{"id":2329,"date":"2024-07-22T22:29:50","date_gmt":"2024-07-22T22:29:50","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=2329"},"modified":"2025-08-15T15:42:24","modified_gmt":"2025-08-15T15:42:24","slug":"systems-of-linear-equations-two-variables-apply-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/systems-of-linear-equations-two-variables-apply-it-1\/","title":{"raw":"Systems of Linear Equations: Two Variables: Apply It 1","rendered":"Systems of Linear Equations: Two Variables: Apply It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find solutions to systems of equations by drawing their graphs and finding where they cross<\/li>\r\n \t<li>Solve systems of equations by replacing one variable with an expression from another equation<\/li>\r\n \t<li>Solve systems of equations by adding equations to eliminate a variable<\/li>\r\n \t<li>Figure out when systems of equations have no solution or infinitely many solutions<\/li>\r\n<\/ul>\r\n<\/section>\r\n<div class=\"page\" title=\"Page 822\">\r\n<div class=\"layoutArea\">\r\n<div class=\"column\">\r\n<h2>Writing and Solving a System of Equations in Two Variables<\/h2>\r\nNow that we've learned all about systems of equations, it's time to put that knowledge to use! Sometimes, we need to develop our own equations to solve real-world problems.\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Follow these steps to create and solve your own system of equations:\r\n<ul>\r\n \t<li><strong>Identify the Situation: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Think about a real-world scenario where you need to find two unknown quantities. This could be anything from planning a party to budgeting for groceries.<\/span><\/li>\r\n \t<li><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Pick Your Variables: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Choose two variables to represent the unknown quantities. Let\u2019s call them [latex]x[\/latex] and [latex]y[\/latex]. Note: You don't always need to pick [latex]x[\/latex] and [latex]y[\/latex].<\/span><\/li>\r\n \t<li><strong>Create Your Equations: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Write two equations based on the scenario. Each equation should represent a different relationship between [latex]x[\/latex] and [latex]y[\/latex].<\/span><\/li>\r\n \t<li><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Choose a Method to Solve: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Decide which method you want to use to solve the system:<\/span>\r\n<ul>\r\n \t<li><strong>Substitution<\/strong>: Solve one equation for one variable, then substitute that into the other equation.<\/li>\r\n \t<li><strong>Elimination<\/strong>: Add or subtract the equations to eliminate one variable.<\/li>\r\n \t<li><strong>Graphing<\/strong>: Draw both equations on a graph and see where they intersect.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Solve the System: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Use your chosen method to find the values of [latex]x[\/latex] and [latex]y[\/latex] <\/span><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">that make both equations true.<\/span><\/li>\r\n \t<li><strong>Interpret the Solution: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Look at the values you found. What do they mean in the context of your scenario? If there is no solution or infinitely many solutions, explain why that happens.<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h3>Using Systems of Equations to Investigate Profits<\/h3>\r\nUsing what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer\u2019s <strong>revenue function<\/strong> is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[\/latex], where [latex]x=[\/latex] quantity and [latex]p=[\/latex] price. The revenue function is shown in orange in the graph below.\r\n\r\nThe <strong>cost function<\/strong> is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The [latex]x[\/latex] -axis represents quantity in hundreds of units. The [latex]y[\/latex]-axis represents either cost or revenue in hundreds of dollars.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183617\/CNX_Precalc_Figure_09_01_0092.jpg\" alt=\"A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis. A line representing cost and a line representing revenue cross at the point (7,33), which is marked break-even. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"488\" height=\"347\" \/> A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis[\/caption]\r\n\r\nThe point at which the two lines intersect is called the <strong>break-even point<\/strong>. We can see from the graph that if [latex]700[\/latex] units are produced, the cost is [latex]$3,300[\/latex] and the revenue is also [latex]$3,300[\/latex]. In other words, the company breaks even if they produce and sell [latex]700[\/latex] units. They neither make money nor lose money.\r\n\r\nThe shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The <strong>profit function<\/strong> is the revenue function minus the cost function, written as [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Given the cost function [latex]C\\left(x\\right)=0.85x+35{,}000[\/latex] and the revenue function [latex]R\\left(x\\right)=1.55x[\/latex], find the break-even point and the profit function.[reveal-answer q=\"569292\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"569292\"]Write the system of equations using [latex]y[\/latex] to replace function notation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} y&amp;=0.85x+35{,}000 \\\\ y&amp;=1.55x \\end{align}[\/latex]<\/p>\r\nSubstitute the expression [latex]0.85x+35{,}000[\/latex] from the first equation into the second equation and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}0.85x+35{,}000=1.55x\\\\ 35{,}000=0.7x\\\\ 50{,}000=x\\end{gathered}[\/latex]<\/p>\r\nThen, we substitute [latex]x=50{,}000[\/latex] into either the cost function or the revenue function.\r\n<p style=\"text-align: center;\">[latex]1.55\\left(50{,}000\\right)=77{,}500[\/latex]<\/p>\r\nThe break-even point is [latex]\\left(50{,}000,77{,}500\\right)[\/latex].\r\n\r\nThe profit function is found using the formula [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}P\\left(x\\right)&amp;=1.55x-\\left(0.85x+35{,}000\\right) \\\\ &amp;=0.7x - 35{,}000 \\end{align}[\/latex]<\/p>\r\nThe profit function is [latex]P\\left(x\\right)=0.7x - 35{,}000[\/latex].\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"400\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183619\/CNX_Precalc_Figure_09_01_0102.jpg\" alt=\"A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"400\" height=\"320\" \/> A graph showing money in dollars on the y axis and quantity on the x axis, C(x), and R(x)[\/caption]\r\n\r\n<strong>Analysis of the Solution<\/strong>\r\n\r\nThe cost to produce [latex]50,000[\/latex] units is [latex]$77,500[\/latex], and the revenue from the sales of [latex]50,000[\/latex] units is also [latex]$77,500[\/latex]. To make a profit, the business must produce and sell more than [latex]50,000[\/latex] units.\r\n\r\nWe see from the graph below that the profit function has a negative value until [latex]x=50{,}000[\/latex], when the graph crosses the [latex]x[\/latex]-axis. Then, the graph emerges into positive [latex]y[\/latex]-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is [latex]0[\/latex]. The area to the left of the break-even point represents operating at a loss.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"398\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183621\/CNX_Precalc_Figure_09_01_0112.jpg\" alt=\"A graph showing dollars profit on the y axis and quantity on the x axis. The profit line crosses the break-even point at fifty thousand, zero. The profit line's equation is P(x)=0.7x-35,000.\" width=\"398\" height=\"276\" \/> A graph showing dollars profit on the y axis and quantity on the x axis and P(x)[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24805[\/ohm2_question]<\/section><section aria-label=\"Try It\">\r\n<h3>Using Systems of Equations in Application<\/h3>\r\n<section class=\"textbox example\" aria-label=\"Example\">The cost of a ticket to the circus is [latex]$25.00[\/latex] for children and [latex]$50.00[\/latex] for adults. On a certain day, attendance at the circus is [latex]2,000[\/latex] and the total gate revenue is [latex]$70,000[\/latex]. How many children and how many adults bought tickets?[reveal-answer q=\"455809\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455809\"]Let [latex]c[\/latex] = the number of children and [latex]a[\/latex] = the number of adults in attendance.The total number of people is [latex]2,000[\/latex]. We can use this to write an equation for the number of people at the circus that day.\r\n<p style=\"text-align: center;\">[latex]c+a=2{,}000[\/latex]<\/p>\r\nThe revenue from all children can be found by multiplying [latex]$25.00[\/latex] by the number of children, [latex]25c[\/latex]. The revenue from all adults can be found by multiplying [latex]$50.00[\/latex] by the number of adults, [latex]50a[\/latex]. The total revenue is [latex]$70,000[\/latex]. We can use this to write an equation for the revenue.\r\n<p style=\"text-align: center;\">[latex]25c+50a=70{,}000[\/latex]<\/p>\r\nWe now have a system of linear equations in two variables.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}c+a=2,000\\\\ 25c+50a=70{,}000\\end{gathered}[\/latex]<\/p>\r\nIn the first equation, the coefficient of both variables is [latex]1[\/latex]. We can quickly solve the first equation for either [latex]c[\/latex] or [latex]a[\/latex]. We will solve for [latex]a[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}c+a=2{,}000\\\\ a=2{,}000-c\\end{gathered}[\/latex]<\/p>\r\nSubstitute the expression [latex]2{,}000-c[\/latex] in the second equation for [latex]a[\/latex] and solve for [latex]c[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 25c+50\\left(2{,}000-c\\right)&amp;=70{,}000 \\\\ 25c+100{,}000 - 50c&amp;=70{,}000 \\\\ -25c&amp;=-30{,}000 \\\\ c&amp;=1{,}200 \\end{align}[\/latex]<\/p>\r\nSubstitute [latex]c=1{,}200[\/latex] into the first equation to solve for [latex]a[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}1{,}200+a&amp;=2{,}000 \\\\ a&amp;=800 \\end{align}[\/latex]<\/p>\r\n<strong>We find that [latex]1,200[\/latex] children and [latex]800[\/latex] adults bought tickets to the circus that day.<\/strong>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of [latex]$20[\/latex], then [latex]59[\/latex] cents a mile. The second, Move It Your Way, charges an up-front fee of [latex]$16[\/latex], then [latex]63[\/latex] cents a mile.[footnote]Rates retrieved Aug 2, 2010 from http:\/\/www.budgettruck.com and http:\/\/www.uhaul.com\/[\/footnote] When will Keep on Trucking, Inc. be the better choice for Jamal?[reveal-answer q=\"869152\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"869152\"]The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.\r\n<table summary=\"Three rows and three columns. In the first column, are the years 1950 and 2000. In the second columns are the house values for Indiana, which are 37700 for 1950 and 94300 for 2000. In the third columns are the house values for Alabama, which are 27100 for 1950 and 85100 for 2000.\">\r\n<tbody>\r\n<tr>\r\n<td>Input<\/td>\r\n<td>[latex]d[\/latex], distance driven in miles<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Outputs<\/td>\r\n<td>[latex]K(d)[\/latex]: cost, in dollars, for renting from Keep on Trucking\r\n[latex]M(d)[\/latex]: cost, in dollars, for renting from Move It Your Way<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Initial Value<\/td>\r\n<td>Up-front fee: [latex]K(0) = 20[\/latex] and [latex]M(0) = 16[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rate of Change<\/td>\r\n<td>[latex]K(d) = $0.59[\/latex]\/mile and [latex]P(d) = $0.63[\/latex]\/mile<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nA linear function is of the form [latex]f\\left(x\\right)=mx+b[\/latex]. Using the rates of change and initial charges, we can write the equations\r\n<p style=\"text-align: center;\">[latex]\\begin{align}K\\left(d\\right)=0.59d+20\\\\ M\\left(d\\right)=0.63d+16\\end{align}[\/latex]<\/p>\r\nUsing these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\\left(d\\right)&lt;M\\left(d\\right)[\/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\\left(d\\right)[\/latex] function is smaller.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19011929\/CNX_Precalc_Figure_02_03_0072.jpg\" alt=\"\" width=\"731\" height=\"340\" \/> A graph of K(d), M(d) and their intersection[\/caption]\r\n\r\nThese graphs are sketched above, with <em>K<\/em>(<em>d<\/em>)\u00a0in blue.\r\n\r\nTo find the intersection, we set the equations equal and solve:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}K\\left(d\\right)&amp;=M\\left(d\\right) \\\\ 0.59d+20&amp;=0.63d+16 \\\\ 4&amp;=0.04d \\\\ 100&amp;=d \\\\ d&amp;=100 \\end{align}[\/latex]<\/p>\r\nThis tells us that the cost from the two companies will be the same if [latex]100[\/latex] miles are driven. Either by looking at the graph, or noting that [latex]K\\left(d\\right)[\/latex]\u00a0is growing at a slower rate, we can conclude that <strong>Keep on Trucking, Inc. will be the cheaper price when more than [latex]100[\/latex] miles are driven<\/strong>, that is [latex]d&gt;100[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">A chemist has [latex]70[\/latex]mL of a [latex]50\\%[\/latex] methane solution. How much of a [latex]80\\%[\/latex] solution must she add so the final solution is [latex]60\\%[\/latex] methane?\r\n[reveal-answer q=\"350379\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"350379\"]We will use the following table to help us solve this mixture problem:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>Amount<\/td>\r\n<td>Part<\/td>\r\n<td>Total<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Start<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Final<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe start with [latex]70[\/latex]mL of solution, and the unknown amount can be [latex]x[\/latex]. The part is the percentages, or concentration of solution [latex]0.5[\/latex] for start, [latex]0.8[\/latex] for add.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>Amount<\/td>\r\n<td>Part<\/td>\r\n<td>Total<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Start<\/td>\r\n<td>[latex]70[\/latex]mL<\/td>\r\n<td>[latex]0.5[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add<\/td>\r\n<td>[latex]x[\/latex]<\/td>\r\n<td>[latex]0.8[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Final<\/td>\r\n<td>[latex]70+x[\/latex]<\/td>\r\n<td>[latex]0.6[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAdd amount column to get final amount. The part for this amount is 0.6 because we want the final solution to be 60% methane.\r\n<table style=\"height: 55px;\">\r\n<tbody>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"height: 15px;\"><\/td>\r\n<td style=\"height: 15px;\">Amount<\/td>\r\n<td style=\"height: 15px;\">Part<\/td>\r\n<td style=\"height: 15px;\">Total<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"height: 15px;\">Start<\/td>\r\n<td style=\"height: 15px;\">[latex]70[\/latex]mL<\/td>\r\n<td style=\"height: 15px;\">[latex]0.5[\/latex]<\/td>\r\n<td style=\"height: 15px;\">[latex]35[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 10px;\">\r\n<td style=\"height: 10px;\">Add<\/td>\r\n<td style=\"height: 10px;\">[latex]x[\/latex]<\/td>\r\n<td style=\"height: 10px;\">[latex]0.8[\/latex]<\/td>\r\n<td style=\"height: 10px;\">[latex]0.8x[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"height: 15px;\">Final<\/td>\r\n<td style=\"height: 15px;\">[latex]70+x[\/latex]<\/td>\r\n<td style=\"height: 15px;\">[latex]0.6[\/latex]<\/td>\r\n<td style=\"height: 15px;\">\u00a0[latex]42+0.6x[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nMultiply amount by part to get total. be sure to distribute on the last row:[latex](70 + x)0.6[\/latex].\r\n\r\nIf we add the start and add entries in the Total column, we get the final equation that represents the total amount and it's concentration.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}35+0.8x&amp; = 42+0.6x \\\\ 0.2x&amp;=7 \\\\ \\frac{0.2}{0.2}x&amp;=\\frac{7}{0.2} \\\\ x&amp;=35 \\end{align}[\/latex]<\/p>\r\n<strong>[latex]35[\/latex]mL of [latex]80\\%[\/latex] solution must be added to [latex]70[\/latex]mL of [latex]50\\%[\/latex] solution to get a [latex]60\\%[\/latex] solution of Methane.<\/strong>\r\n\r\nThe same process can be used if the starting and final amount have a price attached to them, rather than a percentage.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24806[\/ohm2_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24807[\/ohm2_question]<\/section><\/section><\/div>\r\n<\/div>\r\n<\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find solutions to systems of equations by drawing their graphs and finding where they cross<\/li>\n<li>Solve systems of equations by replacing one variable with an expression from another equation<\/li>\n<li>Solve systems of equations by adding equations to eliminate a variable<\/li>\n<li>Figure out when systems of equations have no solution or infinitely many solutions<\/li>\n<\/ul>\n<\/section>\n<div class=\"page\" title=\"Page 822\">\n<div class=\"layoutArea\">\n<div class=\"column\">\n<h2>Writing and Solving a System of Equations in Two Variables<\/h2>\n<p>Now that we&#8217;ve learned all about systems of equations, it&#8217;s time to put that knowledge to use! Sometimes, we need to develop our own equations to solve real-world problems.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Follow these steps to create and solve your own system of equations:<\/p>\n<ul>\n<li><strong>Identify the Situation: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Think about a real-world scenario where you need to find two unknown quantities. This could be anything from planning a party to budgeting for groceries.<\/span><\/li>\n<li><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Pick Your Variables: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Choose two variables to represent the unknown quantities. Let\u2019s call them [latex]x[\/latex] and [latex]y[\/latex]. Note: You don&#8217;t always need to pick [latex]x[\/latex] and [latex]y[\/latex].<\/span><\/li>\n<li><strong>Create Your Equations: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Write two equations based on the scenario. Each equation should represent a different relationship between [latex]x[\/latex] and [latex]y[\/latex].<\/span><\/li>\n<li><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Choose a Method to Solve: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Decide which method you want to use to solve the system:<\/span>\n<ul>\n<li><strong>Substitution<\/strong>: Solve one equation for one variable, then substitute that into the other equation.<\/li>\n<li><strong>Elimination<\/strong>: Add or subtract the equations to eliminate one variable.<\/li>\n<li><strong>Graphing<\/strong>: Draw both equations on a graph and see where they intersect.<\/li>\n<\/ul>\n<\/li>\n<li><strong style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Solve the System: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Use your chosen method to find the values of [latex]x[\/latex] and [latex]y[\/latex] <\/span><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">that make both equations true.<\/span><\/li>\n<li><strong>Interpret the Solution: <\/strong><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">Look at the values you found. What do they mean in the context of your scenario? If there is no solution or infinitely many solutions, explain why that happens.<\/span><\/li>\n<\/ul>\n<\/section>\n<h3>Using Systems of Equations to Investigate Profits<\/h3>\n<p>Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer\u2019s <strong>revenue function<\/strong> is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[\/latex], where [latex]x=[\/latex] quantity and [latex]p=[\/latex] price. The revenue function is shown in orange in the graph below.<\/p>\n<p>The <strong>cost function<\/strong> is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The [latex]x[\/latex] -axis represents quantity in hundreds of units. The [latex]y[\/latex]-axis represents either cost or revenue in hundreds of dollars.<\/p>\n<figure style=\"width: 488px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183617\/CNX_Precalc_Figure_09_01_0092.jpg\" alt=\"A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis. A line representing cost and a line representing revenue cross at the point (7,33), which is marked break-even. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"488\" height=\"347\" \/><figcaption class=\"wp-caption-text\">A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis<\/figcaption><\/figure>\n<p>The point at which the two lines intersect is called the <strong>break-even point<\/strong>. We can see from the graph that if [latex]700[\/latex] units are produced, the cost is [latex]$3,300[\/latex] and the revenue is also [latex]$3,300[\/latex]. In other words, the company breaks even if they produce and sell [latex]700[\/latex] units. They neither make money nor lose money.<\/p>\n<p>The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The <strong>profit function<\/strong> is the revenue function minus the cost function, written as [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Given the cost function [latex]C\\left(x\\right)=0.85x+35{,}000[\/latex] and the revenue function [latex]R\\left(x\\right)=1.55x[\/latex], find the break-even point and the profit function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q569292\">Show Solution<\/button><\/p>\n<div id=\"q569292\" class=\"hidden-answer\" style=\"display: none\">Write the system of equations using [latex]y[\/latex] to replace function notation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} y&=0.85x+35{,}000 \\\\ y&=1.55x \\end{align}[\/latex]<\/p>\n<p>Substitute the expression [latex]0.85x+35{,}000[\/latex] from the first equation into the second equation and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}0.85x+35{,}000=1.55x\\\\ 35{,}000=0.7x\\\\ 50{,}000=x\\end{gathered}[\/latex]<\/p>\n<p>Then, we substitute [latex]x=50{,}000[\/latex] into either the cost function or the revenue function.<\/p>\n<p style=\"text-align: center;\">[latex]1.55\\left(50{,}000\\right)=77{,}500[\/latex]<\/p>\n<p>The break-even point is [latex]\\left(50{,}000,77{,}500\\right)[\/latex].<\/p>\n<p>The profit function is found using the formula [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}P\\left(x\\right)&=1.55x-\\left(0.85x+35{,}000\\right) \\\\ &=0.7x - 35{,}000 \\end{align}[\/latex]<\/p>\n<p>The profit function is [latex]P\\left(x\\right)=0.7x - 35{,}000[\/latex].<\/p>\n<figure style=\"width: 400px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183619\/CNX_Precalc_Figure_09_01_0102.jpg\" alt=\"A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"400\" height=\"320\" \/><figcaption class=\"wp-caption-text\">A graph showing money in dollars on the y axis and quantity on the x axis, C(x), and R(x)<\/figcaption><\/figure>\n<p><strong>Analysis of the Solution<\/strong><\/p>\n<p>The cost to produce [latex]50,000[\/latex] units is [latex]$77,500[\/latex], and the revenue from the sales of [latex]50,000[\/latex] units is also [latex]$77,500[\/latex]. To make a profit, the business must produce and sell more than [latex]50,000[\/latex] units.<\/p>\n<p>We see from the graph below that the profit function has a negative value until [latex]x=50{,}000[\/latex], when the graph crosses the [latex]x[\/latex]-axis. Then, the graph emerges into positive [latex]y[\/latex]-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is [latex]0[\/latex]. The area to the left of the break-even point represents operating at a loss.<\/p>\n<figure style=\"width: 398px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183621\/CNX_Precalc_Figure_09_01_0112.jpg\" alt=\"A graph showing dollars profit on the y axis and quantity on the x axis. The profit line crosses the break-even point at fifty thousand, zero. The profit line's equation is P(x)=0.7x-35,000.\" width=\"398\" height=\"276\" \/><figcaption class=\"wp-caption-text\">A graph showing dollars profit on the y axis and quantity on the x axis and P(x)<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24805\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24805&theme=lumen&iframe_resize_id=ohm24805&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section aria-label=\"Try It\">\n<h3>Using Systems of Equations in Application<\/h3>\n<section class=\"textbox example\" aria-label=\"Example\">The cost of a ticket to the circus is [latex]$25.00[\/latex] for children and [latex]$50.00[\/latex] for adults. On a certain day, attendance at the circus is [latex]2,000[\/latex] and the total gate revenue is [latex]$70,000[\/latex]. How many children and how many adults bought tickets?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q455809\">Show Solution<\/button><\/p>\n<div id=\"q455809\" class=\"hidden-answer\" style=\"display: none\">Let [latex]c[\/latex] = the number of children and [latex]a[\/latex] = the number of adults in attendance.The total number of people is [latex]2,000[\/latex]. We can use this to write an equation for the number of people at the circus that day.<\/p>\n<p style=\"text-align: center;\">[latex]c+a=2{,}000[\/latex]<\/p>\n<p>The revenue from all children can be found by multiplying [latex]$25.00[\/latex] by the number of children, [latex]25c[\/latex]. The revenue from all adults can be found by multiplying [latex]$50.00[\/latex] by the number of adults, [latex]50a[\/latex]. The total revenue is [latex]$70,000[\/latex]. We can use this to write an equation for the revenue.<\/p>\n<p style=\"text-align: center;\">[latex]25c+50a=70{,}000[\/latex]<\/p>\n<p>We now have a system of linear equations in two variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}c+a=2,000\\\\ 25c+50a=70{,}000\\end{gathered}[\/latex]<\/p>\n<p>In the first equation, the coefficient of both variables is [latex]1[\/latex]. We can quickly solve the first equation for either [latex]c[\/latex] or [latex]a[\/latex]. We will solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}c+a=2{,}000\\\\ a=2{,}000-c\\end{gathered}[\/latex]<\/p>\n<p>Substitute the expression [latex]2{,}000-c[\/latex] in the second equation for [latex]a[\/latex] and solve for [latex]c[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 25c+50\\left(2{,}000-c\\right)&=70{,}000 \\\\ 25c+100{,}000 - 50c&=70{,}000 \\\\ -25c&=-30{,}000 \\\\ c&=1{,}200 \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]c=1{,}200[\/latex] into the first equation to solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}1{,}200+a&=2{,}000 \\\\ a&=800 \\end{align}[\/latex]<\/p>\n<p><strong>We find that [latex]1,200[\/latex] children and [latex]800[\/latex] adults bought tickets to the circus that day.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of [latex]$20[\/latex], then [latex]59[\/latex] cents a mile. The second, Move It Your Way, charges an up-front fee of [latex]$16[\/latex], then [latex]63[\/latex] cents a mile.<a class=\"footnote\" title=\"Rates retrieved Aug 2, 2010 from http:\/\/www.budgettruck.com and http:\/\/www.uhaul.com\/\" id=\"return-footnote-2329-1\" href=\"#footnote-2329-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> When will Keep on Trucking, Inc. be the better choice for Jamal?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q869152\">Show Solution<\/button><\/p>\n<div id=\"q869152\" class=\"hidden-answer\" style=\"display: none\">The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.<\/p>\n<table summary=\"Three rows and three columns. In the first column, are the years 1950 and 2000. In the second columns are the house values for Indiana, which are 37700 for 1950 and 94300 for 2000. In the third columns are the house values for Alabama, which are 27100 for 1950 and 85100 for 2000.\">\n<tbody>\n<tr>\n<td>Input<\/td>\n<td>[latex]d[\/latex], distance driven in miles<\/td>\n<\/tr>\n<tr>\n<td>Outputs<\/td>\n<td>[latex]K(d)[\/latex]: cost, in dollars, for renting from Keep on Trucking<br \/>\n[latex]M(d)[\/latex]: cost, in dollars, for renting from Move It Your Way<\/td>\n<\/tr>\n<tr>\n<td>Initial Value<\/td>\n<td>Up-front fee: [latex]K(0) = 20[\/latex] and [latex]M(0) = 16[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Rate of Change<\/td>\n<td>[latex]K(d) = $0.59[\/latex]\/mile and [latex]P(d) = $0.63[\/latex]\/mile<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>A linear function is of the form [latex]f\\left(x\\right)=mx+b[\/latex]. Using the rates of change and initial charges, we can write the equations<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}K\\left(d\\right)=0.59d+20\\\\ M\\left(d\\right)=0.63d+16\\end{align}[\/latex]<\/p>\n<p>Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\\left(d\\right)<M\\left(d\\right)[\/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\\left(d\\right)[\/latex] function is smaller.\n\n\n\n<figure style=\"width: 731px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19011929\/CNX_Precalc_Figure_02_03_0072.jpg\" alt=\"\" width=\"731\" height=\"340\" \/><figcaption class=\"wp-caption-text\">A graph of K(d), M(d) and their intersection<\/figcaption><\/figure>\n<p>These graphs are sketched above, with <em>K<\/em>(<em>d<\/em>)\u00a0in blue.<\/p>\n<p>To find the intersection, we set the equations equal and solve:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}K\\left(d\\right)&=M\\left(d\\right) \\\\ 0.59d+20&=0.63d+16 \\\\ 4&=0.04d \\\\ 100&=d \\\\ d&=100 \\end{align}[\/latex]<\/p>\n<p>This tells us that the cost from the two companies will be the same if [latex]100[\/latex] miles are driven. Either by looking at the graph, or noting that [latex]K\\left(d\\right)[\/latex]\u00a0is growing at a slower rate, we can conclude that <strong>Keep on Trucking, Inc. will be the cheaper price when more than [latex]100[\/latex] miles are driven<\/strong>, that is [latex]d>100[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A chemist has [latex]70[\/latex]mL of a [latex]50\\%[\/latex] methane solution. How much of a [latex]80\\%[\/latex] solution must she add so the final solution is [latex]60\\%[\/latex] methane?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q350379\">Show Solution<\/button><\/p>\n<div id=\"q350379\" class=\"hidden-answer\" style=\"display: none\">We will use the following table to help us solve this mixture problem:<\/p>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>Amount<\/td>\n<td>Part<\/td>\n<td>Total<\/td>\n<\/tr>\n<tr>\n<td>Start<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Add<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Final<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We start with [latex]70[\/latex]mL of solution, and the unknown amount can be [latex]x[\/latex]. The part is the percentages, or concentration of solution [latex]0.5[\/latex] for start, [latex]0.8[\/latex] for add.<\/p>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>Amount<\/td>\n<td>Part<\/td>\n<td>Total<\/td>\n<\/tr>\n<tr>\n<td>Start<\/td>\n<td>[latex]70[\/latex]mL<\/td>\n<td>[latex]0.5[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Add<\/td>\n<td>[latex]x[\/latex]<\/td>\n<td>[latex]0.8[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Final<\/td>\n<td>[latex]70+x[\/latex]<\/td>\n<td>[latex]0.6[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Add amount column to get final amount. The part for this amount is 0.6 because we want the final solution to be 60% methane.<\/p>\n<table style=\"height: 55px;\">\n<tbody>\n<tr style=\"height: 15px;\">\n<td style=\"height: 15px;\"><\/td>\n<td style=\"height: 15px;\">Amount<\/td>\n<td style=\"height: 15px;\">Part<\/td>\n<td style=\"height: 15px;\">Total<\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"height: 15px;\">Start<\/td>\n<td style=\"height: 15px;\">[latex]70[\/latex]mL<\/td>\n<td style=\"height: 15px;\">[latex]0.5[\/latex]<\/td>\n<td style=\"height: 15px;\">[latex]35[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 10px;\">\n<td style=\"height: 10px;\">Add<\/td>\n<td style=\"height: 10px;\">[latex]x[\/latex]<\/td>\n<td style=\"height: 10px;\">[latex]0.8[\/latex]<\/td>\n<td style=\"height: 10px;\">[latex]0.8x[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"height: 15px;\">Final<\/td>\n<td style=\"height: 15px;\">[latex]70+x[\/latex]<\/td>\n<td style=\"height: 15px;\">[latex]0.6[\/latex]<\/td>\n<td style=\"height: 15px;\">\u00a0[latex]42+0.6x[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Multiply amount by part to get total. be sure to distribute on the last row:[latex](70 + x)0.6[\/latex].<\/p>\n<p>If we add the start and add entries in the Total column, we get the final equation that represents the total amount and it&#8217;s concentration.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}35+0.8x& = 42+0.6x \\\\ 0.2x&=7 \\\\ \\frac{0.2}{0.2}x&=\\frac{7}{0.2} \\\\ x&=35 \\end{align}[\/latex]<\/p>\n<p><strong>[latex]35[\/latex]mL of [latex]80\\%[\/latex] solution must be added to [latex]70[\/latex]mL of [latex]50\\%[\/latex] solution to get a [latex]60\\%[\/latex] solution of Methane.<\/strong><\/p>\n<p>The same process can be used if the starting and final amount have a price attached to them, rather than a percentage.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24806\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24806&theme=lumen&iframe_resize_id=ohm24806&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24807\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24807&theme=lumen&iframe_resize_id=ohm24807&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-2329-1\">Rates retrieved Aug 2, 2010 from http:\/\/www.budgettruck.com and http:\/\/www.uhaul.com\/ <a href=\"#return-footnote-2329-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":12,"menu_order":11,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":300,"module-header":"apply_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2329"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2329\/revisions"}],"predecessor-version":[{"id":7854,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2329\/revisions\/7854"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/300"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2329\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=2329"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2329"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=2329"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=2329"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}