How do I solve inequalities where both sides are not equal to zero, and that use absolute values and fractions?

Question

For example something like

$$\dfrac{|x-2|+3}{4-|2x+8|}\geq-5.$$

I understand how to solve inequalities where there is an absolute value, however in fraction form just confuses me.

Answer 1

Hint:

• When $4-|2x+8|>0$ the inequality is always satisfied (LHS is positive).
• When $4-|2x+8|<0$ the inequality is satisfied if and only if $|x-2|+3\leq -5(4-|2x+8|)$ (the inequality reverses when multiplying both sides by a negative number).

Answer 2

Add +5 to both sides and you got 0 on one side. Now add the two terms you got on the left (you may want to search about the least common multiple)

Now the inequality is asking you for what values of x your fraction is greater than or equal to 0.

The answer is: for every value of x that makes numerator and denominator either botth positive or both negative.

Answer 3

The answer by Especially Lime shows a simpler way that works in your particular case, but a general method that always works (but in certain specific cases, including this one, this method will be longer to carry out) is to solve separately in intervals where the absolute value expressions can be written without the use of absolute values.

Note that $|x-2|$ is equal to $-(x-2)$ when $x-2 \leq 0$ and $|x-2|$ is equal to $(x-2)$ when $x-2 \geq 0.$ This is just the definition of absolute value (i.e. $|u|$ equals $-u$ when $u \leq 0$ and $|u|$ equals $u$ when $u \geq 0$; now rewrite this replacing $u$ with $x-2).$ Thus, simplifying a bit, we have $|x-2|$ equal to $-x+2$ when $x \leq 2$ and equal to $x-2$ when $x \geq 2.$ That is, the non-absolute value version of $|x-2|$ changes its form at $x=2.$

In the same way, the non-absolute value version of $|2x + 8|$ changes its form where $2x + 8 = 0,$ or $x = -4.$ For $x \leq -4$ we have $|2x + 8| = -2x - 8$ and for $x \geq -4$ we have $|2x + 8| = 2x + 8.$

Combining the results in the last two paragraphs, we see that the entire fractional expression you're dealing with has three different forms of expression in three intervals, which are determined by the points $x = 2$ and $x = -4,$ the places where the insides of at least one of the absolute value expressions is equal to zero.

Interval 1: $\;x \leq -4$

Replace $|x-2|$ with $(-x+2)$ and replace $|2x + 8|$ with $(-2x - 8).$

$$\frac{(-x + 2) + 3}{4 - (-2x - 8)} \; \geq \; -5$$

$$\frac{-x + 5}{2x + 12} \; \geq \; -5$$

Solve this the usual way, and then include ONLY those values you get that ALSO satisfy $x \leq -4.$

Interval 2: $\; -4 \leq x \leq 2$

Replace $|x-2|$ with $(-x+2)$ and replace $|2x + 8|$ with $(2x + 8).$

$$\frac{(-x + 2) +3}{4 - (2x+8)} \; \geq \; -5$$

$$\frac{-x + 5}{-2x - 4} \; \geq \; -5$$

Solve this the usual way, and then include ONLY those values you get that ALSO satisfy $-4 \leq x \leq 2.$

Interval 3: $\;x \geq 2$

Replace $|x-2|$ with $(x-2)$ and replace $|2x + 8|$ with $(2x + 8).$

$$\frac{(x - 2) + 3}{4 - (2x + 8)} \; \geq \; -5$$

$$\frac{x + 1}{-2x - 4} \; \geq \; -5$$

Solve this the usual way, and then include ONLY those values you get that ALSO satisfy $x \geq 2.$

Final Solution

To obtain the final solution, combine together (i.e. take the union) the solutions in the three intervals.

Although this seems rather lengthy, the same method works for equations as well as inequalities, and it also works when exponential and/or logarithmic and/or trig functions are involved (and other possibilities). Just divide up the number line into intervals in such a way that, within each interval, you can replace ALL of the absolute value expressions with the inside of the absolute value expression or the negative of the inside of the absolute value expression, as appropriate.

A Harder Example

To solve

$$ 2\left | x^2 \; - \; |x^2 - 1|\, \right| \;\; \geq \;\; 1,$$

first observe that the inside of the innermost absolute value, namely $x^2 - 1,$ is equal to $0$ when $x = -1,\,1.$ Now note that $x^2 - 1$ is negative when $-1 < x < 1$ and $x^2 - 1$ is positive when $x < -1$ or $x > 1.$ (Sometimes you might have to work a bit to find this part out. Here, however, it's easy to see if you think about the graph of $y = x^2 - 1,$ which is the graph of $y = x^2$ shifted down by $1$ unit.) Thus, we can remove the inner absolute value by breaking this down into two cases.

Case 1: $\; -1 \leq x \leq 1$

$$ 2\left | x^2 \; - \; (-x^2 + 1)\, \right| \;\; \geq \;\; 1$$

$$ 2\left | 2x^2 - 1 \, \right| \;\; \geq \;\; 1$$

To solve this, note that the inside of the absolute value is equal to $0$ when $2x^2 - 1 = 0,$ or $x = \pm \frac{1}{\sqrt 2}.$ Note that $2x^2 - 1$ is negative when $-\frac{1}{\sqrt 2} < x < \frac{1}{\sqrt 2}$ and $x^2 - 1$ is positive when $x < -\frac{1}{\sqrt 2}$ OR $x > \frac{1}{\sqrt 2}.$ Thus, we can remove the last absolute value by breaking this down into two sub-cases.

Sub-case A: $\;-\frac{1}{\sqrt 2} \leq x \leq \frac{1}{\sqrt 2}$

$$ 2(-2x^2 + 1 ) \;\; \geq \;\; 1$$

$$ x^2 \leq \frac{1}{4}$$

$$-\frac{1}{2} \leq x \leq \frac{1}{2}$$

Since we're in sub-case A, we can only use those values above that also satisfy $-\frac{1}{\sqrt 2} < x < \frac{1}{\sqrt 2}.$ Since $\frac{1}{\sqrt 2}$ is approximately $0.707$ (all we need to know is that it's greater $\frac{1}{2}$ to deduce what follows, and this can be seen by comparing the squares of $\frac{1}{\sqrt 2}$ and $\frac{1}{2},$ or by observing which of the denominators $2$ and $\sqrt 2$ is greater), this means that all of these values qualify for inclusion in Sub-case A. Therefore, the values of $x$ we get from Sub-case A satisfy $-\frac{1}{2} \leq x \leq \frac{1}{2}.$

Sub-case B: $\;x \leq -\frac{1}{\sqrt 2}$ OR $x \geq \frac{1}{\sqrt 2}$

$$ 2(2x^2 - 1 ) \;\; \geq \;\; 1$$

$$ x^2 \geq \frac{3}{4}$$

$$x \leq -\frac{\sqrt 3}{2} \;\; \text{OR} \;\; x \geq \frac{\sqrt 3}{2}$$

Since we're in sub-case B, we can only use those values above that also satisfy $x \leq -\frac{1}{\sqrt 2}$ OR $x \geq \frac{1}{\sqrt 2}.$ Since $\frac{1}{\sqrt 2} < \frac{\sqrt 3}{2}$ (you can see this by squaring each of these positive fractions), and hence $-\frac{\sqrt 3}{2} < -\frac{1}{\sqrt 2},$ it follows that all of the values of $x$ we got from Sub-case B qualify for inclusion in Sub-case B. To see this, it will help to draw a number line, labeling the following points in order from left to right: $-\frac{\sqrt 3}{2},\;$ $-\frac{1}{\sqrt 2},\;$ $0,\;$ $\frac{1}{\sqrt 2},\;$ $\frac{\sqrt 3}{2}.$

Therefore, the values of $x$ we get from combining Sub-case A and Sub-case B are $-\frac{1}{2} \leq x \leq \frac{1}{2}$ along with $x \leq -\frac{\sqrt 3}{2}$ along with $x \geq \frac{\sqrt 3}{2}.$

Therefore, the values of $x$ we get in Case 1 are the combined values of $x$ (just listed) that ALSO satisfy $-1 \leq x \leq 1$ (the original condition we started with for Case 1). This gives $-\frac{1}{2} \leq x \leq \frac{1}{2}$ along with $-1 \leq x \leq -\frac{\sqrt 3}{2}$ along with $\frac{\sqrt 3}{2} \leq x \leq 1.$

Case 2: $\;x \leq -1$ OR $x \geq 1$

$$ \left | x^2 \; - \; (x^2 - 1)\, \right| \;\; \geq \;\; 1$$

$$ \left | 1 \, \right| \; \geq \;\; 1$$

Since this is true for all values of $x,$ we obtain no restriction on $x$ within Case 2, and thus the values of $x$ from Case 2 are just the Case 2 restriction, namely $x \leq -1$ OR $x \geq 1.$

Final Solution

Combining all values of $x$ we got in Case 1 with those we got in Case 2 gives $x \leq -\frac{\sqrt 3}{2}$ OR $-\frac{1}{2} \leq x \leq \frac{1}{2}$ OR $x \geq \frac{\sqrt 3}{2}.$ See this graph for a visualization of what's going on.

Final Thoughts

If you know how to quickly sketch the graph of $y = |f(x)|$ when the graph of $y = f(x)$ is known (leave stuff above $x$-axis alone, reflect about $x$-axis the stuff that's below the $x$-axis), then the solution above can be shortened quite a bit. In particular, it is not difficult to hand-sketch the graph of $y = |2x^2 - 1|$ (graph $y = 2x^2,$ then shift down by $1$ unit, then graph the absolute value of the result). Having done this, you can then solve $|2x^2 - 1| \geq \frac{1}{2}$ after figuring out the coordinates of some appropriate points on your graph.