{"id":1082,"date":"2024-05-04T17:47:26","date_gmt":"2024-05-04T17:47:26","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=1082"},"modified":"2025-08-21T23:15:53","modified_gmt":"2025-08-21T23:15:53","slug":"modeling-with-linear-equations-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/modeling-with-linear-equations-learn-it-3\/","title":{"raw":"Modeling with Linear Equations: Learn It 3","rendered":"Modeling with Linear Equations: Learn It 3"},"content":{"raw":"

Using a Formula to Solve a Real-World Application<\/h2>\r\nMany applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem\u2019s question is answered. Sometimes, these problems involve two equations representing two unknowns, which can be written using one equation in one variable by expressing one unknown in terms of the other.\r\n\r\n
Examples of formulas include the perimeter<\/strong> of a rectangle, [latex]P=2L+2W[\/latex]; the area<\/strong> of a rectangular region, [latex]A=LW[\/latex]; and the volume<\/strong> of a rectangular solid, [latex]V=LWH[\/latex].<\/section>
The\u00a0distance<\/strong> [latex]d[\/latex] covered when traveling at a constant\u00a0rate<\/strong> [latex]r[\/latex] for some\u00a0time\u00a0<\/strong>[latex]t[\/latex] is given by the formula [latex]d=rt[\/latex].<\/section>
It takes Andrew [latex]30[\/latex] minutes to drive to work in the morning. He drives home using the same route, but it takes [latex]10[\/latex] minutes longer, and he averages [latex]10[\/latex] mi\/h less than in the morning. How far does Andrew drive to work?[reveal-answer q=\"599628\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"599628\"]Step 1: Define the Variables\r\n<\/strong>Let:\r\n
    \r\n \t
  • [latex]d[\/latex] be the distance to work (in miles).<\/li>\r\n \t
  • [latex]r[\/latex] be the morning driving speed (in miles per hour).<\/li>\r\n \t
  • Morning trip time: [latex]30[\/latex] minutes.<\/li>\r\n \t
  • Afternoon trip time, which is [latex]10[\/latex] minutes longer: [latex]40[\/latex] minutes.<\/li>\r\n<\/ul>\r\nStep 2: Convert Times to Hours<\/strong>\r\n\r\nSince the rate is in miles per hour, convert the time from minutes to hours:\r\n
      \r\n \t
    • Morning trip time [latex]= \\frac{30}{60} = \\frac{1}{2}[\/latex] hours<\/li>\r\n \t
    • Afternoon trip time [latex]= \\frac{40}{60} = \\frac{2}{3}[\/latex] hours<\/li>\r\n<\/ul>\r\nStep 3: Set Up Equations<\/strong>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      <\/th>\r\n[latex]d[\/latex] (miles)<\/th>\r\n[latex]r[\/latex] (mph)<\/th>\r\n[latex]t[\/latex] (hours)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      Morning Trip<\/strong><\/td>\r\n[latex]d[\/latex]<\/td>\r\n[latex]r[\/latex]<\/td>\r\n[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n
      Afternoon Trip<\/strong><\/td>\r\n[latex]d[\/latex]<\/td>\r\n[latex]r - 10[\/latex]<\/td>\r\n[latex]\\frac{2}{3}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
        \r\n \t
      • Morning trip equation: [latex]d=r(\\frac{1}{2})[\/latex]<\/li>\r\n \t
      • Afternoon trip equation: [latex]d=(r - 10)(\\frac{2}{3})[\/latex]<\/li>\r\n<\/ul>\r\nStep 4: <\/strong>Equate and Solve for [latex]r[\/latex]<\/strong>\r\n\r\nSince the distance ([latex]d[\/latex]) is the same for both trips, set the equations equal to each other:\r\n

        [latex]r(\\frac{1}{2}) =(r - 10)(\\frac{2}{3})[\/latex]<\/p>\r\nSolve for [latex]r[\/latex]:\r\n\r\n[latex]\\begin{align*} \\text{Given equation:} && r\\left(\\frac{1}{2}\\right) &= (r - 10)\\left(\\frac{2}{3}\\right) \\\\ \\text{Distribute the } \\frac{2}{3}: && \\frac{1}{2}r &= \\frac{2}{3}r - \\frac{20}{3} \\\\ \\text{Subtract } \\frac{2}{3}r \\text{ from both sides:} && \\frac{1}{2}r - \\frac{2}{3}r &= -\\frac{20}{3} \\\\ \\text{Combine like terms:} && \\left(\\frac{1}{2} - \\frac{2}{3}\\right)r &= -\\frac{20}{3} \\\\ \\text{Get common denominators for coefficients of } r: && \\left(\\frac{3}{6} - \\frac{4}{6}\\right)r &= -\\frac{20}{3} \\\\ \\text{Simplify:} && -\\frac{1}{6}r &= -\\frac{20}{3} \\\\ \\text{Divide both sides by } -\\frac{1}{6}: && r &= \\left(-\\frac{20}{3}\\right) \\cdot \\left(-6\\right) \\\\ \\text{Simplify:} && r &= 40 \\end{align*}[\/latex]\r\n\r\nStep 5: Calculate distance [latex]d[\/latex] and write your conclusion.<\/strong>\r\n\r\nSubstitute [latex]r = 40[\/latex] back into the morning trip equation:\r\n

        [latex]d = 40(\\frac{1}{2}) = 20 \\text{ miles}[\/latex]<\/p>\r\nThus, Andrew drives [latex]20[\/latex] miles to work.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

        How were the fractions handled in the example above?\r\n<\/strong>\r\n\r\n
        \r\n\r\nRecall that when solving multi-step equations, it is helpful to multiply by the LCD to clear the denominators from the equation. But it is also permissible to use operations on fractions to combine like terms. The example above demonstrates both.[reveal-answer q=\"257523\"]more[\/reveal-answer]\r\n[hidden-answer a=\"257523\"]\r\n
          \r\n \t
        • In the text of the solution when solving for [latex]r[\/latex] the first time, the parentheses were eliminated using the distributive property, then operations on fractions were used to combine like terms.<\/li>\r\n<\/ul>\r\n

          [latex]\\begin{array}{c}r\\left(\\frac{1}{2}\\right)=\\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill \\\\ \\frac{1}{2}r=\\frac{2}{3}r-\\frac{20}{3}\\hfill \\\\ \\frac{1}{2}r-\\frac{2}{3}r=-\\frac{20}{3}\\hfill \\\\ -\\frac{1}{6}r=-\\frac{20}{3}\\hfill \\\\ r=-\\frac{20}{3}\\left(-6\\right)\\hfill \\\\ r=40\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n

            \r\n \t
          • Later, in the analysis of the solution, the LCD between the denominators [latex]2 \\text{ and } 3 [\/latex] was multiplied on both sides of the equation to cancel out the denominators so that operations on fractions were not necessary. Do you see how the LCD wasn't actually multiplied through on both sides, but that the denominators cancelled out, resulting in a linear equation in one variable without denominators?<\/li>\r\n<\/ul>\r\n

            [latex]\\begin{array}{l}r\\left(\\frac{1}{2}\\right)=\\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill \\\\ 6\\times r\\left(\\frac{1}{2}\\right)=6\\times \\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill \\\\ 3r=4\\left(r - 10\\right)\\hfill \\\\ 3r=4r - 40\\hfill \\\\ -r=-40\\hfill \\\\ r=40\\hfill \\end{array}[\/latex]<\/p>\r\nBoth methods are equally correct unless your instructor requires you to specifically demonstrate knowledge of one or the other.\r\n\r\nWhich do you prefer?\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

            How To: Solve Multi-Step Equations<\/strong>\r\n
              \r\n \t
            1. (Optional<\/em>) Multiply to clear any fractions or decimals.<\/li>\r\n \t
            2. Simplify each side by clearing parentheses and combining like terms.<\/li>\r\n \t
            3. Add or subtract to isolate the variable term\u2014you may have to move a term with the variable.<\/li>\r\n \t
            4. Multiply or divide to isolate the variable.<\/li>\r\n \t
            5. Check the solution.<\/li>\r\n<\/ol>\r\n<\/section>
              [ohm2_question hide_question_numbers=1]18939[\/ohm2_question]<\/section>
              The perimeter of a rectangular outdoor patio is [latex]54[\/latex] ft. The length is [latex]3[\/latex] ft. greater than the width. What are the dimensions of the patio?[reveal-answer q=\"852942\"]Show Answer[\/reveal-answer]<\/span>[hidden-answer a=\"852942\"]Given: [latex]P = 54[\/latex]ft. and [latex]L = W+3[\/latex] ft.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"384\"]\"A A rectangle with the length labeled as: L = W + 3 and the width labeled as: W[\/caption]\r\n\r\nThe relationship can be expressed and solved as follows:\r\n
              [latex]\\begin{align*} \\text{Given the perimeter formula:} && P &= 2L + 2W \\\\ \\text{Substitute } P=54 \\text{ and } L = W + 3: && 54 &= 2(W + 3) + 2W \\\\ \\text{Expand and simplify:} && 54 &= 2W + 6 + 2W \\\\ \\text{Combine like terms:} && 54 &= 4W + 6 \\\\ \\text{Solve for } W: && 48 &= 4W \\\\ && W &= 12 \\\\ \\text{Solve for } L: && L &= W + 3 = 12 + 3 = 15 \\end{align*}[\/latex]<\/center>\r\nThus, the dimensions are [latex]L=15[\/latex] ft and [latex]W=12[\/latex] ft.[\/hidden-answer]<\/section>
              [ohm2_question hide_question_numbers=1]18940[\/ohm2_question]<\/section>
              The perimeter of a tablet of graph paper is [latex]48[\/latex] inches[latex]^2[\/latex]. The length is [latex]6[\/latex] inches more than the width. Find the area of the graph paper.\r\n[reveal-answer q=\"81698\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"81698\"]The standard formula for area is [latex]A=LW[\/latex]; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.\r\n[latex]\\\\[\/latex]\r\nWe know that the length is [latex]6[\/latex] inches more than the width, so we can write length as [latex]L=W+6[\/latex].\r\n[latex]\\\\[\/latex]\r\nSubstitute the value of the perimeter and the expression for length into the perimeter formula and find the length.\r\n
              [latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 48=2\\left(W+6\\right)+2W\\hfill \\\\ 48=2W+12+2W\\hfill \\\\ 48=4W+12\\hfill \\\\ 36=4W\\hfill \\\\ 9=W\\hfill \\\\ \\left(9+6\\right)=L\\hfill \\\\ 15=L\\hfill \\end{array}[\/latex]<\/div>\r\nNow, we find the area given the dimensions of [latex]L=15[\/latex] in. and [latex]W=9[\/latex] in.\r\n
              [latex]\\begin{array}{l}A\\hfill&=LW\\hfill \\\\ A\\hfill&=15\\left(9\\right)\\hfill \\\\ \\hfill&=135\\text{ in}^{2}\\hfill \\end{array}[\/latex]<\/div>\r\nThe area is [latex]135[\/latex] inches[latex]^2[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
              [ohm_question hide_question_numbers=1]293768[\/ohm_question]<\/section>
              Find the dimensions of a shipping box given that the length is twice the width, the height is [latex]8[\/latex] inches, and the volume is [latex]1,600[\/latex] in.3<\/sup>.\r\n[reveal-answer q=\"889305\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"889305\"]The formula for the volume of a box is given as [latex]V=LWH[\/latex], the product of length, width, and height. We are given that [latex]L=2W[\/latex], and [latex]H=8[\/latex]. The volume is [latex]1,600[\/latex] cubic inches.\r\n
              [latex]\\begin{array}{l}V=LWH \\\\ 1,600=\\left(2W\\right)W\\left(8\\right) \\\\ 1,600=16{W}^{2} \\\\ 100={W}^{2} \\\\ 10=W \\end{array}[\/latex]<\/div>\r\nThe dimensions are [latex]L=20[\/latex] in., [latex]W=10[\/latex] in., and [latex]H=8[\/latex] in.\r\n\r\nAnalysis of the Solution<\/strong>\r\n\r\nNote that the square root of [latex]{W}^{2}[\/latex] would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><\/section>","rendered":"

              Using a Formula to Solve a Real-World Application<\/h2>\n

              Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem\u2019s question is answered. Sometimes, these problems involve two equations representing two unknowns, which can be written using one equation in one variable by expressing one unknown in terms of the other.<\/p>\n

              Examples of formulas include the perimeter<\/strong> of a rectangle, [latex]P=2L+2W[\/latex]; the area<\/strong> of a rectangular region, [latex]A=LW[\/latex]; and the volume<\/strong> of a rectangular solid, [latex]V=LWH[\/latex].<\/section>\n
              The\u00a0distance<\/strong> [latex]d[\/latex] covered when traveling at a constant\u00a0rate<\/strong> [latex]r[\/latex] for some\u00a0time\u00a0<\/strong>[latex]t[\/latex] is given by the formula [latex]d=rt[\/latex].<\/section>\n
              It takes Andrew [latex]30[\/latex] minutes to drive to work in the morning. He drives home using the same route, but it takes [latex]10[\/latex] minutes longer, and he averages [latex]10[\/latex] mi\/h less than in the morning. How far does Andrew drive to work?<\/p>\n