The value of $y$ varies inversely as $\sqrt x$ and when $x=2$, $y=4$. What is $x$ when $y=1$?
We know that $y$ varies inversely as $\sqrt x$, so we write that as $y=\frac{k}{\sqrt x}$ where $k$ is some constant. We know that when $x=2$, $y=4$, so we can substitute that in to find the value of $k$. We get $4=\frac{k}{\sqrt{2}}$, so if we multiply both sides by $\sqrt{2}$, we get that $k=4\sqrt{2}$ Now we know that $y=\frac{4\sqrt{2}}{\sqrt x}$, so if we want to find $x$ when $y=1$, we can substitute that in. We get $1=\frac{4\sqrt{2}}{\sqrt x}$, so if we multiply both sides by $\sqrt x$ we get $\sqrt x=4\sqrt{2}$ Finally, if we square both sides of that equation, we get that $x=32$. That's correct. So the value of $x$ when $y=1$ is $32$.

