Absolute value inequality is a type of inequality that contains an absolute or mod (modulus) value sign with a variable inside it. The value of the variable represents its distance from the origin, which can be plotted on a number line.
Types
Now, depending on whether the expression inside the modulus is positive or negative, there can be two possible cases:
With < or ≤ symbol
With > or ≥ symbol
Rules for Solving
Inequalities with < or ≤
Suppose we consider the absolute value inequality |x| < 6. Here, the distance from the origin to the variable ‘x’ is less than 6 units.
To solve |x| < 6, we follow the below steps.
Step 1: Splitting the Inequality into Two Cases
The inequality is already simplified. So, |x| < 6 ⇒ -x < 6 and x < 6
Step 2: Determining the Range of the Variable
Using the inversion property in the first inequality to find the value of the variable ‘x,’
-x < 6 ⇒ x > -6
Thus, |x| < 6 ⇒ x > -6 and x < 6, which is written in the set notation as {x | -6 < x < 6} and in the interval notation as (-6, 6)
Step 3: Graphing the Solution
When it is plotted on the number line, we get:
How to Solve Absolute Value Inequalities
In general, for any real numbers x and y, if |x| < y, then -y < x < y. In this case, the solution always shows an intersection between the two.
Solve and Graph the solution set of the inequality |2x – 1| + 5 ≤ 14
Solution:
Given, |2x – 1| + 5 ≤ 14 Step 1: Simplifying Using the subtraction property, we get |2x – 1| + 5 – 5 ≤ 14 – 5 ⇒ |2x – 1| ≤ 9 Step 2: Splitting the Inequality into Two Cases By splitting the above absolute value inequality into two, we get 2x – 1 ≥ -9 and 2x – 1 ≤ 9 Step 3: Determining the Range of the Variable Now, 2x – 1 ≥ -9 and 2x – 1 ≤ 9 ⇒ 2x – 1 + 1 ≥ -9 + 1 (by addition property) and 2x – 1 + 1 ≤ 9 + 1 (by addition property) ⇒ 2x ≥ -8 and 2x ≤ 10 ⇒ x ≥ -4 (by division property) and x ≤ 5 (by division property) ⇒ x ≥ -4 and x ≤ 5 Step 4: Graphing the Solution Here, we observe that the solution is [-4, 5], and the graph is formed with the ‘closed’ circles for weak inequalities.
Solve the inequality|6x| – 3 < 9
Solution:
Given, |6x| – 3 < 9 Step 1: SImplifying Using the addition property, |6x| – 3 + 3 < 9 + 3 ⇒ |6x| < 12 Step 2: Splitting the Inequality into Two Cases By splitting the above absolute value inequality into two separate inequalities, we get 6x > -12 and 6x < 12 Step 3: Determining the Range of the Variable Now, 6x > -12 and 6x < 12 ⇒ x > -2 and x < 2 (by division property) Thus, the solution set is {x | -2 < x < 2} or (-2, 2)
Inequalities With > or ≥
Now, let us solve the inequality |x| > 6.
|x| > 6 means the distance from the origin to the variable ‘x’ is greater than 6 units.
To solve |x| > 6, the steps are:
Step 1: Splitting the Inequality into Two Cases
The inequality is already simplified. So, |x| > 6 ⇒ -x > 6 or x > 6
Step 2: Determining the Range of the Variable
Using the inversion property in the first inequality, we get -x > 6 ⇒ x < -6
Thus, |x| > 6 ⇒ x < -6 or x > 6.
The solution set is written in the set builder notation as {x | x < -6 or x > 6} and in the interval notation as (-∞, -6) ∪ (6, ∞)
Step 3: Graphing the Solution
When it is plotted on the number line, we get
Absolute Value Inequalities 1
In general, for any real numbers x and y, if |x| > y, then x < -y or x > y.
Solve and graph the solution set of the inequality 5|4 – 3x| > 10
Solution:
Given, 5|4 – 3x| > 10 Step 1: Simplifying Using the division property, we get |4 – 2x| > 2 Step 2: Splitting the Inequality into Two Cases By splitting the above absolute value inequality into two separate inequalities, we get 4 – 2x < -12 or 4 – 2x > 12 Step 3: Determining the Range of the Variable Now, 4 – 2x < -12 or 4 – 2x > 12 ⇒ 4 – 2x – 4 < -12 – 4 or 4 – 2x – 4 > 12 – 4 (by subtraction property) ⇒ -2x < -16 or -2x > 8 ⇒ -x < -8 or -x > 4 (by division property) ⇒ x > 8 or x < -4 (by inversion property) Thus, the solution set is {x | x < -4 or x > 8} or (-∞, -4) ∪ (8, ∞) Step 4: Graphing the Solution
Solve and graph the solution set of the inequality: |2x + 5| ≥ 9
Solution:
Here, |2x + 5| ≥ 9 Step 1: Splitting the Inequality into Two Cases By splitting the above inequality into two separate inequalities, we get 2x + 5 ≤ -9 or 2x + 5 ≥ 9 Step 2: Determining the Range of the Variable Now, 2x + 5 ≤ -9 or 2x + 5 ≥ 9 ⇒ 2x +5 – 5 ≤ -9 – 5 (by subtraction property) or 2x + 5 – 5 ≥ 9 – 5 (by subtraction property) ⇒ 2x ≤ -14 or 2x ≥ 4 ⇒ x ≤ -7 (by division property) or x ≥ 2 (by division property) ⇒ x ≤ -7 or x ≥ 2 Step 3: Graphing the Solution Since weak inequalities are used here, circles are ‘closed,’ and the solution in the interval notation is (-∞, -7] ∪ [2, ∞).
