How do you solve # 1/(x-2) + 3/ (x+3) = 4/(x^2 +x-6)#?

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Answer 1 · 2018-03-14T18:50:17

#x = 7/4#

Note that:

#(x-2)(x+3) = x^2+x-6#

So multiplying both sides of the given equation by #x^2+x-6# we get:

#(x+3)+3(x-2) = 4#

which simplifies to:

#4x-3 = 4#

Add #3# to both sides to get:

#4x = 7#

Divide both sides by #4# to get:

#x = 7/4#

Since this value makes none of the denominators #0#, it is a valid solution of the original equation.