# Solving $\left| \frac{-2x-6}{4} \right| \le 5$ for $x$

Say I have a statement like:

$$\left| \frac{-2x-6}{4} \right| \le 5.$$

And I want to find the closed interval form of $x$. i.e. I want to know what the maximum and minimum $x$ can be. How do I do this?

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You can get rid of the - inside of the absolute value and have the inequality $${|2x + 6|\over 4} \le 5$$ or $$|x + 3| < 10.$$ Now you can use the fact that for real numbers $a$ and $b$, we have ${\rm distance}(a,b) = |a - b|$ and observe that $|x + 3| = {\rm distance}(x, -3)$.

You now have the inequality ${\rm distance}(x, -3) < 10$, which yields $-13 < x < 7$. This is a nice geometric way to think about the problem.

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$|z|\le a$ is the same thing as $-|a|\le z\le|a|$. If you take $z$ to be $(-2x-6)/4$, and $a$ to be $5$, that should get you started.

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You can translate the absolute value to two statements: $\frac{-2x-6}{4} \le 5$ and $-\frac{-2x-6}{4} \le 5$. Each one gives one end of the interval. Can you take it from here?

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I think so, am I trying to solve for 'x' in both of those? –  InBetween Jun 15 '11 at 1:22
@InBetween, yes (what then?). –  Gerry Myerson Jun 15 '11 at 1:35
Sweet! Got it, thanks for the help! –  InBetween Jun 15 '11 at 3:02