# How do you solve \frac { x - 1} { 4} - \frac { x } { 3} = \frac { 1} { 12}?

Feb 19, 2017

See the entire solution below:$x = - 4$

#### Explanation:

First, multiply each side of the equation by $\textcolor{red}{12}$ to eliminate the fractions while keeping the equation balanced:

$\textcolor{red}{12} \left(\frac{x - 1}{4} - \frac{x}{3}\right) = \textcolor{red}{12} \times \frac{1}{12}$

$\left(\textcolor{red}{12} \times \frac{x - 1}{4}\right) - \left(\textcolor{red}{12} \times \frac{x}{3}\right) = \cancel{\textcolor{red}{12}} \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{12}}}}$

$\left(\cancel{\textcolor{red}{12}} 3 \times \frac{x - 1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}}\right) - \left(\cancel{\textcolor{red}{12}} 4 \times \frac{x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}\right) = 1$

$3 \left(x - 1\right) - 4 x = 1$

Next, expand the terms in parenthesis:

$\left(3 \times x\right) - \left(3 \times 1\right) - 4 x = 1$

$3 x - 3 - 4 x = 1$

Then, group and combine like terms on the left side of the equation:

$3 x - 4 x - 3 = 1$

$\left(3 - 4\right) x - 3 = 1$

$- 1 x - 3 = 1$

Next, add $\textcolor{red}{3}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced:

$- 1 x - 3 + \textcolor{red}{3} = 1 + \textcolor{red}{3}$

$- 1 x - 0 = 4$

$- 1 x = 4$

Now, multiply each side of the equation by $\textcolor{red}{- 1}$ to solve for $x$ while keeping the equation balanced:

$\textcolor{red}{- 1} \times - 1 x = \textcolor{red}{- 1} \times 4$

$1 x = - 4$

$x = - 4$