How do you solve #\frac { x - 6} { x + 2} + 1= \frac { x - 2} { x + 3}#?

1 Answer
Andy Y.
May 30, 2017

#-8.899# and #0.899#

Explanation:

The first step is to get rid of all the fractions. You can easily do this by every term by the product of every denominator. So you have two denominators: #x+2# and #x+3#.

The product of the denominators is: #(x+2)(x+3)#

Multiplying every single term by this product gives,

#(x+2)(x+3) (x-6)/(x+2)+(x+2)(x+3)(1)=(x+2)(x+3) (x-2)/(x+3)#

Cancel out terms that are in both numerator and denominator:

#cancel((x+2))(x+3) (x-6)/cancel((x+2))+(x+2)(x+3)(1)=(x+2)cancel((x+3)) (x-2)/cancel((x+3))#

This simplifies to

#(x+3) (x-6)+(x+2)(x+3)(1)=(x+2) (x-2)#

On the left hand side, the term #(x+3)# factors out of both terms.

#(x+3)(x-6+x+2)=(x+2)(x-2)#

Simplifying the left even more gives

#(x+3)(2x-4)=(x+2)(x-2)#

Factoring out the left and right gives

#2x^2+8x-12=x^2-4#

#x^2+8x-12=-4#

#x^2+8x-8=0#

This doesn't factor easily, so using the quadratic equation gives

#x=(-8+-sqrt(64-4(1)(-8)))/(2(1))=-4+-sqrt(96)/(2)~~-8.899# and #0.899#

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