Question

How do you solve `(y^2+5y-6)/(y^3-2y^2)=5/y-6/(y^3-2y^2)`?

Answer 1

Restrict the domain to prevent division by 0.
Multiply both sides by a factor that eliminates the denominators.
Solve the resulting quadratic equation.

Explanation:

Given: `(y^2+5y-6)/(y^3-2y^2)=5/y-6/(y^3-2y^2)`

Restrict the domain to prevent division by 0.

`(y^2+5y-6)/(y^3-2y^2)=5/y-6/(y^3-2y^2); y !=0, y!=2`

Multiply both sides by a factor that eliminates the denominators.

`(y^2+5y-6)/(y^3-2y^2)(y^3-2y^2)=5/y(y^3-2y^2)-6/(y^3-2y^2)(y^3-2y^2); y !=0, y!=2`

`(y^2+5y-6)/cancel(y^3-2y^2)cancel(y^3-2y^2)=5/cancely(cancel(y)(y^2-2y))-6/cancel(y^3-2y^2)cancel(y^3-2y^2); y !=0, y!=2`

`y^2+5y-6=5(y^2-2y)-6; y !=0, y!=2`

`y^2+5y-6=5y^2-10y-6; y !=0, y!=2`

`0=4y^2-15y; y !=0, y!=2`

Solve the resulting equation quadratic equation.

`0=4y^2-15y; y !=0, y!=2`

Factor:

`y(4y-15)=0`; y !=0, y=2

Because of the restriction `y!=0`, we discard the `y` factor:

`4y-15=0`

`y = 15/4`