(y^2-4)/(y+3)=2-(y-2)/(y+3)=(2(y+3)-(y-2))/(y+3). The denominators are now the same so, provided y is not -3, we can equate the numerators: y^2-4=2y+6-y+2; y^2-4=y+8; y^2-y-12=0; (y-4)(y+3)=0, so y=4, because y=-3 is not permissible since it would mean dividing by zero (y+3)=0.
Check y=4 in original: (16-4)/7=12/7=2-(2/7)=12/7. Correct!