# How do you solve (x-2)/(x^2-7x+10)=1/(5x-10)-1/(x-5)?

##### 3 Answers
Dec 29, 2015

Simplify by getting everything on the same common denominator and then rearrange to find x

#### Explanation:

$\frac{x - 2}{\left(x - 2\right) \left(x - 5\right)} = \frac{\left(x - 5\right) - \left(5 x - 10\right)}{\left(5 x - 10\right) \left(x - 5\right)}$
(x-2)/((x-2)(x-5)) = (-4x + 5)/(5(x-2)(x-5)
$5 \left(x - 2\right) = \left(5 - 4 x\right)$
$5 x + 4 x = 5 + 10$
$x = \frac{15}{9} = \frac{5}{3}$

Dec 31, 2015

You can simplify the left side.

#### Explanation:

Obtaining the roots for ${x}^{2} - 7 x + 10 = 0$
$\Delta = 49 - 40 = 9$
$x = \frac{7 \pm \sqrt{9}}{2}$ => ${x}_{1} = 2$ and ${x}_{2} = 5$
Then ${x}^{2} - 7 x + 10 = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) = \left(x - 2\right) \left(x - 5\right)$

In the expression:
$\frac{\cancel{x - 2}}{\cancel{\left(x - 2\right)} \left(x - 5\right)} = \frac{1}{5 x - 10} - \frac{1}{x - 5}$
$\frac{2}{x - 5} = \frac{1}{5 x - 10}$
$10 x - 20 = x - 5$
$9 x = 15$ => $x = \frac{5}{3}$

Jan 9, 2016

$x = \frac{5}{3}$

#### Explanation:

Factor each denominator.

$\frac{x - 2}{\left(x - 5\right) \left(x - 2\right)} = \frac{1}{5 \left(x - 2\right)} - \frac{1}{x - 5}$

Simplify the first fraction by dividing $\frac{x - 2}{x - 2} = 1$.

$\frac{1}{x - 5} = \frac{1}{5 \left(x - 2\right)} - \frac{1}{x - 5}$

Add $\frac{1}{x - 5}$ to both sides.

2/(x-5)=1/(5(x-2)

Cross multiply.

$10 \left(x - 2\right) = x - 5$

Simplify and solve.

$10 x - 20 = x - 5$

$9 x - 20 = - 5$

$9 x = 15$

$x = \frac{15}{9}$

$x = \frac{5}{3}$

When solving rational equations such as these, it's always a good idea to go back and check that this answer isn't an invalid answer. Invalidity would occur if $x = \frac{5}{3}$ caused a denominator to equal $0$, but the only two values for $x$ that would cause that here are $2$ and $5$, so this is a valid answer.