How do you solve #\frac { x } { x - 2} - \frac { 3x } { x ^ { 2} - x - 2} = \frac { 6} { x + 1}#?

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Answer 1 · 2018-06-13T05:49:54

#x=6# only

#x/(x-2)-(3x)/(x^2-x-2)=6/(x+1)# where #x!=2#

#x/(x-2)-(3x)/((x-2)(x+1))=6/(x+1)#

#(x(x+1)-3x)/((x-2)(x+1))=(6(x-2))/((x+1)(x-2))#

#x^2+x-3x=6x-12#

#x^2-8x+12=0#

#(x-6)(x-2)=0#

#x=6# or #x=2#

But #x!=2#, so #x=6# only