# How do you solve \frac{3x^2+2x-1}{x^2-1}=-2 using cross multiplication?

Dec 11, 2014

Step 1

Make the right side of the equation a fraction:

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

Step 2

Cross multiply:

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

$- 3 {x}^{2} - 2 x + 1 = - 2 {x}^{2} + 2$

Step 3

Simplify:

$- 2 x + 1 = - 2 {x}^{2} + 3 {x}^{2} + 2$

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

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

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

Step 4

Solve for $x$:

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

We notice that the above represents the expanded form of the perfect square:

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

Because of the perfect square, we know that x has only one value:

$x = - 1$