Question

What is the solution to this problem?

Answer 1

Answer:

$\frac{2y^2-5y-7}{6y^2+10y+4}-\frac{7}{9y^2+6y}=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

Step-by-step explanation:

we are given

$\frac{2y^2-5y-7}{6y^2+10y+4}-\frac{7}{9y^2+6y}$

we have to simplify it

Firstly, we will factor it

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

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

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

$=\frac{\left(y+1\right)\left(2y-7\right)}{2\left(y+1\right)\left(3y+2\right)}-\frac{7}{3y(3y+2)}$

now, we can cancel like terms

$=\frac{\left(2y-7\right)}{2\left(3y+2\right)}-\frac{7}{3y(3y+2)}$

now, we can combine it

$=\frac{3y(2y-7\right)}{6y\left(3y+2\right)}-\frac{7\times 2}{6y(3y+2)}$

$=\frac{3y(2y-7\right)-14}{6y\left(3y+2\right)}$

$=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

so, we get

$=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$