Question

How do you multiply this?​

Answer 1

`(-2)/(y^2+3y)`

Explanation:

You multiply it the same way you multiply any fractions. The numerator of the result is the product of numerators, and the denominator of the result is the product of denominators. Quite often, these problems are arranged so that factors in the numerator cancel factors in the denominator, so it is generally advisable to factor each expression completely at some point in the process.

`(6y+2)/(y^2-9)\cdot(3-y)/(3y^2+y)=(2(3y+1)(3-y))/((y-3)(y+3)(y)(3y+1))\\\\=(-2)/(y(y+3))\cdot((3y+1)(y-3))/((3y+1)(y-3))\\\\=(-2)/(y^2+3y)`