Find the smallest number $y$ such that $y^2 = 81$.
I know that $y^2 = 81$ means that $y$ is either the positive or negative square root of 81. So I need to find the smallest of those two possible values. The positive square root of 81 is 9, and the negative square root of 81 is -9. The smallest of those two numbers is -9, since it is farther to the left on the number line. Therefore, the smallest number $y$ that satisfies the equation is -9. 