{"id":1139,"date":"2024-05-06T20:33:46","date_gmt":"2024-05-06T20:33:46","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=1139"},"modified":"2024-12-03T17:46:53","modified_gmt":"2024-12-03T17:46:53","slug":"linear-inequalities-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/linear-inequalities-learn-it-2\/","title":{"raw":"Linear Inequalities: Learn It 2","rendered":"Linear Inequalities: Learn It 2"},"content":{"raw":"

Using the Properties of Inequalities<\/h2>\r\nWhen we work with inequalities, we can usually treat them similarly to but not exactly as we treat equations. We can use the addition property<\/strong> and the multiplication property<\/strong> to help us solve them. The one exception is when we multiply or divide by a negative number, we must reverse the inequality symbol.\r\n\r\n
\r\n

properties of inequalities<\/h3>\r\n

[latex]\\begin{array}{ll}\\text{Addition Property}\\hfill& \\text{If }a< b,\\text{ then }a+c< b+c.\\hfill \\\\ \\hfill & \\hfill \\\\ \\text{Multiplication Property}\\hfill & \\text{If }a< b\\text{ and }c> 0,\\text{ then }ac< bc.\\hfill \\\\ \\hfill & \\text{If }a< b\\text{ and }c< 0,\\text{ then }ac> bc.\\hfill \\end{array}[\/latex]<\/p>\r\n \r\n\r\nThese properties also apply to [latex]a\\le b[\/latex], [latex]a>b[\/latex], and [latex]a\\ge b[\/latex].\r\n\r\n<\/section>

Illustrate the addition property for inequalities by solving each of the following:\r\n
    \r\n \t
  1. [latex]x - 15<4[\/latex]<\/li>\r\n \t
  2. [latex]6\\ge x - 1[\/latex]<\/li>\r\n \t
  3. [latex]x+7>9[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"105622\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"105622\"]\r\n\r\nThe addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality.\r\n
      \r\n \t
    1. [latex]\\begin{array}{ll} \\text{Given inequality:} & x - 15 < 4 \\\\ \\text{Add 15 to both sides:} & x - 15 + 15 < 4 + 15 \\\\ \\text{Simplify:} & x < 19 \\end{array}[\/latex]<\/center>In interval notation, the solution is expressed as: [latex](-\\infty, 19)[\/latex].<\/li>\r\n \t
    2. [latex]\\begin{array}{ll} \\text{Given inequality:} & 6 \\geq x - 1 \\\\ \\text{Add 1 to both sides:} & 6 + 1 \\geq x - 1 + 1 \\\\ \\text{Simplify:} & 7 \\geq x \\\\ \\text{or equivalently:} & x \\leq 7 \\end{array}[\/latex]<\/center>In interval notation, the solution is expressed as: [latex](-\\infty, 7][\/latex]<\/li>\r\n \t
    3. [latex]\\begin{array}{ll} \\text{Given inequality:} & x + 7 > 9 \\\\ \\text{Subtract 7 from both sides:} & x + 7 - 7 > 9 - 7 \\\\ \\text{Simplify:} & x > 2 \\end{array}[\/latex]<\/center>In interval notation, the solution is expressed as: [latex](2, \\infty)[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      [ohm2_question hide_question_numbers=1]18944[\/ohm2_question]<\/section>
      Illustrate the multiplication property for inequalities by solving each of the following:\r\n
        \r\n \t
      1. [latex]3x<6[\/latex]<\/li>\r\n \t
      2. [latex]-2x - 1\\ge 5[\/latex]<\/li>\r\n \t
      3. [latex]5-x>10[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"725573\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"725573\"]\r\n
          \r\n \t
        1. [latex]\\begin{array}{ll} \\text{Given inequality:} && 3x &< 6 \\\\ \\text{Multiply both sides by} \\frac{1}{3} \\text{or divide by 3:} && x &< \\frac{6}{3} \\\\ \\text{Simplify:} && x &< 2 \\end{array}[\/latex]<\/center>\r\nIn interval notation, the solution is expressed as: [latex](-\\infty, 2)[\/latex]<\/li>\r\n \t
        2. [latex]\\begin{array}{ll} \\text{Given inequality:} && -2x - 1 &\\geq 5 \\\\ \\text{Add 1 to both sides:} && -2x - 1 + 1 &\\geq 5 + 1 \\\\ \\text{Simplify:} && -2x &\\geq 6 \\\\ \\text{Divide both sides by -2 (reverse inequality):} && x &\\leq -3 \\end{array}[\/latex]<\/center>\r\nIn interval notation, the solution is expressed as: [latex](-\\infty, -3][\/latex]<\/li>\r\n \t
        3. [latex]\\begin{array}{ll} \\text{Given inequality:} & 5 - x > 10 \\\\ \\text{Subtract 5 from both sides:} & 5 - x - 5 > 10 - 5 \\\\ \\text{Simplify:} & -x > 5 \\\\ \\text{Multiply both sides by -1 (reverse inequality):} & x < -5 \\end{array}[\/latex]<\/center>\r\nIn interval notation, the solution is: [latex](-\\infty, -5)[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
          [ohm2_question hide_question_numbers=1]18945[\/ohm2_question]<\/section>
          [ohm2_question hide_question_numbers=1]18946[\/ohm2_question]<\/section>We can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable.\r\n\r\n
          Solve the following inequality:
          [latex]13 - 7x\\ge 10x - 4[\/latex]<\/center>[reveal-answer q=\"453286\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"453286\"]Solving this inequality is similar to solving an equation up until the last step.\r\n
          [latex]\\begin{array}{ll}13 - 7x\\ge 10x - 4\\hfill & \\hfill \\\\ 13 - 17x\\ge -4\\hfill & \\text{Move variable terms to one side of the inequality}.\\hfill \\\\ -17x\\ge -17\\hfill & \\text{Isolate the variable term}.\\hfill \\\\ x\\le 1\\hfill & \\text{Dividing both sides by }-17\\text{ reverses the inequality}.\\hfill \\end{array}[\/latex]<\/div>\r\nThe solution set is given by the interval [latex]\\left(-\\infty ,1\\right][\/latex], or all real numbers less than and including [latex]1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
          Solve the following inequality and write the answer in interval notation:
          [latex]-\\dfrac{3}{4}x\\ge -\\dfrac{5}{8}+\\dfrac{2}{3}x[\/latex]<\/center>[reveal-answer q=\"37354\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"37354\"]\r\nWe begin solving in the same way we do when solving an equation.\r\n
          [latex]\\begin{array}{ll}-\\frac{3}{4}x\\ge -\\frac{5}{8}+\\frac{2}{3}x\\hfill & \\hfill \\\\ -\\frac{3}{4}x-\\frac{2}{3}x\\ge -\\frac{5}{8}\\hfill & \\text{Put variable terms on one side}.\\hfill \\\\ -\\frac{9}{12}x-\\frac{8}{12}x\\ge -\\frac{5}{8}\\hfill & \\text{Write fractions with common denominator}.\\hfill \\\\ -\\frac{17}{12}x\\ge -\\frac{5}{8}\\hfill & \\hfill \\\\ x\\le -\\frac{5}{8}\\left(-\\frac{12}{17}\\right)\\hfill & \\text{Multiplying by a negative number reverses the inequality}.\\hfill \\\\ x\\le \\frac{15}{34}\\hfill & \\hfill \\end{array}[\/latex]<\/div>\r\nThe solution set is the interval [latex]\\left(-\\infty ,\\frac{15}{34}\\right][\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
          [ohm_question hide_question_numbers=1]293769[\/ohm_question]<\/section>","rendered":"

          Using the Properties of Inequalities<\/h2>\n

          When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equations. We can use the addition property<\/strong> and the multiplication property<\/strong> to help us solve them. The one exception is when we multiply or divide by a negative number, we must reverse the inequality symbol.<\/p>\n

          \n

          properties of inequalities<\/h3>\n

          [latex]\\begin{array}{ll}\\text{Addition Property}\\hfill& \\text{If }a< b,\\text{ then }a+c< b+c.\\hfill \\\\ \\hfill & \\hfill \\\\ \\text{Multiplication Property}\\hfill & \\text{If }a< b\\text{ and }c> 0,\\text{ then }ac< bc.\\hfill \\\\ \\hfill & \\text{If }a< b\\text{ and }c< 0,\\text{ then }ac> bc.\\hfill \\end{array}[\/latex]<\/p>\n

           <\/p>\n

          These properties also apply to [latex]a\\le b[\/latex], [latex]a>b[\/latex], and [latex]a\\ge b[\/latex].<\/p>\n<\/section>\n

          Illustrate the addition property for inequalities by solving each of the following:<\/p>\n
            \n
          1. [latex]x - 15<4[\/latex]<\/li>\n
          2. [latex]6\\ge x - 1[\/latex]<\/li>\n
          3. [latex]x+7>9[\/latex]<\/li>\n<\/ol>\n