Define absolute value inequalities and draw on a number line

If absolute value of a real number represents its distance from the origin on the number line then absolute value inequalities are type of inequalities that are consisted of absolute values.

If we are trying to solve a simple absolute value equation, the solution is quite simple, it usually has two solutions.

Example 1.

$|x| = 2$

$x_1 = 2$, $x_2 = – 2$

Now we want to find out what happens if we “change our equality sign into an inequality sign”.

By solving any inequality we’ll get a set of solutions as our final solution, which means that this will apply to absolute inequalities as well.

Let’s try to solve example 1. but change the equality sign.

Example 2.

$|x| < 2$

If absolute value represents numbers distance from the origin, this would mean that we are searching for all numbers whose distance from the origin is lesser than two.

We can see the solution for this inequality is the set $x \in <-2, 2>$, but how can we be sure?

First you break down your inequality into two parts:

-first is the part in which your expression in absolute value is positive,

-and second in which that expression is negative.

1. $x ≥ 0$ – if x is greater or equal to zero, we can just “ignore” absolute value sign. We got the inequality $x < 2$.

The solution for this inequality is $x \in [0, 2>$.

2. $x < 0$ – if variable $x$ is lesser than zero, we have to change its sign. We got inequality $– x < 2$. When we solve this simple inequality we get $x > – 2$.

The solution for this inequality is $x \in <- 2, 0>$.

Our final solution will be the union of these two intervals, which means that the final solution is in the form:

$x \in <- 2, 2>$

If we want to draw it on the number line:

Usually you’ll get a whole expression in your inequality. This is solved just like the example 2. Set your grounds first before going any further. Example 2 is basic absolute value inequality task, but using it you can solve any other absolute value task, no matter how much is complicated.

Example 3.

$|x + 2| ≥ 1$
First thing you need to do is find the points where the absolute value will change. You do this by equating the term that is affected by the absolute value with $0$. We get $x = – 2$. This means that $-2$ will be our turning point. From minus infinity to $-2$ our term under the absolute value will be negative, which means that we’ll have to change the sign of every member of that term, and from $2$ to plus infinity it will be positive so we’ll leave it as it is.

Again, we’ll divide it into two parts.

1. For $x + 2 ≥ 0$

$x + 2 ≥ 1$

$x ≥ – 1$

The common solution for these two inequalities is the interval $[-1, +\infty>$.

2. For $x + 2 < 0$

$– x – 2 ≥ 1$

$– x ≥ 3$

$x ≤ – 3$

The common solution for these two inequalities is the interval $<-\infty, – 3]$.

The final solution is the union of solutions of separate parts:

$x \in <-\infty, – 3] \cup [- 1, +\infty>$
This solution can be also written as $\mathbb{R} \ <- 3, – 1>$. This means that our solution is the whole set of real numbers except interval $x \in <- 3, – 1>$.

Example 4.

$|\frac{1}{3}x + 1| – |2x – 2| ≤ 8$
First interval is $<-\infty, – 3>$. We can insert any number from that interval into those two absolute values to see are they positive or negative. If they are positive, we leave them as they are, and if they are negative we change signs of all members of expressions in absolute values. You can insert any number, so we’ll use $x = – 4$.

For the first absolute value $\frac{1}{3}x + 1$ => $\frac{1}{3} * (- 4) + 1 = – \frac{1}{3}$ which is lesser than zero. This means that for the first interval first absolute value will change signs of its terms.

For the second absolute value $2x – 2$ => $– 8 – 2 = – 10$ which is lesser than zero. This means that for the first interval second absolute value will change signs of its terms.

$– \frac{1}{3}x – 1 – (-2x + 2) ≤ 8$

$- \frac{1}{3}x + 2x – 1 – 2 ≤ 8$

$\frac{5}{3}x ≤ 11$

$5x ≤ 33$

$x ≤ \frac{33}{5}$
Second interval is $[-3 , 1>$. From this interval we’ll remove zero.

For the first absolute value $\frac{1}{3}x + 1$ => $\frac{1}{3} \cdot 0 + 1 = 1$ which is greater than zero. This means that for the second interval the first absolute value will not change signs of its terms.

For the second absolute value $2x – 2$ => $2 \cdot 0 – 2 = – 2$ which is lesser than zero. This means that for the second interval second absolute value will change signs of its terms.

$|\frac{1}{3}x + 1| – |2x – 2| ≤ 8$

$\frac{1}{3}x + 1 + 2x – 2 ≤ 8$

$\frac{7}{3}x ≤ 9$

$7x ≤ 27$

$x ≤ \frac{27}{7}$
Third interval is $[1,+\infty>$. From this interval we’ll remove point that represent number two.

For both absolute values the solution will be positive, which means that we leave them as they are.

$|\frac{1}{3}x +1| – |2x – 2| ≤ 8$

$\frac{1}{3}x + 1 -2x + 2 ≤ 8$

$– \frac{5}{3}x ≤ 5$

$– 5x ≤ 15$

$x ≥ – 3$
Note how the intervals are made considering the points. From the left, the first one is always an open interval, the second is half open segment where we include the left point, and second one is half open segment where we include the left point. This means that we will construct segments that always include the left endpoints.

The final solution is the union of these intervals which is, in this case, the whole set of real numbers.

Now, we can solve one more example.

Example 5.

$|\frac{1}{x-2}|\geq 2$

Solution:

$\frac{1}{x-1} \geq 2 /\cdot|x-1|, x\neq 1$

$2|x-1|\leq 1 /:2$

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

$-\frac{1}{2}\leq x-1 \leq \frac{1}{2} /+1$ $, x\neq 1$

$\frac{1}{2}\leq x \leq \frac{3}{2}, x\neq 1$

Absolute value inequalities worksheets

Integers - One or less operations (541.1 KiB, 804 hits)

Integers - More than one operations (656.8 KiB, 783 hits)

Decimals - One or less operations (566.3 KiB, 539 hits)

Decimals - More than one operations (883.6 KiB, 628 hits)

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Fractions - More than one operations (1,009.1 KiB, 658 hits)