How do you solve #\frac { 9} { x ^ { 2} - 2x - 8} + \frac { x } { x + 2} = 2#?

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Answer 1 · 2017-06-21T10:36:15

Solution: #x=5 or x=-5#

#9/(x^2-2x-8)+x/(x+2)=2# or

#9/((x-4)(x+2))+x/(x+2)=2# . Multiplying by #(x-4)(x+2)# on both sides we get ,

#9+x(x-4)=2(x-4)(x+2)# or

#9+x^2-cancel(4x)=2x^2-cancel(4x)-16# or

#2x^2-x^2-16-9=0# or

#x^2-25=0 or (x+5)(x-5)=0#. Either #x+5=0 :. x = -5# or

#x-5=0 :. x = 5#

Solution: #x=5 or x=-5# [Ans]