How do you solve #(x)/(x-2) – (8)/(x+3) = 10/(x^2+x-6)#?

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Answer 1 · 2016-08-29T14:29:01

#x = 3, x!= -3, 2#

Start by factoring everything to see what the denominators are and what we can cancel out.

#=>x/(x - 2) - 8/(x + 3) = 10/((x + 3)(x - 2))#

The common denominator will be #(x + 3)(x - 2)#.

#=>(x(x + 3))/((x + 3)(x - 2)) - (8(x - 2))/((x + 3)(x - 2)) = 10/((x + 3)(x - 2))#

We can now cancel the denominators and solve as a regular quadratic.

#=>x^2 + 3x - 8x + 16 = 10#

#=>x^2 - 5x + 6 = 0#

#=>(x - 3)(x - 2) = 0#

#=>x = 3 and 2#

However, the #x = 2# is extraneous, since it is a restriction on the variable (it makes the denominator equal to #0# and therefore undefined).

Checking in the original equation, #x = 3# works.

Hopefully this helps!