How do you solve #(x^2-3)/(x+2)=(x-3)/2#?

1 Answer
May 31, 2017

Answer:

#x=0,-1#

Explanation:

Solve:

#(x^2-3)/color(blue)((x+2))=(x-3)/color(red)(2)#

Cross multiply.

#color(red)(2)(x^2-3)=color(blue)((x+2))(x-3)#

Expand.

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

FOIL the right side.

#2x^2-6=x^2-3x+2x-6#

Simplify.

#2x^2-6=x^2-x-6#

Add #6# to both sides.

#2x^2-color(red)cancel(color(black)(6))+color(red)cancel(color(black)(6))=x^2-x-color(red)cancel(color(black)(6))+color(red)cancel(color(black)(6))#

Simplify.

#2x^2=x^2-x#

Move all terms to the left side.

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

Simplify.

#x^2+x=0#

Factor out the #x#.

#x(x+1)=0#

Solutions for #x#.

#x=0#

#x=-1#