How do you solve #2x ( x - 3) ^ { - 1} + 4( x + 3) ^ { - 1} = - 24( x ^ { 2} - 9) ^ { - 1}#?

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Answer 1 · 2016-12-10T01:49:07

#x= -2#.

Start by rewriting using positive exponents only, using the rule #a^-n = 1/a^n#.

#(2x)/(x- 3) + 4/(x + 3) = -24/(x^2 - 9)#

Factor the denominators as much as possible to determine the least common denominator (LCD).

#(2x)/(x - 3) + 4/(x + 3) = -24/((x + 3)(x - 3))#

#:.# The LCD is #(x + 3)(x - 3)#. Multiply to make all the denominators equal to this product.

#(2x(x + 3))/((x + 3)(x- 3)) + (4(x - 3))/((x + 3)(x - 3)) = -24/((x + 3)(x - 3))#

We can now eliminate the denominators since everything is equivalent.

#2x^2 + 6x + 4x - 12 = -24#

#2x^2 + 10x + 12 = 0#

#2(x^2 + 5x + 6) = 0#

#(x + 3)(x + 2) = 0#

#x = -3 and -2#

However, our non permissible values are #x != +-3#, so #x = -3# is extraneous. Therefore, the only valid solution is #x = -2#.

Hopefully this helps!