How do you evaluate #\frac { 4} { x ^ { 2} - 25} + \frac { 6} { x ^ { 2} + 6x + 5}#?

1 Answer
Nov 11, 2017
  1. Factor the denominators
  2. Find LCD
  3. Make equivalent fractions with Like denominators
  4. Add like Terms

#(10x - 26)/((x-5)(x+5)(x+1) )#

Explanation:

#4/(x^2 -25) + 6/(x^2 + 6x +5) #

to factor #x^2-25# you need to understand difference of squares where #(x-5)(x+5) = x^2 -25#

#4/((x-5)(x+5)) + 6/((x+1)(x+5)#

our LCD is #(x-5)(x+5)(x+1) #

#4/((x-5)(x+5)) * ((x+1))/((x+1)) = (4(x+1))/((x-5)(x+5)(x+1)# and

#6/((x+1)(x+5)) * ((x-5))/((x-5)) = (6(x-5))/((x-5)(x+5)(x+1))#

#(4(x+1) + 6(x-5))/((x-5)(x+5)(x+1) )#

Distributive Property

#(4x + 4 + 6x - 30)/((x-5)(x+5)(x+1) )#

add like terms

#(10x - 26)/((x-5)(x+5)(x+1) )# Fully simplified