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

Sep 7, 2016

$\implies x = 0 \text{ and } x = 1$

#### Explanation:

Configuring the equation so that both sides end up with the same denominator.

Write $1 \text{ as } 1 = \frac{x}{x}$ giving

$\frac{2}{x - 1} + \frac{x}{x} = \frac{2}{{x}^{2} - x}$

But ${x}^{2} - x = x \left(x - 1\right)$

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

Consider just the left hand side

$\frac{2}{x - 1} + \frac{x}{x} \text{ " ->" } \frac{2 + x \left(x - 1\right)}{x \left(x - 1\right)}$

Putting it all together

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

As the denominators are the same on both sides the equation is equally true if you just equate the numerators to each other and totally forget about the denominators.

$2 + x \left(x - 1\right) = 2$

$x \left(x - 1\right) = 0$

$\implies x = 0 \text{ and } x = 1$

Sep 7, 2016

$x = - 1 , 2$

#### Explanation:

We have: $\frac{2}{x - 1} + 1 = \frac{2}{{x}^{2} - x}$

Let's simplify the left-hand side of the equation:

$\implies \frac{2 + \left(x - 1\right)}{x - 1} = \frac{2}{{x}^{2} - x}$

$\implies \frac{x + 1}{x - 1} = \frac{2}{{x}^{2} - x}$

Then, let's cross-multiply:

$\implies \left(x - 1\right) \left({x}^{2} - x\right) = 2 \left(x - 1\right)$

We can expand the parentheses to get:

$\implies {x}^{3} - 2 {x}^{2} + x = 2 x - 2$

Let's move all terms to the left-hand side to form a cubic equation:

$\implies {x}^{3} - 2 {x}^{2} - x + 2 = 0$

$\implies \left({x}^{3} - 2 {x}^{2}\right) + \left(- x + 2\right) = 0$

Now, let's factorise:

$\implies {x}^{2} \left(x - 2\right) - \left(x - 2\right) = 0$

$\implies \left(x - 2\right) \left({x}^{2} - 1\right) = 0$

The difference of squares can be simplified to:

$\implies \left(x - 2\right) \left(x + 1\right) \left(x - 1\right) = 0$

$\implies x = - 1 , 1 , 2$

However, using $x = 1$ yields an undefined result.

Therefore, the solutions to the equation are $x = - 1$ and $x = 2$.