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#