How do you simplify #(3x)/(x²-3x-18)- ( x-4)/(x-6)#?

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Answer 1 · 2016-02-17T02:06:11

First, factor the first denominator (#x^2 - 3x - 18#) to see what has to be our LCD (Least Common Denominator)

To factor a trinomial of form #ax^2 + bx + c, a = 1#, you must find two numbers that multiply to c and that add to b. These numbers are -6 and +3.

#(3x)/((x - 6)(x + 3))#

The LCD of the expression is (x - 6)(x + 3).

= #(3x)/((x - 6)(x + 3)) - ((x - 4)(x + 3))/((x - 6)(x + 3))#

= #(3x - (x^2 - 4x + 3x - 12))/((x - 6)(x + 3))#

= #(3x - x^2 + 4x - 3x + 12)/((x - 6)(x + 3))#

= #(-x^2 + 4x + 12)/((x - 6)(x + 3))#

= #(-x^2 + 4x + 12)/(x^2 - 3x - 18)#