How do you evaluate #\frac { 1} { x + 1} + \frac { 3} { 3x - 13} = \frac { 2} { x - 7}#?

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Answer 1 · 2017-07-05T06:43:08

#x=3#

First, we will transfer all the terms to one side and then get one fraction:

#1/(x+1)+3/(3x-13)-2/(x-7)=0#

#((3x-13)(x-7)+3(x+1)(x-7)-2(x+1)(3x-13))/((x+1)(3x-13)(x-7))=0#

Then we will expand the numerator:

#cancel(3x^2)-21x-13x+91cancel(+3x^2)-21x+3x-21cancel(-6x^2)+26x-6x+26=0#

and state that

#x!=-1# and #x!=13/3 # and #x!=7#

Let's sum like terms and get:

#-32x+96=0#

and the solution is:

#x=96/32=3#