{"id":830,"date":"2025-07-15T20:11:20","date_gmt":"2025-07-15T20:11:20","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=830"},"modified":"2026-01-12T18:31:23","modified_gmt":"2026-01-12T18:31:23","slug":"absolute-value-functions-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/absolute-value-functions-learn-it-4\/","title":{"raw":"Absolute Value Functions: Learn It 4","rendered":"Absolute Value Functions: Learn It 4"},"content":{"raw":"

Absolute Value Inequalities<\/h2>\r\nAn absolute value inequality<\/strong> is an equation of the form\r\n
[latex]|A| < B,|A|\\le B,|A| > B,\\text{or }|A|\\ge B[\/latex],<\/div>\r\nwhere A<\/em>, and sometimes B<\/em>, represents an algebraic expression dependent on a variable x. <\/em>Solving the inequality means finding the set of all [latex]x[\/latex] -<\/em>values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.\r\n\r\n
\r\n

absolute value inequality<\/h3>\r\nFor an algebraic expression [latex]X[\/latex]\u00a0<\/em>and [latex]k>0[\/latex], an absolute value inequality<\/strong> is an inequality of the form:\r\n
<\/div>\r\n
[latex]\\begin{array}{c} |X| < k \\text{ is equivalent to } -k < X < k \\\\ \\text{or} \\\\ |X| > k \\text{ is equivalent to } X < -k \\text{ or } X > k \\\\ \\end{array}[\/latex]<\/div>\r\n \r\n\r\nThese statements also apply to [latex]|X|\\le k[\/latex] and [latex]|X|\\ge k[\/latex].\r\n\r\n<\/section>There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.\r\n\r\n
Suppose we want to know all possible returns on an investment if we could earn some amount of money within [latex]$200[\/latex] of [latex]$600[\/latex].\r\n\r\n
\r\n\r\nWe can solve algebraically for the set of [latex]x-[\/latex]values such that the distance between [latex]x[\/latex] and [latex]600 [\/latex] is less than [latex]200[\/latex]. We represent the distance between [latex]x[\/latex] and [latex]600 [\/latex] as [latex]|x - 600|[\/latex], and therefore,\r\n\r\n
[latex]|x - 600|\\le 200[\/latex]<\/center>
or<\/center>\r\n
[latex]\\begin{array}{c}-200\\le x - 600\\le 200\\\\ -200+600\\le x - 600+600\\le 200+600\\\\ 400\\le x\\le 800\\end{array}[\/latex]<\/div>\r\nThis means our returns would be between [latex]$400[\/latex] and [latex]$800[\/latex].\r\n\r\n<\/section>To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.\r\n\r\n
Describe all values [latex]x[\/latex] within a distance of [latex]4[\/latex] from the number [latex]5[\/latex].\r\n\r\n
\r\n\r\nWe want the distance between [latex]x[\/latex] and [latex]5[\/latex] to be less than or equal to [latex]4[\/latex]. We can draw a number line to represent the condition to be satisfied.\r\n\r\n\"NumberThe distance from [latex]x[\/latex] to [latex]5[\/latex] can be represented using an absolute value symbol, [latex]|x - 5|[\/latex]. Write the values of [latex]x[\/latex] that satisfy the condition as an absolute value inequality.\r\n
[latex]|x - 5|\\le 4[\/latex]<\/div>\r\nWe need to write two inequalities as there are always two solutions to an absolute value equation.\r\n
[latex]\\begin{array}{lll}x - 5\\le 4\\hfill & \\text{and}\\hfill & x - 5\\ge -4\\hfill \\\\ x\\le 9\\hfill & \\hfill & x\\ge 1\\hfill \\end{array}[\/latex]<\/div>\r\nIf the solution set is [latex]x\\le 9[\/latex] and [latex]x\\ge 1[\/latex], then the solution set is an interval including all real numbers between and including [latex]1[\/latex] and [latex]9[\/latex].\r\n\r\nSo, [latex]|x - 5|\\le 4[\/latex] is equivalent to [latex]\\left[1,9\\right][\/latex] in interval notation.\r\n\r\n<\/section>
Solve the following:
[latex]|x - 1|\\le 3[\/latex]<\/center>[reveal-answer q=\"4865\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"4865\"]\r\n

[latex]\\begin{array}{c}|x - 1|\\le 3\\hfill \\\\ \\hfill \\\\ -3\\le x - 1\\le 3\\hfill \\\\ \\hfill \\\\ -2\\le x\\le 4\\hfill \\\\ \\hfill \\\\ \\left[-2,4\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n

[\/hidden-answer]<\/div>\r\n<\/section>
Given the equation
[latex]y=-\\frac{1}{2}|4x - 5|+3[\/latex],<\/center>determine the [latex]x[\/latex]-values for which the [latex]y[\/latex]-values are negative.[reveal-answer q=\"624558\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"624558\"]We are trying to determine where [latex]y<0[\/latex] which is when [latex]-\\frac{1}{2}|4x - 5|+3<0[\/latex]. We begin by isolating the absolute value.\r\n
[latex]\\begin{array}{ll}-\\frac{1}{2}|4x - 5|< -3\\hfill & \\text{Multiply both sides by -2, and reverse the inequality}.\\hfill \\\\ |4x - 5|> 6\\hfill & \\hfill \\end{array}[\/latex]<\/div>\r\nNext, we solve [latex]|4x - 5|=6[\/latex].\r\n
[latex]\\begin{array}{lll}4x - 5=6\\hfill & \\hfill & 4x - 5=-6\\hfill \\\\ 4x=11\\hfill & \\text{or}\\hfill & 4x=-1\\hfill \\\\ x=\\frac{11}{4}\\hfill & \\hfill & x=-\\frac{1}{4}\\hfill \\end{array}[\/latex]<\/div>\r\nNow, we can examine the graph to observe where the [latex]y[\/latex]-<\/em>values are negative. We observe where the branches are below the [latex]x[\/latex]-<\/em>axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at [latex]x=-\\frac{1}{4}[\/latex] and [latex]x=\\frac{11}{4}[\/latex] and that the graph opens downward.\r\n\r\n\"A\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
[ohm_question hide_question_numbers=1]318772[\/ohm_question]<\/section>
\r\n

Sometimes a picture is worth a thousand words. You can turn a single variable inequality into a two variable inequality and make a graph. The [latex]x[\/latex]-intercepts of the graph will correspond with the solution to the inequality you can find by hand.<\/p>\r\n

When solving [latex]-2|k - 4|\\le -6[\/latex] we can change the variable to [latex]x[\/latex] to make it easier to enter in an online graphing calculator.<\/p>\r\n

To turn [latex]-2|x - 4|\\le -6[\/latex] into a two variable equation, move everything to one side, and place the variable y on the other side like this:<\/p>\r\n

[latex]-2|x - 4|\\le -6[\/latex]\r\n[latex]-2|x - 4|+6\\le y[\/latex]<\/p>\r\n

Now enter this inequality in an online graphing calculator and hover over the [latex]x[\/latex]-intercepts.<\/p>\r\n\r\n<\/section>

[ohm_question hide_question_numbers=1]318773[\/ohm_question]<\/section>","rendered":"

Absolute Value Inequalities<\/h2>\n

An absolute value inequality<\/strong> is an equation of the form<\/p>\n

[latex]|A| < B,|A|\\le B,|A| > B,\\text{or }|A|\\ge B[\/latex],<\/div>\n

where A<\/em>, and sometimes B<\/em>, represents an algebraic expression dependent on a variable x. <\/em>Solving the inequality means finding the set of all [latex]x[\/latex] –<\/em>values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.<\/p>\n

\n

absolute value inequality<\/h3>\n

For an algebraic expression [latex]X[\/latex]\u00a0<\/em>and [latex]k>0[\/latex], an absolute value inequality<\/strong> is an inequality of the form:<\/p>\n

<\/div>\n
[latex]\\begin{array}{c} |X| < k \\text{ is equivalent to } -k < X < k \\\\ \\text{or} \\\\ |X| > k \\text{ is equivalent to } X < -k \\text{ or } X > k \\\\ \\end{array}[\/latex]<\/div>\n

 <\/p>\n

These statements also apply to [latex]|X|\\le k[\/latex] and [latex]|X|\\ge k[\/latex].<\/p>\n<\/section>\n

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.<\/p>\n

Suppose we want to know all possible returns on an investment if we could earn some amount of money within [latex]$200[\/latex] of [latex]$600[\/latex].<\/p>\n
\n

We can solve algebraically for the set of [latex]x-[\/latex]values such that the distance between [latex]x[\/latex] and [latex]600[\/latex] is less than [latex]200[\/latex]. We represent the distance between [latex]x[\/latex] and [latex]600[\/latex] as [latex]|x - 600|[\/latex], and therefore,<\/p>\n

[latex]|x - 600|\\le 200[\/latex]<\/div>\n
or<\/div>\n
[latex]\\begin{array}{c}-200\\le x - 600\\le 200\\\\ -200+600\\le x - 600+600\\le 200+600\\\\ 400\\le x\\le 800\\end{array}[\/latex]<\/div>\n

This means our returns would be between [latex]$400[\/latex] and [latex]$800[\/latex].<\/p>\n<\/section>\n

To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.<\/p>\n

Describe all values [latex]x[\/latex] within a distance of [latex]4[\/latex] from the number [latex]5[\/latex].<\/p>\n
\n

We want the distance between [latex]x[\/latex] and [latex]5[\/latex] to be less than or equal to [latex]4[\/latex]. We can draw a number line to represent the condition to be satisfied.<\/p>\n

\"NumberThe distance from [latex]x[\/latex] to [latex]5[\/latex] can be represented using an absolute value symbol, [latex]|x - 5|[\/latex]. Write the values of [latex]x[\/latex] that satisfy the condition as an absolute value inequality.<\/p>\n

[latex]|x - 5|\\le 4[\/latex]<\/div>\n

We need to write two inequalities as there are always two solutions to an absolute value equation.<\/p>\n

[latex]\\begin{array}{lll}x - 5\\le 4\\hfill & \\text{and}\\hfill & x - 5\\ge -4\\hfill \\\\ x\\le 9\\hfill & \\hfill & x\\ge 1\\hfill \\end{array}[\/latex]<\/div>\n

If the solution set is [latex]x\\le 9[\/latex] and [latex]x\\ge 1[\/latex], then the solution set is an interval including all real numbers between and including [latex]1[\/latex] and [latex]9[\/latex].<\/p>\n

So, [latex]|x - 5|\\le 4[\/latex] is equivalent to [latex]\\left[1,9\\right][\/latex] in interval notation.<\/p>\n<\/section>\n

Solve the following:<\/p>\n
[latex]|x - 1|\\le 3[\/latex]<\/div>\n