{"id":1280,"date":"2021-12-02T19:37:18","date_gmt":"2021-12-03T00:37:18","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/4-3-linear-absolute-value-inequalities\/"},"modified":"2023-08-30T12:53:32","modified_gmt":"2023-08-30T16:53:32","slug":"linear-absolute-value-inequalities","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/linear-absolute-value-inequalities\/","title":{"raw":"4.3 Linear Absolute Value Inequalities","rendered":"4.3 Linear Absolute Value Inequalities"},"content":{"raw":"Absolute 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 [latex]| x | < 4.[\/latex]\r\n\r\nThis means that the unknown [latex]x[\/latex] value is less than 4, so [latex]| x | < 4[\/latex] becomes [latex]x < 4.[\/latex] However, there is more to this with regards to negative values for [latex]x.[\/latex]\r\n\r\n| \u22121 | is a value that is a solution, since 1 < 4.\r\n\r\nHowever, | \u22125 | < 4 is not a solution, since 5 > 4.\r\n\r\nThe boundary of [latex]| x | < 4[\/latex] works out to be between \u22124 and +4.\r\n\r\nThis means that [latex]| x | < 4[\/latex] ends up being bounded as [latex]-4 < x < 4.[\/latex]\r\n\r\nIf the inequality is written as [latex]| x | \\le 4[\/latex], then little changes, except that [latex]x[\/latex] can then equal \u22124 and +4, rather than having to be larger or smaller.\r\n\r\nThis means that [latex]| x | \\le 4[\/latex] ends up being bounded as [latex]-4 \\le x \\le 4.[\/latex]\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 [latex]|x|[\/latex] > [latex]4.[\/latex]\r\n\r\nThis means that the unknown [latex]x[\/latex] value is greater than 4, so [latex]|x|[\/latex] > [latex]4[\/latex] becomes [latex]x[\/latex] > [latex]4.[\/latex] However, the negative values for [latex]x[\/latex] must still be considered.\r\n\r\nThe boundary of [latex]|x|[\/latex] > [latex]4[\/latex] works out to be smaller than \u22124 and larger than +4.\r\n\r\nThis means that [latex]|x|[\/latex] > [latex]4[\/latex] ends up being bounded as [latex]x < -4 \\text{ or } 4 < x.[\/latex]\r\n\r\nIf the inequality is written as [latex]| x | \\ge 4,[\/latex] then little changes, except that [latex]x[\/latex] can then equal \u22124 and +4, rather than having to be larger or smaller.\r\n\r\nThis means that [latex]|x| \\ge 4[\/latex] ends up being bounded as [latex]x \\le -4 \\text{ or } 4 \\le x.[\/latex]\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
[latex]| x | <4 [\/latex]<\/td>\r\n\"x<\/td>\r\n<\/tr>\r\n
[latex]| x | \\le 4[\/latex]<\/td>\r\n\"x<\/td>\r\n<\/tr>\r\n
[latex]| x |[\/latex] > [latex]4[\/latex]<\/td>\r\n\"x<\/td>\r\n<\/tr>\r\n
[latex]| x | \\ge 4[\/latex]<\/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 [latex]-4 - 3 | x | \\ge -16.[\/latex]\r\n\r\nFirst, isolate the inequality:\r\n\r\n[latex]\\begin{array}{rrrrrl}\r\n-4&-&3|x|& \\ge & -16 &\\\\\r\n+4&&&&+4& \\text{add 4 to both sides}\\\\\r\n\\hline\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}[\/latex]\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 [latex]-4 \\le | x | \\le 4.[\/latex]\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: [latex][-4, 4].[\/latex]\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 [latex]| 2x - 4 | \\le 6.[\/latex]\r\n\r\nThis means that the inequality to solve is [latex]-6\\le 2x - 4\\le 6[\/latex]:\r\n

[latex]\\begin{array}{rrrcrrr}\r\n-6&\\le & 2x&-&4&\\le & 6 \\\\\r\n+4&&&+&4&&+4 \\\\\r\n\\hline\r\n\\dfrac{-2}{2}&\\le &&\\dfrac{2x}{2}&&\\le & \\dfrac{10}{2} \\\\ \\\\\r\n-1 &\\le &&x&&\\le & 5\r\n\\end{array}[\/latex]<\/p>\r\n\"\u22121<\/span>\r\n\r\nTo write the solution in interval notation, use the symbols and numbers on the number line: [latex][-1,5].[\/latex]\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 [latex]9 - 2 | 4x + 1 |[\/latex] > [latex]3.[\/latex]\r\n\r\nFirst, isolate the inequality by subtracting 9 from both sides:\r\n

[latex]\\begin{array}{rrrrrrr}9&-&2|4x&+&1&>&3 \\\\ -9&&&&&&-9 \\\\ \\hline &&-2|4x&+&1|&>&-6 \\end{array}[\/latex]<\/p>\r\n

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

[latex]\\begin{array}{rcc}\\dfrac{-2|4x+1|}{-2}&>&\\dfrac{-6}{-2} \\\\ |4x+1|&<&3 \\end{array}[\/latex]<\/p>\r\nAt this point, it is known that the inequality expression is between \u22123 and 3, so [latex]-3 < 4x + 1 < 3.[\/latex]\r\n\r\nAll that is left is to isolate [latex]x[\/latex]. First, subtract 1 from all three parts:\r\n

[latex]\\begin{array}{rrrrrrr}\r\n-3&<&4x&+&1&<&3 \\\\\r\n-1&&&-&1&&-1 \\\\\r\n\\hline\r\n-4&<&&4x&&<&2 \\\\\r\n\\end{array}[\/latex]<\/p>\r\nThen, divide all three parts by 4:\r\n

[latex]\\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}[\/latex]<\/p>\r\n\"\u22121<\/span>\r\n\r\nIn interval notation, this is written as [latex]\\left(-1,\\dfrac{1}{2}\\right).[\/latex]\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

[latex]\\begin{array}{c}| 2x - 4| < -6 \\text{ has no possible solution } (x \\ne \\mathbb{R}) \\\\ \\\\ \\text{and}\\\\ \\\\ |2x-4| > -6 \\text{ has every number as a solution and is written as } (-\\infty, \\infty) \\end{array}[\/latex]<\/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. [latex]| x | < 3[\/latex]<\/li>\r\n \t
  2. [latex]| x | \\le 8[\/latex]<\/li>\r\n \t
  3. [latex]| 2x | < 6[\/latex]<\/li>\r\n \t
  4. [latex]| x + 3 | < 4[\/latex]<\/li>\r\n \t
  5. [latex]| x - 2 | < 6[\/latex]<\/li>\r\n \t
  6. [latex]| x - 8 | < 12[\/latex]<\/li>\r\n \t
  7. [latex]| x - 7 | < 3[\/latex]<\/li>\r\n \t
  8. [latex]| x + 3 | \\le 4[\/latex]<\/li>\r\n \t
  9. [latex]| 3x - 2 | < 9[\/latex]<\/li>\r\n \t
  10. [latex]| 2x + 5 | < 9[\/latex]<\/li>\r\n \t
  11. [latex]1 + 2 | x - 1 | \\le 9[\/latex]<\/li>\r\n \t
  12. [latex]10 - 3 | x - 2 | \\ge 4[\/latex]<\/li>\r\n \t
  13. [latex]6 - | 2x - 5 |[\/latex] > [latex]3[\/latex]<\/li>\r\n \t
  14. [latex]| x |[\/latex] > [latex]5[\/latex]<\/li>\r\n \t
  15. [latex]| 3x |[\/latex] > [latex]5[\/latex]<\/li>\r\n \t
  16. [latex]| x - 4 |[\/latex] > [latex]5[\/latex]<\/li>\r\n \t
  17. [latex]| x + 3 |[\/latex] > [latex]3[\/latex]<\/li>\r\n \t
  18. [latex]| 2x - 4 |[\/latex] > [latex]6[\/latex]<\/li>\r\n \t
  19. [latex]| x - 5 |[\/latex] > [latex]3[\/latex]<\/li>\r\n \t
  20. [latex]3 - | 2 - x | < 1[\/latex]<\/li>\r\n \t
  21. [latex]4 + 3 | x - 1 | < 10[\/latex]<\/li>\r\n \t
  22. [latex]3 - 2 | 3x - 1 | \\ge -7[\/latex]<\/li>\r\n \t
  23. [latex]3 - 2 | x - 5 | \\le -15[\/latex]<\/li>\r\n \t
  24. [latex]4 - 6 | -6 - 3x | \\le -5[\/latex]<\/li>\r\n \t
  25. [latex]-2 - 3 | 4 - 2x | \\ge -8[\/latex]<\/li>\r\n \t
  26. [latex]-3 - 2 | 4x - 5 | \\ge 1[\/latex]<\/li>\r\n \t
  27. [latex]4 - 5 | -2x - 7 | < -1[\/latex]<\/li>\r\n \t
  28. [latex]-2 + 3 | 5 - x | \\le 4[\/latex]<\/li>\r\n \t
  29. [latex]3 - 2 | 4x - 5 | \\ge 1[\/latex]<\/li>\r\n \t
  30. [latex]-2 - 3 | - 3x - 5| \\ge -5[\/latex]<\/li>\r\n \t
  31. [latex]-5 - 2 | 3x - 6 | < -8[\/latex]<\/li>\r\n \t
  32. [latex]6 - 3 | 1 - 4x | < -3[\/latex]<\/li>\r\n \t
  33. [latex]4 - 4 | -2x + 6 |[\/latex] > [latex]-4[\/latex]<\/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 [latex]| x | < 4.[\/latex]\n\nThis means that the unknown [latex]x[\/latex] value is less than 4, so [latex]| x | < 4[\/latex] becomes [latex]x < 4.[\/latex] However, there is more to this with regards to negative values for [latex]x.[\/latex]\n\n| \u22121 | is a value that is a solution, since 1 < 4.\n\nHowever, | \u22125 | < 4 is not a solution, since 5 > 4.\n\nThe boundary of [latex]| x | < 4[\/latex] works out to be between \u22124 and +4.\n\nThis means that [latex]| x | < 4[\/latex] ends up being bounded as [latex]-4 < x < 4.[\/latex]\n\nIf the inequality is written as [latex]| x | \\le 4[\/latex], then little changes, except that [latex]x[\/latex] can then equal \u22124 and +4, rather than having to be larger or smaller.\n\nThis means that [latex]| x | \\le 4[\/latex] ends up being bounded as [latex]-4 \\le x \\le 4.[\/latex]\n\n<\/div>\n<\/div>\n

    \n
    \n

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

    \n

    Consider [latex]|x|[\/latex] > [latex]4.[\/latex]<\/p>\n

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

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

    This means that [latex]|x|[\/latex] > [latex]4[\/latex] ends up being bounded as [latex]x < -4 \\text{ or } 4 < x.[\/latex]\n\nIf the inequality is written as [latex]| x | \\ge 4,[\/latex] then little changes, except that [latex]x[\/latex] can then equal \u22124 and +4, rather than having to be larger or smaller.\n\nThis means that [latex]|x| \\ge 4[\/latex] ends up being bounded as [latex]x \\le -4 \\text{ or } 4 \\le x.[\/latex]\n\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
    [latex]| x | <4[\/latex]<\/td>\n\"x<\/td>\n<\/tr>\n
    [latex]| x | \\le 4[\/latex]<\/td>\n\"x<\/td>\n<\/tr>\n
    [latex]| x |[\/latex] > [latex]4[\/latex]<\/td>\n\"x<\/td>\n<\/tr>\n
    [latex]| x | \\ge 4[\/latex]<\/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 [latex]-4 - 3 | x | \\ge -16.[\/latex]<\/p>\n

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

    [latex]\\begin{array}{rrrrrl} -4&-&3|x|& \\ge & -16 &\\\\ +4&&&&+4& \\text{add 4 to both sides}\\\\ \\hline &&\\dfrac{-3|x|}{-3}& \\ge & \\dfrac{-12}{-3}&\\text{divide by }-3 \\text{ and flip the sense} \\\\ \\\\ &&|x|&\\le & 4 && \\end{array}[\/latex]<\/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 [latex]-4 \\le | x | \\le 4.[\/latex]<\/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: [latex][-4, 4].[\/latex]<\/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 [latex]| 2x - 4 | \\le 6.[\/latex]<\/p>\n

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

    [latex]\\begin{array}{rrrcrrr} -6&\\le & 2x&-&4&\\le & 6 \\\\ +4&&&+&4&&+4 \\\\ \\hline \\dfrac{-2}{2}&\\le &&\\dfrac{2x}{2}&&\\le & \\dfrac{10}{2} \\\\ \\\\ -1 &\\le &&x&&\\le & 5 \\end{array}[\/latex]<\/p>\n

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

    To write the solution in interval notation, use the symbols and numbers on the number line: [latex][-1,5].[\/latex]<\/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 [latex]9 - 2 | 4x + 1 |[\/latex] > [latex]3.[\/latex]<\/p>\n

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

    [latex]\\begin{array}{rrrrrrr}9&-&2|4x&+&1&>&3 \\\\ -9&&&&&&-9 \\\\ \\hline &&-2|4x&+&1|&>&-6 \\end{array}[\/latex]<\/p>\n

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

    [latex]\\begin{array}{rcc}\\dfrac{-2|4x+1|}{-2}&>&\\dfrac{-6}{-2} \\\\ |4x+1|&<&3 \\end{array}[\/latex]<\/p>\n

    At this point, it is known that the inequality expression is between \u22123 and 3, so [latex]-3 < 4x + 1 < 3.[\/latex]\n\nAll that is left is to isolate [latex]x[\/latex]. First, subtract 1 from all three parts:\n\n\n

    [latex]\\begin{array}{rrrrrrr} -3&<&4x&+&1&<&3 \\\\ -1&&&-&1&&-1 \\\\ \\hline -4&<&&4x&&<&2 \\\\ \\end{array}[\/latex]<\/p>\n

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

    [latex]\\begin{array}{rrrrr} \\dfrac{-4}{4}&<&\\dfrac{4x}{4}&<&\\dfrac{2}{4} \\\\ \\\\ -1&<&x&<&\\dfrac{1}{2} \\\\ \\end{array}[\/latex]<\/p>\n

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

    In interval notation, this is written as [latex]\\left(-1,\\dfrac{1}{2}\\right).[\/latex]<\/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

    [latex]\\begin{array}{c}| 2x - 4| < -6 \\text{ has no possible solution } (x \\ne \\mathbb{R}) \\\\ \\\\ \\text{and}\\\\ \\\\ |2x-4| > -6 \\text{ has every number as a solution and is written as } (-\\infty, \\infty) \\end{array}[\/latex]<\/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. [latex]| x | < 3[\/latex]<\/li>\n
    2. [latex]| x | \\le 8[\/latex]<\/li>\n
    3. [latex]| 2x | < 6[\/latex]<\/li>\n
    4. [latex]| x + 3 | < 4[\/latex]<\/li>\n
    5. [latex]| x - 2 | < 6[\/latex]<\/li>\n
    6. [latex]| x - 8 | < 12[\/latex]<\/li>\n
    7. [latex]| x - 7 | < 3[\/latex]<\/li>\n
    8. [latex]| x + 3 | \\le 4[\/latex]<\/li>\n
    9. [latex]| 3x - 2 | < 9[\/latex]<\/li>\n
    10. [latex]| 2x + 5 | < 9[\/latex]<\/li>\n
    11. [latex]1 + 2 | x - 1 | \\le 9[\/latex]<\/li>\n
    12. [latex]10 - 3 | x - 2 | \\ge 4[\/latex]<\/li>\n
    13. [latex]6 - | 2x - 5 |[\/latex] > [latex]3[\/latex]<\/li>\n
    14. [latex]| x |[\/latex] > [latex]5[\/latex]<\/li>\n
    15. [latex]| 3x |[\/latex] > [latex]5[\/latex]<\/li>\n
    16. [latex]| x - 4 |[\/latex] > [latex]5[\/latex]<\/li>\n
    17. [latex]| x + 3 |[\/latex] > [latex]3[\/latex]<\/li>\n
    18. [latex]| 2x - 4 |[\/latex] > [latex]6[\/latex]<\/li>\n
    19. [latex]| x - 5 |[\/latex] > [latex]3[\/latex]<\/li>\n
    20. [latex]3 - | 2 - x | < 1[\/latex]<\/li>\n
    21. [latex]4 + 3 | x - 1 | < 10[\/latex]<\/li>\n
    22. [latex]3 - 2 | 3x - 1 | \\ge -7[\/latex]<\/li>\n
    23. [latex]3 - 2 | x - 5 | \\le -15[\/latex]<\/li>\n
    24. [latex]4 - 6 | -6 - 3x | \\le -5[\/latex]<\/li>\n
    25. [latex]-2 - 3 | 4 - 2x | \\ge -8[\/latex]<\/li>\n
    26. [latex]-3 - 2 | 4x - 5 | \\ge 1[\/latex]<\/li>\n
    27. [latex]4 - 5 | -2x - 7 | < -1[\/latex]<\/li>\n
    28. [latex]-2 + 3 | 5 - x | \\le 4[\/latex]<\/li>\n
    29. [latex]3 - 2 | 4x - 5 | \\ge 1[\/latex]<\/li>\n
    30. [latex]-2 - 3 | - 3x - 5| \\ge -5[\/latex]<\/li>\n
    31. [latex]-5 - 2 | 3x - 6 | < -8[\/latex]<\/li>\n
    32. [latex]6 - 3 | 1 - 4x | < -3[\/latex]<\/li>\n
    33. [latex]4 - 4 | -2x + 6 |[\/latex] > [latex]-4[\/latex]<\/li>\n<\/ol>\n

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