 United States of AmericaPA
High School Core Standards - Algebra I Assessment Anchors

3.11 Absolute value inequalities

Lesson

Inequalities of the form $\left|x\right||x|<b or$\left|x\right|\le b$|x|≤b To say$x\le4$x4 is to say that$x$x can be any number positive or negative that is not greater than four. On a number line we draw$x\le4$x4 as such: If the notation is changed slightly to$|x|\le4$|x|4, then we are restricting$x$x to be any number between$-4$4 and$4$4, inclusive. On a number line we draw$|x|\le4$|x|4 as such: You could check that if$x$x is a number less than$-4$4, then the absolute value of the number is greater than$4$4. It follows that a statement like$|x|\le4$|x|4 is equivalent to a pair of inequalities that can be written$-4\le x\le4$4x4. See the different representations below: Absolute Value Inequality Interval Notation Number Line$\left|x\right|\le4$|x|4$-4\le x\le4$4x4$\left[-4,4\right]$[4,4] $\left|x\right|<2$|x|<2$-22<x<2 $\left(-2,2\right)$(2,2) Inequalities of the form $\left|x\right|>b$|x|>b or $\left|x\right|\ge b$|x|≥b

To solve an inequality (sometimes called an inequation) means to find out the range or ranges of values of the variable for which the statement is true.

Above we looked at $|x|\le4$|x|4, and said this was the same as restricting $x$x to be any number between $-4$4 and $4$4, inclusive.

What about something like $\left|x\right|\ge2$|x|2? Well, any number greater than or equal to $2$2 will satisfy this equation, but also any values less than or equal to $-2$2. This means that this single expression is represented by two disjoint intervals as show below. See the different representations below:

Absolute Value Inequality Interval Notation Number Line
$\left|x\right|\ge2$|x|2 $x\le-2$x2 or $x\ge2$x2 $\left(-\infty,-2\right]\cup\left[2,\infty\right)$(,2][2,) $\left|x\right|>2$|x|>2 $x<-2$x<2 or $x>2$x>2 $\left(-\infty,-2\right)\cup\left(2,\infty\right)$(,2)(2,) Worked examples

Question 1

Solve $\left|x+2\right|>5$|x+2|>5.

Think: It is common practice to consider what happens to the statement in $2$2 parts.  The positive side and the negative side.

Do:

Positive side means to consider and solve $+(x+2)>5$+(x+2)>5

 $+(x+2)$+(x+2) $>$> $5$5 (original positive inequality) $x+2$x+2 $>$> $5$5 (remove unnecessary parentheses) $x$x $>$> $3$3 (subtract $2$2 from both sides)

Negative side means to consider and solve $-(x+2)>5$(x+2)>5

 $-(x+2)$−(x+2) $>$> $5$5 (original positive inequality) $x+2$x+2 $<$< $-5$−5 (divide both sides by $-1$−1, remember this flips the sign) $x$x $<$< $-7$−7 (subtract $2$2 from both sides)

This means that for values of $x$x, that are either greater then $3$3 or less than $-7$7, then the inequality $|x+2|>5$|x+2|>5 will hold true. This graph shows the solution set.  Spot check values inside or outside the range to check.

Alternative approach for Question 1

Another way to think about the previous example is via translations.

Consider first the solution on a number line for $|x|>5$|x|>5.  We would know that this means all values of $x$x on the line that are a distance of greater than $5$5 from zero.  It would look like this. The effect of the $+2$+2, is a horizontal translation of $2$2 units to the left.  Resulting in our final solution: Practice questions

Question 2

Rewrite the inequality $\left|x\right|<2$|x|<2 without using absolute values.

Question 3

Rewrite the inequality $\left|x\right|\ge5$|x|5 without using absolute values.

Question 4

Solve $\left|4x-8\right|+1=13$|4x8|+1=13.

Write both solutions as equations on the same line separated by a comma.

Outcomes

A1.1.3.1.1

Write or solve compound inequalities and/or graph their solution sets on a number line (may include absolute value inequalities).

A1.1.3.1.2

Identify or graph the solution set to a linear inequality on a number line.