How do you solve #\frac{3}{2x-4}-\frac{2}{x^{2}-x-2}=\frac{9}{x+1}#?

2 Answers
Dean R.
Apr 19, 2018

#x=7/3#

Explanation:

First we factor the denominators, then find a common denominator so we can subtract, then cross multiply and solve the resulting polynomial equation.

#frac{3}{2x-4} - frac{2}{x^2 - x -2 } = frac{9}{x+1}#

#frac{3}{2(x-2)} - frac{2}{(x+1)(x-2)} = frac{9}{x+1}#

#frac{3(x+1) - 2(2)}{2(x+1)(x-2)} = frac{9}{x+1}#

# frac{3x-1}{2(x+1)(x-2)} = frac{9}{x+1}#

# (3x-1)(x+1) = 18(x+1)(x-2) #

# (x+1)( (3x - 1) - 18(x-2) ) = 0#

# (x+1)(-15 x + 35) = 0 #

#x = -1 # or #x = 35/15 = 7/3 #

Shiva Prakash M V
Apr 19, 2018

#x=7/3#

Explanation:

#3/(2x-4)-2/(x^2-x-2)=9/(x+1)#

Factorising

#2x-4=2(x-2)#

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

#=x(x-2)+(x-2)#

#=(x+1)(x-2)#

#3/(2x-4)-2/(x^2-x-2)=3/(2(x-2))-2/((x+1)(x-2))#

#3/(2(x-2))-2/((x+1)(x-2))=9/(x+1)#

#3/(2(x-2))-9/(x+1)=2/((x+1)(x-2))#

Multiplying throughout by #(x+1)(x-2)#

#3/2(x+1)-9(x-2)=2#

#3/2x+3/2-9x+18=2#

#(3/2-9)x+(3/2+18-2)=0#

#-15/2x+35/2=0#

#-15x+35=0#

#15x=35#

#x=35/15#

#x=7/3#

Check:

lhs=
#3/(2x-4)-2/(x^2-x-2)=3/(2xx7/3-4)-2/((7/3)^2-7/3-2)#

#=3/(14/3-4)-2/(49/9-7/3-2)=9/(14-12)-18/(49-21-18)#

#=9/2-18/10=45/10-18/10=(45-18)/10=27/10#

#rhs=9/(x+1)=9/(7/3+1)=27/(7+3)=27/10#

#lhs=rhs#

Related questions
Impact of this question
2727 views around the world
You can reuse this answer
Creative Commons License