# How do you solve abs(3x+2)-5<0?

Aug 1, 2016

$\text{ } - \frac{7}{3} < x < 1$

#### Explanation:

$| 3 x + 2 | < 5$

Note that $| 3 x + 2 |$ is always positive so there is a limit to the negative value of $3 x + 2$

Suppose $3 x + 2 = 5 \implies x = 1$

Then the maximum value attributable to $x$ is such that $x < 1$

Suppose $3 x + 2 = - 5 \implies x = - \frac{7}{3}$

So the minimum value of $x$ is such that $- \frac{7}{3} < x$

So $\text{ } - \frac{7}{3} < x < 1$
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$\textcolor{b l u e}{\text{In summery:}}$

solve for $3 x + 2 = | \pm 5 |$ and make the appropriate value of $x$ the $\textcolor{m a \ge n t a}{\underline{\text{not inclusive}}}$ upper and lower bounds. $\text{ } - \frac{7}{3} \textcolor{m a \ge n t a}{<} x \textcolor{m a \ge n t a}{<} 1$

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$\textcolor{b r o w n}{\text{Just for information:}}$
If they had been $\textcolor{b l u e}{\underline{\text{inclusive}}}$ then we would have $- \frac{7}{3} \le x \le 1$