Virginia SOL Algebra 2 - 2020 Edition
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1.11 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

AII.3a

Solve absolute value linear equations and inequalities

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