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

May 7, 2017

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: $\frac{{y}^{2} + 5 y - 6}{{y}^{3} - 2 {y}^{2}} = \frac{5}{y} - \frac{6}{{y}^{3} - 2 {y}^{2}}$

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:

$y \left(4 y - 15\right) = 0$; y !=0, y=2

Because of the restriction $y \ne 0$, we discard the $y$ factor:

$4 y - 15 = 0$

$y = \frac{15}{4}$