How do you solve #6/(x+2)+5/(x-2)=20/(x^2-4)# and check for extraneous solutions?

1 Answer
Mar 4, 2018

Answer:

This equation has no solution.

The "solution" #x=2# that can be obtained by manipulating this equation is extraneous.

Explanation:

Multiply both sides of the equation

# 6/{x+2}+5/{x-2} = 20/{x^2-4}#

by #x^2 -4# to get

# 6(x-2)+5(x+2) = 20 implies 11x -2 = 20 implies 11x = 22#

This seems to imply that the solution is #x=2#.

Note, however, that the original equation is meaningless for #x=2# (as well as #x=-2#). So, this is an extraneous solution. This does not satisfy the original equation, and, indeed, has arisen because we multiplied both sides of the equation by #x^2-4# - which vanishes when #x=2#.