# How do you solve x+ \frac { - 3} { x } = 2?

Mar 27, 2017

$x = - 1 \mathmr{and} 3$

#### Explanation:

$x \times \left(x + - \frac{3}{x} = 2\right)$ is the first step which gives

${x}^{2} - 3 = 2 x$ Now subtract $2 x$ from each side.

${x}^{2} - 2 x - 3 = 2 x - 2 x$ resulting in

${x}^{2} - 2 x - 3 = 0$ break this into two binomials.

$- 3$ means that one of the factors of three must be negative and one must be positive.

$- 2$ means that the negative factor must be larger than the positive factor.

The possible factors of $- 3$ are $1 x \left(- 3\right)$ and $- 1 x \left(+ 3\right)$ using the information above the correct choice is $+ 1 x - 3$ so

$\left(x + 1\right) \times \left(x - 3\right) = 0$ solve for each binomial

$x + 1 = 0$ subtract -1 from each side

$x + 1 - 1 = 0 - 1$ so

$x = - 1$

$x - 3 = 0$ add three to each side

$x - 3 + 3 = 0 + 3$

$x = + 3$

Put in this values to make sure there are no extraneous variables:

$- 1 + \left(- \frac{3}{-} 1\right) = 2$

That works. Now try the other one,

$3 + \left(- \frac{3}{3}\right) = 2$

That works too. So both work!