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

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Answer 1 · 2014-12-11T20:05:30

Step 1

Make the right side of the equation a fraction:

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

Step 2

Cross multiply:

$-1(3x^2+2x-1) = -2(x^2 - 1)$

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

Step 3

Simplify:

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

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

$-2x = x^2 + 1$

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

Step 4

Solve for $x$ :

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

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

$0 = (x + 1)^2$

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

$x = -1$

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