$

$\textcolor{w h i t e}{}$
Are there any other solutions?

Given:

$\frac{3}{5 - x} + \frac{2}{4 - x} = \frac{8}{x + 2}$

Subtract $\frac{3}{5 - x} + \frac{2}{4 - x}$ from both sides to get:

$0 = \frac{8}{x + 2} - \frac{3}{5 - x} - \frac{2}{4 - x}$

$\textcolor{w h i t e}{0} = \frac{8}{x + 2} + \frac{3}{x - 5} + \frac{2}{x - 4}$

Multiply through by $\left(x + 2\right) \left(x - 5\right) \left(x - 4\right)$ to get:

$0 = 8 \left(x - 5\right) \left(x - 4\right) + 3 \left(x + 2\right) \left(x - 4\right) + 2 \left(x + 2\right) \left(x - 5\right)$

$\textcolor{w h i t e}{0} = 8 \left({x}^{2} - 9 x + 20\right) + 3 \left({x}^{2} - 2 x - 8\right) + 2 \left({x}^{2} - 3 x - 10\right)$

$\textcolor{w h i t e}{0} = \left(8 {x}^{2} - 72 x + 160\right) + \left(3 {x}^{2} - 6 x - 24\right) + \left(2 {x}^{2} - 6 x - 20\right)$

$\textcolor{w h i t e}{0} = 13 {x}^{2} - 84 x + 116$

We can solve this quadratic by any method we like, but we already know that $x = 2$ is a zero of this quadratic, so $\left(x - 2\right)$ is a factor:

$13 {x}^{2} - 84 x + 116 = \left(x - 2\right) \left(13 x - 58\right)$

So the other solution is $x = \frac{58}{13}$

Sep 17, 2016

$x = 2$ or $x = \frac{58}{13}$

Explanation:

$\frac{3}{5 - x} + \frac{2}{4 - x} = \frac{8}{x + 2}$ can be simplified as

$\frac{3 \left(4 - x\right) + 2 \left(5 - x\right)}{\left(5 - x\right) \left(4 - x\right)} = \frac{8}{x + 2}$ or

$\frac{12 - 3 x + 10 - 2 x}{5 \left(4 - x\right) - x \left(4 - x\right)} = \frac{8}{x + 2}$ or

$\frac{22 - 5 x}{20 - 5 x - 4 x + {x}^{2}} = \frac{8}{x + 2}$ or

$\frac{22 - 5 x}{20 - 9 x + {x}^{2}} = \frac{8}{x + 2}$ or

$8 \left(20 - 9 x + {x}^{2}\right) = \left(x + 2\right) \left(22 - 5 x\right)$

$160 - 72 x + 8 {x}^{2} = x \left(22 - 5 x\right) + 2 \left(22 - 5 x\right)$ or

$160 - 72 x + 8 {x}^{2} = 22 x - 5 {x}^{2} + 44 - 10 x$ or

$8 {x}^{2} + 5 {x}^{2} - 72 x - 22 x + 10 x + 160 - 44 = 0$ or

$13 {x}^{2} - 84 x + 116 = 0$ or

$13 {x}^{2} - 26 x - 58 x + 116 = 0$ or

$13 x \left(x - 2\right) - 58 \left(x - 2\right) = 0$ or

$\left(13 x - 58\right) \left(x - 2\right) = 0$

Hence $x = 2$ or $x = \frac{58}{13}$