{"id":6715,"date":"2020-10-08T15:19:15","date_gmt":"2020-10-08T15:19:15","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6715"},"modified":"2023-07-28T22:45:18","modified_gmt":"2023-07-28T22:45:18","slug":"2-2-applications-using-linear-inequalities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/2-2-applications-using-linear-inequalities\/","title":{"raw":"2.2: Applications using Linear Inequalities","rendered":"2.2: Applications using Linear Inequalities"},"content":{"raw":"
\r\n

SECTION 2.2 Learning Objective<\/h3>\r\n2.2:\u00a0 Applications Using Linear Inequalities<\/strong>\r\n
    \r\n \t
  • Solve applications using linear inequalities<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n

    Solve applications using linear inequalities<\/h2>\r\nThis section will discuss several types of applications. No matter how simple or complex the <\/span>application is, the following steps will help to translate and solve the problem. <\/span>\r\n
      \r\n \t
    1. Read\u00a0<\/span>through the entire problem\u00a0<\/span><\/strong><\/li>\r\n \t
    2. Organize\u00a0<\/span>the information\u00a0<\/span><\/strong><\/li>\r\n \t
    3. Write the\u00a0<\/span>inequali<\/span>ty\u00a0<\/span><\/strong><\/li>\r\n \t
    4. Solve\u00a0<\/span>the inequality<\/span><\/strong><\/li>\r\n \t
    5. Check\u00a0<\/span>the answer<\/span><\/strong><\/li>\r\n<\/ol>\r\nLook for key <\/span>words and <\/span>phrases. The following table gives some common translations\u00a0<\/span>for <\/span>inequalities<\/span>.<\/span>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Words or Phrase<\/strong><\/td>\r\nSymbol<\/strong><\/td>\r\n<\/tr>\r\n
      Is less than<\/td>\r\n<<\/td>\r\n<\/tr>\r\n
      Is more than<\/td>\r\n><\/td>\r\n<\/tr>\r\n
      At least<\/td>\r\n\u2265<\/td>\r\n<\/tr>\r\n
      At most<\/td>\r\n\u2264<\/td>\r\n<\/tr>\r\n
      Minimum<\/td>\r\n\u2265<\/td>\r\n<\/tr>\r\n
      Maximum<\/td>\r\n\u2264<\/td>\r\n<\/tr>\r\n
      No more than<\/td>\r\n\u2264<\/td>\r\n<\/tr>\r\n
      No less than<\/td>\r\n\u2265<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
      Below is a video example of an application problem involving linear inequalities.<\/div>\r\n
      \r\n\r\n[embed]https:\/\/youtu.be\/AMjQd_3Vw_c[\/embed]\r\n\r\n<\/div>\r\n
      \r\n

      Example 1<\/h3>\r\nMaria\u2019s job provides a yearly base salary of $25,000 plus a commission of 7% of her sales. <\/span>If <\/span>h<\/span>er goal is to earn a yearly income <\/span>of at least\u00a0<\/span>$39,000<\/span>, what will her total sales need to be\u00a0<\/span>to <\/span>achieve that goal<\/span>?<\/span>\r\n\r\n[reveal-answer q=\"719869\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"719869\"]\r\n

      Salary is\u00a0<\/span>$25,000<\/span><\/p>\r\n

      Let x be the total sales, so that Commission is 0.07 of x, or 0.07x<\/span><\/p>\r\n

      Income is at least<\/strong> $39,000 or\u00a0\u2265\u00a0<\/span>$39,000<\/span><\/p>\r\n

      \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[latex]25000 + 0.07x\u2265 39000[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Salary + Commission\u00a0\u2265 Income<\/p>\r\n\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [latex]0.07x\u2265 14000[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Subtract 25,000 from 39,000\r\n

      \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[latex]\\frac{0.07x}{0.07}\u2265\\frac{14000}{0.07}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Divide both sides by 0.07<\/p>\r\n

      [latex]x\u2265200000[\/latex]<\/p>\r\nIf Maria\u2019s total sales are $200,000, she will earn $39,000.\u00a0<\/span>\r\n\r\nIf her total sa<\/span>les are more than $200,000, she will earn more than $39,000.<\/span>\r\n

      Answer<\/span><\/h4>\r\nMaria's total sales must be at least $200,000 in order to earn a yearly income of at least $39,000.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is another similar example:\r\n
      \r\n

      Example 2<\/h3>\r\nLeslie want<\/span>s to rent a car\u00a0<\/span>for the day. The rental company charges a daily rate of $18.95 plus 19 <\/span>cents for each mile driven. If Leslie only has $50 to spend on the car rental, what is the maximum <\/span>number of miles she can drive?\u00a0<\/span>Round the answer to the nearest mile.<\/span>\r\n\r\n[reveal-answer q=\"293419\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"293419\"]\r\n\r\nThe cost pe<\/span>r day is $18.95<\/span>\r\n\r\nLet x be the miles driven, <\/span>so the mileage charge is 0.19[latex]x[\/latex]<\/span>\r\n\r\nLeslie cannot go over $50, so <\/span>\u2264\u00a0<\/span>$50.\u00a0<\/span>\r\n\r\n \r\n\r\nDaily rate + Mileage\u00a0\u2264 $50\r\n

      [latex]18.95 + 0.19x\u226450[\/latex]<\/p>\r\nSubtract 18.95 on both sides\r\n

      [latex] 0.19x\u226431.05[\/latex]<\/p>\r\nDivide by 0.19 on both sides\r\n

      [latex]\\frac{0.19x}{0.19}\u2264\\frac{31.05}{0.19}[\/latex]<\/p>\r\n

      [latex]x\u2264163.4210526...[\/latex]<\/p>\r\nRound to the nearest mile\r\n

      [latex]x\u2264163[\/latex]<\/p>\r\nIf Leslie drives 163 miles, she will spend $50 on the\u00a0<\/span>car rental.\u00a0<\/span>If Leslie drives less than 163 miles, she will spend less than $50 on the car rental.<\/span>\r\n

      Answer<\/span><\/h4>\r\nIn order to spend at most $50 on the car rental, Leslie must drive no more than 163 miles.\r\n\r\n \r\n

      Special Consideration<\/span><\/h4>\r\nWhat if the solution to our inequality had been\u00a0[latex]x\u2264163.8210526...[\/latex]?\u00a0 Normally, in this case we would round our answer to 164.\u00a0 However, let's look at what happens if we substitute 164 for the variable in our inequality.\r\n

      [latex]18.95 + 0.19(164)\u226450[\/latex]<\/p>\r\n

      [latex]18.95 + 31.16\u226450[\/latex]<\/p>\r\n

      [latex]50.11\u226450[\/latex]<\/p>\r\n

      This is a false statement.<\/p>\r\nSubstituting\u00a0[latex]x=164[\/latex] into the inequality results in a false statement.\u00a0 Therefore, 164 is not a solution to the inequality.\u00a0 Because Leslie has a maximum of $50 to spend, we cannot round our solution up or the cost of the car rental will exceed $50.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n

      \r\n

      Example 3<\/h3>\r\nLeslie found a second rental company that charges 75 cents per mile with no additional fee.\u00a0 Recall the first rental company (from Example 2) charged $18.95 plus 19 cents for each mile driven.\u00a0 How many miles can she drive so that the second rental company is a better deal?\u00a0 Round down<\/em> to the nearest whole mile.\r\n\r\n[reveal-answer q=\"213943\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"213943\"]\r\n\r\nThe cost for the first rental company is\r\n

      [latex]18.95+0.19x[\/latex]<\/p>\r\nwhile the cost for the second rental company is\r\n

      [latex].75x[\/latex].<\/p>\r\nTo determine the number of miles Leslie can drive so that the second company is cheaper, we solve the following inequality:\r\n

      [latex]0.75x< 18.95+0.19x[\/latex]<\/p>\r\n

      [latex]\\hspace{-.14in}\\underline{-0.19x \\hspace{.64in}-0.19x}[\/latex]<\/p>\r\n

      [latex]0.56x<18.95[\/latex]<\/p>\r\n

      [latex]\\frac{0.56x}{0.56}<\\frac{18.95}{0.56}[\/latex]<\/p>\r\n

      [latex]x<\\frac{18.95}{0.56}[\/latex]<\/p>\r\nInitially, [latex]\\frac{18.95}{0.56}\\approx 33.8[\/latex].\u00a0 If we round down, we determine that if Leslie drives up to 33 miles, the second rental company will cost less.\r\n

      Answer<\/span><\/h4>\r\nLeslie can drive up to 33 miles.\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\nIn order to do the next example, you need to know how to find the average of a set of numbers.\u00a0 Recall that to find an average, you need to first add all the numbers together.\u00a0 Then you divide that sum by the number of values in your set.\r\n\r\n[embed]https:\/\/youtu.be\/eEY3U3x_THc[\/embed]\r\n
      \r\n

      Example 4<\/h3>\r\nA student has scores of 85, 65, and 82 on his first three tests. He needs an average of at least 80 <\/span>to earn a grade of B in the class. What is the minimum score that the student <\/span>needs on the fourth <\/span>test to ensure a grade of B?<\/span>\r\n\r\n[reveal-answer q=\"452603\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"452603\"]\r\n\r\nRecall that to find an average, add the items and then divide by the number of <\/span>items. Do not forget to include the fourth test score, which we can denote by [latex]x[\/latex].<\/span>\r\n

      [latex]\\frac{85+65+82+x}{4}\u226580[\/latex]<\/p>\r\nMultiply both sides by 4 to clear the fraction\r\n

      [latex]4\\cdot\\frac{85+65+82+x}{4}\u226580\\cdot4[\/latex]<\/p>\r\n

      [latex]85+65+82+x\u2265320[\/latex]<\/p>\r\n

      [latex]232+x\u2265320[\/latex]<\/p>\r\nSubtract 232 from both sides\r\n

      [latex]x\u226588[\/latex]<\/p>\r\nIf the student earns an 88 on <\/span>the fourth test, he will have an average of 80.\u00a0<\/span>If the student earns a score higher than 88, his grade will be higher than 80.<\/span>\r\n

      Answer<\/span><\/h4>\r\nThe student must get a minimum score of 88 on the fourth test to earn an average of at least 80.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n ","rendered":"
      \n

      SECTION 2.2 Learning Objective<\/h3>\n

      2.2:\u00a0 Applications Using Linear Inequalities<\/strong><\/p>\n

        \n
      • Solve applications using linear inequalities<\/li>\n<\/ul>\n<\/div>\n

         <\/p>\n

        Solve applications using linear inequalities<\/h2>\n

        This section will discuss several types of applications. No matter how simple or complex the <\/span>application is, the following steps will help to translate and solve the problem. <\/span><\/p>\n

          \n
        1. Read\u00a0<\/span>through the entire problem\u00a0<\/span><\/strong><\/li>\n
        2. Organize\u00a0<\/span>the information\u00a0<\/span><\/strong><\/li>\n
        3. Write the\u00a0<\/span>inequali<\/span>ty\u00a0<\/span><\/strong><\/li>\n
        4. Solve\u00a0<\/span>the inequality<\/span><\/strong><\/li>\n
        5. Check\u00a0<\/span>the answer<\/span><\/strong><\/li>\n<\/ol>\n

          Look for key <\/span>words and <\/span>phrases. The following table gives some common translations\u00a0<\/span>for <\/span>inequalities<\/span>.<\/span><\/p>\n\n\n\n\n\n\n\n\n\n\n\n
          Words or Phrase<\/strong><\/td>\nSymbol<\/strong><\/td>\n<\/tr>\n
          Is less than<\/td>\n<<\/td>\n<\/tr>\n
          Is more than<\/td>\n><\/td>\n<\/tr>\n
          At least<\/td>\n\u2265<\/td>\n<\/tr>\n
          At most<\/td>\n\u2264<\/td>\n<\/tr>\n
          Minimum<\/td>\n\u2265<\/td>\n<\/tr>\n
          Maximum<\/td>\n\u2264<\/td>\n<\/tr>\n
          No more than<\/td>\n\u2264<\/td>\n<\/tr>\n
          No less than<\/td>\n\u2265<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
          Below is a video example of an application problem involving linear inequalities.<\/div>\n
          \n