# How do you solve the linear equation 1/2 (x-3) = 1/3 (2x+1)?

May 17, 2018

$x = - 11$

#### Explanation:

Simplify to remove fractions.

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

Multiply by 6 through out:

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

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

$3 x - 9 = 4 x + 2$

$3 x - 4 x = 2 + 9$

$- x = 11$

$x = - 11$

$\frac{1}{2} \left(- 11 - 3\right) = \frac{1}{3} \left(2 \times \left(- 11\right) + 1\right)$
$\frac{1}{2} \times - 14 = \frac{1}{3} \left(- 22 + 1\right)$
$- 7 = \frac{1}{3} \times \left(- 21\right)$
$- 7 = - 7$

May 17, 2018

$x = - 11$

#### Explanation:

$\text{eliminate the fractions by multiplying both sides by}$
$\text{the "color(blue)"lowest common multiple of 2 and 3}$

$\text{the lowest common multiple of 2 and 3 is 6}$

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

$\Rightarrow 3 \left(x - 3\right) = 2 \left(2 x + 1\right) \leftarrow \textcolor{b l u e}{\text{no fractions}}$

$\text{distribute brackets on both sides}$

$3 x - 9 = 4 x + 2$

$\text{subtract "3x" from both sides}$

$\Rightarrow - 9 = x + 2$

$\text{subtract 2 from both sides}$

$\Rightarrow - 11 = x \Rightarrow x = - 11$

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

Substitute this value into the equation and if both sides are equal then it is the solution.

$\text{left } = \frac{1}{2} \left(- 14\right) = - 7$

$\text{right } = \frac{1}{3} \left(- 21\right) = - 7$

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