How do you solve #(3x)/(x+1)=12/(x^2-1)+2#?

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Answer 1 · 2017-09-23T12:30:38

Solution: #x=5 , x= -2#

# (3x)/(x+1)= 12/(x^2-1) +2# or

# (3x)/(x+1) -12/(x^2-1) = 2# Multiplying by #(x^2-1)# on both

sides we get # 3x(x-1) -12 = 2 (x^2-1) # or

# 3x^2-3x -12 = 2x^2-2 # or

# x^2-3x = 10 or x^2 -3x +9/4 = 10 +9/4 ; (9/4)# is added

on both side to make L.H.S a square.

# (x-3/2)^2 = 49/4 or x-3/2 = +- sqrt (49/4)# or

#x-3/2 = +- 7/2 :. x =3/2 +- 7/2 or x = 1/2( 3+-7) :.#

#x=5 , x= -2# . Solution: #x=5 , x= -2# [Ans]