The distance between A(-3, 2) and B(1, $y$) is 5 units. Find every possible value of $y$.

$\sqrt{(1-(-3){)}^2+(y-2{)}^2}=5$, so $16+(y-2{)}^2=25$ and $(y-2{)}^2=9$. Hence $y-2=\pm 3$, giving $y=5$ and $y=-1$; both points really are 5 units from A. Keeping only the positive square root loses the second answer, and $y=3$ or 1 comes from writing $(y-2{)}^2=1$.