# How do you solve \frac { 4x + 3} { 15} - \frac { 2x - 3} { 9} = \frac { 6x + 4} { 6} - x?

Aug 5, 2018

$x = 3$

#### Explanation:

Here ,

$\frac{4 x + 3}{15} - \frac{2 x - 3}{9} = \frac{6 x + 4}{6} - x$

$\therefore \frac{4 x + 3}{15} - \frac{2 x - 3}{9} = \frac{\cancel{2} \left(3 x + 2\right)}{\cancel{6}} ^ 3 - x$

$\therefore \frac{4 x + 3}{15} - \frac{2 x - 3}{9} = \frac{3 x + 2}{3} - x$

Multiplying each term of eqn. by $45$

$45 \left(\frac{4 x + 3}{15}\right) - 45 \left(\frac{2 x - 3}{9}\right) = 45 \left(\frac{3 x + 2}{3}\right) - 45 \left(x\right)$

$\therefore 3 \left(4 x + 3\right) - 5 \left(2 x - 3\right) = 15 \left(3 x + 2\right) - 45 x$

$\therefore 12 x + 9 - 10 x + 15 = \cancel{45 x} + 30 - \cancel{45 x}$

$\therefore 2 x + 24 = 30$

Adding $\left(- 24\right)$both sides

$2 x + 24 + \left(- 24\right) = 30 + \left(- 24\right)$

$\therefore 2 x = 6$

$\therefore x = 3$

Aug 5, 2018

$x = 3$

#### Explanation:

$\frac{4 x + 3}{15} - \frac{2 x - 3}{9} = \frac{6 x + 4}{6} - x$

$\frac{3 \cdot \left(4 x + 3\right) - 5 \cdot \left(2 x - 3\right)}{45} = \frac{6 x + 4 - 6 x}{6}$

$\frac{\left(12 x + 9\right) - \left(10 x - 15\right)}{45} = \frac{4}{6}$

$\frac{2 x + 24}{45} = \frac{2}{3}$

$3 \cdot \left(2 x + 24\right) = 2 \cdot 45$

$6 x + 72 = 90$

$6 x = 18$

$x = \frac{18}{6} = 3$