{"id":1751,"date":"2023-10-12T00:32:05","date_gmt":"2023-10-12T00:32:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-models-and-applications\/"},"modified":"2025-11-19T16:54:17","modified_gmt":"2025-11-19T16:54:17","slug":"introduction-models-and-applications","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-models-and-applications\/","title":{"raw":"Models and Applications","rendered":"Models and Applications"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write a linear equation to express the relationship between unknown quantities.<\/li>\r\n \t<li>Write a linear equation that models two different cell phone packages.<\/li>\r\n \t<li>Use a linear model to answer questions.<\/li>\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\nJosh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200338\/CNX_CAT_Figure_02_03_001N.jpg\" alt=\"Many students studying in a large lecture hall \" width=\"488\" height=\"366\" \/> College students taking an exam. Credit: Kevin Dooley[\/caption]\r\n\r\nMany real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce <em>x <\/em>widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.\r\n<h2>Writing a Linear Equation to Solve an Application<\/h2>\r\nTo set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10\/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write [latex]0.10x[\/latex]. This expression represents a variable cost because it changes according to the number of miles driven.\r\n\r\nIf a quantity is independent of a variable, we usually just add or subtract it according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10\/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost [latex]C[\/latex].\r\n<div style=\"text-align: center;\">[latex]C=0.10x+50[\/latex]<\/div>\r\nWhen dealing with real-world applications, there are certain expressions that we can translate directly into math. The table\u00a0lists some common verbal expressions and their equivalent mathematical expressions.\r\n<table summary=\"A table with 8 rows and 2 columns. The entries in the first row are: Verbal and Translation to math operations. The entries in the second row are: One number exceeds another by a and x, x+a. The entries in the third row are: Twice a number and 2x. The entries in the fourth row are: One number is a more than another number and x, x plus a. The entries in the fifth row are: One number is a less than twice another number and x,2 times x minus a. The entries in the sixth row are: The product of a number and a, decreased by b and a times x minus b. The entries in the seventh row are: The quotient of a number and the number plus a is three times the number and x divided by the quantity x plus a equals three times x. The entries in the eighth row are: The product of three times a number and the number decreased by b is c and three times x times the quantity x minus b equals c.\">\r\n<thead>\r\n<tr>\r\n<th>Verbal<\/th>\r\n<th>Translation to Math Operations<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>One number exceeds another by <em>a<\/em><\/td>\r\n<td>[latex]x,\\text{ }x+a[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Twice a number<\/td>\r\n<td>[latex]2x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>One number is <em>a <\/em>more than another number<\/td>\r\n<td>[latex]x,\\text{ }x+a[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>One number is <em>a <\/em>less than twice another number<\/td>\r\n<td>[latex]x,2x-a[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The product of a number and <em>a<\/em>, decreased by <em>b<\/em><\/td>\r\n<td>[latex]ax-b[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The quotient of a number and the number plus <em>a <\/em>is three times the number<\/td>\r\n<td>[latex]\\frac{x}{x+a}=3x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The product of three times a number and the number decreased by <em>b <\/em>is <em>c<\/em><\/td>\r\n<td>[latex]3x\\left(x-b\\right)=c[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a real-world problem, model a linear equation to fit it<\/h3>\r\n<ol>\r\n \t<li>Identify known quantities.<\/li>\r\n \t<li>Assign a variable to represent the unknown quantity.<\/li>\r\n \t<li>If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.<\/li>\r\n \t<li>Write an equation interpreting the words as mathematical operations.<\/li>\r\n \t<li>Solve the equation. Be sure the solution can be explained in words including the units of measure.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Modeling a Linear Equation to Solve an Unknown Number Problem<\/h3>\r\nFind a linear equation to solve for the following unknown quantities: One number exceeds another number by [latex]17[\/latex] and their sum is [latex]31.[\/latex]\r\n\r\nFind the two numbers.\r\n[reveal-answer q=\"760873\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"760873\"]\r\n\r\nLet [latex]x[\/latex] equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as [latex]x+17[\/latex]. The sum of the two numbers is 31. We usually interpret the word <em>is<\/em> as an equal sign.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}x+\\left(x+17\\right)=31\\hfill \\\\ 2x+17=31\\hfill&amp;\\text{Simplify and solve}.\\hfill &amp; \\\\ 2x=14\\hfill \\\\ x=7\\hfill &amp; \\\\ \\hfill &amp; \\\\ x+17=7+17=24\\hfill \\end{array}[\/latex]<\/div>\r\nThe two numbers are [latex]7[\/latex] and [latex]24[\/latex].\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 a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is [latex]36[\/latex], find the numbers.\r\n[reveal-answer q=\"930268\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"930268\"]\r\n\r\n11 and 25[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7647&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=30987&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe>\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=13665&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Setting Up a Linear Equation to Solve a Real-World Application<\/h3>\r\nThere are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05\/min talk-time. Company B charges a monthly service fee of $40 plus $.04\/min talk-time.\r\n<ol>\r\n \t<li>Write a linear equation that models the packages offered by both companies.<\/li>\r\n \t<li>If the average number of minutes used each month is 1,160, which company offers the better plan?<\/li>\r\n \t<li>If the average number of minutes used each month is 420, which company offers the better plan?<\/li>\r\n \t<li>How many minutes of talk-time would yield equal monthly statements from both companies?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"785384\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"785384\"]\r\n<ol>\r\n \t<li>The model for Company <em>A<\/em> can be written as [latex]A=0.05x+34[\/latex]. This includes the variable cost of [latex]0.05x[\/latex] plus the monthly service charge of $34. Company <em>B<\/em>\u2019s package charges a higher monthly fee of $40, but a lower variable cost of [latex]0.04x[\/latex]. Company <em>B<\/em>\u2019s model can be written as [latex]B=0.04x+40[\/latex].<\/li>\r\n \t<li>If the average number of minutes used each month is 1,160, we have the following:\r\n<div>[latex]\\begin{array}{l}\\text{Company }A\\hfill&amp;=0.05\\left(1,160\\right)+34\\hfill \\\\ \\hfill&amp;=58+34\\hfill \\\\ \\hfill&amp;=92\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&amp;=0.04\\left(1,160\\right)+40\\hfill \\\\ \\hfill&amp;=46.4+40\\hfill \\\\ \\hfill&amp;=86.4\\hfill \\end{array}[\/latex]<\/div>\r\nSo, Company <em>B<\/em> offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company <em>A<\/em> when the average number of minutes used each month is 1,160.<\/li>\r\n \t<li>If the average number of minutes used each month is 420, we have the following:\r\n<div>[latex]\\begin{array}{l}\\text{Company }A\\hfill&amp;=0.05\\left(420\\right)+34\\hfill \\\\ \\hfill&amp;=21+34\\hfill \\\\ \\hfill&amp;=55\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&amp;=0.04\\left(420\\right)+40\\hfill \\\\ \\hfill&amp;=16.8+40\\hfill \\\\ \\hfill&amp;=56.8\\hfill \\end{array}[\/latex]<\/div>\r\nIf the average number of minutes used each month is 420, then Company <em>A <\/em>offers a lower monthly cost of $55 compared to Company <em>B<\/em>\u2019s monthly cost of $56.80.<\/li>\r\n \t<li>To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of [latex]\\left(x,y\\right)[\/latex] coordinates: At what point are both the <em>x-<\/em>value and the <em>y-<\/em>value equal? We can find this point by setting the equations equal to each other and solving for <em>x.<\/em>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}0.05x+34=0.04x+40\\hfill \\\\ 0.01x=6\\hfill \\\\ x=600\\hfill \\end{array}[\/latex]<\/div>\r\nCheck the <em>x-<\/em>value in each equation.\r\n<div style=\"text-align: left;\">[latex]\\begin{array}{l}0.05\\left(600\\right)+34=64\\hfill \\\\ 0.04\\left(600\\right)+40=64\\hfill \\end{array}[\/latex]<\/div>\r\nTherefore, a monthly average of 600 talk-time minutes renders the plans equal.<\/li>\r\n<\/ol>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200339\/CNX_CAT_Figure_02_03_002.jpg\" alt=\"Coordinate plane with the x-axis ranging from 0 to 1200 in intervals of 100 and the y-axis ranging from 0 to 90 in intervals of 10. The functions A = 0.05x + 34 and B = 0.04x + 40 are graphed on the same plot\" width=\"731\" height=\"420\" \/>\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 a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company\u2019s monthly expenses?\r\n[reveal-answer q=\"68149\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"68149\"][latex]\r\n\r\nC=2.5x+3,650[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=92426&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"500\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Using Formulas to Solve Problems<\/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. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the <strong>area<\/strong> of a rectangular region, [latex]A=LW[\/latex]; the <strong>perimeter<\/strong> of a rectangle, [latex]P=2L+2W[\/latex]; and the <strong>volume<\/strong> of a rectangular solid, [latex]V=LWH[\/latex]. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.\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 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<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=52436&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\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 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<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7679&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe>\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<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1688&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\"><\/iframe>\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>\r\n<div>\r\n<h2>Key Concepts<\/h2>\r\n<\/div>\r\n<ul>\r\n \t<li>A linear equation can be used to solve for an unknown in a number problem.<\/li>\r\n \t<li>Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities.<\/li>\r\n \t<li>There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the [latex]d=rt[\/latex] formula.<\/li>\r\n \t<li>Many geometry problems are solved using the perimeter formula [latex]P=2L+2W[\/latex], the area formula [latex]A=LW[\/latex], or the volume formula [latex]V=LWH[\/latex].<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165131990658\" class=\"definition\">\r\n \t<dt><strong>area<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165131990661\">in square units, the area formula used in this section is used to find the area of any two-dimensional rectangular region: [latex]A=LW[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165132943522\" class=\"definition\">\r\n \t<dt><strong>perimeter<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165132943525\">in linear units, the perimeter formula is used to find the linear measurement, or outside length and width, around a two-dimensional regular object; for a rectangle: [latex]P=2L+2W[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165132943528\" class=\"definition\">\r\n \t<dt><strong>volume<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165134297639\">in cubic units, the volume measurement includes length, width, and depth: [latex]V=LWH[\/latex]<\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write a linear equation to express the relationship between unknown quantities.<\/li>\n<li>Write a linear equation that models two different cell phone packages.<\/li>\n<li>Use a linear model to answer questions.<\/li>\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>Josh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer.<\/p>\n<div style=\"width: 498px\" class=\"wp-caption aligncenter\"><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\/25200338\/CNX_CAT_Figure_02_03_001N.jpg\" alt=\"Many students studying in a large lecture hall\" width=\"488\" height=\"366\" \/><\/p>\n<p class=\"wp-caption-text\">College students taking an exam. Credit: Kevin Dooley<\/p>\n<\/div>\n<p>Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce <em>x <\/em>widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.<\/p>\n<h2>Writing a Linear Equation to Solve an Application<\/h2>\n<p>To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10\/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write [latex]0.10x[\/latex]. This expression represents a variable cost because it changes according to the number of miles driven.<\/p>\n<p>If a quantity is independent of a variable, we usually just add or subtract it according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10\/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost [latex]C[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]C=0.10x+50[\/latex]<\/div>\n<p>When dealing with real-world applications, there are certain expressions that we can translate directly into math. The table\u00a0lists some common verbal expressions and their equivalent mathematical expressions.<\/p>\n<table summary=\"A table with 8 rows and 2 columns. The entries in the first row are: Verbal and Translation to math operations. The entries in the second row are: One number exceeds another by a and x, x+a. The entries in the third row are: Twice a number and 2x. The entries in the fourth row are: One number is a more than another number and x, x plus a. The entries in the fifth row are: One number is a less than twice another number and x,2 times x minus a. The entries in the sixth row are: The product of a number and a, decreased by b and a times x minus b. The entries in the seventh row are: The quotient of a number and the number plus a is three times the number and x divided by the quantity x plus a equals three times x. The entries in the eighth row are: The product of three times a number and the number decreased by b is c and three times x times the quantity x minus b equals c.\">\n<thead>\n<tr>\n<th>Verbal<\/th>\n<th>Translation to Math Operations<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>One number exceeds another by <em>a<\/em><\/td>\n<td>[latex]x,\\text{ }x+a[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Twice a number<\/td>\n<td>[latex]2x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>One number is <em>a <\/em>more than another number<\/td>\n<td>[latex]x,\\text{ }x+a[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>One number is <em>a <\/em>less than twice another number<\/td>\n<td>[latex]x,2x-a[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The product of a number and <em>a<\/em>, decreased by <em>b<\/em><\/td>\n<td>[latex]ax-b[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The quotient of a number and the number plus <em>a <\/em>is three times the number<\/td>\n<td>[latex]\\frac{x}{x+a}=3x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The product of three times a number and the number decreased by <em>b <\/em>is <em>c<\/em><\/td>\n<td>[latex]3x\\left(x-b\\right)=c[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox\">\n<h3>How To: Given a real-world problem, model a linear equation to fit it<\/h3>\n<ol>\n<li>Identify known quantities.<\/li>\n<li>Assign a variable to represent the unknown quantity.<\/li>\n<li>If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.<\/li>\n<li>Write an equation interpreting the words as mathematical operations.<\/li>\n<li>Solve the equation. Be sure the solution can be explained in words including the units of measure.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Modeling a Linear Equation to Solve an Unknown Number Problem<\/h3>\n<p>Find a linear equation to solve for the following unknown quantities: One number exceeds another number by [latex]17[\/latex] and their sum is [latex]31.[\/latex]<\/p>\n<p>Find the two numbers.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q760873\">Show Solution<\/span><\/p>\n<div id=\"q760873\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]x[\/latex] equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as [latex]x+17[\/latex]. The sum of the two numbers is 31. We usually interpret the word <em>is<\/em> as an equal sign.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}x+\\left(x+17\\right)=31\\hfill \\\\ 2x+17=31\\hfill&\\text{Simplify and solve}.\\hfill & \\\\ 2x=14\\hfill \\\\ x=7\\hfill & \\\\ \\hfill & \\\\ x+17=7+17=24\\hfill \\end{array}[\/latex]<\/div>\n<p>The two numbers are [latex]7[\/latex] and [latex]24[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is [latex]36[\/latex], find the numbers.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q930268\">Show Solution<\/span><\/p>\n<div id=\"q930268\" class=\"hidden-answer\" style=\"display: none\">\n<p>11 and 25<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7647&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=30987&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=13665&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Setting Up a Linear Equation to Solve a Real-World Application<\/h3>\n<p>There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05\/min talk-time. Company B charges a monthly service fee of $40 plus $.04\/min talk-time.<\/p>\n<ol>\n<li>Write a linear equation that models the packages offered by both companies.<\/li>\n<li>If the average number of minutes used each month is 1,160, which company offers the better plan?<\/li>\n<li>If the average number of minutes used each month is 420, which company offers the better plan?<\/li>\n<li>How many minutes of talk-time would yield equal monthly statements from both companies?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q785384\">Show Solution<\/span><\/p>\n<div id=\"q785384\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The model for Company <em>A<\/em> can be written as [latex]A=0.05x+34[\/latex]. This includes the variable cost of [latex]0.05x[\/latex] plus the monthly service charge of $34. Company <em>B<\/em>\u2019s package charges a higher monthly fee of $40, but a lower variable cost of [latex]0.04x[\/latex]. Company <em>B<\/em>\u2019s model can be written as [latex]B=0.04x+40[\/latex].<\/li>\n<li>If the average number of minutes used each month is 1,160, we have the following:\n<div>[latex]\\begin{array}{l}\\text{Company }A\\hfill&=0.05\\left(1,160\\right)+34\\hfill \\\\ \\hfill&=58+34\\hfill \\\\ \\hfill&=92\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&=0.04\\left(1,160\\right)+40\\hfill \\\\ \\hfill&=46.4+40\\hfill \\\\ \\hfill&=86.4\\hfill \\end{array}[\/latex]<\/div>\n<p>So, Company <em>B<\/em> offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company <em>A<\/em> when the average number of minutes used each month is 1,160.<\/li>\n<li>If the average number of minutes used each month is 420, we have the following:\n<div>[latex]\\begin{array}{l}\\text{Company }A\\hfill&=0.05\\left(420\\right)+34\\hfill \\\\ \\hfill&=21+34\\hfill \\\\ \\hfill&=55\\hfill \\\\ \\hfill \\\\ \\text{Company }B\\hfill&=0.04\\left(420\\right)+40\\hfill \\\\ \\hfill&=16.8+40\\hfill \\\\ \\hfill&=56.8\\hfill \\end{array}[\/latex]<\/div>\n<p>If the average number of minutes used each month is 420, then Company <em>A <\/em>offers a lower monthly cost of $55 compared to Company <em>B<\/em>\u2019s monthly cost of $56.80.<\/li>\n<li>To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of [latex]\\left(x,y\\right)[\/latex] coordinates: At what point are both the <em>x-<\/em>value and the <em>y-<\/em>value equal? We can find this point by setting the equations equal to each other and solving for <em>x.<\/em>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}0.05x+34=0.04x+40\\hfill \\\\ 0.01x=6\\hfill \\\\ x=600\\hfill \\end{array}[\/latex]<\/div>\n<p>Check the <em>x-<\/em>value in each equation.<\/p>\n<div style=\"text-align: left;\">[latex]\\begin{array}{l}0.05\\left(600\\right)+34=64\\hfill \\\\ 0.04\\left(600\\right)+40=64\\hfill \\end{array}[\/latex]<\/div>\n<p>Therefore, a monthly average of 600 talk-time minutes renders the plans equal.<\/li>\n<\/ol>\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\/25200339\/CNX_CAT_Figure_02_03_002.jpg\" alt=\"Coordinate plane with the x-axis ranging from 0 to 1200 in intervals of 100 and the y-axis ranging from 0 to 90 in intervals of 10. The functions A = 0.05x + 34 and B = 0.04x + 40 are graphed on the same plot\" width=\"731\" height=\"420\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company\u2019s monthly expenses?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q68149\">Show Solution<\/span><\/p>\n<div id=\"q68149\" class=\"hidden-answer\" style=\"display: none\">[latex]C=2.5x+3,650[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=92426&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"500\"><\/iframe><\/p>\n<\/div>\n<h2>Using Formulas to Solve Problems<\/h2>\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. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the <strong>area<\/strong> of a rectangular region, [latex]A=LW[\/latex]; the <strong>perimeter<\/strong> of a rectangle, [latex]P=2L+2W[\/latex]; and the <strong>volume<\/strong> of a rectangular solid, [latex]V=LWH[\/latex]. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.<\/p>\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 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=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=52436&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/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 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=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7679&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/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=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1688&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\"><\/iframe><\/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<div>\n<h2>Key Concepts<\/h2>\n<\/div>\n<ul>\n<li>A linear equation can be used to solve for an unknown in a number problem.<\/li>\n<li>Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities.<\/li>\n<li>There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the [latex]d=rt[\/latex] formula.<\/li>\n<li>Many geometry problems are solved using the perimeter formula [latex]P=2L+2W[\/latex], the area formula [latex]A=LW[\/latex], or the volume formula [latex]V=LWH[\/latex].<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165131990658\" class=\"definition\">\n<dt><strong>area<\/strong><\/dt>\n<dd id=\"fs-id1165131990661\">in square units, the area formula used in this section is used to find the area of any two-dimensional rectangular region: [latex]A=LW[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165132943522\" class=\"definition\">\n<dt><strong>perimeter<\/strong><\/dt>\n<dd id=\"fs-id1165132943525\">in linear units, the perimeter formula is used to find the linear measurement, or outside length and width, around a two-dimensional regular object; for a rectangle: [latex]P=2L+2W[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165132943528\" class=\"definition\">\n<dt><strong>volume<\/strong><\/dt>\n<dd id=\"fs-id1165134297639\">in cubic units, the volume measurement includes length, width, and depth: [latex]V=LWH[\/latex]<\/dd>\n<\/dl>\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-1751\">\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 7647, 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 30987, 13665. <strong>Authored by<\/strong>: James Sousa. <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 92426. <strong>Authored by<\/strong>: Michael Jenck. <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":708740,"menu_order":12,"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 al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 52436\",\"author\":\"Edward Wicks\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 7647, 7679\",\"author\":\"Tyler Wallace\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC- BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 30987, 13665\",\"author\":\"James Sousa\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC- BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 92426\",\"author\":\"Michael Jenck\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 1688\",\"author\":\"WebWork-Rochester\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1751","chapter","type-chapter","status-publish","hentry"],"part":1736,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1751","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/users\/708740"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1751\/revisions"}],"predecessor-version":[{"id":2565,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1751\/revisions\/2565"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/parts\/1736"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1751\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/media?parent=1751"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1751"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/contributor?post=1751"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/license?post=1751"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}