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

##### 1 Answer
Feb 1, 2017

$x \ge 15$

#### Explanation:

To get rid of the denominators, you could multiply by $2$ or $3$, but this would only take care of one problem. Because of this, its typically best to multiply both sides by (2 and 3's) least common factor, which is $6$:

$\frac{6 \left(x - 1\right)}{2} \ge \frac{6 x}{3} + 12$

Since $6 > 0$, the inequality remains the same. If we multiplied (or divided) by a negative number, it would be reversed. Then, we can do some simplifications on the fractions, because

$\frac{6}{2} = 3$ and $\frac{6}{3} = 2$.

$3 \left(x - 1\right) \ge 2 x + 12$

$3 x - 3 \ge 2 x + 12$

$3 x - 2 x \cancel{- 3} \cancel{+ 3} \ge \cancel{2 x} \cancel{- 2 x} + 12 + 3$

$x \ge 15$