How do you solve #2/(x+1) + 5/(x-2)=-2#?

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Answer 1 · 2016-07-24T16:50:34

#x=1/2# or #x=-3#

#2/(x+1)+5/(x-2)=-2#

#hArr(2(x-2)+5(x+1))/((x+1)(x-2))=-2#

#hArr2(x-2)+5(x+1)=-2(x+1)(x-2)#

#hArr2x-4+5x+5=-2(x^2-2x+x-2)#

#hArr7x+1=-2x^2+2x+4#

#hArr2x^2+5x-3=0#

#hArr2x^2+6x-x-3=0#

#hArr2x(x+3)-1(x+3)=0#

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

Hence either #2x-1=0# i.e. #x=1/2#

or #x+3=0# i.e. #x=-3#