3x-6y/2x+6y * x^2+5xy+6y^2/6x^2-24y^2

Question

divide the fractions

Answer 1

factor as much as possible

(3x-6y)/(2x+6y) = 3(x-2y)/2(x+3y) = numerator over denominator

x^2 + 5xy + 6y^2 = (x+3y)(x+2y) = numerator

6x^2 -24y^2 = 6(x^2-4y^2) = 3(2)(x-2y)(x+2y) = denominator

notice the factors that are the same and cancel them if one is in the numerator and the same is in the denominator 3's cancel, x-2y cancels, x+2y cancels

that leaves only 1/(2)(2) = 1/4

Answer 2

So after rewriting the expression with parenthesis you will find the solution of the expression:(3x-6y) / (2x+6y) * (x²+5xy+6y²) / (6x²-24y²) is 1/4.

1. Before doing any multiplication factor all the numerators and denominators of the fractions in involved.
2. Now you have: 3(x-2y) / 2(x+3y) * (x+3y)(x+2y) / 6(x + 2y)(x - 2y)
3. Now cancel out the common factors 3(x-2y) / 2(x+3y) × (x+3y)(x+2y) / 6(x + 2y)(x - 2y)
4. We're not done yet: Look at what we have now: 3/2 *1/6 which is the same as 3/2 * 1/ (3*2). 
5. So 3/2 * 1/ (3*2) = 1/2 * 1/2 = 1/4. TADA!

Hope that helps.