# How do you solve \frac { 2x } { 5} - \frac { x + 18} { 6} = 23+ \frac { x } { 30}?

Dec 22, 2016

$x = 130$

#### Explanation:

Rearrange the equation. all $x$'s on one side.

$\frac{2 x}{5} - \frac{x + 18}{6} - \frac{x}{30} = 23$

Equalize the denominators to $30$

$\frac{2 x}{5} \cdot \frac{6}{6} - \frac{x + 18}{6} \cdot \frac{5}{5} - \frac{x}{30} = 23$

$\frac{12 x}{30} - \frac{5 \left(x + 18\right)}{30} - \frac{x}{30} = 23$

$\frac{12 x - 5 x - 90 - x}{30} = 23$

Multiply both sides by $30$

$12 x - 5 x - 90 - x = 690$

$6 x = 690 + 90$

$6 x = 780$

$x = \frac{780}{6}$

$x = 130$

Dec 23, 2016

$x = 130$

#### Explanation:

$\frac{2 x}{5} - \frac{x + 18}{6} = 23 + \frac{x}{30}$

When you have an equation which has fractions in it, you can get rid of the fractions immediately.

Multiply each term by the LCM of the denominators and cancel.

In this case the LCM = color(blue)(30

$\frac{\textcolor{b l u e}{30 \times} 2 x}{5} - \frac{\textcolor{b l u e}{30 \times} \left(x + 18\right)}{6} = \textcolor{b l u e}{30 \times} 23 + \frac{\textcolor{b l u e}{30 \times} x}{30}$

Now cancel all the denominators:

$\frac{\textcolor{b l u e}{{\cancel{30}}^{6} \times} 2 x}{\cancel{5}} - \frac{\textcolor{b l u e}{{\cancel{30}}^{5} \times} \left(x + 18\right)}{\cancel{6}} = \textcolor{b l u e}{30 \times} 23 + \frac{\textcolor{b l u e}{\cancel{30} \times} x}{\cancel{30}}$

$6 \times 2 x - 5 \left(x + 18\right) = 30 \times 23 + x \text{ } \leftarrow$ no fractions!! Simplify

$12 x - 5 x - 90 = 690 + x \text{ } \leftarrow$ now solve the equation

$12 x - 5 x - x = 690 + 90 \text{ } \leftarrow$ re-arrange the terms

$6 x = 780 \text{ } \leftarrow \div 6$ on both sides

$x = 130$