Solve for yyy: yy−2+3y+2=8y2−4\frac{y}{y-2} + \frac{3}{y+2} = \frac{8}{y^2-4}y−2y+y+23=y2−48.
y=2y=2y=2
y=−2y=-2y=−2
y=1y=1y=1
No solution