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

Aug 17, 2017

$x = \frac{1}{2} \left(1 \pm \sqrt{3}\right)$

#### Explanation:

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

Cross multiply

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

$2 {x}^{2} - 2 x - 1 = 0$

For a quadratic equation of the form: $a {x}^{2} + b x + c$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In our case: $a = 2 , b = - 2 , c = - 1$

$\therefore x = \frac{+ 2 \pm \sqrt{4 + 4 \cdot 2 \cdot 1}}{2 \cdot 2}$

$= \frac{+ 2 \pm \sqrt{12}}{4}$

$= \frac{2 \pm 2 \sqrt{3}}{4}$

$= \frac{1}{2} \left(1 \pm \sqrt{3}\right)$

Aug 17, 2017

Multiply through by the denominator of the right-hand side and then solve the resulting quadratic equation.

$x = 1.366$ or $- 0.366$

#### Explanation:

$\frac{2 x}{1}$ is just $2 x$, so

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

Multiply both sides by $\left(x - 1\right)$

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

$2 {x}^{2} - 2 x = 1$

$2 {x}^{2} - 2 x - 1 = 0$

Now we can use the quadratic formula (or any other method) to solve this quadratic equation.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{2 \pm \sqrt{4 + 8}}{4}$

$x = 1.366$ or $- 0.366$