Isolating quantities

Isolating a variable is about undoing whatever has been done to it.
Work backward through the operations, doing the same step to both sides each time.
Keep going until the variable stands alone.

This works for formulas too.
From $V = \pi r^2 h$, to solve for $r^2$, divide both sides by $\pi h$, giving $r^2 = \frac{V}{\pi h}$.
Then take the square root to get $r$.

When the variable appears on both sides, first gather it on one side.
From $kx = m + x$, subtract $x$ to get $kx - x = m$.
Factor out $x$: $x(k - 1) = m$, so $x = \frac{m}{k - 1}$.

Undo the operations in the reverse of the order they were applied.
In $50 + 3h$, the $3$ multiplies $h$ first and then $50$ is added, so undo the addition first, then the multiplication.
From $230 = 50 + 3h$, subtract $50$, then divide by $3$, giving $h = 60$.

Solve $40 = 5 + 7t$ for $t$.
Subtract $5$ from both sides: $35 = 7t$.
Divide by $7$: $t = 5$.

The base area of a cone is $\pi r^2$. Given $\pi r^2 = 2304\pi$, find $r$.
Divide both sides by $\pi$: $r^2 = 2304$.
Take the square root: $r = \sqrt{2304} = 48$.

Solve $m^2 x^2 - 2 = x^2$ for $m^2$.
Add $2$ to both sides: $m^2 x^2 = x^2 + 2$.
Divide by $x^2$: $m^2 = \frac{x^2 + 2}{x^2} = 1 + \frac{2}{x^2}$.