{"id":6605,"date":"2020-10-01T17:00:18","date_gmt":"2020-10-01T17:00:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6605"},"modified":"2021-06-01T22:25:56","modified_gmt":"2021-06-01T22:25:56","slug":"1-7-applications-using-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/1-7-applications-using-linear-equations\/","title":{"raw":"1.7: Applications Using Linear Equations","rendered":"1.7: Applications Using Linear Equations"},"content":{"raw":"
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Module 1 Learning Objectives<\/h3>\r\n1.7: Applications Using Linear Equations<\/strong>\r\n
    \r\n \t
  • Direct Translation - Translate to an algebraic equation and solve<\/li>\r\n \t
  • Consecutive Integers - Use an algebraic equation to find consecutive integers, consecutive even integers, or consecutive odd integers when given the sum<\/li>\r\n \t
  • Perimeter - Use an algebraic equation to find the dimensions of a rectangle when given the perimeter<\/li>\r\n \t
  • Comparison Applications - Given information about how two unknowns are related, use an algebraic equation to solve for the unknowns<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n

    Define a Process for Problem Solving<\/h2>\r\nThe power of algebra is how it can help you model real situations in order to answer questions about them.\r\n\r\nHere are some\u00a0steps to translate problem situations into algebraic equations you can solve. Not every<\/em> word problem fits perfectly into these steps, but they will help you get started.\r\n
      \r\n \t
    1. Read and understand the problem.<\/li>\r\n \t
    2. Determine the constants and variables in the problem.<\/li>\r\n \t
    3. Translate words into algebraic expressions and equations.<\/li>\r\n \t
    4. Write an equation to represent the problem.<\/li>\r\n \t
    5. Solve the equation.<\/li>\r\n \t
    6. Check and interpret your answer. Sometimes writing a sentence helps.<\/li>\r\n<\/ol>\r\n

      <\/h2>\r\n

      Direct Translation<\/h2>\r\nWord problems can be tricky. Often it takes a bit of practice to convert an English sentence into a mathematical sentence, which is one of the first steps to solving word problems. In the table below, words or phrases commonly associated with mathematical operators are categorized. Word problems often contain these or similar words, so it's good to see what mathematical operators are associated with them.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Addition [latex]+[\/latex]<\/th>\r\nSubtraction [latex]-[\/latex]<\/th>\r\nMultiplication [latex]\\times[\/latex]<\/th>\r\nDivision[latex]\\div[\/latex]<\/th>\r\nVariable ?<\/th>\r\nEquals [latex]=[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      More than<\/td>\r\nLess than<\/td>\r\nDouble<\/td>\r\nRatio<\/td>\r\nA number<\/td>\r\nIs<\/td>\r\n<\/tr>\r\n
      Together<\/td>\r\nIn the past<\/td>\r\nProduct<\/td>\r\nQuotient<\/td>\r\nOften, a value for which no information is given.<\/td>\r\nThe same as<\/td>\r\n<\/tr>\r\n
      Sum<\/td>\r\nSlower than<\/td>\r\nTimes<\/td>\r\nPer<\/td>\r\nAfter how many hours?<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      Total<\/td>\r\nThe remainder of<\/td>\r\nOf<\/td>\r\n<\/td>\r\nHow much will it cost?<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      In the future<\/td>\r\nDifference<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
      Faster than<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSome examples follow:\r\n
        \r\n \t
      • [latex]x\\text{ is }5[\/latex] \u00a0becomes [latex]x=5[\/latex]<\/li>\r\n \t
      • Three more than a number becomes [latex]x+3[\/latex]<\/li>\r\n \t
      • Four less than a number becomes [latex]x-4[\/latex]<\/li>\r\n \t
      • Double the cost becomes [latex]2\\cdot\\text{ cost }[\/latex]<\/li>\r\n \t
      • Groceries and gas together for the week cost $250 means [latex]\\text{ groceries }+\\text{ gas }=250[\/latex]<\/li>\r\n \t
      • The difference of 9 and a number becomes [latex]9-x[\/latex]. Notice how 9 is first in the sentence and the expression<\/li>\r\n<\/ul>\r\nLet's practice translating a few more English phrases into algebraic\u00a0expressions.\r\n
        \r\n

        Example 1<\/h3>\r\nTranslate each phrase in the table into algebraic expressions:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        \u00a0some number<\/td>\r\n\u00a0the sum of the number and 3<\/td>\r\n\u00a0twice the sum of the number and 3<\/td>\r\n<\/tr>\r\n
        \u00a0a length<\/td>\r\n\u00a0double the length<\/td>\r\n\u00a0double the length, decreased by 6<\/td>\r\n<\/tr>\r\n
        \u00a0a cost<\/td>\r\n\u00a0the difference of the cost and 20<\/td>\r\n\u00a02 times the difference of the cost and 20<\/td>\r\n<\/tr>\r\n
        \u00a0some quantity<\/td>\r\n\u00a0the difference of 5 and the quantity<\/td>\r\n\u00a0\u00a0the difference of 5 and the quantity, divided by 2<\/td>\r\n<\/tr>\r\n
        \u00a0an amount of time<\/td>\r\n\u00a0triple the amount of time<\/td>\r\n\u00a0triple the amount of time, increased by 5<\/td>\r\n<\/tr>\r\n
        \u00a0a distance<\/td>\r\n\u00a0the sum of [latex]-4[\/latex] and the distance<\/td>\r\n\u00a0the sum of [latex]-4[\/latex] and the twice the distance<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"790402\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"790402\"]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        \u00a0[latex]a[\/latex]<\/td>\r\n\u00a0[latex]a+3[\/latex]<\/td>\r\n\u00a0[latex]2\\left(a+3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        \u00a0[latex]l[\/latex]<\/td>\r\n\u00a0[latex]2l[\/latex]<\/td>\r\n\u00a0[latex]2l-6[\/latex]<\/td>\r\n<\/tr>\r\n
        \u00a0[latex]c[\/latex]<\/td>\r\n\u00a0\u00a0[latex]c-20[\/latex]<\/td>\r\n\u00a0[latex]2\\left(c-20\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        \u00a0[latex]q[\/latex]<\/td>\r\n\u00a0[latex]5-q[\/latex]<\/td>\r\n\u00a0[latex]\\frac{5-q}{2}[\/latex]<\/td>\r\n<\/tr>\r\n
        \u00a0[latex]t[\/latex]<\/td>\r\n\u00a0[latex]3t[\/latex]<\/td>\r\n\u00a0[latex]3t+5[\/latex]<\/td>\r\n<\/tr>\r\n
        \u00a0[latex]d[\/latex]<\/td>\r\n\u00a0[latex]-4+d[\/latex]<\/td>\r\n\u00a0[latex]-4+2d[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this example video, we show how to translate more words into mathematical expressions.\r\n\r\nhttps:\/\/youtu.be\/uD_V5t-6Kzs\r\n\r\n \r\n
        \r\n

        Example 2<\/h3>\r\nTwenty-eight\u00a0less than five times a certain number is 232. What is the number?\r\n\r\n[reveal-answer q=\"720402\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"720402\"]\r\n\r\nFollowing the steps provided:\r\n
          \r\n \t
        1. Read and understand:<\/strong> we are looking for a number.<\/li>\r\n \t
        2. Constants and variables:<\/strong> 28 and 232 are constants, \"a certain number\" is our variable because we don't know its value, and we are asked to find it. We will call it x.<\/em><\/li>\r\n \t
        3. Translate:\u00a0<\/strong>five times a certain number translates to [latex]5x[\/latex]\r\nTwenty-eight\u00a0less than five times a certain number translates to\u00a0[latex]5x-28[\/latex] because subtraction is built backward.\r\nis 232 translates to [latex]=232[\/latex] because \"is\" is associated with equals.<\/li>\r\n \t
        4. Write an equation:<\/strong>\u00a0[latex]5x-28=232[\/latex]<\/li>\r\n \t
        5. Solve the equation using what you know about solving linear equations:<\/strong>\r\n

          [latex]\\begin{array}{r}5x-28=232\\\\5x=260\\\\x=52\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

        6. Check and interpret:<\/strong> We can substitute 52 for x.\r\n

          [latex]\\begin{array}{r}5\\left(52\\right)-28=232\\\\5\\left(52\\right)=260\\\\260=260\\end{array}[\/latex].<\/p>\r\nTRUE!<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video that follows, we show another example of how to translate a sentence into a mathematical expression using a problem solving method.\r\n\r\nhttps:\/\/youtu.be\/izIIqOztUyI\r\n\r\nIn the last direct translation example, we see what language would require us to utilize parentheses.\r\n

          \r\n

          Example 3<\/h3>\r\nThree times the difference of a number and 4 is 18.\u00a0 Find the number.\r\n\r\n[reveal-answer q=\"924521\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924521\"]\r\n\r\nTranslate:<\/strong>\r\n\r\nWhen there is a word that indicates multiplication just prior to a word that means either addition or subtraction, parentheses are needed.\r\n\r\nTherefore, \"three times the difference of a number and 4\" translates into [latex]3(x-4)[\/latex].\u00a0 This is because three is being multiplied by the entire subtraction expression and not just [latex]x[\/latex].\r\n\r\nWrite an equation:<\/strong>\r\n\r\nTranslating the entire sentence results in the following equation:\u00a0 [latex]3(x-4)=18[\/latex]\r\n\r\nSolve the equation:<\/strong>\r\n

          [latex]3(x-4)=18[\/latex]<\/p>\r\n

          [latex]3x-12=18[\/latex]<\/p>\r\n

          [latex]\\underline{\\hspace{.32in}+12 \\hspace{.1in}+12}[\/latex]<\/p>\r\n

          [latex]\\hspace{.05in}3x\\hspace{.42in}=30[\/latex]<\/p>\r\n

          [latex]\\frac{3x}{3}=\\frac{30}{3}[\/latex]<\/p>\r\n

          [latex]x=10[\/latex]<\/p>\r\nCheck:<\/strong>\r\n

          [latex]3(x-4)=18[\/latex]<\/p>\r\n

          [latex]3(10-4)=18[\/latex]<\/p>\r\n

          [latex]3(6)=18[\/latex]<\/p>\r\n

          [latex]18=18[\/latex]<\/p>\r\n

          TRUE<\/p>\r\n\r\n

          Answer<\/span><\/h4>\r\nThe number is 10.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n

          Consecutive Integers<\/h2>\r\n

          Consecutive Integers<\/h3>\r\nAnother type of number problem involves consecutive numbers. Consecutive numbers are numbers that come one after the other, such as 3, 4, 5. If we are looking for several consecutive numbers it is important to first identify what they look like with variables before we set up the equation.\r\n\r\nFor example, let's say I want to know the next consecutive integer after 4. In mathematical terms, we would add 1 to 4 to get 5. We can generalize this idea as follows: the consecutive integer of any number, x<\/em>, is [latex]x+1[\/latex]. If we continue this pattern we can define any number of consecutive integers from any starting point. The following table shows how to describe four consecutive integers using algebraic notation.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
          First<\/td>\r\n[latex]x[\/latex]<\/td>\r\n<\/tr>\r\n
          Second<\/td>\r\n[latex]x+1[\/latex]<\/td>\r\n<\/tr>\r\n
          Third<\/td>\r\n[latex]x+2[\/latex]<\/td>\r\n<\/tr>\r\n
          Fourth<\/td>\r\n\u00a0[latex]x+3[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe apply the idea of consecutive integers to solving a word problem in the following example.\r\n
          \r\n

          Example 4<\/h3>\r\nThe sum of three consecutive integers is 93. What are the integers?\r\n\r\n[reveal-answer q=\"120402\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"120402\"]\r\nFollowing the steps provided:\r\n
            \r\n \t
          1. Read and understand:<\/strong>\u00a0We are looking for three numbers, and we know they are consecutive integers.<\/li>\r\n \t
          2. Constants and Variables:\u00a0<\/strong>93 is a constant.\r\nThe first integer we will call x<\/em>.\r\nSecond: [latex]x+1[\/latex]\r\nThird: [latex]x+2[\/latex]<\/li>\r\n \t
          3. Translate:\u00a0<\/strong>The sum of three consecutive integers translates to [latex]x+\\left(x+1\\right)+\\left(x+2\\right)[\/latex], based on how we defined the first, second, and third integers. Notice how we placed parentheses around the second and third integers. This is just to make each integer more distinct. is 93<\/em> translates to [latex]=93[\/latex] because is<\/em> is associated with equals.<\/li>\r\n \t
          4. Write an equation:<\/strong>\u00a0[latex]x+\\left(x+1\\right)+\\left(x+2\\right)=93[\/latex]<\/li>\r\n \t
          5. Solve the equation using what you know about solving linear equations:\u00a0<\/strong>We can't simplify within each set of parentheses, and we don't need to use the distributive property so we can rewrite the equation without parentheses.\r\n

            [latex]x+x+1+x+2=93[\/latex]<\/p>\r\nCombine like terms, simplify, and solve.\r\n

            [latex]\\begin{array}{r}x+x+1+x+2=93\\\\3x+3 = 93\\\\\\underline{-3\\,\\,\\,\\,\\,-3}\\\\3x=90\\\\\\frac{3x}{3}=\\frac{90}{3}\\\\x=30\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

          6. Check and Interpret:<\/strong> Okay, we have found a value for x<\/em>. We were asked to find the value of three consecutive integers, so we need to do a couple more steps. Remember how we defined our variables: The first integer we will call [latex]x[\/latex], [latex]x=30[\/latex]\r\nSecond: [latex]x+1[\/latex] so [latex]30+1=31[\/latex]\r\nThird: [latex]x+2[\/latex] so [latex]30+2=32[\/latex] The three consecutive integers whose sum is [latex]93[\/latex] are [latex]30\\text{, }31\\text{, and }32[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show another example of a consecutive integer problem.\r\nhttps:\/\/youtu.be\/S5HZy3jKodg\r\n

            Consecutive Even or Odd Integers<\/h3>\r\nThe following are examples of consecutive odd<\/strong> integers:<\/span><\/span>\r\n\r\n7 and 9<\/strong> are two consecutive odd integers. -19 and -17<\/strong> are a different set of two consecutive odd integers.<\/span><\/span>\r\n\r\nNotice that 7+2 = 9 and -19+2 = -17<\/span><\/span>.<\/span><\/span>\r\n\r\nIn order to go from one odd integer to the next consecutive odd integer you need to add 2.<\/span><\/span>\r\n\r\nThe following are examples of consecutive even<\/strong> integers:\r\n\r\n10 and 12<\/strong> are two consecutive even integers.\u00a0 -44 and -42<\/strong> are a different set of two consecutive even integers\r\n\r\nNotice that 10+2 = 12 and -44+2 = -42.\r\n\r\nIn order to go from one even integer to the next consecutive even integer you need to add 2.\r\n
            \r\n\r\nBased on these examples, we can use the following labels when we are asked to find the following:\r\n\r\nConsecutive Integers:\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[latex]x, x+1, x+2, ...[\/latex]\r\n\r\nConsecutive Odd Integers:\u00a0\u00a0\u00a0\u00a0\u00a0[latex]x, x+2, x+4, ...[\/latex]\r\n\r\nConsecutive Even Integers:\u00a0 \u00a0 [latex]x, x+2, x+4, ...[\/latex]\r\n\r\n<\/div>\r\n \r\n
            \r\n

            Example 5<\/h3>\r\nThe sum of two consecutive even integers is -74.\u00a0Find the integers.\r\n\r\n[reveal-answer q=\"651614\"]Show Answer[\/reveal-answer]\r\n\r\n[hidden-answer a=\"651614\"]\r\n\r\nLet\u00a0[latex]x= [\/latex] the first even integer.\u00a0\u00a0\u00a0\u00a0\u00a0 Then [latex]x+2=[\/latex] the next consecutive even integer.\r\n

            [latex]\\begin{array}{r}x+(x+2)=-74\\\\2x+2 = -74\\\\\\underline{-2\\,\\,\\,\\,\\,-2}\\\\2x=-76\\\\\\frac{2x}{2}=\\frac{-76}{2}\\\\x=-38\\\\x+2=-36\\end{array}[\/latex]<\/p>\r\nThe integers are -38 and -36.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

            \r\n

            Example 6<\/h3>\r\nThe sum of three consecutive odd integers is -15.\u00a0List the integers from smallest to largest.\r\n\r\n[reveal-answer q=\"713731\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"713731\"]\r\n\r\nLet\u00a0[latex]x= [\/latex] the first odd integer. Then [latex]x+2[\/latex] and\u00a0 [latex]x+4[\/latex] would represent the next two consecutive odd integers.\r\n

            [latex]\\begin{array}{r}x+(x+2)+(x+4)=-15\\\\3x+6 = -15\\\\\\underline{-6\\,\\,\\,\\,\\,-6}\\\\3x=-21\\\\\\frac{3x}{3}=\\frac{-21}{3}\\\\x=-7\\\\x+2=-5\\\\x+4=-3\\end{array}[\/latex]<\/p>\r\nThe integers are -7, -5 and -3\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n

            Perimeter<\/h2>\r\nPerimeter is the distance around an object. For example, consider a rectangle with a length of 8 and a width of 3. There are two lengths and two widths in a rectangle (opposite sides), so we add [latex]8+8+3+3=22[\/latex]. Since\u00a0there are two lengths and two widths in a rectangle, you may find the perimeter of a rectangle using\u00a0the formula [latex]{P}=2\\left({L}\\right)+2\\left({W}\\right)[\/latex] where\r\n\r\nL = Length\r\n\r\nW = Width\r\n\r\nIn the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution.\r\n
            \r\n

            Example 7<\/h3>\r\nYou want to make another garden box the same size as the one you already have. You write down the dimensions of the box and go to the lumber store to buy some boards. When you get there you realize you didn't write down the width dimension\u2014only the perimeter and length. You want the exact dimensions so you can have the store cut the lumber for you.\r\n\r\nHere is what you have written down:\r\n\r\nPerimeter = 16.4 feet\r\nLength = 4.7 feet\r\n\r\nCan you find the dimensions you need to have your boards cut at the lumber store? If so, how many boards do you need and what lengths should they be?\r\n\r\n[reveal-answer q=\"719712\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"719712\"]\r\n\r\nRead and Understand:<\/strong>\u00a0We know perimeter = 16.4 feet and length = 4.7 feet, and we want to find width.\r\n\r\nDefine and Translate:<\/strong>\r\n\r\nDefine the known and unknown dimensions:\r\n\r\nw = width\r\n\r\np = 16.4\r\n\r\nl = 4.7\r\n\r\nWrite and Solve:<\/strong>\r\n\r\nFirst we will substitute the dimensions we know into the formula for perimeter:\r\n

            [latex]\\begin{array}{l}\\,\\,\\,\\,\\,P=2{W}+2{L}\\\\\\\\16.4=2\\left(w\\right)+2\\left(4.7\\right)\\end{array}[\/latex]<\/p>\r\nThen we will isolate w<\/em> to find the unknown width.\r\n

            [latex]\\begin{array}{l}16.4=2\\left(w\\right)+2\\left(4.7\\right)\\\\16.4=2{w}+9.4\\\\\\underline{-9.4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-9.4}\\\\\\,\\,\\,\\,\\,\\,\\,7=2\\left(w\\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{7}{2}=\\frac{2\\left(w\\right)}{2}\\\\\\,\\,\\,\\,3.5=w\\end{array}[\/latex]<\/p>\r\nWrite the width as a decimal to make cutting the boards easier and replace the units on the measurement, or you won't get the right size of board!\r\n\r\nCheck and Interpret:<\/strong>\r\n\r\nIf we replace the width we found, [latex]w=3.5\\text{ feet }[\/latex] into the formula for perimeter with the dimensions we wrote down, we can check our work:\r\n

            [latex]\\begin{array}{l}\\,\\,\\,\\,\\,{P}=2\\left({L}\\right)+2\\left({W}\\right)\\\\\\\\{16.4}=2\\left({4.7}\\right)+2\\left({3.5}\\right)\\\\\\\\{16.4}=9.4+7\\\\\\\\{16.4}=16.4\\end{array}[\/latex]<\/p>\r\nOur calculation for width checks out. We need to ask for 2 boards cut to 3.5 feet and 2 boards cut to 4.7 feet so we can make the new garden box.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\nThe perimeter of any shape can be found by adding the lengths of all the sides.\u00a0 The first step in each of the following examples is to define the variable and which side length it represents.\u00a0 Then label each side of the figure in terms of the same variable.\u00a0 Once all the sides of the figure are labeled, the sum of the lengths of the sides should equal the given perimeter.\r\n

            \r\n

            Example 8<\/h3>\r\nA rectangular room is 7 meters longer than it is wide, and its perimeter is 62 meters.\u00a0 Find the dimensions of the room.\r\n\r\n[reveal-answer q=\"957209\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"957209\"]\r\n\r\nLet\u00a0[latex]x= [\/latex] the width of the rectangle. Then [latex]x+7[\/latex] would represent how long the rectangle is.\r\n\r\nRemember that opposite sides of a rectangle are equal in length.\r\n\r\n\"Rectangle\r\n\r\n \r\n

            [latex]\\begin{array}{r}x+(x+7)+x + (x+7)=62\\\\4x+14 = 62\\\\\\underline{-14\\,\\,-14}\\\\4x=48\\\\\\frac{4x}{4}=\\frac{48}{4}\\\\x=12\\\\x+7=19\\end{array}[\/latex]<\/p>\r\nThe width of the room is 12 meters.\r\n\r\nThe length the room is 19 meters.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

            \r\n

            Example 9<\/h3>\r\nOne side of a triangle is twice as long as the shortest side, and the third side is four times as long as the shortest side.\u00a0 The perimeter is 63 feet.\u00a0 Find the dimensions of the triangle.\r\n[reveal-answer q=\"773270\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"773270\"]\r\n\r\nLet\u00a0[latex]x= [\/latex] the length of the shortest side.\r\n\r\n\"Triangle\r\n

            [latex]\\begin{array}{r}x+2x+4x =63\\\\7x= 63\\\\\\frac{7x}{7}=\\frac{63}{7}\\\\x=9\\\\2x=18\\\\4x=36\\end{array}[\/latex]<\/p>\r\nThe lengths of the sides of the triangle are 9 ft., 18 ft., and 36 ft.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

            Comparison Applications<\/h2>\r\nThe first step in these problems is to define the variable and which unknown in the problem it represents.\u00a0 Then define the other unknown(s) in the problem in terms of that same variable.\u00a0 Next, read the problem carefully to create an equation.\u00a0 Solve the equation and be sure to label the answers.\r\n
            \r\n

            Example 10<\/h3>\r\nA sofa and a loveseat together cost $630. The cost of the sofa is twice the cost of the loveseat.\u00a0 How much do they each cost?\r\n[reveal-answer q=\"23572\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"23572\"]\r\n\r\nLet\u00a0[latex]L= [\/latex] the cost of the loveseat. Let\u00a0[latex]2L= [\/latex] the cost of the sofa. (Notice we use\u00a0[latex]2L [\/latex] here to indicate \"twice the cost\" and NOT\u00a0[latex]L+2 [\/latex].)\r\n

            Cost of the loveseat + Cost of the sofa\u00a0 = 630<\/p>\r\n

            [latex]\\begin{array}{r}L + 2L =630\\\\3L= 630\\\\\\frac{3L}{3}=\\frac{630}{3}\\\\L=210\\\\2L=420\\end{array}[\/latex]<\/p>\r\nThe loveseat costs $210, and the sofa costs $420.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

            \r\n

            Example 11<\/h3>\r\nA bag is filled with green and blue marbles. There are 111 marbles in the bag.\u00a0 If there are 17 more green marbles than blue marbles, find the number of green marbles and the number of blue marbles in the bag.\r\n\r\n[reveal-answer q=\"79127\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"79127\"]\r\n\r\nLet\u00a0[latex]B= [\/latex] the number of blue marbles. Let\u00a0[latex]B+17= [\/latex] the number of green marbles\r\n

            Number of blue marbles + Number of green marbles\u00a0 = 111<\/p>\r\n

            [latex]\\begin{array}{r}B + (B+17) =111\\\\2B+17= 111\\\\\\underline{-17\\,\\,\\,-17}\\\\2B= 94\\\\\\frac{2B}{2}=\\frac{94}{2}\\\\B=47\\\\B+17=64\\end{array}[\/latex]<\/p>\r\nThere are 47 blue marbles and 64 green marbles.\r\n\r\nNotice how the final answers adds up to the total, 47 + 64 = 111 marbles.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

            \r\n

            Example 12<\/h3>\r\nA 12-foot board is cut into two pieces.\u00a0 One piece is 4 feet shorter than the other piece. How long are the pieces?\r\n\r\n[reveal-answer q=\"53825\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"53825\"]\r\n\r\nLet\u00a0[latex]x= [\/latex] the length of the longer piece. Then\u00a0[latex]x-4= [\/latex] the length of the shorter piece.\r\n

            Length of longer piece + Length of shorter piece = 12<\/p>\r\n

            [latex]\\begin{array}{r}x + (x-4) =12\\\\2x-4= 12\\\\\\underline{+4\\,\\,\\,+4}\\\\2x= 16\\\\\\frac{2x}{2}=\\frac{16}{2}\\\\x=8\\\\x-4=4\\end{array}[\/latex]<\/p>\r\nThe longer piece is 8 feet and the shorter piece is 4 feet.\r\n\r\nAgain, notice how the final answers add up to the total, 8 feet + 4 feet = 12 feet.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n ","rendered":"

            \n

            Module 1 Learning Objectives<\/h3>\n

            1.7: Applications Using Linear Equations<\/strong><\/p>\n

              \n
            • Direct Translation – Translate to an algebraic equation and solve<\/li>\n
            • Consecutive Integers – Use an algebraic equation to find consecutive integers, consecutive even integers, or consecutive odd integers when given the sum<\/li>\n
            • Perimeter – Use an algebraic equation to find the dimensions of a rectangle when given the perimeter<\/li>\n
            • Comparison Applications – Given information about how two unknowns are related, use an algebraic equation to solve for the unknowns<\/li>\n<\/ul>\n<\/div>\n

               <\/p>\n

              Define a Process for Problem Solving<\/h2>\n

              The power of algebra is how it can help you model real situations in order to answer questions about them.<\/p>\n

              Here are some\u00a0steps to translate problem situations into algebraic equations you can solve. Not every<\/em> word problem fits perfectly into these steps, but they will help you get started.<\/p>\n

                \n
              1. Read and understand the problem.<\/li>\n
              2. Determine the constants and variables in the problem.<\/li>\n
              3. Translate words into algebraic expressions and equations.<\/li>\n
              4. Write an equation to represent the problem.<\/li>\n
              5. Solve the equation.<\/li>\n
              6. Check and interpret your answer. Sometimes writing a sentence helps.<\/li>\n<\/ol>\n

                <\/h2>\n

                Direct Translation<\/h2>\n

                Word problems can be tricky. Often it takes a bit of practice to convert an English sentence into a mathematical sentence, which is one of the first steps to solving word problems. In the table below, words or phrases commonly associated with mathematical operators are categorized. Word problems often contain these or similar words, so it’s good to see what mathematical operators are associated with them.<\/p>\n\n\n\n\n\n\n\n\n\n\n
                Addition [latex]+[\/latex]<\/th>\nSubtraction [latex]-[\/latex]<\/th>\nMultiplication [latex]\\times[\/latex]<\/th>\nDivision[latex]\\div[\/latex]<\/th>\nVariable ?<\/th>\nEquals [latex]=[\/latex]<\/th>\n<\/tr>\n<\/thead>\n
                More than<\/td>\nLess than<\/td>\nDouble<\/td>\nRatio<\/td>\nA number<\/td>\nIs<\/td>\n<\/tr>\n
                Together<\/td>\nIn the past<\/td>\nProduct<\/td>\nQuotient<\/td>\nOften, a value for which no information is given.<\/td>\nThe same as<\/td>\n<\/tr>\n
                Sum<\/td>\nSlower than<\/td>\nTimes<\/td>\nPer<\/td>\nAfter how many hours?<\/td>\n<\/td>\n<\/tr>\n
                Total<\/td>\nThe remainder of<\/td>\nOf<\/td>\n<\/td>\nHow much will it cost?<\/td>\n<\/td>\n<\/tr>\n
                In the future<\/td>\nDifference<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
                Faster than<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                Some examples follow:<\/p>\n

                  \n
                • [latex]x\\text{ is }5[\/latex] \u00a0becomes [latex]x=5[\/latex]<\/li>\n
                • Three more than a number becomes [latex]x+3[\/latex]<\/li>\n
                • Four less than a number becomes [latex]x-4[\/latex]<\/li>\n
                • Double the cost becomes [latex]2\\cdot\\text{ cost }[\/latex]<\/li>\n
                • Groceries and gas together for the week cost $250 means [latex]\\text{ groceries }+\\text{ gas }=250[\/latex]<\/li>\n
                • The difference of 9 and a number becomes [latex]9-x[\/latex]. Notice how 9 is first in the sentence and the expression<\/li>\n<\/ul>\n

                  Let’s practice translating a few more English phrases into algebraic\u00a0expressions.<\/p>\n

                  \n

                  Example 1<\/h3>\n

                  Translate each phrase in the table into algebraic expressions:<\/p>\n\n\n\n\n\n\n\n\n
                  \u00a0some number<\/td>\n\u00a0the sum of the number and 3<\/td>\n\u00a0twice the sum of the number and 3<\/td>\n<\/tr>\n
                  \u00a0a length<\/td>\n\u00a0double the length<\/td>\n\u00a0double the length, decreased by 6<\/td>\n<\/tr>\n
                  \u00a0a cost<\/td>\n\u00a0the difference of the cost and 20<\/td>\n\u00a02 times the difference of the cost and 20<\/td>\n<\/tr>\n
                  \u00a0some quantity<\/td>\n\u00a0the difference of 5 and the quantity<\/td>\n\u00a0\u00a0the difference of 5 and the quantity, divided by 2<\/td>\n<\/tr>\n
                  \u00a0an amount of time<\/td>\n\u00a0triple the amount of time<\/td>\n\u00a0triple the amount of time, increased by 5<\/td>\n<\/tr>\n
                  \u00a0a distance<\/td>\n\u00a0the sum of [latex]-4[\/latex] and the distance<\/td>\n\u00a0the sum of [latex]-4[\/latex] and the twice the distance<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                  Show Solution<\/span><\/p>\n
                  \n\n\n\n\n\n\n\n\n
                  \u00a0[latex]a[\/latex]<\/td>\n\u00a0[latex]a+3[\/latex]<\/td>\n\u00a0[latex]2\\left(a+3\\right)[\/latex]<\/td>\n<\/tr>\n
                  \u00a0[latex]l[\/latex]<\/td>\n\u00a0[latex]2l[\/latex]<\/td>\n\u00a0[latex]2l-6[\/latex]<\/td>\n<\/tr>\n
                  \u00a0[latex]c[\/latex]<\/td>\n\u00a0\u00a0[latex]c-20[\/latex]<\/td>\n\u00a0[latex]2\\left(c-20\\right)[\/latex]<\/td>\n<\/tr>\n
                  \u00a0[latex]q[\/latex]<\/td>\n\u00a0[latex]5-q[\/latex]<\/td>\n\u00a0[latex]\\frac{5-q}{2}[\/latex]<\/td>\n<\/tr>\n
                  \u00a0[latex]t[\/latex]<\/td>\n\u00a0[latex]3t[\/latex]<\/td>\n\u00a0[latex]3t+5[\/latex]<\/td>\n<\/tr>\n
                  \u00a0[latex]d[\/latex]<\/td>\n\u00a0[latex]-4+d[\/latex]<\/td>\n\u00a0[latex]-4+2d[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n

                  In this example video, we show how to translate more words into mathematical expressions.<\/p>\n