{"id":1377,"date":"2019-07-29T23:25:07","date_gmt":"2019-07-29T23:25:07","guid":{"rendered":"https:\/\/opentextbc.ca\/businesstechnicalmath\/chapter\/solve-applications-with-systems-of-equations\/"},"modified":"2021-08-31T21:21:08","modified_gmt":"2021-08-31T21:21:08","slug":"solve-applications-with-systems-of-equations","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/businesstechnicalmath\/chapter\/solve-applications-with-systems-of-equations\/","title":{"raw":"4.4. Solve Applications with Systems of Equations","rendered":"4.4. Solve Applications with Systems of Equations"},"content":{"raw":"[latexpage]\n
\n

Learning Objectives<\/p>\n\n<\/header>\n

\n\nBy the end of this section it is expected that you will be able to:\n
    \n \t
  • Translate to a system of equations<\/li>\n \t
  • Solve direct translation applications<\/li>\n \t
  • Solve geometry applications<\/li>\n \t
  • Solve uniform motion applications<\/li>\n<\/ul>\n<\/div>\n<\/div>\n

    Previously in this chapter we solved several applications with systems of linear equations. In this section, we\u2019ll look at some specific types of applications that relate two quantities. We\u2019ll translate the words into linear equations, decide which is the most convenient method to use, and then solve them.<\/p>\n

    We will use our Problem Solving Strategy for Systems of Linear Equations.<\/p>\n\n

    \n
    \n
    \n

    Use a problem solving strategy for systems of linear equations.<\/p>\n\n<\/header>\n

    \n
      \n \t
    1. Read<\/strong> the problem. Make sure all the words and ideas are understood.<\/li>\n \t
    2. Identify<\/strong> what we are looking for.<\/li>\n \t
    3. Name<\/strong> what we are looking for. Choose variables to represent those quantities.<\/li>\n \t
    4. Translate<\/strong> into a system of equations.<\/li>\n \t
    5. Solve<\/strong> the system of equations using good algebra techniques.<\/li>\n \t
    6. Check<\/strong> the answer in the problem and make sure it makes sense.<\/li>\n \t
    7. Answer<\/strong> the question with a complete sentence.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      Translate to a System of Equations<\/h1>\n

      Many of the problems we solved in earlier applications related two quantities.<\/p>\n

      Let\u2019s see how we can translate these problems into a system of equations with two variables. We\u2019ll focus on Steps 1 through 4 of our Problem Solving Strategy.<\/p>\n\n

      \n

      EXAMPLE 1<\/p>\n\n<\/header>\n

      \n
      How to Translate to a System of Equations<\/div>\n
      \n
      \n

      Translate to a system of equations:<\/p>\n

      The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.<\/p>\n\n<\/div>\n

      \n
      <\/div>\n
      Solution<\/strong><\/div>\n
      <\/div>\n\"This<\/span>\"The<\/span>\"The<\/span>\"The<\/span>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      TRY IT 1<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations:<\/p>\n

      The sum of two numbers is negative twenty-three. One number is 7 less than the other. Find the numbers.<\/p>\n\n<\/div>\n

      Show answer<\/summary>\n

      \\(\\left\\{\\begin{array}{c}m+n=-23\\hfill \\\\ m=n-7\\hfill \\end{array}\\)<\/p>\n\n<\/details><\/div>\n<\/div>\n<\/div>\n

      We\u2019ll do another example where we stop after we write the system of equations.<\/p>\n\n

      \n

      EXAMPLE 2<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations:<\/p>\n

      A married couple together earns \\$110,000 a year. The wife earns \\$16,000 less than twice what her husband earns. What does the husband earn?<\/p>\n\n<\/div>\n

      \n
      <\/div>\n
      Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n
      We are looking for the amount that the husband and wife each earn.<\/td>\nLet \\(h=\\) the amount the husband earns.\u00a0\n<\/span>\\(w=\\) the amount the wife earns.<\/td>\n<\/tr>\n
      Translate.<\/td>\nA married couple together earns \\$110,000.<\/td>\n<\/tr>\n
      <\/td>\n\\(w+h=110,000\\)<\/td>\n<\/tr>\n
      <\/td>\nThe wife earns \\$16,000 less than twice what husband earns.<\/td>\n<\/tr>\n
      <\/td>\n\\(w=2h-16,000\\)<\/td>\n<\/tr>\n
      The system of equations is:<\/td>\n\\(\\left\\{\\begin{array}{c}w+h=110,000\\hfill \\\\ w=2h-16,000\\hfill \\end{array}\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      TRY IT 2<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations:<\/p>\n

      A couple has a total household income of \\$84,000. The husband earns \\$18,000 less than twice what the wife earns. How much does the wife earn?<\/p>\n\n<\/div>\n

      Show answer<\/summary>\n

      \\(\\left\\{\\begin{array}{c}w+h=84,000\\hfill \\\\ h=2w-18,000\\hfill \\end{array}\\)<\/p>\n\n<\/details><\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Solve Direct Translation Applications<\/h1>\n

      We set up, but did not solve, the systems of equations in examples 1 and 2. Now we\u2019ll translate a situation to a system of equations and then solve it.<\/p>\n\n

      \n

      EXAMPLE 3<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      Devon is 26 years older than his son Cooper. The sum of their ages is 50. Find their ages.<\/p>\n\n<\/div>\n

      \n
      <\/div>\n
      Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
      Step 1. Read<\/strong>\u00a0the problem.<\/td>\n<\/td>\n<\/tr>\n
      Step 2. Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the ages of Devon and Cooper.<\/td>\n<\/tr>\n
      Step 3. Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \\(d=\\) Devon\u2019s age.\n<\/span>\\(\\phantom{\\rule{1.5em}{0ex}}c=\\) Cooper\u2019s age<\/td>\n<\/tr>\n
      Step 4. Translate<\/strong>\u00a0into a system of equations.<\/td>\nDevon is 26 years older than Cooper.<\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\nThe sum of their ages is 50.<\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      The system is:<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Step 5. Solve<\/strong>\u00a0the system of equations.\n<\/span>\n<\/span>Solve by substitution.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Substitute\u00a0c<\/em>\u00a0+ 26 into the second equation.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Solve for\u00a0c<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Substitute\u00a0c<\/em>\u00a0= 12 into the first equation and then solve for\u00a0d<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Step 6. Check<\/strong>\u00a0the answer in the problem.<\/td>\nIs Devon\u2019s age 26 more than Cooper\u2019s?\n<\/span>Yes, 38 is 26 more than 12.\n<\/span>Is the sum of their ages 50?\n<\/span>Yes, 38 plus 12 is 50.<\/td>\n<\/tr>\n
      Step 7. Answer<\/strong>\u00a0the question.<\/td>\nDevon is 38 and Cooper is 12 years old.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      TRY IT 3<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      Ali is 12 years older than his youngest sister, Jameela. The sum of their ages is 40. Find their ages.<\/p>\n\n<\/div>\n

      Show answer<\/summary>\n

      Ali is 28 and Jameela is 16.<\/p>\n\n<\/details><\/div>\n<\/div>\n<\/div>\n

      \n

      EXAMPLE 4<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories does she burn for each minute of circuit training?<\/p>\n\n<\/div>\n

      \n
      <\/div>\n
      Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
      Step 1. Read<\/strong>\u00a0the problem.<\/td>\n<\/td>\n<\/tr>\n
      Step 2. Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the number of\n<\/span>calories burned each minute on the\n<\/span>elliptical trainer and each minute of\n<\/span>circuit training.<\/td>\n<\/tr>\n
      Step 3. Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \\(e=\\) number of calories burned per minute on the elliptical trainer.\n<\/span>\\(\\phantom{\\rule{1.5em}{0ex}}c=\\) number of calories burned per minute while circuit training<\/td>\n<\/tr>\n
      Step 4. Translate<\/strong>\u00a0into a system of equations.<\/td>\n10 minutes on the elliptical and circuit\n<\/span>training for 20 minutes, burned\n<\/span>278 calories<\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n20 minutes on the elliptical and\n<\/span>30 minutes of circuit training burned\n<\/span>473 calories<\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      The system is:<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Step 5. Solve<\/strong>\u00a0the system of equations.<\/td>\n<\/td>\n<\/tr>\n
      Multiply the first equation by \u22122 to get opposite coefficients of\u00a0e<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Simplify and add the equations.\n<\/span>\n<\/span>Solve for\u00a0c<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Substitute\u00a0c<\/em>\u00a0= 8.3 into one of the original equations to solve for\u00a0e<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Step 6. Check<\/strong>\u00a0the answer in the problem.<\/td>\nCheck the math on your own.<\/td>\n<\/tr>\n
      \".\"<\/span><\/td>\n<\/td>\n<\/tr>\n
      Step 7. Answer<\/strong>\u00a0the question.<\/td>\nJenna burns 8.3 calories per minute\n<\/span>circuit training and 11.2 calories per\n<\/span>minute while on the elliptical trainer.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      TRY IT 4<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?<\/p>\n\n<\/div>\n

      Show answer<\/summary>\n

      Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.<\/p>\n\n<\/details><\/div>\n<\/div>\n<\/div>\n

      Solve Geometry Applications<\/h1>\n<\/div>\n
      \n

      We solved geometry applications using properties of triangles and rectangles. Now we\u2019ll add to our list some properties of angles.<\/p>\n

      The measures of two complementary angles<\/span> add to 90 degrees. The measures of two supplementary angles<\/span> add to 180 degrees.<\/p>\n\n

      \n
      \n
      \n

      Complementary and Supplementary Angles<\/p>\n\n<\/header>\n

      \n

      Two angles are complementary<\/strong> if the sum of the measures of their angles is 90 degrees.<\/p>\n

      Two angles are supplementary<\/strong> if the sum of the measures of their angles is 180 degrees.<\/p>\n\n<\/div>\n<\/div>\nIf two angles are complementary, we say that <\/span>one angle is the complement of the other.<\/em>\n\n<\/div>\n<\/div>\n

      If two angles are supplementary, we say that one angle is the supplement of the other.<\/em><\/p>\n\n

      \n

      EXAMPLE 5<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      The difference of two complementary angles is 26 degrees. Find the measures of the angles.<\/p>\n\n<\/div>\n

      \n
      <\/div>\n
      Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n
      Step 1. Read<\/strong>\u00a0the problem.<\/td>\n<\/td>\n<\/tr>\n
      Step 2. Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the measure of each angle.<\/td>\n<\/tr>\n
      Step 3. Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \\(x=\\) the measure of the first angle \\(x=\\).\u00a0\n<\/span>\\(m=\\) the measure of the second angle.<\/td>\n<\/tr>\n
      Step 4. Translate<\/strong>\u00a0into a system of equations.<\/td>\nThe angles are complementary.\u00a0\n<\/span>\\(x+y=90\\)<\/td>\n<\/tr>\n
      <\/td>\nThe difference of the two angles is 26 degrees.\u00a0\n<\/span>\\(x-y=26\\)<\/td>\n<\/tr>\n
      The system is<\/td>\n\\(\\left\\{\\begin{array}{c}x+y=90\\hfill \\\\ x-y=26\\hfill \\end{array}\\)<\/td>\n<\/tr>\n
      Step 5. Solve<\/strong>\u00a0the system of equations by elimination.<\/td>\n\\(\\begin{array}{c}\\text{}{\\left\\{\\begin{array}{l}x+y=90\\hfill \\\\ x-y=26\\hfill \\end{array}}\\hfill \\\\ 2x\\phantom{\\rule{1.5em}{0ex}}=116\\hfill \\end{array}\\)<\/td>\n<\/tr>\n
      Substitute \\(x=58\\) into the first equation.<\/td>\n\\(\\begin{array}{ccc}\\hfill x+y& =\\hfill & 90\\hfill \\\\ \\hfill 58+y& =\\hfill & 90\\hfill \\\\ \\hfill y& =\\hfill & 32\\hfill \\end{array}\\)<\/td>\n<\/tr>\n
      Step 6. Check<\/strong>\u00a0the answer in the problem.\n<\/span>\\(\\begin{array}{c}\\hfill 58+32\\phantom{\\rule{0.2em}{0ex}}=\\phantom{\\rule{0.2em}{0ex}}90\u2713\\hfill \\\\ \\hfill 58-32\\phantom{\\rule{0.2em}{0ex}}=\\phantom{\\rule{0.2em}{0ex}}26\u2713\\hfill \\end{array}\\)<\/td>\n<\/td>\n<\/tr>\n
      Step 7. Answer<\/strong>\u00a0the question.<\/td>\nThe angle measures are 58 degrees and 42 degrees.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      TRY IT 5<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      The difference of two complementary angles is 20 degrees. Find the measures of the angles.<\/p>\n\n<\/div>\n

      Show answer<\/summary>\n

      The angle measures are 55 degrees and 35 degrees.<\/p>\n\n<\/details><\/div>\n<\/div>\n<\/div>\n

      \n

      EXAMPLE 6<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.<\/p>\n\n<\/div>\n

      \n
      <\/div>\n
      Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
      Step 1. Read<\/strong>\u00a0the problem.<\/td>\n<\/td>\n<\/tr>\n
      Step 2. Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the measure of each angle.<\/td>\n<\/tr>\n
      Step 3. Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \\(x=\\) the measure of the first angle.\n<\/span>\\(\\phantom{\\rule{1.5em}{0ex}}y=\\) the measure of the second angle<\/td>\n<\/tr>\n
      Step 4. Translate<\/strong>\u00a0into a system of equations.<\/td>\nThe angles are supplementary.<\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\nThe larger angle is twelve less than five times the smaller angle<\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      The system is:\n<\/span>\n<\/span>\n<\/span>Step 5. Solve<\/strong>\u00a0the system of equations substitution.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Substitute 5x<\/em>\u00a0\u2212 12 for\u00a0y<\/em>\u00a0in the first equation.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Solve for\u00a0x<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Substitute 32 for in the second equation, then solve for\u00a0y<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Step 6. Check<\/strong>\u00a0the answer in the problem.\n<\/span>\n<\/span>\\(\\begin{array}{ccc}\\hfill 32+158& =\\hfill & 180\\phantom{\\rule{0.2em}{0ex}}\u2713\\hfill \\\\ \\hfill 5 \\cdot 32-12& =\\hfill & 147\\phantom{\\rule{0.2em}{0ex}}\u2713\\hfill \\end{array}\\)<\/td>\n<\/td>\n<\/tr>\n
      Step 7. Answer<\/strong>\u00a0the question.<\/td>\nThe angle measures are 148 and 32.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      TRY IT 6<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.<\/p>\n\n<\/div>\n

      Show answer<\/summary>\n

      The angle measures are 42 degrees and 138 degrees.<\/p>\n\n<\/details><\/div>\n<\/div>\n<\/div>\n

      \n

      EXAMPLE 7<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      Randall has 125 feet of fencing to enclose the rectangular part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.<\/p>\n\n<\/div>\n

      \n
      <\/div>\n
      Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
      Step 1. Read<\/strong>\u00a0the problem.<\/td>\n<\/td>\n<\/tr>\n
      Step 2. Identify<\/strong>\u00a0what you are looking for.<\/td>\nWe are looking for the length and width.<\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Step 3. Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \\(L=\\) the length of the fenced yard.\n<\/span>\\(\\phantom{\\rule{1.3em}{0ex}}W=\\) the width of the fenced yard<\/td>\n<\/tr>\n
      Step 4. Translate<\/strong>\u00a0into a system of equations.<\/td>\nOne length and two widths equal 125.<\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\nThe length will be 5 feet more than four times the width.<\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      The system is:\n<\/span>\n<\/span>Step 5. Solve<\/strong>\u00a0the system of equations by substitution.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Substitute\u00a0L<\/em>\u00a0= 4W<\/em>\u00a0+ 5 into the first\n<\/span>equation, then solve for\u00a0W<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Substitute 20 for\u00a0W<\/em>\u00a0in the second\n<\/span>equation, then solve for\u00a0L<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Step 6. Check<\/strong>\u00a0the answer in the problem.\n<\/span>\n<\/span>\\(\\begin{array}{ccc}\\hfill 20+28+20& =\\hfill & 125\\phantom{\\rule{0.2em}{0ex}}\u2713\\hfill \\\\ \\hfill 85& =\\hfill & 4 \\cdot 20+5\\phantom{\\rule{0.2em}{0ex}}\u2713\\hfill \\end{array}\\)<\/td>\n<\/td>\n<\/tr>\n
      Step 7. Answer<\/strong>\u00a0the equation.<\/td>\nThe length is 85 feet and the width is 20 feet.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      TRY IT 7<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      Mario wants to put a rectangular fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.<\/p>\n\n<\/div>\n

      Show answer<\/summary>\n

      The length is 60 feet and the width is 35 feet.<\/p>\n\n<\/details><\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Solve Uniform Motion Applications<\/h1>\n

      We used a table to organize the information in uniform motion problems when we introduced them earlier. We\u2019ll continue using the table here. The basic equation was D<\/em> = rt<\/em> where D<\/em> is the distance travelled, r<\/em> is the rate, and t<\/em> is the time.<\/p>\n

      Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.<\/p>\n\n

      \n

      EXAMPLE 8<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?<\/p>\n\n<\/div>\n

      \n
      <\/div>\n
      Solution<\/strong><\/div>\n
      <\/div>\n

      A diagram is useful in helping us visualize the situation.\n<\/span><\/p>\n\"This<\/span>\n\n \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
      Identify and name<\/strong>\u00a0what we are looking for.\n<\/span>A chart will help us organize the data.\n<\/span>We know the rates of both Joni and Kelly, and so\n<\/span>we enter them in the chart.<\/td>\n<\/td>\n<\/tr>\n
      We are looking for the length of time Kelly,\n<\/span>k<\/em>, and Joni,\u00a0j<\/em>, will each drive.\n<\/span>Since \\(D=r \\cdot t\\) we can fill in the Distance column.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Translate<\/strong>\u00a0into a system of equations.\n<\/span>To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So, \\(65j=78k.\\)\n<\/span>\n<\/span>Also, since Kelly left later, her time will be \\(\\frac{1}{2}\\) hour less than Joni\u2019s time.\n<\/span>\n<\/span>So, \\(k=j-\\frac{1}{2}.\\)<\/td>\n\u00a0<\/strong><\/td>\n<\/tr>\n
      Now we have the system.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Solve<\/strong>\u00a0the system of equations by substitution.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Substitute \\(k=j-\\frac{1}{2}\\) into the second equation, then solve for\u00a0j<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      To find Kelly\u2019s time, substitute\u00a0j<\/em>\u00a0= 3 into the first equation, then solve for\u00a0k<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Check<\/strong>\u00a0the answer in the problem.\n<\/span>\u2003\u2003Joni 3 hours (65 mph) = 195 miles.\n<\/span>\u2003\u2003Kelly \\(2\\frac{1}{2}\\) hours (78 mph) = 195 miles.\n<\/span>\u2003\u2003Yes, they will have traveled the same distance\n<\/span>when they meet.<\/td>\n<\/td>\n<\/tr>\n
      Answer<\/strong>\u00a0the question.<\/td>\nKelly will catch up to Joni in \\(2\\frac{1}{2}\\) hours.\n<\/span>By then, Joni will have traveled 3 hours.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      TRY IT 8<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve: Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?<\/p>\n\n<\/div>\n

      Show answer<\/summary>\n

      It will take Clark 4 hours to catch Mitchell.<\/p>\n\n<\/details><\/div>\n<\/div>\n<\/div>\n

      Many real-world applications of uniform motion arise because of the effects of currents\u2014of water or air\u2014on the actual speed of a vehicle. Cross-country airplane flights generally take longer going west than going east because of the prevailing wind currents.<\/p>\n

      Let\u2019s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.<\/p>\n

      The images below show how a river current affects the speed at which a boat is actually travelling. We\u2019ll call the speed of the boat in still water b<\/em> and the speed of the river current c<\/em>.<\/p>\n

      The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat\u2019s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b<\/em> + c<\/em>.<\/p>\n\n

      \"This<\/span><\/div>\n

      The boat is going upstream, opposite to the river current. The current is going against the boat, so the boat\u2019s actual speed is slower than its speed in still water. The actual speed of the boat is \\(b-c\\).<\/p>\n\n

      \"This<\/span><\/div>\n

      We\u2019ll put some numbers to this situation in the next example.<\/p>\n\n

      \n

      EXAMPLE 9<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.<\/p>\n\n<\/div>\n

      \n
      <\/div>\n
      Solution<\/strong><\/div>\n \n

      Read<\/strong>\u00a0the problem.<\/p>\n

      This is a uniform motion problem and a picture will help us visualize the situation.\n<\/span><\/p>\n\"This<\/span>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
      Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the speed of the ship\n<\/span>in still water and the speed of the current.<\/td>\n<\/tr>\n
      Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \\(s=\\) the rate of the ship in still water.\n<\/span>\\(\\phantom{\\rule{1.5em}{0ex}}c=\\) the rate of the current<\/td>\n<\/tr>\n
      A chart will help us organize the information.\n<\/span>The ship goes downstream and then upstream.\n<\/span>Going downstream, the current helps the\n<\/span>ship; therefore, the ship\u2019s actual rate is\u00a0s<\/em>\u00a0+\u00a0c<\/em>.\n<\/span>Going upstream, the current slows the ship;\n<\/span>therefore, the actual rate is\u00a0s<\/em>\u00a0\u2212\u00a0c<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Downstream it takes 4 hours.\n<\/span>Upstream it takes 5 hours.\n<\/span>Each way the distance is 60 miles.<\/td>\n<\/td>\n<\/tr>\n
      Translate<\/strong>\u00a0into a system of equations.\n<\/span>Since rate times time is distance, we can\n<\/span>write the system of equations.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Solve<\/strong>\u00a0the system of equations.\n<\/span>Distribute to put both equations in standard\n<\/span>form, then solve by elimination.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Multiply the top equation by 5 and the bottom equation by 4.\n<\/span>Add the equations, then solve for\u00a0s<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Substitute\u00a0s<\/em>\u00a0= 13.5 into one of the original equations.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Check<\/strong>\u00a0the answer in the problem.\n<\/span>\n<\/span>\u2003The downstream rate would be\n<\/span>\u2003\u200313.5 + 1.5 = 15 mph.\n<\/span>\u2003In 4 hours the ship would travel\n<\/span>\u2003\u2003\u2003\u200315 \\cdot 4 = 60 miles.\n<\/span>\u2003The upstream rate would be\n<\/span>\u2003\u200313.5 \u2212 1.5 = 12 mph.\n<\/span>\u2003In 5 hours the ship would travel\n<\/span>\u2003\u2003\u2003\u200312 \\cdot 5 = 60 miles.\n<\/span><\/td>\n<\/td>\n<\/tr>\n
      Answer<\/strong>\u00a0the question.<\/td>\nThe rate of the ship is 13.5 mph and\n<\/span>the rate of the current is 1.5 mph.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      TRY IT 9<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve: Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.<\/p>\n\n<\/div>\n

      Show answer<\/summary>\n

      The speed of the canoe is 7 mph and the speed of the current is 1 mph.<\/p>\n\n<\/details><\/div>\n<\/div>\n<\/div>\n

      Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We\u2019ll see this in the next example. A wind current in the same direction as the plane is flying is called a tailwind<\/em>. A wind current blowing against the direction of the plane is called a headwind<\/em>.<\/p>\n\n

      \n

      EXAMPLE 10<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve:<\/p>\n

      A private jet can fly 1095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.<\/p>\n\n<\/div>\n

      \n
      <\/div>\n
      Solution<\/strong><\/div>\n \n

      Read<\/strong>\u00a0the problem.<\/p>\n

      This is a uniform motion problem and a picture will help us visualize.\n<\/span><\/p>\n\"This<\/span>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
      Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the speed of the jet\n<\/span>in still air and the speed of the wind.<\/td>\n<\/tr>\n
      Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \\(j=\\) the speed of the jet in still air.\n<\/span>\\(\\phantom{\\rule{1.4em}{0ex}}w=\\) the speed of the wind<\/td>\n<\/tr>\n
      A chart will help us organize the information.\n<\/span>The jet makes two trips-one in a tailwind\n<\/span>and one in a headwind.\n<\/span>In a tailwind, the wind helps the jet and so\n<\/span>the rate is\u00a0j<\/em>\u00a0+\u00a0w<\/em>.\n<\/span>In a headwind, the wind slows the jet and\n<\/span>so the rate is\u00a0j<\/em>\u00a0\u2212\u00a0w<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Each trip takes 3 hours.\n<\/span>In a tailwind the jet flies 1095 miles.\n<\/span>In a headwind the jet flies 987 miles.<\/td>\n<\/td>\n<\/tr>\n
      Translate<\/strong>\u00a0into a system of equations.\n<\/span>Since rate times time is distance, we get the\n<\/span>system of equations.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Solve<\/strong>\u00a0the system of equations.\n<\/span>Distribute, then solve by elimination.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Add, and solve for\u00a0j<\/em>.\n<\/span>\n<\/span>Substitute\u00a0j<\/em>\u00a0= 347 into one of the original\n<\/span>equations, then solve for\u00a0w<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
      Check<\/strong>\u00a0the answer in the problem.\n<\/span>\n<\/span>\u2003With the tailwind, the actual rate of the\n<\/span>\u2003jet would be\n<\/span>\u2003\u2003\u2003347 + 18 = 365 mph.\n<\/span>\u2003In 3 hours the jet would travel\n<\/span>\u2003\u2003\u2003\u2003365 \\cdot 3 = 1095 miles.\n<\/span>\u2003Going into the headwind, the jet\u2019s actual\n<\/span>\u2003rate would be\n<\/span>\u2003\u2003\u2003347 \u2212 18 = 329 mph.\n<\/span>\u2003In 3 hours the jet would travel\n<\/span>\u2003\u2003\u2003\u2003329 \\cdot 3 = 987 miles.<\/td>\n<\/td>\n<\/tr>\n
      Answer<\/strong>\u00a0the question.<\/td>\nThe rate of the jet is 347 mph and the\n<\/span>rate of the wind is 18 mph.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      TRY IT 10<\/p>\n\n<\/header>\n

      \n
      \n

      Translate to a system of equations and then solve: A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1025 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.<\/p>\n\n<\/div>\n

      Show answer<\/summary>\n

      The speed of the jet is 235 mph and the speed of the wind is 30 mph.<\/p>\n\n<\/details><\/div>\n<\/div>\n<\/div>\n

      \n

      Glossary<\/h1>\n
      <\/dl>\n
      \n \t
      \n
      \n
      \n \t
      complementary angles<\/dt>\n \t
      Two angles are complementary if the sum of the measures of their angles is 90 degrees.<\/dd>\n<\/dl>\n
      \n \t
      supplementary angles<\/dt>\n \t
      Two angles are supplementary if the sum of the measures of their angles is 180 degrees.<\/dd>\n<\/dl>\n<\/div><\/dd>\n<\/dl>\n

      4.4 Exercise Set<\/h1>\n

      In the following exercises, translate to a system of equations and solve the system.<\/p>\n\n

        \n \t
      1. The sum of two numbers is fifteen. One number is three less than the other. Find the numbers.<\/li>\n \t
      2. The sum of two numbers is negative thirty. One number is five times the other. Find the numbers.<\/li>\n \t
      3. Twice a number plus three times a second number is twenty-two. Three times the first number plus four times the second is thirty-one. Find the numbers.<\/li>\n \t
      4. Three times a number plus three times a second number is fifteen. Four times the first plus twice the second number is fourteen. Find the numbers.<\/li>\n \t
      5. A married couple together earn ?75,000. The husband earns ?15,000 more than five times what his wife earns. What does the wife earn?<\/li>\n \t
      6. Daniela invested a total of ?50,000, some in a certificate of deposit (CD) and the remainder in bonds. The amount invested in bonds was ?5000 more than twice the amount she put into the CD. How much did she invest in each account?<\/li>\n \t
      7. In her last two years in college, Marlene received? 42,000 in loans. The first year she received a loan that was ?6,000 less than three times the amount of the second year\u2019s loan. What was the amount of her loan for each year?<\/li>\n<\/ol>\n
        \n
        In the following exercises, translate to a system of equations and solve.<\/span><\/div>\n
          \n \t
        1. Alyssa is twelve years older than her sister, Bethany. The sum of their ages is forty-four. Find their ages.<\/li>\n \t
        2. The age of Noelle\u2019s dad is six less than three times Noelle\u2019s age. The sum of their ages is seventy-four. Find their ages.<\/li>\n \t
        3. Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?<\/li>\n \t
        4. Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?<\/li>\n \t
        5. Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for \\$116. Lisa bought two notebooks and three thumb dives for \\$68. Find the cost of each notebook and each thumb drive.<\/li>\n<\/ol>\n<\/div>\n

          In the following exercises, translate to a system of equations and solve.<\/p>\n\n

            \n \t
          1. The difference of two complementary angles is 30 degrees. Find the measures of the angles.<\/li>\n \t
          2. The difference of two supplementary angles is 70 degrees. Find the measures of the angles.<\/li>\n \t
          3. The difference of two supplementary angles is 8 degrees. Find the measures of the angles.<\/li>\n \t
          4. The difference of two complementary angles is 55 degrees. Find the measures of the angles.<\/li>\n \t
          5. Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.<\/li>\n \t
          6. Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.<\/li>\n \t
          7. Wayne is hanging a string of lights 45 feet long around the three sides of his rectangular patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.<\/li>\n \t
          8. A frame around a rectangular family portrait has a perimeter of 60 inches. The length is fifteen less than twice the width. Find the length and width of the frame.<\/li>\n<\/ol>\n

            In the following exercises, translate to a system of equations and solve.<\/p>\n\n

              \n \t
            1. Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah\u2019s sister to catch up to Sarah?<\/li>\n \t
            2. At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy\u2019s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy\u2019s friend to catch up to Lucy?<\/li>\n \t
            3. The Jones family took a 12 mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.<\/li>\n \t
            4. A motor boat traveled 18 miles down a river in two hours but going back upstream, it took 4.5 hours due to the current. Find the rate of the motor boat in still water and the rate of the current. (Round to the nearest hundredth.).<\/li>\n \t
            5. A small jet can fly 1,072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.<\/li>\n \t
            6. A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.\u00a0<\/span><\/li>\n \t
            7. At a school concert, 425 tickets were sold. Student tickets cost \\$5 each and adult tickets cost \\$8 each. The total receipts for the concert were \\$2,851. Solve the system \\(\\left\\{\\begin{array}{c}s+a=425\\hfill \\\\ 5s+8a=2,851\\hfill \\end{array}\\) to find \\(s\\), the number of student tickets and \\(a\\), the number of adult tickets.<\/li>\n<\/ol>\n<\/div>\n
              \n

              Answers:<\/h1>\n
                \n \t
              1. The numbers are 6 and 9.<\/li>\n \t
              2. The numbers are \u22125 and \u221225.<\/li>\n \t
              3. The numbers are 5 and 4.<\/li>\n \t
              4. The numbers are 2 and 3.<\/li>\n \t
              5. \\$10,000<\/li>\n \t
              6. She put \\$15,000 into a CD and \\$35,000 in bonds.<\/li>\n \t
              7. The amount of the first year\u2019s loan was \\$30,000 and the amount of the second year\u2019s loan was \\$12,000.<\/li>\n \t
              8. Bethany is 16 years old and Alyssa is 28 years old.<\/li>\n \t
              9. Noelle is 20 years old and her dad is 54 years old.<\/li>\n \t
              10. The small container holds 20 gallons and the large container holds 30 gallons.<\/li>\n \t
              11. There were 10 calories burned jogging and 10 calories burned cycling.<\/li>\n \t
              12. Notebooks are \\$4 and thumb drives are \\$20.<\/li>\n \t
              13. The measures are 60 degrees and 30 degrees.<\/li>\n \t
              14. The measures are 125 degrees and 55 degrees.<\/li>\n \t
              15. 94 degrees and 86 degrees<\/li>\n \t
              16. 72.5 degrees and 17.5 degrees<\/li>\n \t
              17. The measures are 44 degrees and 136 degrees.<\/li>\n \t
              18. The measures are 34 degrees and 56 degrees.<\/li>\n \t
              19. The width is 10 feet and the length is 25 feet.<\/li>\n \t
              20. The width is 15 feet and the length is 15 feet.<\/li>\n \t
              21. It took Sarah\u2019s sister 12 hours.<\/li>\n \t
              22. It took Lucy\u2019s friend 2 hours.<\/li>\n \t
              23. The canoe rate is 5 mph and the current rate is 1 mph.<\/li>\n \t
              24. The boat rate is 6.5 mph and the current rate is 2.5 mph.<\/li>\n \t
              25. The jet rate is 240 mph and the wind speed is 28 mph.<\/li>\n \t
              26. The jet rate is 415 mph and the wind speed is 19 mph.<\/li>\n \t
              27. \\(s=183,a=242\\)<\/li>\n<\/ol>\n<\/div>\n<\/div>","rendered":"
                \n
                \n

                Learning Objectives<\/p>\n<\/header>\n

                \n

                By the end of this section it is expected that you will be able to:<\/p>\n

                  \n
                • Translate to a system of equations<\/li>\n
                • Solve direct translation applications<\/li>\n
                • Solve geometry applications<\/li>\n
                • Solve uniform motion applications<\/li>\n<\/ul>\n<\/div>\n<\/div>\n

                  Previously in this chapter we solved several applications with systems of linear equations. In this section, we\u2019ll look at some specific types of applications that relate two quantities. We\u2019ll translate the words into linear equations, decide which is the most convenient method to use, and then solve them.<\/p>\n

                  We will use our Problem Solving Strategy for Systems of Linear Equations.<\/p>\n

                  \n
                  \n
                  \n
                  \n

                  Use a problem solving strategy for systems of linear equations.<\/p>\n<\/header>\n

                  \n
                    \n
                  1. Read<\/strong> the problem. Make sure all the words and ideas are understood.<\/li>\n
                  2. Identify<\/strong> what we are looking for.<\/li>\n
                  3. Name<\/strong> what we are looking for. Choose variables to represent those quantities.<\/li>\n
                  4. Translate<\/strong> into a system of equations.<\/li>\n
                  5. Solve<\/strong> the system of equations using good algebra techniques.<\/li>\n
                  6. Check<\/strong> the answer in the problem and make sure it makes sense.<\/li>\n
                  7. Answer<\/strong> the question with a complete sentence.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
                    \n

                    Translate to a System of Equations<\/h1>\n

                    Many of the problems we solved in earlier applications related two quantities.<\/p>\n

                    Let\u2019s see how we can translate these problems into a system of equations with two variables. We\u2019ll focus on Steps 1 through 4 of our Problem Solving Strategy.<\/p>\n

                    \n
                    \n

                    EXAMPLE 1<\/p>\n<\/header>\n

                    \n
                    How to Translate to a System of Equations<\/div>\n
                    \n
                    \n

                    Translate to a system of equations:<\/p>\n

                    The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.<\/p>\n<\/div>\n

                    \n
                    <\/div>\n
                    Solution<\/strong><\/div>\n
                    <\/div>\n

                    \"This<\/span>\"The<\/span>\"The<\/span>\"The<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                    \n
                    \n

                    TRY IT 1<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations:<\/p>\n

                    The sum of two numbers is negative twenty-three. One number is 7 less than the other. Find the numbers.<\/p>\n<\/div>\n

                    \n
                    \nShow answer<\/summary>\n

                    \"\left\{\begin{array}{c}m+n=-23\hfill \\ m=n-7\hfill \end{array}\"<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n

                    We\u2019ll do another example where we stop after we write the system of equations.<\/p>\n

                    \n
                    \n

                    EXAMPLE 2<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations:<\/p>\n

                    A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?<\/p>\n<\/div>\n

                    \n
                    <\/div>\n
                    Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n
                    We are looking for the amount that the husband and wife each earn.<\/td>\nLet \"h=\" the amount the husband earns.\u00a0
                    \n<\/span>\"w=\" the amount the wife earns.<\/td>\n<\/tr>\n
                    Translate.<\/td>\nA married couple together earns $110,000.<\/td>\n<\/tr>\n
                    <\/td>\n\"w+h=110,000\"<\/td>\n<\/tr>\n
                    <\/td>\nThe wife earns $16,000 less than twice what husband earns.<\/td>\n<\/tr>\n
                    <\/td>\n\"w=2h-16,000\"<\/td>\n<\/tr>\n
                    The system of equations is:<\/td>\n\"\left\{\begin{array}{c}w+h=110,000\hfill \\ w=2h-16,000\hfill \end{array}\"<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
                    \n
                    \n

                    TRY IT 2<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations:<\/p>\n

                    A couple has a total household income of $84,000. The husband earns $18,000 less than twice what the wife earns. How much does the wife earn?<\/p>\n<\/div>\n

                    \n
                    \nShow answer<\/summary>\n

                    \"\left\{\begin{array}{c}w+h=84,000\hfill \\ h=2w-18,000\hfill \end{array}\"<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                    \n

                    Solve Direct Translation Applications<\/h1>\n

                    We set up, but did not solve, the systems of equations in examples 1 and 2. Now we\u2019ll translate a situation to a system of equations and then solve it.<\/p>\n

                    \n
                    \n

                    EXAMPLE 3<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    Devon is 26 years older than his son Cooper. The sum of their ages is 50. Find their ages.<\/p>\n<\/div>\n

                    \n
                    <\/div>\n
                    Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
                    Step 1. Read<\/strong>\u00a0the problem.<\/td>\n<\/td>\n<\/tr>\n
                    Step 2. Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the ages of Devon and Cooper.<\/td>\n<\/tr>\n
                    Step 3. Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \"d=\" Devon\u2019s age.
                    \n<\/span>\"\phantom{\rule{1.5em}{0ex}}c=\" Cooper\u2019s age<\/td>\n<\/tr>\n
                    Step 4. Translate<\/strong>\u00a0into a system of equations.<\/td>\nDevon is 26 years older than Cooper.<\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\nThe sum of their ages is 50.<\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    The system is:<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Step 5. Solve<\/strong>\u00a0the system of equations.
                    \n<\/span>
                    \n<\/span>Solve by substitution.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Substitute\u00a0c<\/em>\u00a0+ 26 into the second equation.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Solve for\u00a0c<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Substitute\u00a0c<\/em>\u00a0= 12 into the first equation and then solve for\u00a0d<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Step 6. Check<\/strong>\u00a0the answer in the problem.<\/td>\nIs Devon\u2019s age 26 more than Cooper\u2019s?
                    \n<\/span>Yes, 38 is 26 more than 12.
                    \n<\/span>Is the sum of their ages 50?
                    \n<\/span>Yes, 38 plus 12 is 50.<\/td>\n<\/tr>\n
                    Step 7. Answer<\/strong>\u00a0the question.<\/td>\nDevon is 38 and Cooper is 12 years old.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
                    \n
                    \n

                    TRY IT 3<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    Ali is 12 years older than his youngest sister, Jameela. The sum of their ages is 40. Find their ages.<\/p>\n<\/div>\n

                    \n
                    \nShow answer<\/summary>\n

                    Ali is 28 and Jameela is 16.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n

                    \n
                    \n

                    EXAMPLE 4<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories does she burn for each minute of circuit training?<\/p>\n<\/div>\n

                    \n
                    <\/div>\n
                    Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
                    Step 1. Read<\/strong>\u00a0the problem.<\/td>\n<\/td>\n<\/tr>\n
                    Step 2. Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the number of
                    \n<\/span>calories burned each minute on the
                    \n<\/span>elliptical trainer and each minute of
                    \n<\/span>circuit training.<\/td>\n<\/tr>\n
                    Step 3. Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \"e=\" number of calories burned per minute on the elliptical trainer.
                    \n<\/span>\"\phantom{\rule{1.5em}{0ex}}c=\" number of calories burned per minute while circuit training<\/td>\n<\/tr>\n
                    Step 4. Translate<\/strong>\u00a0into a system of equations.<\/td>\n10 minutes on the elliptical and circuit
                    \n<\/span>training for 20 minutes, burned
                    \n<\/span>278 calories<\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n20 minutes on the elliptical and
                    \n<\/span>30 minutes of circuit training burned
                    \n<\/span>473 calories<\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    The system is:<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Step 5. Solve<\/strong>\u00a0the system of equations.<\/td>\n<\/td>\n<\/tr>\n
                    Multiply the first equation by \u22122 to get opposite coefficients of\u00a0e<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Simplify and add the equations.
                    \n<\/span>
                    \n<\/span>Solve for\u00a0c<\/em>.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Substitute\u00a0c<\/em>\u00a0= 8.3 into one of the original equations to solve for\u00a0e<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Step 6. Check<\/strong>\u00a0the answer in the problem.<\/td>\nCheck the math on your own.<\/td>\n<\/tr>\n
                    \".\"<\/span><\/td>\n<\/td>\n<\/tr>\n
                    Step 7. Answer<\/strong>\u00a0the question.<\/td>\nJenna burns 8.3 calories per minute
                    \n<\/span>circuit training and 11.2 calories per
                    \n<\/span>minute while on the elliptical trainer.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
                    \n
                    \n

                    TRY IT 4<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?<\/p>\n<\/div>\n

                    \n
                    \nShow answer<\/summary>\n

                    Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n

                    Solve Geometry Applications<\/h1>\n<\/div>\n
                    \n

                    We solved geometry applications using properties of triangles and rectangles. Now we\u2019ll add to our list some properties of angles.<\/p>\n

                    The measures of two complementary angles<\/span> add to 90 degrees. The measures of two supplementary angles<\/span> add to 180 degrees.<\/p>\n

                    \n
                    \n
                    \n
                    \n

                    Complementary and Supplementary Angles<\/p>\n<\/header>\n

                    \n

                    Two angles are complementary<\/strong> if the sum of the measures of their angles is 90 degrees.<\/p>\n

                    Two angles are supplementary<\/strong> if the sum of the measures of their angles is 180 degrees.<\/p>\n<\/div>\n<\/div>\n

                    If two angles are complementary, we say that <\/span>one angle is the complement of the other.<\/em><\/p>\n<\/div>\n<\/div>\n

                    If two angles are supplementary, we say that one angle is the supplement of the other.<\/em><\/p>\n

                    \n
                    \n

                    EXAMPLE 5<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    The difference of two complementary angles is 26 degrees. Find the measures of the angles.<\/p>\n<\/div>\n

                    \n
                    <\/div>\n
                    Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n
                    Step 1. Read<\/strong>\u00a0the problem.<\/td>\n<\/td>\n<\/tr>\n
                    Step 2. Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the measure of each angle.<\/td>\n<\/tr>\n
                    Step 3. Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \"x=\" the measure of the first angle \"x=\".\u00a0
                    \n<\/span>\"m=\" the measure of the second angle.<\/td>\n<\/tr>\n
                    Step 4. Translate<\/strong>\u00a0into a system of equations.<\/td>\nThe angles are complementary.\u00a0
                    \n<\/span>\"x+y=90\"<\/td>\n<\/tr>\n
                    <\/td>\nThe difference of the two angles is 26 degrees.\u00a0
                    \n<\/span>\"x-y=26\"<\/td>\n<\/tr>\n
                    The system is<\/td>\n\"\left\{\begin{array}{c}x+y=90\hfill \\ x-y=26\hfill \end{array}\"<\/td>\n<\/tr>\n
                    Step 5. Solve<\/strong>\u00a0the system of equations by elimination.<\/td>\n\"\begin{array}{c}\text{}{\left\{\begin{array}{l}x+y=90\hfill \\ x-y=26\hfill \end{array}}\hfill \\ 2x\phantom{\rule{1.5em}{0ex}}=116\hfill \end{array}\"<\/td>\n<\/tr>\n
                    Substitute \"x=58\" into the first equation.<\/td>\n\"\begin{array}{ccc}\hfill x+y& =\hfill & 90\hfill \\ \hfill 58+y& =\hfill & 90\hfill \\ \hfill y& =\hfill & 32\hfill \end{array}\"<\/td>\n<\/tr>\n
                    Step 6. Check<\/strong>\u00a0the answer in the problem.
                    \n<\/span>\"\begin{array}{c}\hfill 58+32\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}90✓\hfill \\ \hfill 58-32\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}26✓\hfill \end{array}\"<\/td>\n                    		<\/td>\n<\/tr>\n
                    Step 7. Answer<\/strong>\u00a0the question.<\/td>\nThe angle measures are 58 degrees and 42 degrees.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
                    \n
                    \n

                    TRY IT 5<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    The difference of two complementary angles is 20 degrees. Find the measures of the angles.<\/p>\n<\/div>\n

                    \n
                    \nShow answer<\/summary>\n

                    The angle measures are 55 degrees and 35 degrees.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n

                    \n
                    \n

                    EXAMPLE 6<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.<\/p>\n<\/div>\n

                    \n
                    <\/div>\n
                    Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
                    Step 1. Read<\/strong>\u00a0the problem.<\/td>\n<\/td>\n<\/tr>\n
                    Step 2. Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the measure of each angle.<\/td>\n<\/tr>\n
                    Step 3. Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \"x=\" the measure of the first angle.
                    \n<\/span>\"\phantom{\rule{1.5em}{0ex}}y=\" the measure of the second angle<\/td>\n<\/tr>\n
                    Step 4. Translate<\/strong>\u00a0into a system of equations.<\/td>\nThe angles are supplementary.<\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\nThe larger angle is twelve less than five times the smaller angle<\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    The system is:
                    \n<\/span>
                    \n<\/span>
                    \n<\/span>Step 5. Solve<\/strong>\u00a0the system of equations substitution.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Substitute 5x<\/em>\u00a0\u2212 12 for\u00a0y<\/em>\u00a0in the first equation.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Solve for\u00a0x<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Substitute 32 for in the second equation, then solve for\u00a0y<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Step 6. Check<\/strong>\u00a0the answer in the problem.
                    \n<\/span>
                    \n<\/span>\"\begin{array}{ccc}\hfill 32+158& =\hfill & 180\phantom{\rule{0.2em}{0ex}}✓\hfill \\ \hfill 5 \cdot 32-12& =\hfill & 147\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\"<\/td>\n                    		<\/td>\n<\/tr>\n
                    Step 7. Answer<\/strong>\u00a0the question.<\/td>\nThe angle measures are 148 and 32.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
                    \n
                    \n

                    TRY IT 6<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.<\/p>\n<\/div>\n

                    \n
                    \nShow answer<\/summary>\n

                    The angle measures are 42 degrees and 138 degrees.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n

                    \n
                    \n

                    EXAMPLE 7<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    Randall has 125 feet of fencing to enclose the rectangular part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.<\/p>\n<\/div>\n

                    \n
                    <\/div>\n
                    Solution<\/strong><\/div>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
                    Step 1. Read<\/strong>\u00a0the problem.<\/td>\n<\/td>\n<\/tr>\n
                    Step 2. Identify<\/strong>\u00a0what you are looking for.<\/td>\nWe are looking for the length and width.<\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Step 3. Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \"L=\" the length of the fenced yard.
                    \n<\/span>\"\phantom{\rule{1.3em}{0ex}}W=\" the width of the fenced yard<\/td>\n<\/tr>\n
                    Step 4. Translate<\/strong>\u00a0into a system of equations.<\/td>\nOne length and two widths equal 125.<\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\nThe length will be 5 feet more than four times the width.<\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    The system is:
                    \n<\/span>
                    \n<\/span>Step 5. Solve<\/strong>\u00a0the system of equations by substitution.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Substitute\u00a0L<\/em>\u00a0= 4W<\/em>\u00a0+ 5 into the first
                    \n<\/span>equation, then solve for\u00a0W<\/em>.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Substitute 20 for\u00a0W<\/em>\u00a0in the second
                    \n<\/span>equation, then solve for\u00a0L<\/em>.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Step 6. Check<\/strong>\u00a0the answer in the problem.
                    \n<\/span>
                    \n<\/span>\"\begin{array}{ccc}\hfill 20+28+20& =\hfill & 125\phantom{\rule{0.2em}{0ex}}✓\hfill \\ \hfill 85& =\hfill & 4 \cdot 20+5\phantom{\rule{0.2em}{0ex}}✓\hfill \end{array}\"<\/td>\n                    		<\/td>\n<\/tr>\n
                    Step 7. Answer<\/strong>\u00a0the equation.<\/td>\nThe length is 85 feet and the width is 20 feet.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
                    \n
                    \n

                    TRY IT 7<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    Mario wants to put a rectangular fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.<\/p>\n<\/div>\n

                    \n
                    \nShow answer<\/summary>\n

                    The length is 60 feet and the width is 35 feet.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                    \n

                    Solve Uniform Motion Applications<\/h1>\n

                    We used a table to organize the information in uniform motion problems when we introduced them earlier. We\u2019ll continue using the table here. The basic equation was D<\/em> = rt<\/em> where D<\/em> is the distance travelled, r<\/em> is the rate, and t<\/em> is the time.<\/p>\n

                    Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.<\/p>\n

                    \n
                    \n

                    EXAMPLE 8<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?<\/p>\n<\/div>\n

                    \n
                    <\/div>\n
                    Solution<\/strong><\/div>\n
                    <\/div>\n

                    A diagram is useful in helping us visualize the situation.
                    \n<\/span><\/p>\n

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

                     <\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
                    Identify and name<\/strong>\u00a0what we are looking for.
                    \n<\/span>A chart will help us organize the data.
                    \n<\/span>We know the rates of both Joni and Kelly, and so
                    \n<\/span>we enter them in the chart.<\/td>\n                    		<\/td>\n<\/tr>\n
                    We are looking for the length of time Kelly,
                    \n<\/span>k<\/em>, and Joni,\u00a0j<\/em>, will each drive.
                    \n<\/span>Since \"D=r \cdot t\" we can fill in the Distance column.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Translate<\/strong>\u00a0into a system of equations.
                    \n<\/span>To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So, \"65j=78k.\"
                    \n<\/span>
                    \n<\/span>Also, since Kelly left later, her time will be \"\frac{1}{2}\" hour less than Joni\u2019s time.
                    \n<\/span>
                    \n<\/span>So, \"k=j-\frac{1}{2}.\"<\/td>\n                    		\u00a0<\/strong><\/td>\n<\/tr>\n
                    Now we have the system.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Solve<\/strong>\u00a0the system of equations by substitution.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Substitute \"k=j-\frac{1}{2}\" into the second equation, then solve for\u00a0j<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    To find Kelly\u2019s time, substitute\u00a0j<\/em>\u00a0= 3 into the first equation, then solve for\u00a0k<\/em>.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Check<\/strong>\u00a0the answer in the problem.
                    \n<\/span>\u2003\u2003Joni 3 hours (65 mph) = 195 miles.
                    \n<\/span>\u2003\u2003Kelly \"2\frac{1}{2}\" hours (78 mph) = 195 miles.
                    \n<\/span>\u2003\u2003Yes, they will have traveled the same distance
                    \n<\/span>when they meet.<\/td>\n                    		<\/td>\n<\/tr>\n
                    Answer<\/strong>\u00a0the question.<\/td>\nKelly will catch up to Joni in \"2\frac{1}{2}\" hours.
                    \n<\/span>By then, Joni will have traveled 3 hours.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
                    \n
                    \n

                    TRY IT 8<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve: Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?<\/p>\n<\/div>\n

                    \n
                    \nShow answer<\/summary>\n

                    It will take Clark 4 hours to catch Mitchell.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n

                    Many real-world applications of uniform motion arise because of the effects of currents\u2014of water or air\u2014on the actual speed of a vehicle. Cross-country airplane flights generally take longer going west than going east because of the prevailing wind currents.<\/p>\n

                    Let\u2019s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.<\/p>\n

                    The images below show how a river current affects the speed at which a boat is actually travelling. We\u2019ll call the speed of the boat in still water b<\/em> and the speed of the river current c<\/em>.<\/p>\n

                    The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat\u2019s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b<\/em> + c<\/em>.<\/p>\n

                    \"This<\/span><\/div>\n

                    The boat is going upstream, opposite to the river current. The current is going against the boat, so the boat\u2019s actual speed is slower than its speed in still water. The actual speed of the boat is \"b-c\".<\/p>\n

                    \"This<\/span><\/div>\n

                    We\u2019ll put some numbers to this situation in the next example.<\/p>\n

                    \n
                    \n

                    EXAMPLE 9<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.<\/p>\n<\/div>\n

                    \n
                    <\/div>\n
                    Solution<\/strong><\/div>\n

                     <\/p>\n

                    Read<\/strong>\u00a0the problem.<\/p>\n

                    This is a uniform motion problem and a picture will help us visualize the situation.
                    \n<\/span><\/p>\n

                    \"This<\/span><\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
                    Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the speed of the ship
                    \n<\/span>in still water and the speed of the current.<\/td>\n<\/tr>\n
                    Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \"s=\" the rate of the ship in still water.
                    \n<\/span>\"\phantom{\rule{1.5em}{0ex}}c=\" the rate of the current<\/td>\n<\/tr>\n
                    A chart will help us organize the information.
                    \n<\/span>The ship goes downstream and then upstream.
                    \n<\/span>Going downstream, the current helps the
                    \n<\/span>ship; therefore, the ship\u2019s actual rate is\u00a0s<\/em>\u00a0+\u00a0c<\/em>.
                    \n<\/span>Going upstream, the current slows the ship;
                    \n<\/span>therefore, the actual rate is\u00a0s<\/em>\u00a0\u2212\u00a0c<\/em>.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Downstream it takes 4 hours.
                    \n<\/span>Upstream it takes 5 hours.
                    \n<\/span>Each way the distance is 60 miles.<\/td>\n                    		<\/td>\n<\/tr>\n
                    Translate<\/strong>\u00a0into a system of equations.
                    \n<\/span>Since rate times time is distance, we can
                    \n<\/span>write the system of equations.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Solve<\/strong>\u00a0the system of equations.
                    \n<\/span>Distribute to put both equations in standard
                    \n<\/span>form, then solve by elimination.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Multiply the top equation by 5 and the bottom equation by 4.
                    \n<\/span>Add the equations, then solve for\u00a0s<\/em>.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Substitute\u00a0s<\/em>\u00a0= 13.5 into one of the original equations.<\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Check<\/strong>\u00a0the answer in the problem.
                    \n<\/span>
                    \n<\/span>\u2003The downstream rate would be
                    \n<\/span>\u2003\u200313.5 + 1.5 = 15 mph.
                    \n<\/span>\u2003In 4 hours the ship would travel
                    \n<\/span>\u2003\u2003\u2003\u200315 \\cdot 4 = 60 miles.
                    \n<\/span>\u2003The upstream rate would be
                    \n<\/span>\u2003\u200313.5 \u2212 1.5 = 12 mph.
                    \n<\/span>\u2003In 5 hours the ship would travel
                    \n<\/span>\u2003\u2003\u2003\u200312 \\cdot 5 = 60 miles.
                    \n<\/span><\/td>\n                    		<\/td>\n<\/tr>\n
                    Answer<\/strong>\u00a0the question.<\/td>\nThe rate of the ship is 13.5 mph and
                    \n<\/span>the rate of the current is 1.5 mph.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
                    \n
                    \n

                    TRY IT 9<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve: Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.<\/p>\n<\/div>\n

                    \n
                    \nShow answer<\/summary>\n

                    The speed of the canoe is 7 mph and the speed of the current is 1 mph.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n

                    Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We\u2019ll see this in the next example. A wind current in the same direction as the plane is flying is called a tailwind<\/em>. A wind current blowing against the direction of the plane is called a headwind<\/em>.<\/p>\n

                    \n
                    \n

                    EXAMPLE 10<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve:<\/p>\n

                    A private jet can fly 1095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.<\/p>\n<\/div>\n

                    \n
                    <\/div>\n
                    Solution<\/strong><\/div>\n

                     <\/p>\n

                    Read<\/strong>\u00a0the problem.<\/p>\n

                    This is a uniform motion problem and a picture will help us visualize.
                    \n<\/span><\/p>\n

                    \"This<\/span><\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
                    Identify<\/strong>\u00a0what we are looking for.<\/td>\nWe are looking for the speed of the jet
                    \n<\/span>in still air and the speed of the wind.<\/td>\n<\/tr>\n
                    Name<\/strong>\u00a0what we are looking for.<\/td>\nLet \"j=\" the speed of the jet in still air.
                    \n<\/span>\"\phantom{\rule{1.4em}{0ex}}w=\" the speed of the wind<\/td>\n<\/tr>\n
                    A chart will help us organize the information.
                    \n<\/span>The jet makes two trips-one in a tailwind
                    \n<\/span>and one in a headwind.
                    \n<\/span>In a tailwind, the wind helps the jet and so
                    \n<\/span>the rate is\u00a0j<\/em>\u00a0+\u00a0w<\/em>.
                    \n<\/span>In a headwind, the wind slows the jet and
                    \n<\/span>so the rate is\u00a0j<\/em>\u00a0\u2212\u00a0w<\/em>.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Each trip takes 3 hours.
                    \n<\/span>In a tailwind the jet flies 1095 miles.
                    \n<\/span>In a headwind the jet flies 987 miles.<\/td>\n                    		<\/td>\n<\/tr>\n
                    Translate<\/strong>\u00a0into a system of equations.
                    \n<\/span>Since rate times time is distance, we get the
                    \n<\/span>system of equations.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Solve<\/strong>\u00a0the system of equations.
                    \n<\/span>Distribute, then solve by elimination.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    Add, and solve for\u00a0j<\/em>.
                    \n<\/span>
                    \n<\/span>Substitute\u00a0j<\/em>\u00a0= 347 into one of the original
                    \n<\/span>equations, then solve for\u00a0w<\/em>.<\/td>\n                    		\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    <\/td>\n\".\"<\/span><\/td>\n<\/tr>\n
                    Check<\/strong>\u00a0the answer in the problem.
                    \n<\/span>
                    \n<\/span>\u2003With the tailwind, the actual rate of the
                    \n<\/span>\u2003jet would be
                    \n<\/span>\u2003\u2003\u2003347 + 18 = 365 mph.
                    \n<\/span>\u2003In 3 hours the jet would travel
                    \n<\/span>\u2003\u2003\u2003\u2003365 \\cdot 3 = 1095 miles.
                    \n<\/span>\u2003Going into the headwind, the jet\u2019s actual
                    \n<\/span>\u2003rate would be
                    \n<\/span>\u2003\u2003\u2003347 \u2212 18 = 329 mph.
                    \n<\/span>\u2003In 3 hours the jet would travel
                    \n<\/span>\u2003\u2003\u2003\u2003329 \\cdot 3 = 987 miles.<\/td>\n                    		<\/td>\n<\/tr>\n
                    Answer<\/strong>\u00a0the question.<\/td>\nThe rate of the jet is 347 mph and the
                    \n<\/span>rate of the wind is 18 mph.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n
                    \n
                    \n

                    TRY IT 10<\/p>\n<\/header>\n

                    \n
                    \n

                    Translate to a system of equations and then solve: A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1025 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.<\/p>\n<\/div>\n

                    \n
                    \nShow answer<\/summary>\n

                    The speed of the jet is 235 mph and the speed of the wind is 30 mph.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n

                    \n

                    Glossary<\/h1>\n
                    <\/dl>\n
                    \n
                    \n
                    \n
                    \n
                    complementary angles<\/dt>\n
                    Two angles are complementary if the sum of the measures of their angles is 90 degrees.<\/dd>\n<\/dl>\n
                    \n
                    supplementary angles<\/dt>\n
                    Two angles are supplementary if the sum of the measures of their angles is 180 degrees.<\/dd>\n<\/dl>\n<\/div>\n<\/dd>\n<\/dl>\n

                    4.4 Exercise Set<\/h1>\n

                    In the following exercises, translate to a system of equations and solve the system.<\/p>\n

                      \n
                    1. The sum of two numbers is fifteen. One number is three less than the other. Find the numbers.<\/li>\n
                    2. The sum of two numbers is negative thirty. One number is five times the other. Find the numbers.<\/li>\n
                    3. Twice a number plus three times a second number is twenty-two. Three times the first number plus four times the second is thirty-one. Find the numbers.<\/li>\n
                    4. Three times a number plus three times a second number is fifteen. Four times the first plus twice the second number is fourteen. Find the numbers.<\/li>\n
                    5. A married couple together earn ?75,000. The husband earns ?15,000 more than five times what his wife earns. What does the wife earn?<\/li>\n
                    6. Daniela invested a total of ?50,000, some in a certificate of deposit (CD) and the remainder in bonds. The amount invested in bonds was ?5000 more than twice the amount she put into the CD. How much did she invest in each account?<\/li>\n
                    7. In her last two years in college, Marlene received? 42,000 in loans. The first year she received a loan that was ?6,000 less than three times the amount of the second year\u2019s loan. What was the amount of her loan for each year?<\/li>\n<\/ol>\n
                      \n
                      In the following exercises, translate to a system of equations and solve.<\/span><\/div>\n
                        \n
                      1. Alyssa is twelve years older than her sister, Bethany. The sum of their ages is forty-four. Find their ages.<\/li>\n
                      2. The age of Noelle\u2019s dad is six less than three times Noelle\u2019s age. The sum of their ages is seventy-four. Find their ages.<\/li>\n
                      3. Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?<\/li>\n
                      4. Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?<\/li>\n
                      5. Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $116. Lisa bought two notebooks and three thumb dives for $68. Find the cost of each notebook and each thumb drive.<\/li>\n<\/ol>\n<\/div>\n

                        In the following exercises, translate to a system of equations and solve.<\/p>\n

                          \n
                        1. The difference of two complementary angles is 30 degrees. Find the measures of the angles.<\/li>\n
                        2. The difference of two supplementary angles is 70 degrees. Find the measures of the angles.<\/li>\n
                        3. The difference of two supplementary angles is 8 degrees. Find the measures of the angles.<\/li>\n
                        4. The difference of two complementary angles is 55 degrees. Find the measures of the angles.<\/li>\n
                        5. Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.<\/li>\n
                        6. Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.<\/li>\n
                        7. Wayne is hanging a string of lights 45 feet long around the three sides of his rectangular patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.<\/li>\n
                        8. A frame around a rectangular family portrait has a perimeter of 60 inches. The length is fifteen less than twice the width. Find the length and width of the frame.<\/li>\n<\/ol>\n

                          In the following exercises, translate to a system of equations and solve.<\/p>\n

                            \n
                          1. Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah\u2019s sister to catch up to Sarah?<\/li>\n
                          2. At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy\u2019s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy\u2019s friend to catch up to Lucy?<\/li>\n
                          3. The Jones family took a 12 mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.<\/li>\n
                          4. A motor boat traveled 18 miles down a river in two hours but going back upstream, it took 4.5 hours due to the current. Find the rate of the motor boat in still water and the rate of the current. (Round to the nearest hundredth.).<\/li>\n
                          5. A small jet can fly 1,072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.<\/li>\n
                          6. A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.\u00a0<\/span><\/li>\n
                          7. At a school concert, 425 tickets were sold. Student tickets cost $5 each and adult tickets cost $8 each. The total receipts for the concert were $2,851. Solve the system \"\left\{\begin{array}{c}s+a=425\hfill \\ 5s+8a=2,851\hfill \end{array}\" to find \"s\", the number of student tickets and \"a\", the number of adult tickets.<\/li>\n<\/ol>\n<\/div>\n
                            \n

                            Answers:<\/h1>\n
                              \n
                            1. The numbers are 6 and 9.<\/li>\n
                            2. The numbers are \u22125 and \u221225.<\/li>\n
                            3. The numbers are 5 and 4.<\/li>\n
                            4. The numbers are 2 and 3.<\/li>\n
                            5. $10,000<\/li>\n
                            6. She put $15,000 into a CD and $35,000 in bonds.<\/li>\n
                            7. The amount of the first year\u2019s loan was $30,000 and the amount of the second year\u2019s loan was $12,000.<\/li>\n
                            8. Bethany is 16 years old and Alyssa is 28 years old.<\/li>\n
                            9. Noelle is 20 years old and her dad is 54 years old.<\/li>\n
                            10. The small container holds 20 gallons and the large container holds 30 gallons.<\/li>\n
                            11. There were 10 calories burned jogging and 10 calories burned cycling.<\/li>\n
                            12. Notebooks are $4 and thumb drives are $20.<\/li>\n
                            13. The measures are 60 degrees and 30 degrees.<\/li>\n
                            14. The measures are 125 degrees and 55 degrees.<\/li>\n
                            15. 94 degrees and 86 degrees<\/li>\n
                            16. 72.5 degrees and 17.5 degrees<\/li>\n
                            17. The measures are 44 degrees and 136 degrees.<\/li>\n
                            18. The measures are 34 degrees and 56 degrees.<\/li>\n
                            19. The width is 10 feet and the length is 25 feet.<\/li>\n
                            20. The width is 15 feet and the length is 15 feet.<\/li>\n
                            21. It took Sarah\u2019s sister 12 hours.<\/li>\n
                            22. It took Lucy\u2019s friend 2 hours.<\/li>\n
                            23. The canoe rate is 5 mph and the current rate is 1 mph.<\/li>\n
                            24. The boat rate is 6.5 mph and the current rate is 2.5 mph.<\/li>\n
                            25. The jet rate is 240 mph and the wind speed is 28 mph.<\/li>\n
                            26. The jet rate is 415 mph and the wind speed is 19 mph.<\/li>\n
                            27. \"s=183,a=242\"<\/li>\n<\/ol>\n<\/div>\n<\/div>\n","protected":false},"author":125,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["lynn-marecek","maryanne-anthony-smith"],"pb_section_license":""},"chapter-type":[],"contributor":[63,64],"license":[],"class_list":["post-1377","chapter","type-chapter","status-publish","hentry","contributor-lynn-marecek","contributor-maryanne-anthony-smith"],"part":1094,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/pressbooks\/v2\/chapters\/1377","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/wp\/v2\/users\/125"}],"version-history":[{"count":1,"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/pressbooks\/v2\/chapters\/1377\/revisions"}],"predecessor-version":[{"id":1378,"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/pressbooks\/v2\/chapters\/1377\/revisions\/1378"}],"part":[{"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/pressbooks\/v2\/parts\/1094"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/pressbooks\/v2\/chapters\/1377\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/wp\/v2\/media?parent=1377"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/pressbooks\/v2\/chapter-type?post=1377"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/wp\/v2\/contributor?post=1377"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/businesstechnicalmath\/wp-json\/wp\/v2\/license?post=1377"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}