# How do you solve and write the following in interval notation: (x-5)/3 - (4x+3)/6> -3?

Jul 22, 2016

$x \in \left(- \infty , \frac{5}{2}\right)$

#### Explanation:

The first thing to do here is get rid of the denominators by multiplying the first fraction by $1 = \frac{2}{2}$ and $- 3$, which can be thought of as a fraction with the denominator equal to $1$, by $12 = \frac{6}{6}$.

The inequality becomes

$\frac{x - 5}{3} \cdot \frac{2}{2} - \frac{4 x + 3}{6} > - 3 \cdot \frac{6}{6}$

$\frac{2 \left(x - 5\right)}{6} - \frac{4 x + 3}{6} > - \frac{18}{6}$

This will be equivalent to

$2 \left(x - 5\right) - 4 x - 3 > - 18$

Expand the parentheses and group like terms to find

$2 x - 10 - 4 x - 3 > - 18$

$- 2 x - 13 > - 18$

$- 2 x > - 5$

Now, you must divide both sides of the inequality by $- 2$ to isolate $x$. Do not forget that multiplying or dividing by a negative number changes the sign of the inequality!

In this case, you have

$> \to <$

and

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}} x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 2}}}} < \frac{- 5}{- 2}$

$x < \frac{5}{2}$

This tells you that any value of $x \in \mathbb{R}$ that is smaller than $\frac{5}{2}$ will be a solution to the given inequality. Mind you, $\frac{5}{2}$ is not part of the solution interval.

In interval notation, this can be written as

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x \in \left(- \infty , \frac{5}{2}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

graph{x<5/2 [-10, 10, -5, 5]}