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

Question

Eva Adamson · Accepted Answer

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

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#

Savannah Anderson · Answer

undefined To solve the equation (y^2+5y-6)/(y^3-2y^2)=5/y-6/(y^3-2y^2), we can start by simplifying both sides of the equation.

First, we can factor the numerator and denominator of the left side of the equation:
(y^2+5y-6)/(y^3-2y^2) = [(y+6)(y-1)] / [y^2(y-2)]

Next, we can factor the numerator and denominator of the right side of the equation:
5/y-6/(y^3-2y^2) = 5/y - [(y-2)(y^2)] / [y^2(y-2)]

Now, we can rewrite the equation with the simplified expressions:
[(y+6)(y-1)] / [y^2(y-2)] = 5/y - [(y-2)(y^2)] / [y^2(y-2)]

To eliminate the denominators, we can multiply both sides of the equation by y^2(y-2):
(y+6)(y-1) = 5(y-2) - (y-2)(y^2)

Expanding and simplifying the equation:
y^2 + 5y - 6 = 5y - 10 - y^3 + 2y^2

Rearranging the terms:
y^3 - 2y^2 + y^2 + 5y - 5y + 6 + 10 = 0

Combining like terms:
y^3 - y^2 + 16 = 0

This is a cubic equation. To find the solutions, we can use various methods such as factoring, synthetic division, or numerical methods like Newton's method.