{"id":100,"date":"2023-06-21T13:22:33","date_gmt":"2023-06-21T13:22:33","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/use-formulas-to-solve-problems\/"},"modified":"2023-09-07T15:49:07","modified_gmt":"2023-09-07T15:49:07","slug":"use-formulas-to-solve-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/use-formulas-to-solve-problems\/","title":{"raw":"\u25aa   Using Formulas to Solve Problems","rendered":"\u25aa   Using Formulas to Solve Problems"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Set up a linear equation involving distance, rate, and time.<\/li>\r\n \t<li>Find the dimensions of a rectangle given the area.<\/li>\r\n \t<li>Find the dimensions of a box given information about its side lengths.<\/li>\r\n<\/ul>\r\n<\/div>\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. Examples of formulas include the <strong>perimeter<\/strong> of a rectangle, [latex]P=2L+2W[\/latex]; the <strong>area<\/strong> of a rectangular region, [latex]A=LW[\/latex]; and the <strong>volume<\/strong> of a rectangular solid, [latex]V=LWH[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Recall the relationship between distance, rate and time<\/h3>\r\nThe\u00a0<strong>distance<\/strong> [latex]d[\/latex] covered when traveling at a constant\u00a0<strong>rate<\/strong> [latex]r[\/latex] for some\u00a0<strong>time\u00a0<\/strong>[latex]t[\/latex] is given by the formula [latex]d=rt[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Application Using a Formula<\/h3>\r\nIt takes Andrew 30 minutes to drive to work in the morning. He drives home using the same route, but it takes 10 minutes longer, and he averages 10 mi\/h less than in the morning. How far does Andrew drive to work?\r\n[reveal-answer q=\"993244\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"993244\"]\r\n\r\nThis is a distance problem, so we can use the formula [latex]d=rt[\/latex], where distance equals rate multiplied by time. Note that when rate is given in mi\/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.\r\n\r\nFirst, we identify the known and unknown quantities. Andrew\u2019s morning drive to work takes 30 min, or [latex]\\frac{1}{2}[\/latex] h at rate [latex]r[\/latex]. His drive home takes 40 min, or [latex]\\frac{2}{3}[\/latex] h, and his speed averages 10 mi\/h less than the morning drive. Both trips cover distance [latex]d[\/latex]. A table, such as the one below, is often helpful for keeping track of information in these types of problems.\r\n<table summary=\"A table with 3 rows and 4 columns. The first entry in the first row is blank, the rest are: d, r, and t. The entries in the second row are: To Work, d, r, and . The entries in the third row are: To Home, d, r 10, and 2\/3.\">\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[latex]d[\/latex]<\/th>\r\n<th>[latex]r[\/latex]<\/th>\r\n<th>[latex]t[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td><strong>To Work<\/strong><\/td>\r\n<td>[latex]d[\/latex]<\/td>\r\n<td>[latex]r[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>To Home<\/strong><\/td>\r\n<td>[latex]d[\/latex]<\/td>\r\n<td>[latex]r - 10[\/latex]<\/td>\r\n<td>[latex]\\frac{2}{3}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWrite two equations, one for each trip.\r\n<div>[latex]\\begin{array}{ll}d=r\\left(\\frac{1}{2}\\right)\\hfill &amp; \\text{To work}\\hfill \\\\ d=\\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill &amp; \\text{To home}\\hfill \\end{array}[\/latex]<\/div>\r\nAs both equations equal the same distance, we set them equal to each other and solve for <em>r<\/em>.\r\n<div style=\"text-align: center;\">[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]<\/div>\r\nWe have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi\/h. Now we can answer the question. Substitute the rate back into either equation and solve for <em>d.<\/em>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d\\hfill&amp;=40\\left(\\frac{1}{2}\\right)\\hfill \\\\ \\hfill&amp;=20\\hfill \\end{array}[\/latex]<\/div>\r\nThe distance between home and work is 20 mi.\r\n<h4>Analysis of the Solution<\/h4>\r\nNote that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for [latex]r[\/latex].\r\n<div style=\"text-align: center;\">[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]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>How were the fractions handled in the example above?<\/h3>\r\nRecall that when solving multi-step equations, it is helpful to multiply by the LCD to clear the denominators from the equation.\r\n\r\nBut it is also permissible to use operations on fractions to combine like terms. The example above demonstrates both.\r\n\r\n[reveal-answer q=\"257523\"]more[\/reveal-answer]\r\n[hidden-answer a=\"257523\"]\r\n<ul>\r\n \t<li>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<p style=\"text-align: center;\">[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<ul>\r\n \t<li>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<p style=\"text-align: center;\">[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<p style=\"padding-left: 30px;\"><strong>How To: Solve Multi-Step Equations<\/strong>\r\n1. (Optional) Multiply to clear any fractions or decimals.<\/p>\r\n<p style=\"padding-left: 30px;\">2. Simplify each side by clearing parentheses and combining like terms.<\/p>\r\n<p style=\"padding-left: 30px;\">3. Add or subtract to isolate the variable term\u2014you may have to move a term with the variable.<\/p>\r\n<p style=\"padding-left: 30px;\">4. Multiply or divide to isolate the variable.<\/p>\r\n<p style=\"padding-left: 30px;\">5. Check the solution.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nOn Saturday morning, it took Jennifer 3.6 hours to drive to her mother\u2019s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 hours to return home. Her speed was 5 mi\/h slower on Sunday than on Saturday. What was her speed on Sunday?\r\n[reveal-answer q=\"929719\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"929719\"]\r\n\r\n45 [latex]\\frac{\\text{mi}}{\\text{h}}[\/latex][\/hidden-answer]\r\n\r\n[ohm_question height=\"220\"]52436[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Perimeter Problem<\/h3>\r\nThe 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?\r\n[reveal-answer q=\"815614\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"815614\"]\r\n\r\nThe perimeter formula is standard: [latex]P=2L+2W[\/latex]. We have two unknown quantities, length and width. However, we can write the length in terms of the width as [latex]L=W+3[\/latex]. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as shown below.\r\n\r\n&nbsp;\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200341\/CNX_CAT_Figure_02_03_003.jpg\" alt=\"A rectangle with the length labeled as: L = W + 3 and the width labeled as: W.\" width=\"487\" height=\"166\" \/>\r\n\r\nNow we can solve for the width and then calculate the length.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 54=2\\left(W+3\\right)+2W\\hfill \\\\ 54=2W+6+2W\\hfill \\\\ 54=4W+6\\hfill \\\\ 48=4W\\hfill \\\\ 12=W\\hfill \\\\ \\left(12+3\\right)=L\\hfill \\\\ 15=L\\hfill \\end{array}[\/latex]<\/div>\r\nThe dimensions are [latex]L=15[\/latex] ft and [latex]W=12[\/latex] ft.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Evaluating a variable for an expression<\/h3>\r\nIn the example above, the length was written in terms of the width in order to substitute it for the variable [latex]L[\/latex] to be able to write one equation in one variable.\r\n\r\nThis is a powerful technique in algebra and should be practiced until it becomes familiar, so it's worth taking a closer look.\r\n\r\n[reveal-answer q=\"866900\"]more[\/reveal-answer]\r\n[hidden-answer a=\"866900\"]\r\n\r\nWe had [latex]P = 2L + 2W[\/latex], which is one equation in two variables. Given the value for [latex]P[\/latex], we don't have enough information to solve for either [latex]L[\/latex] or [latex]W[\/latex].\r\n\r\nBut we do know that [latex]L=W+3[\/latex]. We can use this to substitute for the [latex]L[\/latex] in the original equation. Be careful to retain the original statement exactly, substituting only the value for [latex]L[\/latex]. A handy way to handle the substitution is with a set of parentheses.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 54=2\\left(L\\right)+2W \\,\\,\\,\\,\\,\\,\\,\\, \\text{ wrap the } L \\text{ to see where to make the substitution} \\hfill \\\\ 54=2\\left(W+3\\right)+2W \\,\\,\\,\\,\\, \\text{ drop the expression for } L \\text{ into the parentheses}\\hfill \\\\ 54=2W+6+2W\\hfill \\\\ 54=4W+6\\end{array}[\/latex]<\/p>\r\nNow we have one equation in one variable. We can use the properties of equality to isolate the variable [latex]W[\/latex] on one side, with its numerical value on the other.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the dimensions of a rectangle given that the perimeter is [latex]110[\/latex] cm. and the length is 1 cm. more than twice the width.\r\n[reveal-answer q=\"447289\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"447289\"]\r\n\r\nL = 37 cm, W = 18 cm[\/hidden-answer]\r\n\r\n[ohm_question height=\"240\"]7679[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Area Problem<\/h3>\r\nThe perimeter of a tablet of graph paper is 48 in<sup>2<\/sup>. The length is [latex]6[\/latex] in. 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\"]\r\n\r\nThe 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\r\nWe know that the length is 6 in. more than the width, so we can write length as [latex]L=W+6[\/latex]. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.\r\n<div style=\"text-align: center;\">[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<div style=\"text-align: center;\">[latex]\\begin{array}{l}A\\hfill&amp;=LW\\hfill \\\\ A\\hfill&amp;=15\\left(9\\right)\\hfill \\\\ \\hfill&amp;=135\\text{ in}^{2}\\hfill \\end{array}[\/latex]<\/div>\r\nThe area is [latex]135[\/latex] in<sup>2<\/sup>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft<sup>2<\/sup> of new carpeting should be ordered?\r\n[reveal-answer q=\"646790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"646790\"]\r\n\r\n250 ft<sup>2<\/sup>[\/hidden-answer]\r\n\r\n[ohm_question height=\"190\"]1688[\/ohm_question]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Volume Problem<\/h3>\r\nFind 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 1,600 in.<sup>3<\/sup>.\r\n[reveal-answer q=\"889305\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"889305\"]\r\n\r\nThe 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<div style=\"text-align: center;\">[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<h4>Analysis of the Solution<\/h4>\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<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Set up a linear equation involving distance, rate, and time.<\/li>\n<li>Find the dimensions of a rectangle given the area.<\/li>\n<li>Find the dimensions of a box given information about its side lengths.<\/li>\n<\/ul>\n<\/div>\n<p>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. Examples of formulas include the <strong>perimeter<\/strong> of a rectangle, [latex]P=2L+2W[\/latex]; the <strong>area<\/strong> of a rectangular region, [latex]A=LW[\/latex]; and the <strong>volume<\/strong> of a rectangular solid, [latex]V=LWH[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Recall the relationship between distance, rate and time<\/h3>\n<p>The\u00a0<strong>distance<\/strong> [latex]d[\/latex] covered when traveling at a constant\u00a0<strong>rate<\/strong> [latex]r[\/latex] for some\u00a0<strong>time\u00a0<\/strong>[latex]t[\/latex] is given by the formula [latex]d=rt[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Application Using a Formula<\/h3>\n<p>It takes Andrew 30 minutes to drive to work in the morning. He drives home using the same route, but it takes 10 minutes longer, and he averages 10 mi\/h less than in the morning. How far does Andrew drive to work?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q993244\">Show Solution<\/span><\/p>\n<div id=\"q993244\" class=\"hidden-answer\" style=\"display: none\">\n<p>This is a distance problem, so we can use the formula [latex]d=rt[\/latex], where distance equals rate multiplied by time. Note that when rate is given in mi\/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.<\/p>\n<p>First, we identify the known and unknown quantities. Andrew\u2019s morning drive to work takes 30 min, or [latex]\\frac{1}{2}[\/latex] h at rate [latex]r[\/latex]. His drive home takes 40 min, or [latex]\\frac{2}{3}[\/latex] h, and his speed averages 10 mi\/h less than the morning drive. Both trips cover distance [latex]d[\/latex]. A table, such as the one below, is often helpful for keeping track of information in these types of problems.<\/p>\n<table summary=\"A table with 3 rows and 4 columns. The first entry in the first row is blank, the rest are: d, r, and t. The entries in the second row are: To Work, d, r, and . The entries in the third row are: To Home, d, r 10, and 2\/3.\">\n<thead>\n<tr>\n<th><\/th>\n<th>[latex]d[\/latex]<\/th>\n<th>[latex]r[\/latex]<\/th>\n<th>[latex]t[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><strong>To Work<\/strong><\/td>\n<td>[latex]d[\/latex]<\/td>\n<td>[latex]r[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>To Home<\/strong><\/td>\n<td>[latex]d[\/latex]<\/td>\n<td>[latex]r - 10[\/latex]<\/td>\n<td>[latex]\\frac{2}{3}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Write two equations, one for each trip.<\/p>\n<div>[latex]\\begin{array}{ll}d=r\\left(\\frac{1}{2}\\right)\\hfill & \\text{To work}\\hfill \\\\ d=\\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill & \\text{To home}\\hfill \\end{array}[\/latex]<\/div>\n<p>As both equations equal the same distance, we set them equal to each other and solve for <em>r<\/em>.<\/p>\n<div style=\"text-align: center;\">[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]<\/div>\n<p>We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi\/h. Now we can answer the question. Substitute the rate back into either equation and solve for <em>d.<\/em><\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d\\hfill&=40\\left(\\frac{1}{2}\\right)\\hfill \\\\ \\hfill&=20\\hfill \\end{array}[\/latex]<\/div>\n<p>The distance between home and work is 20 mi.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for [latex]r[\/latex].<\/p>\n<div style=\"text-align: center;\">[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]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>How were the fractions handled in the example above?<\/h3>\n<p>Recall that when solving multi-step equations, it is helpful to multiply by the LCD to clear the denominators from the equation.<\/p>\n<p>But it is also permissible to use operations on fractions to combine like terms. The example above demonstrates both.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q257523\">more<\/span><\/p>\n<div id=\"q257523\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>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>\n<\/ul>\n<p style=\"text-align: center;\">[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>\n<ul>\n<li>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&#8217;t actually multiplied through on both sides, but that the denominators cancelled out, resulting in a linear equation in one variable without denominators?<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[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>\n<p>Both methods are equally correct unless your instructor requires you to specifically demonstrate knowledge of one or the other.<\/p>\n<p>Which do you prefer?<\/p>\n<p style=\"padding-left: 30px;\"><strong>How To: Solve Multi-Step Equations<\/strong><br \/>\n1. (Optional) Multiply to clear any fractions or decimals.<\/p>\n<p style=\"padding-left: 30px;\">2. Simplify each side by clearing parentheses and combining like terms.<\/p>\n<p style=\"padding-left: 30px;\">3. Add or subtract to isolate the variable term\u2014you may have to move a term with the variable.<\/p>\n<p style=\"padding-left: 30px;\">4. Multiply or divide to isolate the variable.<\/p>\n<p style=\"padding-left: 30px;\">5. Check the solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>On Saturday morning, it took Jennifer 3.6 hours to drive to her mother\u2019s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 hours to return home. Her speed was 5 mi\/h slower on Sunday than on Saturday. What was her speed on Sunday?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q929719\">Show Solution<\/span><\/p>\n<div id=\"q929719\" class=\"hidden-answer\" style=\"display: none\">\n<p>45 [latex]\\frac{\\text{mi}}{\\text{h}}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm52436\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=52436&theme=oea&iframe_resize_id=ohm52436&show_question_numbers\" width=\"100%\" height=\"220\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Perimeter Problem<\/h3>\n<p>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?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q815614\">Show Solution<\/span><\/p>\n<div id=\"q815614\" class=\"hidden-answer\" style=\"display: none\">\n<p>The perimeter formula is standard: [latex]P=2L+2W[\/latex]. We have two unknown quantities, length and width. However, we can write the length in terms of the width as [latex]L=W+3[\/latex]. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as shown below.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200341\/CNX_CAT_Figure_02_03_003.jpg\" alt=\"A rectangle with the length labeled as: L = W + 3 and the width labeled as: W.\" width=\"487\" height=\"166\" \/><\/p>\n<p>Now we can solve for the width and then calculate the length.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 54=2\\left(W+3\\right)+2W\\hfill \\\\ 54=2W+6+2W\\hfill \\\\ 54=4W+6\\hfill \\\\ 48=4W\\hfill \\\\ 12=W\\hfill \\\\ \\left(12+3\\right)=L\\hfill \\\\ 15=L\\hfill \\end{array}[\/latex]<\/div>\n<p>The dimensions are [latex]L=15[\/latex] ft and [latex]W=12[\/latex] ft.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Evaluating a variable for an expression<\/h3>\n<p>In the example above, the length was written in terms of the width in order to substitute it for the variable [latex]L[\/latex] to be able to write one equation in one variable.<\/p>\n<p>This is a powerful technique in algebra and should be practiced until it becomes familiar, so it&#8217;s worth taking a closer look.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q866900\">more<\/span><\/p>\n<div id=\"q866900\" class=\"hidden-answer\" style=\"display: none\">\n<p>We had [latex]P = 2L + 2W[\/latex], which is one equation in two variables. Given the value for [latex]P[\/latex], we don&#8217;t have enough information to solve for either [latex]L[\/latex] or [latex]W[\/latex].<\/p>\n<p>But we do know that [latex]L=W+3[\/latex]. We can use this to substitute for the [latex]L[\/latex] in the original equation. Be careful to retain the original statement exactly, substituting only the value for [latex]L[\/latex]. A handy way to handle the substitution is with a set of parentheses.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 54=2\\left(L\\right)+2W \\,\\,\\,\\,\\,\\,\\,\\, \\text{ wrap the } L \\text{ to see where to make the substitution} \\hfill \\\\ 54=2\\left(W+3\\right)+2W \\,\\,\\,\\,\\, \\text{ drop the expression for } L \\text{ into the parentheses}\\hfill \\\\ 54=2W+6+2W\\hfill \\\\ 54=4W+6\\end{array}[\/latex]<\/p>\n<p>Now we have one equation in one variable. We can use the properties of equality to isolate the variable [latex]W[\/latex] on one side, with its numerical value on the other.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the dimensions of a rectangle given that the perimeter is [latex]110[\/latex] cm. and the length is 1 cm. more than twice the width.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q447289\">Show Solution<\/span><\/p>\n<div id=\"q447289\" class=\"hidden-answer\" style=\"display: none\">\n<p>L = 37 cm, W = 18 cm<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm7679\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7679&theme=oea&iframe_resize_id=ohm7679&show_question_numbers\" width=\"100%\" height=\"240\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Area Problem<\/h3>\n<p>The perimeter of a tablet of graph paper is 48 in<sup>2<\/sup>. The length is [latex]6[\/latex] in. more than the width. Find the area of the graph paper.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q81698\">Show Solution<\/span><\/p>\n<div id=\"q81698\" class=\"hidden-answer\" style=\"display: none\">\n<p>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.<\/p>\n<p>We know that the length is 6 in. more than the width, so we can write length as [latex]L=W+6[\/latex]. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.<\/p>\n<div style=\"text-align: center;\">[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>\n<p>Now, we find the area given the dimensions of [latex]L=15[\/latex] in. and [latex]W=9[\/latex] in.<\/p>\n<div style=\"text-align: center;\">[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>\n<p>The area is [latex]135[\/latex] in<sup>2<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft<sup>2<\/sup> of new carpeting should be ordered?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q646790\">Show Solution<\/span><\/p>\n<div id=\"q646790\" class=\"hidden-answer\" style=\"display: none\">\n<p>250 ft<sup>2<\/sup><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm1688\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1688&theme=oea&iframe_resize_id=ohm1688&show_question_numbers\" width=\"100%\" height=\"190\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Volume Problem<\/h3>\n<p>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 1,600 in.<sup>3<\/sup>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q889305\">Show Solution<\/span><\/p>\n<div id=\"q889305\" class=\"hidden-answer\" style=\"display: none\">\n<p>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.<\/p>\n<div style=\"text-align: center;\">[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>\n<p>The dimensions are [latex]L=20[\/latex] in., [latex]W=10[\/latex] in., and [latex]H=8[\/latex] in.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Note 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.<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-100\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 52436. <strong>Authored by<\/strong>: Edward Wicks. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><li>Question ID 7679. <strong>Authored by<\/strong>: Tyler Wallace. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><li>Question ID 1688. <strong>Authored by<\/strong>: WebWork-Rochester. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":14,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et 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