(5y/(y^2-3y))-(7/(3-y))

Solution for (5y/(y^2-3y))-(7/(3-y)) equation:

D( y )

y^2-(3*y) = 0

3-y = 0

y^2-(3*y) = 0

y^2-(3*y) = 0

y^2-3*y = 0

y^2-3*y = 0

DELTA = (-3)^2-(0*1*4)

DELTA = 9

DELTA > 0

y = (9^(1/2)+3)/(1*2) or y = (3-9^(1/2))/(1*2)

y = 3 or y = 0

3-y = 0

3-y = 0

3-y = 0 // - 3

-y = -3 // * -1

y = 3

y in (-oo:0) U (0:3) U (3:+oo)

(5*y)/(y^2-(3*y))-(7/(3-y)) = 0

(5*y)/(y^2-3*y)-7*(3-y)^-1 = 0

(5*y)/(y^2-3*y)-7/(3-y) = 0

y^2-3*y = 0

y^2-3*y = 0

y*(y-3) = 0

y-3 = 0 // + 3

y = 3

y*(y-3) = 0

(5*y)/(y*(y-3))-7/(3-y) = 0

(5*y*(3-y))/(y*(y-3)*(3-y))+(-7*y*(y-3))/(y*(y-3)*(3-y)) = 0

5*y*(3-y)-7*y*(y-3) = 0

36*y-12*y^2 = 0

36*y-12*y^2 = 0

12*y*(3-y) = 0

3-y = 0 // - 3

-y = -3 // * -1

y = 3

12*y*(y-3) = 0

(12*y*(y-3))/(y*(y-3)*(3-y)) = 0

(12*y*(y-3))/(y*(y-3)*(3-y)) = 0 // * y*(y-3)*(3-y)

12*y*(y-3) = 0

( 12*y )

12*y = 0 // : 12

y = 0

( y-3 )

y-3 = 0 // + 3

y = 3

y in { 0} 

y in { 3} 

y belongs to the empty set