How do you solve #(9/x)+9/(x-2) = 12#?

1 Answer
Dec 5, 2016

Answer:

#x = 1/2# and #x = 3#

Explanation:

First, you can get each fraction over a common denominator, in this case #x(x - 2)#, so the fractions can be added:

#((x-2)/(x-2))(9/x) + (x/x)(9/(x-2)) = 12#

#(9(x-2) + 9x)/(x(x-2)) = 12#

We can now expand and add the numerator:

#(9x-18 + 9x)/(x(x-2)) = 12#

#(18x-18)/(x(x-2)) = 12#

We can now eliminate the fraction through multiplication on each side of the equation to keep the equation balanced:

#(18x-18)/(x^2-2x) = 12#

#(x^2 - 2x)(18x - 18)/(x^2-2x) = 12(x^2 - 2x)#

#cancel((x^2 - 2x))(18x - 18)/cancel((x^2-2x)) = 12x^2 - 24x#

#18x - 18 = 12x^2 - 24x#

We can now make a single quadratic and factor:

#18x - 18 - 18x + 18 = 12x^2 - 24x - 18x + 18#

#12x^2 - 42x + 18 = 0#

#(6x - 3)(2x - 6) = 0#

We can now solve each factor for #0#:

#6x - 3 = 0#

#6x - 3 + 3 = 0 + 3#

#6x = 3#

#(6x)/6 = 3/6#

#x = 1/2#

and

#2x - 6 = 0#

#2x - 6 + 6 = 0 + 6#

#2x = 6#

#(2x)/2 = 6/2#

#x = 3#