I'm preparing for the ACTM State contest, and I stumbled across a problem asking for the solution to an inequality that contained an absolute value. I'm not very familiar with how to solve equations with absolute values, so can anyone explain it to me?

Find the solution for the inequality ${|2x-4|\over (x+3)}\leq 0$.

  • $\begingroup$ absolute value symbol is just shift and the \ key to get | $\endgroup$ – Heavenly96 Apr 15 '17 at 3:22
  • $\begingroup$ Hint: $\,\frac{|2x-4|}{x+3}$ equals $0$ iff $x=2\,$, else it has the same sign as the denominator $x+3\,$. $\endgroup$ – dxiv Apr 15 '17 at 3:36
  • $\begingroup$ The inequality is true if $x=2$. If $x\neq 2$ then $|2x-4|>0$ and with this, we must have $x+3<0$, that is, $x<-3$. So, the solution set is $$ (−\infty,−3)\cup\{2\}.$$ $\endgroup$ – Juniven Apr 15 '17 at 3:46

$$\frac{|2x-4|}{x+3} \le 0$$

We can easily see that if $x=2$, the inequality will become zero, as the numerator also cancels out to zero. This is a solution, but this isn't the end - we're also looking for negative solutions.

So if we know that $x=2$ produces $0$, then anything less than $x=2$ will produce a numerator that is negative - however, this isn't possible since there are absolute value signs around the numerator. This means that the numerator will never be negative - so we look at the denominator ($x+3$).

We're looking for values of $x+3$ that are less than or equal to zero. Let's set up an equation:

$x+3 \le 0$

We can easily solve this to $x \le -3$, but this is wrong. We have to remember the fact that since this is the denominator of a fraction, it cannot equal zero lest the value be undefined. We can see this happen when $x = - 3$, so $ x \neq 3 $ so we remove that from $x \le - 3$, leaving us with $x < - 3$.

Final answer: $x < -3, x =2$.

Commonly when solving equations with absolute values in them a method is to square both sides, making the absolute value signs disappear. In this case: $$\frac{(2x-4)^2}{(x+3)^2} \le 0$$

One should try and avoid this outcome unless one knows for sure that both sides are not negative because in doing so, half of the solution is eliminated.

  • $\begingroup$ when squaring both sides of the equation You cannot square inequalities (unless both sides are known to be non-negative), and the last line in your post shows precisely what can go wrong if you do. If you meant it as a "don't do that" counterexample, then you should make that plenty clear in the answer. $\endgroup$ – dxiv Apr 15 '17 at 3:54
  • $\begingroup$ Thanks I tweaked the language in the final part to make that sentiment of "don't do that" clearer $\endgroup$ – zeroseventwoeight Apr 15 '17 at 4:18

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