How do you factor #2/3 = 2 - (5x-3)/(x-1)#?

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Answer 1 · 2016-08-15T19:18:18

#11x - 5= 0# does not factor, but can be solved to give #x = 5/11#

We will need to simplify first. This is not in any recognisable form for factoring.

#2/3 = 2 - (5x-3)/(x-1)" "color(blue)(xx 3(x-1))#

Get rid of the denominators:

#2/cancel3color(blue)(xx cancel3(x-1)) = 2color(blue)(xx 3(x-1)) - (5x-3)/cancel((x-1)) color(blue)(xx 3cancel((x-1))#

#2(x-1) = 6(x-1) - 3(5x-3)#

#2x-2 = 6x-6 -15x+9#

#2x-6x+15x = -6+9+2#

This simplifies to a simple linear equation and does not factorise.

#11x = 5#