Question

(2)/(y+3)+(3)/(y-4)=(5)/(y+6)

Answer 1

(6)/(3y+9+2y-8)=5/(y+6)
6/(5y+1) = 5/(y+6)
6(y+6) = 5(5y+1)
6y + 36 = 25y + 5
31 = 19y
Y = 31/19

You can check my maths but the theory should be fine.

Answer 2

2/(y + 3) + 3/(y - 4) = 5/(y + 6)

To solve first find the LCD. The LCD would be the product of the binomial denominators (y + 3) (y - 4) ( y + 6)
Once you multiply through by the LCD, the fractions will clear.
2(y-4)(y+6) + 3(y+3)(y + 6) = 5(y + 3)(y - 4)
2(y^2 +2y -24 +3(y^2 + 9y + 18) = 5(y^2 -y - 12)
2y^2 +4y -48 +3y^2 + 27y + 54 = 5y^2 -5y - 60
31y + 6 = -5y - 60
36y = -66
y = -11/6