Alg 2 Question?

Question

How do I find all values of y satisfying the equation

(12) / (y^2 - 25) = - (3y) / (y + 5)

(If there is more than one solution, separate them with commas.)

Answer 1

Alg 2 Question?
How do I find all values of y satisfying the equation

(12) / (y^2 - 25) = - (3y) / (y + 5)

4(y+5)(y-5)=-(y+5)
y+5=0 or
4y-20=1,y=21/4

Answer 2

(12) / (y^2 - 25) = - (3y) / (y + 5)
Cross multipl getting:
12(y+5) = (y^2-25)(-3y)
12y+60 = -3y^3 +75y
3y^3 - 63y +60 = 0
y^3 -21y +20 = 0
(y-1)(y+5)(y-4) = 0
y = 1, 4, -5

Answer 3

(12) / (y^2 - 25) = - (3y) / (y + 5)

factorize:
(12) / (y+5)(y-5) = - (3y) / (y + 5)

cancel:
(12) / (y-5) = - (3y)

rearange:
12 = - (3y)(y-5)
0 = -(3y^2-15y) - 12
3y^2-15y + 12 = 0

factorize again:
(3y-3)(y-4) = 0
y = 0+3/3 = 1
y = 0+4 = 4

Answer 4

12/(y^2-25)=-3y/(y+5)

Before solving the problem, put the condition that:
y^2-25<>0 and y+5<>0
this means y<>5, y<>-5

12/[(y-5)(y+5)]=-3y/(y+5)
4/(y-5)=-y
y^2-5y+4=0
so: y1=1, y2=4