# How do you solve and write the following in interval notation: 4+abs(3x+2) <=5?

Jan 6, 2017

$x \in \left[- 1 , - \frac{1}{3}\right]$

#### Explanation:

$4 + | 3 x + 2 | \le 5$

$\implies | 3 x + 2 | \le 1$

The absolute value function $| a |$ is defined as

$| a | = \left\{\begin{matrix}a \mathmr{if} a \ge 0 \\ - a \mathmr{if} a < 0\end{matrix}\right.$

We will use this definition to consider all possible cases.

Case 1: $3 x + 2 \ge 0$

Note that for this to occur, we must have $3 x \ge - 2 \implies x \ge - \frac{2}{3}$

Now, by the definition of the absolute value function, we have $| 3 x + 2 | = 3 x + 2$ in this case. Substituting that into our initial inequality, we get

$3 x + 2 \le 1$

$\implies 3 x \le - 1$

$\implies \frac{3 x}{3} \le \frac{- 1}{3}$

Recall that when multiplying or dividing by a positive number, the direction of the inequality stays the same, and when multiplying or dividing by a negative number, we change the direction of the inequality

$\implies x \le - \frac{1}{3}$

So, together with the initial restriction $x \implies - \frac{2}{3}$, we have

$- \frac{2}{3} \le x \le - \frac{1}{3}$

or, in interval notation, $x \in \left[- \frac{2}{3} , - \frac{1}{3}\right]$

Case 2: $3 x + 2 < 0$

As above, we first solve for $x$ to find that this case occurs when $3 x < - 2 \implies x < - \frac{2}{3}$

By the definition of the absolute value function, in this case we have $| 3 x + 2 | = - \left(3 x + 2\right)$. Again, we substitute this into the initial inequality.

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

$\implies 3 x + 2 \ge - 1$

We multiplied by $- 1$ in the prior step, and so reversed the direction of the inequality.

$\implies 3 x \ge - 3$

$\implies x \ge - 1$

Together with the initial restriction of $x < - \frac{2}{3}$ in this case, we have

$- 1 \le x < - \frac{2}{3}$

or, in interval notation, $x \in \left[- 1 , - \frac{2}{3}\right)$

If we look at where the intervals of the solutions of each case lay, we find that they meet at and include $x = - \frac{2}{3}$, and so putting them together, we get

$- 1 \le x \le - \frac{1}{3}$

or, in interval notation, $x \in \left[- 1 , - \frac{1}{3}\right]$