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

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Answer 1 · 2016-08-09T18:46:21

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

In this equation, which has fractions, we may not cross multiply because there are two terms on the left side.

However we can get rid of the denominators by multiplying through by the LCM of the denominators which is #color(red)((x+1)(x-2))#

#color(red)(cancel((x+1))(x-2)xx)2/cancel((x+1)) + color(red)((x+1)cancel((x-2))xx)5/cancel(x-2)=-2color(red)((x+1)(x-2))#

This leaves us with:

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

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

#7x+1 = -2x^2+2x+4#

#2x^2 +5x -3=0 " make the quadratic" = 0#

#(2x-1)(x+3) = 0" factorise"#

#2x -1 =0" " rArr x = 1/2#

#x+3 = 0 " "rArr x = -3#