{"id":370,"date":"2016-06-01T20:49:49","date_gmt":"2016-06-01T20:49:49","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=370"},"modified":"2023-11-08T13:18:38","modified_gmt":"2023-11-08T13:18:38","slug":"applications_of_linear_equations","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/chapter\/applications_of_linear_equations\/","title":{"raw":"1.5 - Applications of Linear Equations","rendered":"1.5 &#8211; Applications of Linear Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>(1.5.1) - Set up a linear equation to solve an application\r\n<ul>\r\n \t<li>Translate words into algebraic expressions and equations<\/li>\r\n \t<li>Solve an application using a formula<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>(1.5.2) - Solve\u00a0distance, rate, and time problems<\/li>\r\n \t<li>(1.5.3) - Solve area and perimeter problems<\/li>\r\n \t<li>(1.5.4) - Rearrange formulas to isolate specific variables<\/li>\r\n \t<li>(1.5.5) - Solve temperature conversion problems<\/li>\r\n<\/ul>\r\n<\/div>\r\nMany real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus an additional charge if you exceed your data plan;\u00a0a car rental company charges a daily fee plus an amount per mile driven; Chipotle offers a base price for your burrito plus additional charges for extra toppings, like guacamole and sour cream. 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<h1>(1.5.1) - Set Up a Linear Equation to Solve an Application<\/h1>\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 use an example of a car rental company. The company charges $0.10\/mi in addition to a flat rate. 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\n<h3>Translate words into algebraic expressions and equations<\/h3>\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<\/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]. Find the two numbers.\r\n\r\n[reveal-answer q=\"460075\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"460075\"]\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}{l}x+\\left(x+17\\right)\\hfill&amp;=31\\hfill \\\\ 2x+17\\hfill&amp;=31\\hfill&amp;\\text{Simplify and solve}.\\hfill \\\\ 2x\\hfill&amp;=14\\hfill \\\\ x\\hfill&amp;=7\\hfill \\\\ \\hfill \\\\ x+17\\hfill&amp;=7+17\\hfill \\\\ \\hfill&amp;=24\\hfill \\end{array}[\/latex]<\/div>\r\n<div>The two numbers are [latex]7[\/latex] and [latex]24[\/latex].<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of how to translate an expression in english into a mathematical equation that can then be solved.\r\nhttps:\/\/youtu.be\/CGC_SwHfwMk\r\nIn the next example we will write equations that will help us compare cell phone plans.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/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=\"14868\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"14868\"]\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>[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>[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[caption id=\"\" align=\"aligncenter\" width=\"731\"]<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\" \/> <b>Figure 2<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows another example of writing two equations that will allow you to compare two different cell phone plans.\r\nhttps:\/\/youtu.be\/Q5hlC_VPKGM\r\n<h3>Solve an Application Using a Formula<\/h3>\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<h1>(1.5.2) - Solve\u00a0distance, rate, and time problems<\/h1>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIt takes Andrew 30 min to drive to work in the morning. He drives home using the same route, but it takes 10 min longer, and he averages 10 mi\/h less than in the morning. How far does Andrew drive to work?\r\n[reveal-answer q=\"61824\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"61824\"]\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 style=\"text-align: center\">[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}{l}r\\left(\\frac{1}{2}\\right)\\hfill&amp;=\\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill \\\\ \\frac{1}{2}r\\hfill&amp;=\\frac{2}{3}r-\\frac{20}{3}\\hfill \\\\ \\frac{1}{2}r-\\frac{2}{3}r\\hfill&amp;=-\\frac{20}{3}\\hfill \\\\ -\\frac{1}{6}r\\hfill&amp;=-\\frac{20}{3}\\hfill \\\\ r\\hfill&amp;=-\\frac{20}{3}\\left(-6\\right)\\hfill \\\\ r\\hfill&amp;=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.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].\r\n<div style=\"text-align: center\">\r\n\r\n[latex]\\begin{array}{l}r\\left(\\frac{1}{2}\\right)\\hfill&amp;=\\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill \\\\ 6\\times r\\left(\\frac{1}{2}\\right)\\hfill&amp; =6\\times \\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill \\\\ 3r\\hfill&amp; =4\\left(r - 10\\right)\\hfill \\\\ 3r\\hfill&amp; =4r - 40\\hfill \\\\ -r\\hfill&amp; =-40\\hfill \\\\ r\\hfill&amp; =40\\hfill \\end{array}[\/latex]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h1>(1.5.3) - Solve area and perimeter problems<\/h1>\r\nIn the next example, we will find the lengths of a rectangular field given it's perimeter and a relationship between it's side lengths.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/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=\"918170\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"918170\"]\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.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<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\" \/> <b>Figure 3<\/b>[\/caption]\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.[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows an example of finding the dimensions of a rectangular field given it's perimeter.\r\nhttps:\/\/youtu.be\/VyK-HQr02iQ\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe perimeter of a tablet of graph paper is 48 in. The length is [latex]6[\/latex] in. more than the width. Find the area of the graph paper.\r\n\r\n[reveal-answer q=\"859423\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"859423\"]\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\\hfill&amp;=2L+2W\\hfill \\\\ 48\\hfill&amp;=2\\left(W+6\\right)+2W\\hfill \\\\ 48\\hfill&amp;=2W+12+2W\\hfill \\\\ 48\\hfill&amp;=4W+12\\hfill \\\\ 36\\hfill&amp;=4W\\hfill \\\\ 9\\hfill&amp;=W\\hfill \\\\ \\left(9+6\\right)\\hfill&amp;=L\\hfill \\\\ 15\\hfill&amp;=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]\r\nThe area is [latex]135[\/latex] in<sup>2<\/sup>.[\/hidden-answer]<\/div>\r\n<\/div>\r\nThe following video shows another example of finding the area\u00a0of a rectangle given it's perimeter and the relationship between it's side lengths.\r\n\r\nhttps:\/\/youtu.be\/zUlU64Umnq4\r\n<h1>(1.5.4) - Rearrange formulas to isolate specific variables<\/h1>\r\nSometimes, it is easier to isolate the variable you you are solving for when you are using a formula. This is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly. In the next examples, we will use algebraic properties to isolate a variable in a formula.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIsolate the term containing the variable, <i>w, <\/i>from the formula for the perimeter of a rectangle<em>: \u00a0<\/em>\r\n<p style=\"text-align: center\">[latex]{P}=2\\left({L}\\right)+2\\left({W}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"967601\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"967601\"]\r\n\r\nFirst, isolate the term with\u00a0<i>w<\/i> by subtracting 2<em>l<\/em> from both sides of the equation.\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,p\\,=\\,\\,\\,\\,2l+2w\\\\\\underline{\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,}\\\\p-2l=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2w\\end{array}[\/latex]<i><\/i><\/p>\r\nNext, clear the coefficient of <i>w <\/i>by dividing both sides of the equation by 2.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{l}\\underline{p-2l}=\\underline{2w}\\\\\\,\\,\\,\\,\\,\\,2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\\\ \\,\\,\\,\\frac{p-2l}{2}\\,\\,=\\,\\,w\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\frac{p-2l}{2}\\end{array}[\/latex]<\/p>\r\nYou can rewrite the equation so the isolated variable is on the left side.\r\n<p style=\"text-align: center\">[latex]w=\\frac{p-2l}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse the multiplication and division properties of equality to isolate the variable <em>b<\/em>\u00a0given [latex]A=\\frac{1}{2}bh[\/latex]\r\n\r\n[reveal-answer q=\"291790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"291790\"]\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,A=\\frac{1}{2}bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\frac{1}{2}bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{h}=\\frac{bh}{h}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{h}=\\frac{b\\cancel{h}}{\\cancel{h}}\\end{array}[\/latex]<\/p>\r\nWrite the equation with the desired variable on the left-hand side as a matter of convention:\r\n<p style=\"text-align: center\">[latex]b=\\frac{2A}{h}[\/latex]\r\n[\/hidden-answer]<\/p>\r\nUse the multiplication and division properties of equality to isolate the variable <em>h\u00a0<\/em>given [latex]A=\\frac{1}{2}bh[\/latex]\r\n[reveal-answer q=\"595790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"595790\"]\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,A=\\frac{1}{2}bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\frac{1}{2}bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{b}=\\frac{bh}{b}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{b}=\\frac{h\\cancel{b}}{\\cancel{b}}\\end{array}[\/latex]<\/p>\r\nWrite the equation with the desired variable on the left-hand side as a matter of convention:\r\n<p style=\"text-align: center\">[latex]h=\\frac{2A}{b}[\/latex]\r\n[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<h1>(1.5.5) - Solve temperature conversion problems<\/h1>\r\nLet\u2019s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.\r\n<p style=\"text-align: center\">[latex]C=\\left(F--32\\right)\\cdot \\frac{5}{9}[\/latex]<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGiven a temperature of [latex]12^{\\circ}{C}[\/latex], find the equivalent in [latex]{}^{\\circ}{F}[\/latex].\r\n[reveal-answer q=\"594254\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"594254\"]\r\n\r\nSubstitute the given temperature in[latex]{}^{\\circ}{C}[\/latex]\u00a0into the conversion formula:\r\n<p style=\"text-align: center\">[latex]12=\\left(F-32\\right)\\cdot \\frac{5}{9}[\/latex]<\/p>\r\nIsolate the variable F to obtain the equivalent temperature.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}12=\\left(F-32\\right)\\cdot \\frac{5}{9}\\\\\\\\\\left(\\frac{9}{5}\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\left(\\frac{108}{5}\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\21.6=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{+32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+32}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\53.6={}^{\\circ}{F}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAs with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.\r\n\r\n[latex]C=\\left(F--32\\right)\\cdot \\frac{5}{9}[\/latex]\r\n\r\n[reveal-answer q=\"591790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"591790\"]\r\n\r\nTo isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex] \\displaystyle \\frac{9}{5}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\\\\\,\\,\\,\\,\\left(\\frac{9}{5}\\right)C=\\left(F-32\\right)\\left(\\frac{5}{9}\\right)\\left(\\frac{9}{5}\\right)\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{9}{5}C=F-32\\end{array}[\/latex]<\/p>\r\nAdd 32 to both sides.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\frac{9}{5}\\,C+32=F-32+32\\\\\\\\\\frac{9}{5}\\,C+32=F\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]F=\\frac{9}{5}C+32[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2><\/h2>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>(1.5.1) &#8211; Set up a linear equation to solve an application\n<ul>\n<li>Translate words into algebraic expressions and equations<\/li>\n<li>Solve an application using a formula<\/li>\n<\/ul>\n<\/li>\n<li>(1.5.2) &#8211; Solve\u00a0distance, rate, and time problems<\/li>\n<li>(1.5.3) &#8211; Solve area and perimeter problems<\/li>\n<li>(1.5.4) &#8211; Rearrange formulas to isolate specific variables<\/li>\n<li>(1.5.5) &#8211; Solve temperature conversion problems<\/li>\n<\/ul>\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 an additional charge if you exceed your data plan;\u00a0a car rental company charges a daily fee plus an amount per mile driven; Chipotle offers a base price for your burrito plus additional charges for extra toppings, like guacamole and sour cream. 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<h1>(1.5.1) &#8211; Set Up a Linear Equation to Solve an Application<\/h1>\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 use an example of a car rental company. The company charges $0.10\/mi in addition to a flat rate. 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<h3>Translate words into algebraic expressions and equations<\/h3>\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<\/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]. Find the two numbers.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q460075\">Show Solution<\/span><\/p>\n<div id=\"q460075\" 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}{l}x+\\left(x+17\\right)\\hfill&=31\\hfill \\\\ 2x+17\\hfill&=31\\hfill&\\text{Simplify and solve}.\\hfill \\\\ 2x\\hfill&=14\\hfill \\\\ x\\hfill&=7\\hfill \\\\ \\hfill \\\\ x+17\\hfill&=7+17\\hfill \\\\ \\hfill&=24\\hfill \\end{array}[\/latex]<\/div>\n<div>The two numbers are [latex]7[\/latex] and [latex]24[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show another example of how to translate an expression in english into a mathematical equation that can then be solved.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-1\" title=\"Write and Solve  Linear Equation - Number Problem with Given Sum\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/CGC_SwHfwMk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\nIn the next example we will write equations that will help us compare cell phone plans.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/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=\"q14868\">Show Solution<\/span><\/p>\n<div id=\"q14868\" 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>[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>[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<div style=\"width: 741px\" 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\/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<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows another example of writing two equations that will allow you to compare two different cell phone plans.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-2\" title=\"Write Linear Equations to Model and Compare Cell Phone Plans with Data Usage\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Q5hlC_VPKGM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Solve an Application Using a Formula<\/h3>\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<h1>(1.5.2) &#8211; Solve\u00a0distance, rate, and time problems<\/h1>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>It takes Andrew 30 min to drive to work in the morning. He drives home using the same route, but it takes 10 min 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=\"q61824\">Show Solution<\/span><\/p>\n<div id=\"q61824\" 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 style=\"text-align: center\">[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}{l}r\\left(\\frac{1}{2}\\right)\\hfill&=\\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill \\\\ \\frac{1}{2}r\\hfill&=\\frac{2}{3}r-\\frac{20}{3}\\hfill \\\\ \\frac{1}{2}r-\\frac{2}{3}r\\hfill&=-\\frac{20}{3}\\hfill \\\\ -\\frac{1}{6}r\\hfill&=-\\frac{20}{3}\\hfill \\\\ r\\hfill&=-\\frac{20}{3}\\left(-6\\right)\\hfill \\\\ r\\hfill&=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.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\">\n<p>[latex]\\begin{array}{l}r\\left(\\frac{1}{2}\\right)\\hfill&=\\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill \\\\ 6\\times r\\left(\\frac{1}{2}\\right)\\hfill& =6\\times \\left(r - 10\\right)\\left(\\frac{2}{3}\\right)\\hfill \\\\ 3r\\hfill& =4\\left(r - 10\\right)\\hfill \\\\ 3r\\hfill& =4r - 40\\hfill \\\\ -r\\hfill& =-40\\hfill \\\\ r\\hfill& =40\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h1>(1.5.3) &#8211; Solve area and perimeter problems<\/h1>\n<p>In the next example, we will find the lengths of a rectangular field given it&#8217;s perimeter and a relationship between it&#8217;s side lengths.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/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=\"q918170\">Show Solution<\/span><\/p>\n<div id=\"q918170\" 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.<\/p>\n<div style=\"width: 497px\" 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\/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 class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\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><\/div>\n<\/div>\n<\/div>\n<p>The following video shows an example of finding the dimensions of a rectangular field given it&#8217;s perimeter.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  Find the Dimensions and Area of a Field Given the Perimeter\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/VyK-HQr02iQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The perimeter of a tablet of graph paper is 48 in. 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=\"q859423\">Show Solution<\/span><\/p>\n<div id=\"q859423\" 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\\hfill&=2L+2W\\hfill \\\\ 48\\hfill&=2\\left(W+6\\right)+2W\\hfill \\\\ 48\\hfill&=2W+12+2W\\hfill \\\\ 48\\hfill&=4W+12\\hfill \\\\ 36\\hfill&=4W\\hfill \\\\ 9\\hfill&=W\\hfill \\\\ \\left(9+6\\right)\\hfill&=L\\hfill \\\\ 15\\hfill&=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]<br \/>\nThe area is [latex]135[\/latex] in<sup>2<\/sup>.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows another example of finding the area\u00a0of a rectangle given it&#8217;s perimeter and the relationship between it&#8217;s side lengths.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex: Find the Area of a Rectangle Given the Perimeter\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/zUlU64Umnq4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(1.5.4) &#8211; Rearrange formulas to isolate specific variables<\/h1>\n<p>Sometimes, it is easier to isolate the variable you you are solving for when you are using a formula. This is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly. In the next examples, we will use algebraic properties to isolate a variable in a formula.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Isolate the term containing the variable, <i>w, <\/i>from the formula for the perimeter of a rectangle<em>: \u00a0<\/em><\/p>\n<p style=\"text-align: center\">[latex]{P}=2\\left({L}\\right)+2\\left({W}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q967601\">Show Solution<\/span><\/p>\n<div id=\"q967601\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, isolate the term with\u00a0<i>w<\/i> by subtracting 2<em>l<\/em> from both sides of the equation.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,p\\,=\\,\\,\\,\\,2l+2w\\\\\\underline{\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,}\\\\p-2l=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2w\\end{array}[\/latex]<i><\/i><\/p>\n<p>Next, clear the coefficient of <i>w <\/i>by dividing both sides of the equation by 2.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{l}\\underline{p-2l}=\\underline{2w}\\\\\\,\\,\\,\\,\\,\\,2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\\\ \\,\\,\\,\\frac{p-2l}{2}\\,\\,=\\,\\,w\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\frac{p-2l}{2}\\end{array}[\/latex]<\/p>\n<p>You can rewrite the equation so the isolated variable is on the left side.<\/p>\n<p style=\"text-align: center\">[latex]w=\\frac{p-2l}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use the multiplication and division properties of equality to isolate the variable <em>b<\/em>\u00a0given [latex]A=\\frac{1}{2}bh[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q291790\">Show Solution<\/span><\/p>\n<div id=\"q291790\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,A=\\frac{1}{2}bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\frac{1}{2}bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{h}=\\frac{bh}{h}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{h}=\\frac{b\\cancel{h}}{\\cancel{h}}\\end{array}[\/latex]<\/p>\n<p>Write the equation with the desired variable on the left-hand side as a matter of convention:<\/p>\n<p style=\"text-align: center\">[latex]b=\\frac{2A}{h}[\/latex]\n<\/div>\n<\/div>\n<p>Use the multiplication and division properties of equality to isolate the variable <em>h\u00a0<\/em>given [latex]A=\\frac{1}{2}bh[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q595790\">Show Solution<\/span><\/p>\n<div id=\"q595790\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,A=\\frac{1}{2}bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\frac{1}{2}bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{b}=\\frac{bh}{b}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{b}=\\frac{h\\cancel{b}}{\\cancel{b}}\\end{array}[\/latex]<\/p>\n<p>Write the equation with the desired variable on the left-hand side as a matter of convention:<\/p>\n<p style=\"text-align: center\">[latex]h=\\frac{2A}{b}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<h1>(1.5.5) &#8211; Solve temperature conversion problems<\/h1>\n<p>Let\u2019s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.<\/p>\n<p style=\"text-align: center\">[latex]C=\\left(F--32\\right)\\cdot \\frac{5}{9}[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Given a temperature of [latex]12^{\\circ}{C}[\/latex], find the equivalent in [latex]{}^{\\circ}{F}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q594254\">Show Solution<\/span><\/p>\n<div id=\"q594254\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the given temperature in[latex]{}^{\\circ}{C}[\/latex]\u00a0into the conversion formula:<\/p>\n<p style=\"text-align: center\">[latex]12=\\left(F-32\\right)\\cdot \\frac{5}{9}[\/latex]<\/p>\n<p>Isolate the variable F to obtain the equivalent temperature.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}12=\\left(F-32\\right)\\cdot \\frac{5}{9}\\\\\\\\\\left(\\frac{9}{5}\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\left(\\frac{108}{5}\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\21.6=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{+32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+32}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\53.6={}^{\\circ}{F}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>As with the other formulas we have worked with, we could have isolated the variable F first, then substituted in the given temperature in Celsius.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.<\/p>\n<p>[latex]C=\\left(F--32\\right)\\cdot \\frac{5}{9}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q591790\">Show Solution<\/span><\/p>\n<div id=\"q591790\" class=\"hidden-answer\" style=\"display: none\">\n<p>To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex]\\displaystyle \\frac{9}{5}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\\\\\,\\,\\,\\,\\left(\\frac{9}{5}\\right)C=\\left(F-32\\right)\\left(\\frac{5}{9}\\right)\\left(\\frac{9}{5}\\right)\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{9}{5}C=F-32\\end{array}[\/latex]<\/p>\n<p>Add 32 to both sides.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\frac{9}{5}\\,C+32=F-32+32\\\\\\\\\\frac{9}{5}\\,C+32=F\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]F=\\frac{9}{5}C+32[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h2><\/h2>\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-370\">\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>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@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/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@3.278:1\/Preface<\/li><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><li>Write and Solve Linear Equation - Number Problem with Given Sum. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/CGC_SwHfwMk\">https:\/\/youtu.be\/CGC_SwHfwMk<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Write Linear Equations to Model and Compare Cell Phone Plans with Data Usage. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Provided by<\/strong>: https:\/\/youtu.be\/Q5hlC_VPKGM. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><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@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/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@3.278:1\/Preface<\/li><li>Ex: Find the Dimensions and Area of a Field Given the Perimeter. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/VyK-HQr02iQ\">https:\/\/youtu.be\/VyK-HQr02iQ<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Find the Area of a Rectangle Given the Perimeter. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/zUlU64Umnq4\">https:\/\/youtu.be\/zUlU64Umnq4<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Find the Volume of a Right Circular Cylinder Formed from a Given Rectangle. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/KS1pGO_g3vM\">https:\/\/youtu.be\/KS1pGO_g3vM<\/a>. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Write and Solve Linear Equation - 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