Question

How do you evaluate `frac { y + 3} { y ^ { 2} } \div \frac { y - 9} { 6y }`?

Answer 1

`(6y+18)/(y^2-9y)`

Explanation:

Law:
`a/b:c/d=(a/b)/(c/d)=(a*d)/(b*c)`

`(y+3)/y^2:(y-9)/(6y)=`
`=((y+3)/y^2)/((y-9)/(6y))=`
`=((y+3)*(6y))/((y^2)*(y-9))=`
`=((y+3)*(6))/((y)*(y-9))=`
`=(6y+18)/(y^2-9y)`