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

Jun 17, 2017

$x = - 5$ or $x = 7$

#### Explanation:

Note : when you multiply something with a fraction, you multiply to the numerator.

First, let's simplify the LHS by multiplying $\frac{x - 5}{x - 1}$ with $2 x + 7$

$\frac{x - 5 \cdot 2 x + 7}{x - 1}$

$\frac{2 {x}^{2} - 10 x + 7 x - 35}{x - 1}$

So we have

$x = \frac{2 {x}^{2} - 3 x - 35}{x - 1}$

Then let's simplify the RHS by multiplying $x$ with $x - 1$.

$\frac{x}{1} = \frac{2 {x}^{2} - 3 x - 35}{x - 1}$ since $x = \frac{x}{1}$,

Then

$x \cdot x - 1 = 2 {x}^{2} - 3 x - 35$

Then ${x}^{2} - x = 2 {x}^{2} - 3 x - 35$

$0 = {x}^{2} - 2 x - 35$

What we just did here is called cross multiplying.

Now, we have a quadratic equation.
We can factorise this to

$\left(x + 5\right) \left(x - 7\right) = 0$

Set each factor equal to $0$, since if either $\left(x + 5\right)$ or $\left(x - 7\right)$ was 0, the equation would hold, because $0$ multiplied by any number will always be 0.

$x + 5 = 0$
$x = - 5$

$x - 7 = 0$
$x = 7$