3 is supposed to be the denominator of 2
6 is supposed to be the denominator of 5
Complex Fractions
_1_ - _2_
x^2 3
____________
_1_ - _5_
x 6
2 answers
So I see it as
(1/x^2 - 2/3) / (1/x - 5/6)
= (3-2x^2)/(3x^2) / (6x)/(6-5x)
= (3 - 2x^2)/(3x^2) (6x)/(6-5x)
= 2(3-2x^2)/(x(6-5x))
or
= 2(2x^2 - 3)/(x(5x - 6))
(1/x^2 - 2/3) / (1/x - 5/6)
= (3-2x^2)/(3x^2) / (6x)/(6-5x)
= (3 - 2x^2)/(3x^2) (6x)/(6-5x)
= 2(3-2x^2)/(x(6-5x))
or
= 2(2x^2 - 3)/(x(5x - 6))
2 answers