{"id":493,"date":"2019-04-29T13:59:50","date_gmt":"2019-04-29T17:59:50","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=493"},"modified":"2019-11-26T13:22:38","modified_gmt":"2019-11-26T18:22:38","slug":"4-3-linear-absolute-value-inequalities","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/4-3-linear-absolute-value-inequalities\/","title":{"raw":"4.3 Linear Absolute Value Inequalities","rendered":"4.3 Linear Absolute Value Inequalities"},"content":{"raw":"[latexpage]\r\n\r\nAbsolute values are positive magnitudes, which means that they represent the positive value of any number.\r\n\r\nFor instance, | \u22125 | and | +5 | are the same, with both having the same value of 5, and | \u221299 | and | +99 | both share the same value of 99.\r\n\r\nWhen used in inequalities, absolute values become a boundary limit to a number.\r\n
\r\n

Example 4.3.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nConsider \\(| x | < 4.\\)\r\n\r\nThis means that the unknown \\(x\\) value is less than 4, so \\(| x | < 4\\) becomes \\(x < 4.\\) However, there is more to this with regards to negative values for \\(x.\\)\r\n\r\n| \u22121 | is a value that is a solution, since 1 <\u00a0 4.\r\n\r\nHowever, | \u22125 | < 4 is not a solution, since 5\u00a0 >\u00a0 4.\r\n\r\nThe boundary of \\(| x | < 4\\) works out to be between \u22124 and +4.\r\n\r\nThis means that \\(| x | < 4\\) ends up being bounded as \\(-4 <\u00a0 x\u00a0 < 4.\\)\r\n\r\nIf the inequality is written as \\(| x | \\le 4\\), then little changes, except that \\(x\\) can then equal \u22124 and +4, rather than having to be larger or smaller.\r\n\r\nThis means that \\(| x | \\le 4\\) ends up being bounded as \\(-4 \\le\u00a0 x\u00a0 \\le 4.\\)\r\n\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 4.3.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nConsider \\(|x| > 4.\\)\r\n\r\nThis means that the unknown \\(x\\) value is greater than 4, so \\(|x| > 4\\) becomes \\(x > 4.\\) However, the negative values for \\(x\\) must still be considered.\r\n\r\nThe boundary of \\(|x| > 4\\) works out to be smaller than \u22124 and larger than +4.\r\n\r\nThis means that \\(|x| > 4\\) ends up being bounded as \\(x < -4\u00a0 \\text{ or }\u00a0 4 < x.\\)\r\n\r\nIf the inequality is written as \\(| x | \\ge 4,\\) then little changes, except that \\(x\\) can then equal \u22124 and +4, rather than having to be larger or smaller.\r\n\r\nThis means that \\(|x| \\ge 4\\) ends up being bounded as\u00a0 \\(x \\le -4\u00a0 \\text{ or }\u00a0 4 \\le x.\\)\r\n\r\n<\/div>\r\n<\/div>\r\nWhen drawing the boundaries for inequalities on a number line graph, use the following conventions:\r\n

For \u2264 or \u2265, use [brackets] as boundary limits.\"Blank<\/p>\r\n

For < or >, use (parentheses) as boundary limits. \"Blank<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Equation<\/th>\r\nNumber Line<\/th>\r\n<\/tr>\r\n
\\(| x | <4 \\)<\/td>\r\n\"x<\/td>\r\n<\/tr>\r\n
\\(| x | \\le 4\\)<\/td>\r\n\"x<\/td>\r\n<\/tr>\r\n
\\(| x | > 4\\)<\/td>\r\n\"x<\/td>\r\n<\/tr>\r\n
\\(| x | \\ge 4\\)<\/td>\r\n\"x<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen an inequality has an absolute value, isolate the absolute value first in order to graph a solution and\/or write it in interval notation. The following examples will illustrate isolating and solving an inequality with an absolute value.\r\n
\r\n

Example 4.3.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve, graph, and give interval notation for the inequality\u00a0 \\(-4\u00a0 - 3 | x | \\ge\u00a0 -16.\\)\r\n\r\nFirst, isolate the inequality:\r\n\r\n\\[\\begin{array}{rrrrrl}\r\n-4&-&3|x|& \\ge & -16 &\\\\\r\n+4&&&&+4& \\text{add 4 to both sides}\\\\\r\n\\midrule\r\n&&\\dfrac{-3|x|}{-3}& \\ge & \\dfrac{-12}{-3}&\\text{divide by }-3 \\text{ and flip the sense} \\\\ \\\\\r\n&&|x|&\\le & 4 &&\r\n\\end{array}\\]\r\n\r\nAt this point, it is known that the inequality is bounded by 4. Specifically, it is between \u22124 and 4.\r\n\r\nThis means that \\(-4 \\le | x | \\le 4.\\)\r\n\r\nThis solution on a number line looks like:\r\n\r\n\"\u22124\r\n\r\nTo write the solution in interval notation, use the symbols and numbers on the number line: \\([-4, 4].\\)\r\n\r\n<\/div>\r\n<\/div>\r\nOther examples of absolute value inequalities result in an algebraic expression that is bounded by an inequality.\r\n
\r\n

Example 4.3.4<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve, graph, and give interval notation for the inequality \\(| 2x - 4 | \\le\u00a0 6.\\)\r\n\r\nThis means that the inequality to solve is \\(-6\\le 2x - 4\\le 6\\):\r\n

\\(\\begin{array}{rrrcrrr}\r\n-6&\\le & 2x&-&4&\\le & 6 \\\\\r\n+4&&&+&4&&+4 \\\\\r\n\\midrule\r\n\\dfrac{-2}{2}&\\le &&\\dfrac{2x}{2}&&\\le & \\dfrac{10}{2} \\\\ \\\\\r\n-1 &\\le &&x&&\\le & 5\r\n\\end{array}\\)<\/p>\r\n\"\u22121<\/span>\r\n\r\nTo write the solution in interval notation, use the symbols and numbers on the number line: \\([-1,5].\\)\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 4.3.5<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve, graph, and give interval notation for the inequality \\(9\u00a0 -\u00a0 2 | 4x + 1 |\u00a0 > 3.\\)\r\n\r\nFirst, isolate the inequality by subtracting 9 from both sides:\r\n\r\n\\[\\begin{array}{rrrrrrr}\r\n9&-&2|4x&+&1|&>&3 \\\\\r\n-9&&&&&&-9 \\\\\r\n\\midrule\r\n&&-2|4x&+&1|&>&-6 \\\\\r\n\\end{array}\\]\r\n

Divide both sides by \u22122 and flip the sense:<\/p>\r\n\\[\\begin{array}{rrr}\r\n\\dfrac{-2|4x+1|}{-2}&>&\\dfrac{-6}{-2} \\\\ \\\\\r\n|4x+1|&<& 3\r\n\\end{array}\\]\r\n\r\nAt this point, it is known that the inequality expression is between \u22123 and 3, so \\(-3\u00a0 <\u00a0 4x + 1\u00a0 <\u00a0 3.\\)\r\n\r\nAll that is left is to isolate \\(x\\). First, subtract 1 from all three parts:\r\n\r\n\\[\\begin{array}{rrrrrrr}\r\n-3&<&4x&+&1&<&3 \\\\\r\n-1&&&-&1&&-1 \\\\\r\n\\midrule\r\n-4&<&&4x&&<&2 \\\\\r\n\\end{array}\\]\r\n\r\nThen, divide all three parts by 4:\r\n\r\n\\[\\begin{array}{rrrrr}\r\n\\dfrac{-4}{4}&<&\\dfrac{4x}{4}&<&\\dfrac{2}{4} \\\\ \\\\\r\n-1&<&x&<&\\dfrac{1}{2} \\\\\r\n\\end{array}\\]\r\n\r\n\"\u22121<\/span>\r\n\r\nIn interval notation, this is written as \\(\\left(-1,\\dfrac{1}{2}\\right).\\)\r\n\r\n<\/div>\r\n<\/div>\r\nIt is important to remember when solving these equations that the absolute value is always positive. If given an absolute value that is less than a negative number, there will be no solution because absolute value will always be positive, i.e., greater than a negative. Similarly, if absolute value is greater than a negative, the answer will be all real numbers.\r\n\r\nThis means that:\r\n

\\(\\begin{array}{c}\r\n| 2x - 4 | <\u00a0 -6 \\text{ has no possible solution } (x \\ne \\mathbb{R}) \\\\ \\\\\r\n\\text{and} \\\\ \\\\\r\n| 2x - 4 | >\u00a0 -6 \\text{ has every number as a solution and is written as } (-\\infty, \\infty)\r\n\\end{array}\\)<\/p>\r\nNote: since infinity can never be reached, use parentheses instead of brackets when writing infinity (positive or negative) in interval notation.\r\n

Questions<\/h1>\r\nFor questions 1 to 33, solve each inequality, graph its solution, and give interval notation.\r\n
    \r\n \t
  1. \\(| x | < 3\\)<\/li>\r\n \t
  2. \\(| x | \\le 8\\)<\/li>\r\n \t
  3. \\(| 2x | < 6\\)<\/li>\r\n \t
  4. \\(| x + 3 | < 4\\)<\/li>\r\n \t
  5. \\(| x - 2 | < 6\\)<\/li>\r\n \t
  6. \\(| x - 8 | < 12\\)<\/li>\r\n \t
  7. \\(| x - 7 | < 3\\)<\/li>\r\n \t
  8. \\(| x + 3 | \\le 4\\)<\/li>\r\n \t
  9. \\(| 3x - 2 | < 9\\)<\/li>\r\n \t
  10. \\(| 2x + 5 | < 9\\)<\/li>\r\n \t
  11. \\(1 + 2 | x - 1 | \\le 9\\)<\/li>\r\n \t
  12. \\(10 - 3 | x - 2 | \\ge 4\\)<\/li>\r\n \t
  13. \\(6 -\u00a0 | 2x - 5 |\u00a0 > 3\\)<\/li>\r\n \t
  14. \\(| x | > 5\\)<\/li>\r\n \t
  15. \\(| 3x |\u00a0 > 5\\)<\/li>\r\n \t
  16. \\(| x - 4 | > 5\\)<\/li>\r\n \t
  17. \\(| x + 3 | > 3\\)<\/li>\r\n \t
  18. \\(| 2x - 4 | > 6\\)<\/li>\r\n \t
  19. \\(| x - 5 | > 3\\)<\/li>\r\n \t
  20. \\(3 -\u00a0 | 2 - x | < 1\\)<\/li>\r\n \t
  21. \\(4 + 3 | x - 1 |\u00a0 < 10\\)<\/li>\r\n \t
  22. \\(3 - 2 | 3x - 1 | \\ge -7\\)<\/li>\r\n \t
  23. \\(3 - 2 | x - 5 | \\le -15\\)<\/li>\r\n \t
  24. \\(4 - 6 | -6 - 3x | \\le -5\\)<\/li>\r\n \t
  25. \\(-2 - 3 | 4 - 2x | \\ge -8\\)<\/li>\r\n \t
  26. \\(-3 - 2 | 4x - 5 | \\ge 1\\)<\/li>\r\n \t
  27. \\(4 - 5 | -2x - 7 | < -1\\)<\/li>\r\n \t
  28. \\(-2 + 3 | 5 - x | \\le 4\\)<\/li>\r\n \t
  29. \\(3 - 2 | 4x - 5 | \\ge 1\\)<\/li>\r\n \t
  30. \\(-2 - 3 | - 3x - 5| \\ge\u00a0 -5\\)<\/li>\r\n \t
  31. \\(-5 - 2 | 3x - 6 | < -8\\)<\/li>\r\n \t
  32. \\(6 - 3 | 1 - 4x | < -3\\)<\/li>\r\n \t
  33. \\(4 - 4 | -2x + 6 | > -4\\)<\/li>\r\n<\/ol>\r\nAnswer Key 4.3<\/a>","rendered":"

    Absolute values are positive magnitudes, which means that they represent the positive value of any number.<\/p>\n

    For instance, | \u22125 | and | +5 | are the same, with both having the same value of 5, and | \u221299 | and | +99 | both share the same value of 99.<\/p>\n

    When used in inequalities, absolute values become a boundary limit to a number.<\/p>\n

    \n
    \n

    Example 4.3.1<\/p>\n<\/header>\n

    \n

    Consider \"| x | < 4.\"<\/p>\n

    This means that the unknown \"x\" value is less than 4, so \"| x | < 4\" becomes \"x < 4.\" However, there is more to this with regards to negative values for \"x.\"<\/p>\n

    | \u22121 | is a value that is a solution, since 1 <\u00a0 4.<\/p>\n

    However, | \u22125 | < 4 is not a solution, since 5\u00a0 >\u00a0 4.<\/p>\n

    The boundary of \"| x | < 4\" works out to be between \u22124 and +4.<\/p>\n

    This means that \"| x | < 4\" ends up being bounded as \"-4 < x < 4.\"<\/p>\n

    If the inequality is written as \"| x | \le 4\", then little changes, except that \"x\" can then equal \u22124 and +4, rather than having to be larger or smaller.<\/p>\n

    This means that \"| x | \le 4\" ends up being bounded as \"-4 \le x \le 4.\"<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 4.3.2<\/p>\n<\/header>\n

    \n

    Consider \"|x| > 4.\"<\/p>\n

    This means that the unknown \"x\" value is greater than 4, so \"|x| > 4\" becomes \"x > 4.\" However, the negative values for \"x\" must still be considered.<\/p>\n

    The boundary of \"|x| > 4\" works out to be smaller than \u22124 and larger than +4.<\/p>\n

    This means that \"|x| > 4\" ends up being bounded as \"x < -4 \text{ or } 4 < x.\"<\/p>\n

    If the inequality is written as \"| x | \ge 4,\" then little changes, except that \"x\" can then equal \u22124 and +4, rather than having to be larger or smaller.<\/p>\n

    This means that \"|x| \ge 4\" ends up being bounded as\u00a0 \"x \le -4 \text{ or } 4 \le x.\"<\/p>\n<\/div>\n<\/div>\n

    When drawing the boundaries for inequalities on a number line graph, use the following conventions:<\/p>\n

    For \u2264 or \u2265, use [brackets] as boundary limits.\"Blank<\/p>\n

    For < or >, use (parentheses) as boundary limits. \"Blank<\/p>\n\n\n\n\n\n\n\n
    Equation<\/th>\nNumber Line<\/th>\n<\/tr>\n
    \"| x | <4\"<\/td>\n\"x<\/td>\n<\/tr>\n
    \"| x | \le 4\"<\/td>\n\"x<\/td>\n<\/tr>\n
    \"| x | > 4\"<\/td>\n\"x<\/td>\n<\/tr>\n
    \"| x | \ge 4\"<\/td>\n\"x<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    When an inequality has an absolute value, isolate the absolute value first in order to graph a solution and\/or write it in interval notation. The following examples will illustrate isolating and solving an inequality with an absolute value.<\/p>\n

    \n
    \n

    Example 4.3.3<\/p>\n<\/header>\n

    \n

    Solve, graph, and give interval notation for the inequality\u00a0 \"-4 - 3 | x | \ge -16.\"<\/p>\n

    First, isolate the inequality:<\/p>\n

      <\/span>   <\/span>\"\[\begin{array}{rrrrrl}<\/p>\n

    At this point, it is known that the inequality is bounded by 4. Specifically, it is between \u22124 and 4.<\/p>\n

    This means that \"-4 \le | x | \le 4.\"<\/p>\n

    This solution on a number line looks like:<\/p>\n

    \"\u22124<\/p>\n

    To write the solution in interval notation, use the symbols and numbers on the number line: \"[-4, 4].\"<\/p>\n<\/div>\n<\/div>\n

    Other examples of absolute value inequalities result in an algebraic expression that is bounded by an inequality.<\/p>\n

    \n
    \n

    Example 4.3.4<\/p>\n<\/header>\n

    \n

    Solve, graph, and give interval notation for the inequality \"| 2x - 4 | \le 6.\"<\/p>\n

    This means that the inequality to solve is \"-6\le 2x - 4\le 6\":<\/p>\n

    \"\begin{array}{rrrcrrr}<\/p>\n

    \"\u22121<\/span><\/p>\n

    To write the solution in interval notation, use the symbols and numbers on the number line: \"[-1,5].\"<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 4.3.5<\/p>\n<\/header>\n

    \n

    Solve, graph, and give interval notation for the inequality \"9 - 2 | 4x + 1 | > 3.\"<\/p>\n

    First, isolate the inequality by subtracting 9 from both sides:<\/p>\n

      <\/span>   <\/span>\"\[\begin{array}{rrrrrrr}<\/p>\n

    Divide both sides by \u22122 and flip the sense:<\/p>\n

      <\/span>   <\/span>\"\[\begin{array}{rrr}<\/p>\n

    At this point, it is known that the inequality expression is between \u22123 and 3, so \"-3 < 4x + 1 < 3.\"<\/p>\n

    All that is left is to isolate \"x\". First, subtract 1 from all three parts:<\/p>\n

      <\/span>   <\/span>\"\[\begin{array}{rrrrrrr}<\/p>\n

    Then, divide all three parts by 4:<\/p>\n

      <\/span>   <\/span>\"\[\begin{array}{rrrrr}<\/p>\n

    \"\u22121<\/span><\/p>\n

    In interval notation, this is written as \"\left(-1,\dfrac{1}{2}\right).\"<\/p>\n<\/div>\n<\/div>\n

    It is important to remember when solving these equations that the absolute value is always positive. If given an absolute value that is less than a negative number, there will be no solution because absolute value will always be positive, i.e., greater than a negative. Similarly, if absolute value is greater than a negative, the answer will be all real numbers.<\/p>\n

    This means that:<\/p>\n

    \"\begin{array}{c}<\/p>\n

    Note: since infinity can never be reached, use parentheses instead of brackets when writing infinity (positive or negative) in interval notation.<\/p>\n

    Questions<\/h1>\n

    For questions 1 to 33, solve each inequality, graph its solution, and give interval notation.<\/p>\n

      \n
    1. \"| x | < 3\"<\/li>\n
    2. \"| x | \le 8\"<\/li>\n
    3. \"| 2x | < 6\"<\/li>\n
    4. \"| x + 3 | < 4\"<\/li>\n
    5. \"| x - 2 | < 6\"<\/li>\n
    6. \"| x - 8 | < 12\"<\/li>\n
    7. \"| x - 7 | < 3\"<\/li>\n
    8. \"| x + 3 | \le 4\"<\/li>\n
    9. \"| 3x - 2 | < 9\"<\/li>\n
    10. \"| 2x + 5 | < 9\"<\/li>\n
    11. \"1 + 2 | x - 1 | \le 9\"<\/li>\n
    12. \"10 - 3 | x - 2 | \ge 4\"<\/li>\n
    13. \"6 - | 2x - 5 | > 3\"<\/li>\n
    14. \"| x | > 5\"<\/li>\n
    15. \"| 3x | > 5\"<\/li>\n
    16. \"| x - 4 | > 5\"<\/li>\n
    17. \"| x + 3 | > 3\"<\/li>\n
    18. \"| 2x - 4 | > 6\"<\/li>\n
    19. \"| x - 5 | > 3\"<\/li>\n
    20. \"3 - | 2 - x | < 1\"<\/li>\n
    21. \"4 + 3 | x - 1 | < 10\"<\/li>\n
    22. \"3 - 2 | 3x - 1 | \ge -7\"<\/li>\n
    23. \"3 - 2 | x - 5 | \le -15\"<\/li>\n
    24. \"4 - 6 | -6 - 3x | \le -5\"<\/li>\n
    25. \"-2 - 3 | 4 - 2x | \ge -8\"<\/li>\n
    26. \"-3 - 2 | 4x - 5 | \ge 1\"<\/li>\n
    27. \"4 - 5 | -2x - 7 | < -1\"<\/li>\n
    28. \"-2 + 3 | 5 - x | \le 4\"<\/li>\n
    29. \"3 - 2 | 4x - 5 | \ge 1\"<\/li>\n
    30. \"-2 - 3 | - 3x - 5| \ge -5\"<\/li>\n
    31. \"-5 - 2 | 3x - 6 | < -8\"<\/li>\n
    32. \"6 - 3 | 1 - 4x | < -3\"<\/li>\n
    33. \"4 - 4 | -2x + 6 | > -4\"<\/li>\n<\/ol>\n

      Answer Key 4.3<\/a><\/p>\n","protected":false},"author":540,"menu_order":13,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-493","chapter","type-chapter","status-publish","hentry"],"part":363,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/493","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/540"}],"version-history":[{"count":19,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/493\/revisions"}],"predecessor-version":[{"id":3566,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/493\/revisions\/3566"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/363"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/493\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=493"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=493"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=493"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=493"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}