Lesson Explainer: One-Variable Absolute Value Inequalities | Nagwa Lesson Explainer: One-Variable Absolute Value Inequalities | Nagwa

Lesson Explainer: One-Variable Absolute Value Inequalities Mathematics • Second Year of Secondary School

In this explainer, we will learn how to solve one-variable inequalities that contain absolute values.

Equations with absolute values cannot be solved in the same way as linear equations. Similarly, inequalities with absolute values require a specific method to be solved.

Definition: The Absolute Value

The absolute value of any number 𝑥 is defined algebraically as follows: |𝑥|=𝑥𝑥≥0,−𝑥𝑥<0.ifif

From the definition of absolute value, we see that there are always two numbers with the same nonzero absolute value, while only 0 has an absolute value of 0. For instance, the solutions to |𝑥|=4 are both numbers at a distance 4 from 0 located on either side of 0, that is, 4 and −4. Algebraically, we can transform this equation into two equations, namely, 𝑥=4 for 𝑥≥0 and −𝑥=4 for 𝑥<0.

Absolute value inequalities are inequalities involving the absolute value of a number expressed in terms of the unknown, 𝑥. The simplest absolute value inequalities are of the form |𝑥|<𝑐,|𝑥|≤𝑐,|𝑥|>𝑐,|𝑥|≥𝑐, where 𝑐 is a constant.

Consider, for example, |𝑥|≤4. Using the definition of the absolute value of a number as being the distance between 0 and the number, the solution to this inequality is the set of numbers that are located at a distance from 0 less than or equal to 4. We can represent this on the number line.

This set of numbers is [−4,4]. Alternatively, we can write −4≤𝑥≤4.

It is worth noting that if 𝑐 is negative, then there are no solutions to the inequality, since the absolute value of a number is always nonnegative.

A slightly more complex form of absolute value inequality is |𝑥−𝑎|<𝑐,|𝑥−𝑎|≤𝑐,|𝑥−𝑎|>𝑐,|𝑥−𝑎|≥𝑐, where 𝑐 is a constant.

Instead of having the absolute value of 𝑥, we have here the absolute value of 𝑥−𝑎. Let us call this number 𝑛. So, 𝑛=𝑥−𝑎, which can be rearranged to 𝑎+𝑛=𝑥. Therefore, 𝑛 is the number that, added to 𝑎, gives 𝑥.

Let us have a cursor located at 𝑎 on a number line; moving it by 𝑛 will position the cursor at 𝑥. If 𝑛 is positive, then 𝑥 is greater than 𝑎 (it is located on the right with respect to 𝑎), and if 𝑛 is negative, then 𝑥 is less than 𝑎 (it is located on the left with respect to 𝑎).

The absolute value of 𝑛, that is, |𝑥−𝑎|, can thus be interpreted as the distance between 𝑥 and 𝑎. Consider, for instance, the inequality |𝑥−2|≤3.

Its solution set is the set of numbers that are located at a distance from number 2 less than or equal to 3. The two numbers that are at a distance of exactly 3 from 2 are −1 and 5. Therefore, all numbers between −1 and 5 are located at a distance from 2 of maximum 3. The solution set is [−1,5].

Let us now reverse the inequality, so |𝑥−2|>3. The solution to this inequality is the set of numbers that are at a distance from 2 greater than 3. These are all the numbers that are less than −1 or greater than 5. In set notation, that is ]−∞,−1[∪]5,+∞[.

Let us summarize our findings.

Standard Result: Solution Set of a Simple Absolute Value Inequality

The solution set of absolute value inequalities of the form |𝑥−𝑎|≤𝑐 (with 𝑐≥0) is an interval centered at 𝑎 with length 2𝑐: [𝑎−𝑐,𝑎+𝑐].

Similarly, |𝑥−𝑎|<𝑐 (with 𝑐≥0) has the solution set ]𝑎−𝑐,𝑎+𝑐[.

The solution set of absolute value inequalities of the form |𝑥−𝑎|>𝑐 is complementary to the solution set of the reversed inequality |𝑥−𝑎|≤𝑐 described above: ℝ−[𝑎−𝑐,𝑎+𝑐]=]−∞,𝑎−𝑐[∪]𝑎+𝑐,+∞[.

Similarly, |𝑥−𝑎|≥𝑐 has the solution set ℝ−]𝑎−𝑐,𝑎+𝑐[=]−∞,𝑎−𝑐]∪[𝑎+𝑐,+∞[.

This can be represented on a number line.

This type of inequalities corresponds to real-life situations of tolerance to a specific measurable attribute of an object. For instance, imagine a carpenter who cuts pieces of wood to a length of 2.54 m with a tolerance of 1 cm. This means that the length does not need to be exactly 2.54 m but can be up to 1 cm larger or shorter than 2.54 m. Therefore, any piece with a length between 2.53 m and 2.55 m has the required length; we say that these lengths are within the tolerance range.

Let us see in our first example how such a situation is described with an absolute value inequality.

Example 1: Solving Word Problems by Finding the Boundaries of the Solution Set of an Absolute Value Inequality

A factory produces cans with weight 𝑥 grams. To control the production quality, the cans are only allowed to be sold if |𝑥−183|≤6. Determine the heaviest and the lightest weight of a can that can be sold.

Answer

Let us first interpret the given inequality, |𝑥−183|≤6. As 𝑥 is the weight of a can in grams, 𝑥−183 represents the difference between the can’s actual weight and the weight of 183 g. |𝑥−183| is then the difference between the can’s actual weight and 183 g. The inequality |𝑥−183|≤6 means that this difference can be up to 6 grams in either direction; that is, it means that the weight of a can be up to 6 grams heavier or lighter than 183 g.

Therefore, the heaviest possible weight is given by 183+6=189,g and the lightest possible weight is given by 183−6=177.g

In the previous example, we have dealt with a situation where a factory aims to produce cans with a weight of 183 g; this weight is then called the nominal weight. However, as it is probably difficult to produce cans with a specific weight, some variation around the nominal weight is allowed, here 6 grams; we say that there is a tolerance of 6 grams. For each can, the difference between its weight and the nominal weight is called the deviation from the nominal weight.

Let us now solve with our next example the inverse problem to our previous example, namely, write an absolute value inequality to describe an interval.

Example 2: Forming Absolute Value Inequalities in a Word Problem

Given that students’ grades in an exam range from 69 to 93, write an absolute value inequality to express the range of grades.

Answer

The given range can be first written as a closed interval, [69,93]. Recall that the solution set of an absolute value inequality of the form |𝑥−𝑎|≤𝑐 is a closed interval centered at 𝑎 with length 2𝑐. We can therefore find this inequality by finding the center, 𝑎, of [69,93] and its half-length, 𝑐.

The length of [69,93] is given by 2𝑐=93−69=24.

Its center can be found either by using the half-length (𝑐=12) and adding it to the lower boundary or subtracting it from the higher boundary: 𝑎=69+12=93−12=81, or by working out the midpoint between 69 and 93 (which is the mean of the two values): 𝑎=12(69+93)=12×162=81.

Thus, we can express the range from 69 to 93 as the absolute value inequality |𝑥−81|≤12.

In the next example, we will see how some complex inequalities are equivalent to a simple absolute value inequality.

Example 3: Rewriting an Inequality as an Absolute Value Inequality

Fill in the blank: The solution set in ℝ of the inequality √4𝑥−8𝑥+4≤8 equals .

  1. {−3,5}
  2. ℝ−[−3,5]
  3. ]−3,5[
  4. [−3,5]
  5. ℝ−]−3,5[

Answer

To start with, let us remember that a square root is always nonnegative. Therefore, we have 0≤√4𝑥−8𝑥+4≤8.

We can now square each side of the inequality, leading to 0≤4𝑥−8𝑥+4≤64.

Dividing each side by 4 gives us 0≤𝑥−2𝑥+1≤16, and factoring 𝑥−2𝑥+1 then leads to 0≤(𝑥−1)≤16.

Taking the square root of each side finally gives us 0≤|𝑥−1|≤4.

Recall that the solution set of an absolute value inequality of the form |𝑥−𝑎|≤𝑐 is a closed interval centered at 𝑎 with length 2𝑐, that is, [𝑎−𝑐,𝑎+𝑐].

Therefore, it is here [−3,5], which is option D.

In the previous example, graphing the parabola 𝑦=𝑥−4𝑥+4 and the horizontal line 𝑦=16 allows us to visualize that, given the line of symmetry of a parabola, the solution set of an inequality of the form 𝑎𝑥+𝑏𝑥+𝑐≤𝑑 is always centered at the 𝑥-coordinate of the vertex of the parabola.

It is worth noting that the inequality 0≤√4𝑥−16𝑥+16≤8 could have been solved as well by finding the boundaries of the interval, that is, the 𝑥-coordinates of the points of intersections of the parabola with the line 𝑦=16. These are the solutions of the equation: 𝑥−4𝑥+4=16,𝑥−4𝑥−12=0, that is, −2 and 6.

As the coefficient of the 𝑥-term in 𝑥−4𝑥+4 is positive, the parabola 𝑦=𝑥−4𝑥+4 opens upward, which means, as it is clear from looking at the graph, that the solution set of the inequality 𝑥−4𝑥+4≤16 is [−2,6].

If the parabola opened downward and still crossed the line 𝑦=16 at the same points, it would be below the line for all 𝑥 in the set ℝ−]−2,6[.

We are now going to learn how we can solve such inequalities graphically and algebraically. These methods will then allow us to solve more complex inequalities. Let us first recap how to plot a graph of the absolute value function. To do this, we can complete a table of value for 𝑦=|𝑥|:

𝑥−3−2−10123
𝑦3210123

Then, we can plot the coordinates on a pair of coordinate axes to draw the graph:

The ability to apply the definition of the absolute value and plot absolute value graphs is very useful when solving absolute value inequalities. So, practicing these skills is very important.

Consider the inequality |𝑥+1|≤3.

Let us solve it graphically first. So, on the same set of axes, we plot the graphs 𝑦=|𝑥+1| and 𝑦=3:

From this graph, we can see that the red graph, 𝑦=|𝑥+1|, is less than or equal to 3 when 𝑥 is greater than or equal to −4 and less than or equal to 2. So, the solution to the inequality is −4≤𝑥≤2.

The solution set is [−4,2].

Note how the graph 𝑦=|𝑥+1| is symmetric with respect to 𝑥=−1, which is consistent with the interpretation of |𝑥+1|=|𝑥−(−1)| as the distance between 𝑥 and −1 as we have learned earlier.

Now, we can also see from the graph how we might approach solving this inequality graphically. The red graph contains part of each of the graphs of 𝑦=𝑥+1 and 𝑦=−(𝑥+1). So, solving |𝑥+1|≤3 is equivalent to solving the set of compound inequalities 𝑥+1≤3 and −(𝑥+1)≤3. By multiplying each side of the latter inequality by −1, we find 𝑥+1≥−3.

Hence, |𝑥+1|≤3 is equivalent to −3≤𝑥+1≤3.

Let us solve it algebraically by first subtracting 1 from the three terms: −3−1≤𝑥+1−1≤3−1−4≤𝑥≤2.

This agrees with our initial findings from inspecting the graphs.

Both methods are equally acceptable for solving absolute value inequalities, but it is worth practicing both, particularly solving graphically, as this helps you to visualize the solution. It is also worth practicing giving your answer in various forms, including as simplified inequalities on number lines and as intervals.

Let us look at a couple more examples.

Example 4: Solving Absolute Value Inequalities

Find the solution set of the inequality |𝑥+4|<9.

Answer

We will solve this question first by using a graphical approach and then by using an algebraic approach. To solve the inequality graphically, we need to plot the graphs 𝑦=|𝑥+4| and 𝑦=9 on the same set of axes.

To plot the graph 𝑦=|𝑥+4|, we first solve 𝑥+4=0, finding that 𝑥=−4. When 𝑥≥−4, 𝑥+4≥0; therefore, |𝑥+4|=𝑥+4 and so our graph is the same as 𝑦=𝑥+4 in this region. For 𝑥<−4, 𝑥+4<0; thus, |𝑥+4|=−(𝑥+4) and the graph 𝑦=|𝑥+4| is the same as the graph 𝑦=−𝑥−4 in this region. Plotting these two lines along with 𝑦=9 on a graph, we get the following.

We observe that the two graphs intersect at (−13,9) and (5,9) and that the graph 𝑦=|𝑥+4| is below the line 𝑦=9 for −13<𝑥<5. Therefore, we conclude that the solution to the inequality is −13<𝑥<5.

The question, however, asks for the solution set of the inequality, which would be written as ]−13,5[.

If we want to solve the inequality algebraically, we rewrite |𝑥+4|<9 as the compound inequality: −9<𝑥+4<9.

Subtracting 4 from each side gives −13<𝑥<5.

Hence, the solution set is ]−13,5[.

Let us look now at an example where the inequality needs rearranging before being solved as we have just done.

Example 5: Solving Absolute Value Inequalities

Find algebraically the solution set of the inequality |7−𝑥|+3≤−6.

Answer

Notice here that the question explicitly asks us to calculate the solution set algebraically; however, for the benefit of explaining the solution, we will also present the graph. If we start by subtracting 3 from each side of the inequality, we get |7−𝑥|≤−9.

Now, the left-hand side of the inequality is an absolute value which is always greater than or equal to zero, and the right-hand side is a negative number, so there is no solution as the left-hand side can never be less than or equal to the right-hand side. This can be seen clearly by drawing the graphs of 𝑦=−9 and 𝑦=|7−𝑥| on the same set of axes:

The red graph here is obviously never less than the blue graph. Therefore, the solution set for the inequality is the empty set, ∅.

In our final example, we are going to solve algebraically a more complex absolute value inequality.

Example 6: Solving Absolute Value Inequalities Algebraically

Find algebraically the solution set of the inequality |𝑥−3|+|𝑥−5|>6.

Answer

We have here an inequality with two absolute value terms, |𝑥−3| and |𝑥−5|. Applying the formal definition of the absolute value for each term gives |𝑥−3|=𝑥−3𝑥−3≥0,𝑥≥3,−(𝑥−3)𝑥−3<0,𝑥<3,iforifor and |𝑥−5|=𝑥−5𝑥−5≥0,𝑥≥5,−(𝑥−5)𝑥−5<0,𝑥<5,iforifor

We see that we need to divide ℝ into two intervals for each absolute value term, so that ℝ is split into 3 intervals in total: ]−∞,3[∪[3,5[∪[5,+∞[. To avoid mistakes, let us use a table to write the value of each absolute value term for each interval and thus rewrite our inequality for each interval.

]−∞,3[[3,5[[5,+∞[
|𝑥−3|−(𝑥−3)𝑥−3𝑥−3
|𝑥−5|−(𝑥−5)−(𝑥−5)𝑥−5
|𝑥−3|+|𝑥−5|>6−(𝑥−3)−(𝑥−5)>6𝑥−3−(𝑥−5)>6𝑥−3+𝑥−5>6

We now have an inequality to solve for each of the three intervals.

For 𝑥<3, we have −(𝑥−3)−(𝑥−5)>6.

Expanding the parentheses gives −𝑥+3−𝑥+5>6, which simplifies to −2𝑥+8>6.

Subtracting 8 from each side yields −2𝑥>−2.

And, finally, multiplying each side by −12 gives 𝑥<1.

For 3≤𝑥<5, we have 𝑥−3−(𝑥−5)>6.

Expanding the parentheses gives 𝑥−3−𝑥+5>6, which simplifies to 2>6.

This inequality is not true. Therefore, 𝑥 cannot be in the interval [3,5[.

For 𝑥≥5, we have 𝑥−3+𝑥−5>6, which simplifies to 2𝑥−8>6.

Adding 8 to each side yields 2𝑥>14.

And, finally, dividing each side by 2 gives 𝑥>7.

Combining our 3 solutions, we find that 𝑥<1𝑥>7,or which corresponds to the solution set ]−∞,1[∪]7,+∞[=ℝ−[1,7].

Let us now summarize what we have learned in this explainer.

Key Points

  • The absolute value of a number can be interpreted as its distance from zero.
  • Absolute value inequalities of the form |𝑥−𝑎|<𝑐 or |𝑥−𝑎|≤𝑐 (with 𝑐≥0) can be solved algebraically by writing them as a compound inequality of the form −𝑐<𝑥−𝑎<𝑐 or −𝑐≤𝑥−𝑎≤𝑐.
  • Absolute value inequalities of the form |𝑥−𝑎|>𝑐 (or |𝑥−𝑎|≥𝑐) are the complementary inequalities of |𝑥−𝑎|<𝑐 or |𝑥−𝑎|≤𝑐, which means that their solution sets are complementary.
  • The solution set of absolute value inequalities of the form |𝑥−𝑎|≤𝑐 (with 𝑐≥0) is an interval centered at 𝑎 with length 2𝑐: [𝑎−𝑐,𝑎+𝑐]; similarly, |𝑥−𝑎|<𝑐 (with 𝑐≥0) has the solution set ]𝑎−𝑐,𝑎+𝑐[.
  • The solution set of absolute value inequalities of the form |𝑥−𝑎|>𝑐 is ℝ−[𝑎−𝑐,𝑎+𝑐] and that of |𝑥−𝑎|≥𝑐 is ℝ−]𝑎−𝑐,𝑎+𝑐[.
  • Absolute value inequalities of the form |𝑥−𝑎|<𝑐 (or any other inequality symbol) can be solved graphically by graphing the corresponding absolute value function 𝑦=|𝑥−𝑎| and the line 𝑦=𝑐 and inspecting for which 𝑥-values the absolute value function is below (for inequalities with < or ≤) or above (for inequalities with > or ≥) the line 𝑦=𝑐.
  • More complex absolute value inequalities can be solved algebraically by splitting ℝ into intervals in which the signs of the numbers inside the absolute value bars of all absolute value terms in the inequality do not change. The inequalities can then be rewritten for each of these intervals using the formal definition of the absolute value and solved separately. The final solution is obtained by combining all solutions.

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