Simplify: [(x2 - 2x - 3)/x2 - x - 2)]/[(x2 - 4x + 3)/(x2 + 5x - 6)]?

Answer 1

simplify each part if possible. (IN ths case yes)

x^2-2x-3 = (x-3)(x+1)
x^2-x-2 = (x-2)(x+1)

so numerator = (x^2-2x-3)/(x^2-x-2) = (x-3)/(x-2)

x^2-4x+3 = (x-1)(x-3)
x^2+5x-6 = (x+6)(x-1)

denominator = (x-1)(x-3)/((x+6)(x-1) = (x-3)/(x+6)

resullt = ((x-3)/(x-2))/((x-3)(x+6)) = (x+6)/(x-2)
'

Answer 2

[(x2 - 2x - 3)/x2 - x - 2)]/[(x2 - 4x + 3)/(x2 + 5x - 6)] = [(-x^3 - x^2 -2x -3)/x^2]/[(x - 3)/x + 6)] = -(x + 6)(x^3 + x^2 + 2x + 3)/[x^2(x-3)] [(x-3)(x+1)/(x-2)(x+1)]/[(x - 3)/x + 6)] = [(x-3)/(x-2)]/[(x - 3)/x + 6)] = (x+6)/(x-2)

Answer 3

[(x^2-2x-3)/(x^2-x-2)]/
[(x^2-4x+3)/(x^2+5x-6)]
=[(x-3)(x+1)/{(x+1)(x-2)}]/
[(x-3)(x-1)/{x-1)(x+6)}]

using the standard rule that for
x^2+bx+c the integers p and q to be found such that p*q = c and p+q = b then x^2+bx+c = (x+p)(x+q)
e.g x^2-2x-3 b=-2 and c=-3 p and q are -3 and +1 since
(-3)*(+1) = -3 and (-3) + (+1) = -2

ok let us go back and simplfy
= [(x-3)/(x-2)]/[(x-3)/(x+6)]..
= (x+6)/(x-2)

Answer 4

Dividing by a fraction a/b means multiplying by b/a, so factoring up,
[(x-3)(x+1) / (x-2)(x+1)] / [(x-1)(x-3) / (x-1)(x+6)]
= (x-3)/(x-2) . (x+6)/(x-3)
= (x+6)/(x-2)

Answer 5

Factor:

[(x-3)(x+1)/(x-2)(x+1)] / [(x-3)(x-1)/(x+6)(x-1)]

If we cancel similar terms, we are left with this:

(x+6) / (x-2)

Answer 6

(x+6)/(x-2)
Conditions apply while cancelling.
All the conditions applied here are x is not in
{-1,+1,3,2,-6}