# How do you solve 1/6-x/2=(x-5)/3?

Jan 27, 2017

see the entire solution process below:

#### Explanation:

First, multiple each side of the equation by $\textcolor{red}{6}$ (the lowest common denominator of all the fractions) to eliminate the fractions while keeping the equation balanced:

$\textcolor{red}{6} \left(\frac{1}{6} - \frac{x}{2}\right) = \textcolor{red}{6} \times \frac{x - 5}{3}$

$\left(\textcolor{red}{6} \times \frac{1}{6}\right) - \left(\textcolor{red}{6} \times \frac{x}{2}\right) = \cancel{\textcolor{red}{6}} 2 \times \frac{x - 5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}$

$\left(\cancel{\textcolor{red}{6}} \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}}\right) - \left(\cancel{\textcolor{red}{6}} 3 \times \frac{x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right) = 2 \left(x - 5\right)$

$1 - 3 x = 2 x - 10$

Next, add $\textcolor{red}{3 x}$ and $\textcolor{b l u e}{10}$ to each side of the equation to isolate the $x$ term while keeping the equation balanced:

$1 - 3 x + \textcolor{red}{3 x} + \textcolor{b l u e}{10} = 2 x - 10 + \textcolor{red}{3 x} + \textcolor{b l u e}{10}$

$1 + \textcolor{b l u e}{10} - 3 x + \textcolor{red}{3 x} = 2 x + \textcolor{red}{3 x} - 10 + \textcolor{b l u e}{10}$

$11 - 0 = 5 x - 0$

$11 = 5 x$

Now, divide each side of the equation by $\textcolor{red}{5}$ to solve for $x$ while keeping the equation balanced:

$\frac{11}{\textcolor{red}{5}} = \frac{5 x}{\textcolor{red}{5}}$

$\frac{11}{5} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} x}{\cancel{\textcolor{red}{5}}}$

$\frac{11}{5} = x$

$x = \frac{11}{5}$

Jan 27, 2017

$x = \frac{11}{5}$

#### Explanation:

Eliminate the fractions in the equation by multiplying ALL terms on both sides of the equation by the $\textcolor{b l u e}{\text{lowest common multiple}}$ ( LCM ) of the denominators 6, 2 and 3

The LCM of 6 , 2 and 3 is 6

Hence multiply All terms by 6

$\left({\cancel{6}}^{1} \times \frac{1}{\cancel{6}} ^ 1\right) - \left({\cancel{6}}^{3} \times \frac{x}{\cancel{2}} ^ 1\right) = \left({\cancel{6}}^{2} \times \frac{x - 5}{\cancel{3}} ^ 1\right)$

$\Rightarrow 1 - 3 x = 2 \left(x - 5\right) \leftarrow \text{ no fractions}$

distribute the bracket on the right side.

$\Rightarrow 1 - 3 x = 2 x - 10$

subtract 2x from both sides.

$1 - 3 x - 2 x = \cancel{2 x} \cancel{- 2 x} - 10$

$\Rightarrow 1 - 5 x = - 10$

subtract 1 from both sides.

$\cancel{1} \cancel{- 1} - 5 x = - 10 - 1$

$\Rightarrow - 5 x = - 11$

To solve for x, divide both sides by - 5

$\frac{\cancel{- 5} x}{\cancel{- 5}} = \frac{- 11}{- 5}$

$\Rightarrow x = \frac{11}{5}$

$\textcolor{b l u e}{\text{As a check}}$

substitute this value into the equation and if the left side equals the right side then it is the solution.

$\text{left side } = \frac{1}{6} - \frac{\frac{11}{5}}{2} = \frac{1}{6} - \frac{11}{10} = \frac{5}{30} - \frac{33}{30} = - \frac{28}{30} = - \frac{14}{15}$

$\text{right side} = \frac{\frac{11}{5} - 5}{3} = \frac{\frac{11}{5} - \frac{25}{5}}{3} = \frac{- \frac{14}{5}}{3} = - \frac{14}{15}$

$\Rightarrow x = \frac{11}{5} \text{ is the solution}$