# How do you solve for l in S= (pi r^2 )+ (pi r l)?

Feb 26, 2017

See the entire solution process below:

#### Explanation:

First, factor out a $\pi r$ from each term in the right side of the equation:

$S = \pi r \left(r + l\right)$

Next, divide each side of the equation by $\textcolor{red}{\pi r}$ to eliminate the coefficient while keeping the equation balanced:

$\frac{S}{\textcolor{red}{\pi r}} = \frac{\pi r \left(r + l\right)}{\textcolor{red}{\pi r}}$

$\frac{S}{\pi r} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\pi r}}} \left(r + l\right)}{\cancel{\textcolor{red}{\pi r}}}$

$\frac{S}{\pi r} = r + l$

Now, subtract $\textcolor{red}{r}$ from each side of the equation to solve for $l$:

$- \textcolor{red}{r} + \frac{S}{\pi r} = - \textcolor{red}{r} + r + l$

$- r + \frac{S}{\pi r} = 0 + l$

$- r + \frac{S}{\pi r} = l$

$l = - r + \frac{S}{\pi r}$

If you want the right side of the equation to be over a common denominator you can multiply $- r$ by $\frac{\pi r}{\pi r}$ giving:

$l = \left(\frac{\pi r}{\pi r} \times - r\right) + \frac{S}{\pi r}$

$l = \frac{- \pi {r}^{2}}{\pi r} + \frac{S}{\pi r}$

$l = \frac{- \pi {r}^{2} + S}{\pi r}$

$l = \frac{S - \pi {r}^{2}}{\pi r}$

The solution is: $l = - r + \frac{S}{\pi r}$ or $l = \frac{S - \pi {r}^{2}}{\pi r}$