# How do you solve \frac { 1- x } { 3} = \frac { x } { 2} - \frac { x + 1} { 6}?

Jul 20, 2017

$x = \frac{3}{4}$

#### Explanation:

$\frac{1 - x}{3} = \frac{x}{2} - \frac{x + 1}{6}$

We first multiply all terms by the LCM of the denominators to remove the fractions. The LCM is $6$.

$\left[6 \times \frac{1 - x}{3}\right] = \left[6 \times \frac{x}{2}\right] - \left[6 \times \frac{x + 1}{6}\right]$

$\left[2 \cancel{6} \times \frac{1 - x}{1 \cancel{3}}\right] = \left[3 \cancel{6} \times \frac{x}{1 \cancel{2}}\right] - \left[1 \cancel{6} \times \frac{x + 1}{1 \cancel{6}}\right]$

$2 \left(1 - x\right) = 3 x - \left(x + 1\right)$

Simplify both sides. On the right side, the opening of the bracket will convert the plus sign to a minus sign since the product of a positive and a negative is a negative.

$2 - 2 x = 3 x - x - 1$

$2 - 2 x = 2 x - 1$

Add $1$ to both sides.

$2 - 2 x + 1 = 2 x - 1 + 1$

$3 - 2 x = 2 x$

Add $2 x$ to both sides.

$3 - 2 x + 2 x = 2 x + 2 x$

$3 = 4 x$

Divide both sides by $4$.

$\frac{3}{4} = \frac{4 x}{4}$

$\frac{3}{4} = \frac{1 \cancel{4} x}{1 \cancel{4}}$

$\frac{3}{4} = x$ or $x = \frac{3}{4}$