# How do you solve for each of the variable 5L= 10L- 15D+ 5?

Jul 30, 2015

You take turns isolating them on one side of the equation.

#### Explanation:

Your starting equation looks like this

$5 L = 10 L - 15 D + 5$

To solve this equation for $L$, get all the terms that contain this variable on one side of the equation.

In your case, you can do this by adding $- 10 L$ to both sides of the equation

$5 L - 10 L = \textcolor{red}{\cancel{\textcolor{b l a c k}{10 L}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{10 L}}} - 15 D + 5$

$- 5 L = - 15 D + 5$

Now divide both sides of the equation by $- 5$ to get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 5}}} L}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 5}}}} = \frac{- 15 D}{- 5} + \frac{5}{- 5}$

$L = \textcolor{g r e e n}{3 D - 1}$

Now do the same for $D$. Notice that you can write

$5 L = 10 L + 5 - 15 D$, so you could add $- 10 L - 5$ to both sides of the equation to get

$5 L - 10 L - 5 = \textcolor{red}{\cancel{\textcolor{b l a c k}{- 10 L - 5}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{10 L + 5}}} - 15 D$

This is equivalent to

$- 15 D = - 5 L - 5$

Finally, divide both sides of the equation by $- 15$ to get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 15}}} D}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 15}}}} = \frac{- 5 L}{- 15} - \frac{5}{- 15}$

$D = \frac{1}{3} L + \frac{1}{3}$, or

$D = \textcolor{g r e e n}{\frac{1}{3} \left(L + 1\right)}$