Question

What is the solution to this problem?

Answer 1

`(2y^2-5y-7)/(6y^2+10y+4)-(7)/(9y^2+6y)=(6y^2-21y-14)/(6y\left(3y+2\right))`

Explanation:

we are given

`(2y^2-5y-7)/(6y^2+10y+4)-(7)/(9y^2+6y)`

we have to simplify it

Firstly, we will factor it

`2y^2-5y-7=\left(y+1\right)\left(2y-7\right)`

`6y^2+10y+4=2\left(y+1\right)\left(3y+2\right)`

`9y^2+6y=3y(3y+2)`

`=(\left(y+1\right)\left(2y-7\right))/(2\left(y+1\right)\left(3y+2\right))-(7)/(3y(3y+2))`

now, we can cancel like terms

`=(\left(2y-7\right))/(2\left(3y+2\right))-(7)/(3y(3y+2))`

now, we can combine it

`=(3y(2y-7\right))/(6y\left(3y+2\right))-(7* 2)/(6y(3y+2))`

`=(3y(2y-7\right)-14)/(6y\left(3y+2\right))`

`=(6y^2-21y-14)/(6y\left(3y+2\right))`

so, we get

`=(6y^2-21y-14)/(6y\left(3y+2\right))`