{"id":375,"date":"2015-10-26T17:34:34","date_gmt":"2015-10-26T17:34:34","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=375"},"modified":"2015-11-12T18:38:00","modified_gmt":"2015-11-12T18:38:00","slug":"setting-up-a-linear-equation-to-solve-a-real-world-application","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/setting-up-a-linear-equation-to-solve-a-real-world-application\/","title":{"raw":"Setting up a Linear Equation to Solve a Real-World Application","rendered":"Setting up a Linear Equation to Solve a Real-World Application"},"content":{"raw":"To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10\/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write [latex]0.10x[\/latex]. This expression represents a variable cost because it changes according to the number of miles driven.\r\n\r\nIf a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10\/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost [latex]C[\/latex].\r\n
[latex]C=0.10x+50[\/latex]<\/div>\r\nWhen dealing with real-world applications, there are certain expressions that we can translate directly into math. The table\u00a0lists some common verbal expressions and their equivalent mathematical expressions.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Verbal<\/th>\r\nTranslation to Math Operations<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
One number exceeds another by a<\/em><\/td>\r\n[latex]x,\\text{ }x+a[\/latex]<\/td>\r\n<\/tr>\r\n
Twice a number<\/td>\r\n[latex]2x[\/latex]<\/td>\r\n<\/tr>\r\n
One number is a <\/em>more than another number<\/td>\r\n[latex]x,\\text{ }x+a[\/latex]<\/td>\r\n<\/tr>\r\n
One number is a <\/em>less than twice another number<\/td>\r\n[latex]x,2x-a[\/latex]<\/td>\r\n<\/tr>\r\n
The product of a number and a<\/em>, decreased by b<\/em><\/td>\r\n[latex]ax-b[\/latex]<\/td>\r\n<\/tr>\r\n
The quotient of a number and the number plus a <\/em>is three times the number<\/td>\r\n[latex]\\frac{x}{x+a}=3x[\/latex]<\/td>\r\n<\/tr>\r\n
The product of three times a number and the number decreased by b <\/em>is c<\/em><\/td>\r\n[latex]3x\\left(x-b\\right)=c[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
\r\n

How To: Given a real-world problem, model a linear equation to fit it.<\/h3>\r\n
    \r\n\t
  1. Identify known quantities.<\/li>\r\n\t
  2. Assign a variable to represent the unknown quantity.<\/li>\r\n\t
  3. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.<\/li>\r\n\t
  4. Write an equation interpreting the words as mathematical operations.<\/li>\r\n\t
  5. Solve the equation. Be sure the solution can be explained in words, including the units of measure.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
    \r\n

    Example 1: Modeling a Linear Equation to Solve an Unknown Number Problem<\/h3>\r\nFind a linear equation to solve for the following unknown quantities: One number exceeds another number by [latex]17[\/latex] and their sum is [latex]31[\/latex]. Find the two numbers.\r\n\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nLet [latex]x[\/latex] equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as [latex]x+17[\/latex]. The sum of the two numbers is 31. We usually interpret the word is<\/em> as an equal sign.\r\n
    [latex]\\begin{array}{l}x+\\left(x+17\\right)\\hfill&=31\\hfill \\\\ 2x+17\\hfill&=31\\hfill&\\text{Simplify and solve}.\\hfill \\\\ 2x\\hfill&=14\\hfill \\\\ x\\hfill&=7\\hfill \\\\ \\hfill \\\\ x+17\\hfill&=7+17\\hfill \\\\ \\hfill&=24\\hfill \\end{array}[\/latex]<\/div>\r\nThe two numbers are [latex]7[\/latex] and [latex]24[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Try It 1<\/h3>\r\nFind a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is [latex]36[\/latex], find the numbers.\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n
    \r\n

    Example 2: Setting Up a Linear Equation to Solve a Real-World Application<\/h3>\r\nThere are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05\/min talk-time. Company B charges a monthly service fee of $40 plus $.04\/min talk-time.\r\n
      \r\n\t
    1. Write a linear equation that models the packages offered by both companies.<\/li>\r\n\t
    2. If the average number of minutes used each month is 1,160, which company offers the better plan?<\/li>\r\n\t
    3. If the average number of minutes used each month is 420, which company offers the better plan?<\/li>\r\n\t
    4. How many minutes of talk-time would yield equal monthly statements from both companies?<\/li>\r\n<\/ol>\r\n<\/div>\r\n
      \r\n

      Solution<\/h3>\r\n
        \r\n\t
      1. The model for Company A<\/em> can be written as [latex]A=0.05x+34[\/latex]. This includes the variable cost of [latex]0.05x[\/latex] plus the monthly service charge of $34. Company B<\/em>\u2019s package charges a higher monthly fee of $40, but a lower variable cost of [latex]0.04x[\/latex]. Company B<\/em>\u2019s model can be written as [latex]B=0.04x+\\$40[\/latex].<\/li>\r\n\t
      2. If the average number of minutes used each month is 1,160, we have the following:\r\n
        [latex]\\begin{array}{l}\\text{Company }A\\hfill&=0.05\\left(1,160\\right)+34\\hfill \\\\ \\hfill&=58+34\\hfill \\\\ \\hfill&=92\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&=0.04\\left(1,160\\right)+40\\hfill \\\\ \\hfill&=46.4+40\\hfill \\\\ \\hfill&=86.4\\hfill \\end{array}[\/latex]<\/div>\r\nSo, Company B<\/em> offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company A<\/em> when the average number of minutes used each month is 1,160.<\/li>\r\n\t
      3. If the average number of minutes used each month is 420, we have the following:\r\n
        [latex]\\begin{array}{l}\\text{Company }A\\hfill&=0.05\\left(420\\right)+34\\hfill \\\\ \\hfill&=21+34\\hfill \\\\ \\hfill&=55\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&=0.04\\left(420\\right)+40\\hfill \\\\ \\hfill&=16.8+40\\hfill \\\\ \\hfill&=56.8\\hfill \\end{array}[\/latex]<\/div>\r\nIf the average number of minutes used each month is 420, then Company A <\/em>offers a lower monthly cost of $55 compared to Company B<\/em>\u2019s monthly cost of $56.80.<\/li>\r\n\t
      4. To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of [latex]\\left(x,y\\right)[\/latex] coordinates: At what point are both the x-<\/em>value and the y-<\/em>value equal? We can find this point by setting the equations equal to each other and solving for x.<\/em>\r\n
        [latex]\\begin{array}{l}0.05x+34=0.04x+40\\hfill \\\\ 0.01x=6\\hfill \\\\ x=600\\hfill \\end{array}[\/latex]<\/div>\r\nCheck the x-<\/em>value in each equation.\r\n
        [latex]\\begin{array}{l}0.05\\left(600\\right)+34=64\\hfill \\\\ 0.04\\left(600\\right)+40=64\\hfill \\end{array}[\/latex]<\/div>\r\nTherefore, a monthly average of 600 talk-time minutes renders the plans equal.<\/li>\r\n<\/ol>\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"Coordinate Figure 2<\/b>[\/caption]\r\n\r\n<\/div>\r\n
        \r\n

        Try It 2<\/h3>\r\nFind a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company\u2019s monthly expenses?\r\n\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

        To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10\/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write [latex]0.10x[\/latex]. This expression represents a variable cost because it changes according to the number of miles driven.<\/p>\n

        If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10\/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost [latex]C[\/latex].<\/p>\n

        [latex]C=0.10x+50[\/latex]<\/div>\n

        When dealing with real-world applications, there are certain expressions that we can translate directly into math. The table\u00a0lists some common verbal expressions and their equivalent mathematical expressions.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
        Verbal<\/th>\nTranslation to Math Operations<\/th>\n<\/tr>\n<\/thead>\n
        One number exceeds another by a<\/em><\/td>\n[latex]x,\\text{ }x+a[\/latex]<\/td>\n<\/tr>\n
        Twice a number<\/td>\n[latex]2x[\/latex]<\/td>\n<\/tr>\n
        One number is a <\/em>more than another number<\/td>\n[latex]x,\\text{ }x+a[\/latex]<\/td>\n<\/tr>\n
        One number is a <\/em>less than twice another number<\/td>\n[latex]x,2x-a[\/latex]<\/td>\n<\/tr>\n
        The product of a number and a<\/em>, decreased by b<\/em><\/td>\n[latex]ax-b[\/latex]<\/td>\n<\/tr>\n
        The quotient of a number and the number plus a <\/em>is three times the number<\/td>\n[latex]\\frac{x}{x+a}=3x[\/latex]<\/td>\n<\/tr>\n
        The product of three times a number and the number decreased by b <\/em>is c<\/em><\/td>\n[latex]3x\\left(x-b\\right)=c[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
        \n

        How To: Given a real-world problem, model a linear equation to fit it.<\/h3>\n
          \n
        1. Identify known quantities.<\/li>\n
        2. Assign a variable to represent the unknown quantity.<\/li>\n
        3. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.<\/li>\n
        4. Write an equation interpreting the words as mathematical operations.<\/li>\n
        5. Solve the equation. Be sure the solution can be explained in words, including the units of measure.<\/li>\n<\/ol>\n<\/div>\n
          \n

          Example 1: Modeling a Linear Equation to Solve an Unknown Number Problem<\/h3>\n

          Find a linear equation to solve for the following unknown quantities: One number exceeds another number by [latex]17[\/latex] and their sum is [latex]31[\/latex]. Find the two numbers.<\/p>\n<\/div>\n

          \n

          Solution<\/h3>\n

          Let [latex]x[\/latex] equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as [latex]x+17[\/latex]. The sum of the two numbers is 31. We usually interpret the word is<\/em> as an equal sign.<\/p>\n

          [latex]\\begin{array}{l}x+\\left(x+17\\right)\\hfill&=31\\hfill \\\\ 2x+17\\hfill&=31\\hfill&\\text{Simplify and solve}.\\hfill \\\\ 2x\\hfill&=14\\hfill \\\\ x\\hfill&=7\\hfill \\\\ \\hfill \\\\ x+17\\hfill&=7+17\\hfill \\\\ \\hfill&=24\\hfill \\end{array}[\/latex]<\/div>\n

          The two numbers are [latex]7[\/latex] and [latex]24[\/latex].<\/p>\n<\/div>\n

          \n

          Try It 1<\/h3>\n

          Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is [latex]36[\/latex], find the numbers.<\/p>\n

          Solution<\/a><\/p>\n<\/div>\n

          \n

          Example 2: Setting Up a Linear Equation to Solve a Real-World Application<\/h3>\n

          There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05\/min talk-time. Company B charges a monthly service fee of $40 plus $.04\/min talk-time.<\/p>\n

            \n
          1. Write a linear equation that models the packages offered by both companies.<\/li>\n
          2. If the average number of minutes used each month is 1,160, which company offers the better plan?<\/li>\n
          3. If the average number of minutes used each month is 420, which company offers the better plan?<\/li>\n
          4. How many minutes of talk-time would yield equal monthly statements from both companies?<\/li>\n<\/ol>\n<\/div>\n
            \n

            Solution<\/h3>\n
              \n
            1. The model for Company A<\/em> can be written as [latex]A=0.05x+34[\/latex]. This includes the variable cost of [latex]0.05x[\/latex] plus the monthly service charge of $34. Company B<\/em>\u2019s package charges a higher monthly fee of $40, but a lower variable cost of [latex]0.04x[\/latex]. Company B<\/em>\u2019s model can be written as [latex]B=0.04x+\\$40[\/latex].<\/li>\n
            2. If the average number of minutes used each month is 1,160, we have the following:\n
              [latex]\\begin{array}{l}\\text{Company }A\\hfill&=0.05\\left(1,160\\right)+34\\hfill \\\\ \\hfill&=58+34\\hfill \\\\ \\hfill&=92\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&=0.04\\left(1,160\\right)+40\\hfill \\\\ \\hfill&=46.4+40\\hfill \\\\ \\hfill&=86.4\\hfill \\end{array}[\/latex]<\/div>\n

              So, Company B<\/em> offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company A<\/em> when the average number of minutes used each month is 1,160.<\/li>\n

            3. If the average number of minutes used each month is 420, we have the following:\n
              [latex]\\begin{array}{l}\\text{Company }A\\hfill&=0.05\\left(420\\right)+34\\hfill \\\\ \\hfill&=21+34\\hfill \\\\ \\hfill&=55\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&=0.04\\left(420\\right)+40\\hfill \\\\ \\hfill&=16.8+40\\hfill \\\\ \\hfill&=56.8\\hfill \\end{array}[\/latex]<\/div>\n

              If the average number of minutes used each month is 420, then Company A <\/em>offers a lower monthly cost of $55 compared to Company B<\/em>\u2019s monthly cost of $56.80.<\/li>\n

            4. To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of [latex]\\left(x,y\\right)[\/latex] coordinates: At what point are both the x-<\/em>value and the y-<\/em>value equal? We can find this point by setting the equations equal to each other and solving for x.<\/em>\n
              [latex]\\begin{array}{l}0.05x+34=0.04x+40\\hfill \\\\ 0.01x=6\\hfill \\\\ x=600\\hfill \\end{array}[\/latex]<\/div>\n

              Check the x-<\/em>value in each equation.<\/p>\n

              [latex]\\begin{array}{l}0.05\\left(600\\right)+34=64\\hfill \\\\ 0.04\\left(600\\right)+40=64\\hfill \\end{array}[\/latex]<\/div>\n

              Therefore, a monthly average of 600 talk-time minutes renders the plans equal.<\/li>\n<\/ol>\n

              \"Coordinate<\/p>\n

              Figure 2<\/b><\/p>\n<\/div>\n<\/div>\n

              \n

              Try It 2<\/h3>\n

              Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company\u2019s monthly expenses?<\/p>\n

              Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

              \n\t\t\t

              Candela Citations<\/h3>\n\t\t\t\t\t
              \n\t\t\t\t\t\t
              \n\t\t\t\t\t\t\t
              CC licensed content, Specific attribution<\/div>