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

Question

Abigail Allen · Accepted Answer

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)#

Ethan Adkins · Answer

undefined To simplify the expression (3x)/(x²-3x-18)- (x-4)/(x-6), we need to find a common denominator and combine the fractions. The common denominator is (x-6)(x+3).

Multiplying the first fraction by (x-6)/(x-6) and the second fraction by (x+3)/(x+3), we get:

[(3x)(x-6)]/[(x-6)(x+3)] - [(x-4)(x+3)]/[(x-6)(x+3)]

Expanding and simplifying, we have:

(3x^2 - 18x - (x^2 - x - 12))/(x^2 - 3x - 18)

Combining like terms, we get:

(2x^2 - 17x + 12)/(x^2 - 3x - 18)

Therefore, the simplified expression is (2x^2 - 17x + 12)/(x^2 - 3x - 18).