How do you combine #(3x-2)*[x/(3x^2+x-2)+2/(x+1)]#?

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Answer 1 · 2016-10-01T15:41:23

#(3x-2)*[x/(3x^2+x-2)+2/(x+1)]=(7x-4)/(x+1)#

Let us first factorize #3x^2+x-2#

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

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

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

Hence #(3x-2)*[x/(3x^2+x-2)+2/(x+1)]#

= #(3x-2)*[x/((3x-2)(x+1))+2/(x+1)]#

= #(3x-2)xx x/((3x-2)(x+1))+(3x-2)xx2/(x+1)#

= #cancel(3x-2)xx x/(cancel((3x-2))(x+1))+(2(3x-2))/(x+1)#

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

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

= #(x+6x-4)/(x+1)#

= #(7x-4)/(x+1)#