Question 74460
the numerator is:
[(2(x+3)-3(x-3)]/(x-3)(x+3)
(2x+6-3x+9)/(x-3)(x+3)
(-x+15)/(x-3)(x+3)
now the denominator
[5(x-3)-4(x^2-9)]/(x^2-9)(x-3)
(5x-15-4x^2+36)/(x-3)(x+3)(x-3)
(-4x^2+5x+21)/(x-3)(x+3)(x-3)
(-x+3)(4x+7)/(x-3)(x+3)(x-3)
-----------------------------------------
now we invert the denominator & multiply
(-x+15)/(x-3)(x+3)*(x-3)(x+3)(x-3)/(-x+3)(4x+7) now we cancel out all common terms in the numerator & the denominator:
(-x+15)(x-3)/(-x+3)(4x+7)