How do you solve #\frac { x } { x + 6} - \frac { 9} { x - 3} = \frac { - 13x + 2} { x ^ { 2} + 3x - 18}#?

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Answer 1 · 2017-03-11T12:42:38

#x_1=-8 and x_2=7#

Since

#x^2+3x-18=(x+6)(x-3)#

you get the equivalent equation:

#x/(x+6)-9/(x-3)-(-13x+2)/((x+6)(x-3))=0#

#(x(x-3)-9(x+6)-(-13x+2))/((x+6)(x-3))=0#

#x^2-3x-9x-54+13x-2=0 and x!=-6 and x!=3#

#x^2+x-56=0#

#x=(-1+-sqrt(1-4*(-56)))/2#

#x=(-1+-15)/2#

#x_1=-8 and x_2=7#