
Steve S. answered 04/07/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
f(x) = 6/(x^2-9) – 9/(x^2–6x+9)
f(x) = 6/((x-3)(x+3)) – 9/((x–3)(x–3))
As in all fraction addition/subtraction problems, we need the same denominator before we can subtract:
f(x) = 6(x-3)/((x-3)(x-3)(x+3))
f(x) = 6/((x-3)(x+3)) – 9/((x–3)(x–3))
As in all fraction addition/subtraction problems, we need the same denominator before we can subtract:
f(x) = 6(x-3)/((x-3)(x-3)(x+3))
– 9(x+3)/((x–3)(x–3)(x+3))
f(x) = (6(x-3) – 9(x+3))/((x–3)(x–3)(x+3))
f(x) = (6x-18 – 9x-27))/((x–3)(x–3)(x+3))
f(x) = (-3x-45))/((x–3)(x–3)(x+3))
f(x) = -3(x+15))/((x+3)(x–3)^2)
f(x) = (6(x-3) – 9(x+3))/((x–3)(x–3)(x+3))
f(x) = (6x-18 – 9x-27))/((x–3)(x–3)(x+3))
f(x) = (-3x-45))/((x–3)(x–3)(x+3))
f(x) = -3(x+15))/((x+3)(x–3)^2)

Steve S.
Strictly PEMDAS. Parentheses used to contain quantities like numerators and denominators. / is divide.
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04/07/14
Dan M.
04/07/14