How do you solve #(x-2)/(x-3)+(x-3)/(x-2)=(2x^2)/(x^2-5x+6)#?

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Answer 1 · 2016-11-23T14:44:24

#x = 13/10#

First, you must get each fraction over a common denominator which is #x^2 - 5x + 6#:

#(x - 2)/(x - 2) (x - 2)/(x - 3) + (x - 3)/(x - 3) (x - 3)/(x - 2) = (2x^2)/(x^2 - 5x + 6)#

#(x^2 - 4x + 4)/(x^2 - 5x + 6) + (x^2 - 6x + 9)/(x^2 - 5x + 6) = (2x^2)/(x^2 - 5x + 6)#

Next, add the fractions on the left side of the equation:

#((x^2 - 4x + 4) + (x^2 - 6x + 9))/(x^2 - 5x + 6) = (2x^2)/(x^2 - 5x + 6)#

#(2x^2 - 10x + 13)/(x^2 - 5x + 6) = (2x^2)/(x^2 - 5x + 6)#

We can now multiply each side of the equation by #x^2 - 5x + 6# to eliminate the fraction:

#(x^2 - 5x + 6) (2x^2 - 10x + 13)/(x^2 - 5x + 6) = (x^2 - 5x + 6) (2x^2)/(x^2 - 5x + 6)#

#cancel((x^2 - 5x + 6)) (2x^2 - 10x + 13)/cancel((x^2 - 5x + 6)) = cancel((x^2 - 5x + 6)) (2x^2)/cancel((x^2 - 5x + 6))#

#2x^2 - 10x + 13 = 2x^2#

We can now solve for #x#:

#2x^2 - 10x + 13 - 2x^2= 2x^2 - 2x^2#

#-10x + 13 = 0#

#-10x + 13 - 13 = 0 - 13#

#-10x = -13#

#(-10x)/-10 = (-13)/(-10)#

#x = 13/10#