{"id":308,"date":"2023-06-21T13:22:54","date_gmt":"2023-06-21T13:22:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/make-sense-of-the-solution-to-a-system-of-linear-equation\/"},"modified":"2023-07-31T17:21:15","modified_gmt":"2023-07-31T17:21:15","slug":"make-sense-of-the-solution-to-a-system-of-linear-equation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/make-sense-of-the-solution-to-a-system-of-linear-equation\/","title":{"raw":"\u25aa   Classify Solutions to Systems","rendered":"\u25aa   Classify Solutions to Systems"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine whether a solution reveals that a system has one, many, or no solutions<\/li>\r\n \t<li>Express the general solution of a dependent system<\/li>\r\n \t<li>Interpret the solution to a system of equations that represents profits for a business<\/li>\r\n \t<li>Write a system of equations that models a given situation<\/li>\r\n<\/ul>\r\n<\/div>\r\nNow that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an <strong>inconsistent system<\/strong> consists of parallel lines that have the same slope but different [latex]y[\/latex] -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as [latex]12=0[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Inconsistent System of Equations<\/h3>\r\nSolve the following system of equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}&amp;x=9 - 2y \\\\ &amp;x+2y=13 \\end{gathered}[\/latex]<\/p>\r\n[reveal-answer q=\"888134\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"888134\"]\r\n\r\nWe can approach this problem in two ways. Because one equation is already solved for [latex]x[\/latex], the most obvious step is to use substitution.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&amp;=13 \\\\ \\left(9 - 2y\\right)+2y&amp;=13 \\\\ 9+0y&amp;=13 \\\\ 9&amp;=13 \\end{align}[\/latex]<\/p>\r\nClearly, this statement is a contradiction because [latex]9\\ne 13[\/latex]. Therefore, the system has no solution.\r\n\r\nThe second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=9 - 2y \\\\ 2y=-x+9 \\\\ y=-\\frac{1}{2}x+\\frac{9}{2} \\end{gathered}[\/latex]<\/p>\r\nWe then convert the second equation expressed to slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+2y=13 \\\\ 2y=-x+13 \\\\ y=-\\frac{1}{2}x+\\frac{13}{2} \\end{gathered}[\/latex]<\/p>\r\nComparing the equations, we see that they have the same slope but different <em>y<\/em>-intercepts. Therefore, the lines are parallel and do not intersect.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y=-\\frac{1}{2}x+\\frac{9}{2} \\\\ y=-\\frac{1}{2}x+\\frac{13}{2} \\end{gathered}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWriting the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183613\/CNX_Precalc_Figure_09_01_0072.jpg\" alt=\"A graph of two parallel lines. The first line's equation is y equals negative one-half x plus 13 over two. The second line's equation is y equals negative one-half x plus 9 over two.\" width=\"488\" height=\"297\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video, see another example of using substitution to solve a system that has no solution.\r\n\r\nhttps:\/\/youtu.be\/kTtKfh5gFUc\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the following system of equations in two variables.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2y - 2x=2\\\\ 2y - 2x=6\\end{gathered}[\/latex]<\/p>\r\n[reveal-answer q=\"681787\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"681787\"]\r\n\r\nNo solution. It is an inconsistent system.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[ohm_question height=\"220\"]29699[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Expressing the Solution of a System of Dependent Equations Containing Two Variables<\/h2>\r\nRecall that a <strong>dependent system<\/strong> of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as [latex]0=0[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nWriting a general solution for a dependent system takes practice to feel natural. Work through the following example on paper, then apply the information in Writing the General Solution below.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Solution to a Dependent System of Linear Equations<\/h3>\r\nFind a solution to the system of equations using the <strong>addition method<\/strong>.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+3y=2\\\\ 3x+9y=6\\end{gathered}[\/latex]<\/p>\r\n[reveal-answer q=\"390828\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"390828\"]\r\n\r\nWith the addition method, we want to eliminate one of the variables by adding the equations. In this case, let\u2019s focus on eliminating [latex]x[\/latex]. If we multiply both sides of the first equation by [latex]-3[\/latex], then we will be able to eliminate the [latex]x[\/latex] -variable.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+3y&amp;=2 \\\\ \\left(-3\\right)\\left(x+3y\\right)&amp;=\\left(-3\\right)\\left(2\\right) \\\\ -3x - 9y&amp;=-6 \\end{align}[\/latex]<\/p>\r\nNow add the equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \u22123x\u22129y&amp;=\u22126 \\\\ +3x+9y&amp;=6 \\\\ \\hline 0&amp;=0 \\end{align}[\/latex]<\/p>\r\nWe can see that there will be an infinite number of solutions that satisfy both equations.\r\n<h4>Analysis of the Solution<\/h4>\r\nIf we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let\u2019s look at what happens when we convert the system to slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\begin{gathered}x+3y=2 \\\\ 3y=-x+2 \\\\ y=-\\frac{1}{3}x+\\frac{2}{3} \\end{gathered} \\hspace{2cm} \\begin{gathered} 3x+9y=6 \\\\9y=-3x+6 \\\\ y=-\\frac{3}{9}x+\\frac{6}{9} \\\\ y=-\\frac{1}{3}x+\\frac{2}{3} \\end{gathered}\\end{align}[\/latex]<\/p>\r\nLook at the graph below. Notice the results are the same. The general solution to the system is [latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183615\/CNX_Precalc_Figure_09_01_0082.jpg\" alt=\"A graph of two lines that overlap each other. The first line's equation is x+3y=2. The second line's equation is 3x-9y=6.\" width=\"487\" height=\"366\" \/>\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Writing the general solution<\/h3>\r\nIn the previous example, we presented an analysis of the solution to the following system of equations:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+3y=2\\\\ 3x+9y=6\\end{gathered}[\/latex]<\/p>\r\nAfter a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as\u00a0[latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex]. Why would we write the solution this way? In some ways, this representation tells us a lot. \u00a0It tells us that <em>x<\/em> can be anything, <em>x<\/em> is <em>x<\/em>. \u00a0It also tells us that <em>y<\/em> is going to depend on <em>x<\/em>, just like when we write a function rule. \u00a0In this case, depending on what you put in for <em>x<\/em>, <em>y<\/em> will be defined in terms of <em>x<\/em> as [latex]-\\frac{1}{3}x+\\frac{2}{3}[\/latex].\r\n\r\nIn other words, there are infinitely many (<em>x<\/em>,<em>y<\/em>) pairs that will satisfy this system of equations, and they all fall on the line\u00a0[latex]f(x)-\\frac{1}{3}x+\\frac{2}{3}[\/latex].\r\n\r\n<\/div>\r\nWatch the next video for an example of a dependent system.\r\n\r\nhttps:\/\/youtu.be\/Pcqb109yK5Q\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the following system of equations in two variables.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y - 2x=5 \\\\ -3y+6x=-15 \\end{gathered}[\/latex]<\/p>\r\n[reveal-answer q=\"218404\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"218404\"]\r\n\r\nThe system is dependent so there are infinitely many solutions of the form [latex]\\left(x,2x+5\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[ohm_question height=\"290\"]15665[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse an online graphing calculator to write three different systems:\r\n<ul>\r\n \t<li>A system of equations with one solution<\/li>\r\n \t<li>A system of equations with no solutions<\/li>\r\n \t<li>A system of equations with infinitely many solutions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2><strong>Using Systems of Equations to Investigate Profits<\/strong><\/h2>\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 <em>y<\/em>-axis represents either cost or revenue in hundreds of dollars.\r\n\r\n<img class=\"aligncenter\" 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\" \/>\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 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 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<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Break-Even Point and the Profit Function Using Substitution<\/h3>\r\nGiven 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.\r\n\r\n[reveal-answer q=\"569292\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"569292\"]\r\n\r\nWrite 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<h4>Analysis of the Solution<\/h4>\r\nThe cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.\r\n\r\n<img class=\"aligncenter\" 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=\"487\" height=\"390\" \/>\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 <em>x<\/em>-axis. Then, the graph emerges into positive <em>y<\/em>-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 0. The area to the left of the break-even point represents operating at a loss.\r\n\r\n<img class=\"aligncenter\" 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=\"731\" height=\"507\" \/>\u00a0[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will see an example of how to find the break-even point for a small sno-cone business.\r\nhttps:\/\/youtu.be\/qey3FmE8saQ\r\n<h2>Writing a System of Linear Equations Given a Situation<\/h2>\r\nIt is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.<\/h3>\r\n<ol>\r\n \t<li>Identify the input and output of each linear model.<\/li>\r\n \t<li>Identify the slope and <em>y<\/em>-intercept of each linear model.<\/li>\r\n \t<li>Find the solution by setting the two linear functions equal to another and solving for <em>x<\/em>, or find the point of intersection on a graph.<\/li>\r\n<\/ol>\r\n<\/div>\r\nNow let's practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nWork through the following examples on paper perhaps more than once to be sure you understand how the situations can be translated into a mathematical model. They use techniques you used earlier in the course to write models for real-world situations.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing and Solving a System of Equations in Two Variables<\/h3>\r\nThe cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?\r\n\r\n[reveal-answer q=\"455809\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455809\"]\r\n\r\nLet <em>c<\/em> = the number of children and <em>a<\/em> = the number of adults in attendance.\r\n\r\nThe total number of people is 2,000. 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 $25.00 by the number of children, [latex]25c[\/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[\/latex]. The total revenue is $70,000. 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 1. 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\nWe find that 1,200 children and 800 adults bought tickets to the circus that day.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nSee the video below for another example of translating a given situation into a system of two linear equations.\r\n\r\nhttps:\/\/youtu.be\/euh9ksWrq0A\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nMeal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets?\r\n\r\n[reveal-answer q=\"454145\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"454145\"]\r\n\r\n700 children, 950 adults\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[ohm_question height=\"245\"]23774[\/ohm_question]\r\n\r\n<\/div>\r\nSometimes, a system of equations can inform a decision. \u00a0In our next example, we help answer the question, \"Which truck rental company will give the best value?\"\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Building a System of Linear Models to Choose a Truck Rental Company<\/h3>\r\nJamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 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?\r\n\r\n[reveal-answer q=\"869152\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"869152\"]\r\n\r\nThe 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><em>d<\/em>, distance driven in miles<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Outputs<\/td>\r\n<td><em>K<\/em>(<em>d<\/em>): cost, in dollars, for renting from Keep on Trucking<em>M<\/em>(<em>d<\/em>) 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: <em>K<\/em>(0) = 20 and <em>M<\/em>(0) = 16<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rate of Change<\/td>\r\n<td><em>K<\/em>(<em>d<\/em>) = $0.59\/mile and <em>P<\/em>(<em>d<\/em>) = $0.63\/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<img class=\"aligncenter\" 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\" width=\"731\" height=\"340\" \/>\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 100 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 Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d&gt;100[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe applications for systems seems almost endless, but we will just show one more. The next few problems involve mixtures.\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nThese examples are challenging. Work through them on paper perhaps more than once or twice to get a feel for how the situations are being translated into mathematical models.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solve a Chemical Mixture Problem<\/h3>\r\nA chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane?\r\n[reveal-answer q=\"350379\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"350379\"]\r\n\r\nWe 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 70 mL of solution, and the unknown amount can be <em>x<\/em>. The part is the percentages, or concentration of solution 0.5 for start, 0.8 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>70mL<\/td>\r\n<td>0.5<\/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>0.8<\/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>0.6<\/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;\">70mL<\/td>\r\n<td style=\"height: 15px;\">0.5<\/td>\r\n<td style=\"height: 15px;\">35<\/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;\">0.8<\/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;\">0.6<\/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\n35mL of 80% solution must be added to 70mL of 50% solution to get a 60% solution of Methane.\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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try IT<\/h3>\r\n[ohm_question height=\"260\"]8589[\/ohm_question]\r\n\r\n[ohm_question height=\"230\"]2239[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine whether a solution reveals that a system has one, many, or no solutions<\/li>\n<li>Express the general solution of a dependent system<\/li>\n<li>Interpret the solution to a system of equations that represents profits for a business<\/li>\n<li>Write a system of equations that models a given situation<\/li>\n<\/ul>\n<\/div>\n<p>Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an <strong>inconsistent system<\/strong> consists of parallel lines that have the same slope but different [latex]y[\/latex] -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as [latex]12=0[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Inconsistent System of Equations<\/h3>\n<p>Solve the following system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}&x=9 - 2y \\\\ &x+2y=13 \\end{gathered}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q888134\">Show Solution<\/span><\/p>\n<div id=\"q888134\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can approach this problem in two ways. Because one equation is already solved for [latex]x[\/latex], the most obvious step is to use substitution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&=13 \\\\ \\left(9 - 2y\\right)+2y&=13 \\\\ 9+0y&=13 \\\\ 9&=13 \\end{align}[\/latex]<\/p>\n<p>Clearly, this statement is a contradiction because [latex]9\\ne 13[\/latex]. Therefore, the system has no solution.<\/p>\n<p>The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=9 - 2y \\\\ 2y=-x+9 \\\\ y=-\\frac{1}{2}x+\\frac{9}{2} \\end{gathered}[\/latex]<\/p>\n<p>We then convert the second equation expressed to slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+2y=13 \\\\ 2y=-x+13 \\\\ y=-\\frac{1}{2}x+\\frac{13}{2} \\end{gathered}[\/latex]<\/p>\n<p>Comparing the equations, we see that they have the same slope but different <em>y<\/em>-intercepts. Therefore, the lines are parallel and do not intersect.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y=-\\frac{1}{2}x+\\frac{9}{2} \\\\ y=-\\frac{1}{2}x+\\frac{13}{2} \\end{gathered}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183613\/CNX_Precalc_Figure_09_01_0072.jpg\" alt=\"A graph of two parallel lines. The first line's equation is y equals negative one-half x plus 13 over two. The second line's equation is y equals negative one-half x plus 9 over two.\" width=\"488\" height=\"297\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next video, see another example of using substitution to solve a system that has no solution.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Solve a System of Equations Using Substitution - No Solution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/kTtKfh5gFUc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the following system of equations in two variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2y - 2x=2\\\\ 2y - 2x=6\\end{gathered}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q681787\">Show Solution<\/span><\/p>\n<div id=\"q681787\" class=\"hidden-answer\" style=\"display: none\">\n<p>No solution. It is an inconsistent system.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm29699\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29699&theme=oea&iframe_resize_id=ohm29699&show_question_numbers\" width=\"100%\" height=\"220\"><\/iframe><\/p>\n<\/div>\n<h2>Expressing the Solution of a System of Dependent Equations Containing Two Variables<\/h2>\n<p>Recall that a <strong>dependent system<\/strong> of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as [latex]0=0[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Writing a general solution for a dependent system takes practice to feel natural. Work through the following example on paper, then apply the information in Writing the General Solution below.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Solution to a Dependent System of Linear Equations<\/h3>\n<p>Find a solution to the system of equations using the <strong>addition method<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+3y=2\\\\ 3x+9y=6\\end{gathered}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q390828\">Show Solution<\/span><\/p>\n<div id=\"q390828\" class=\"hidden-answer\" style=\"display: none\">\n<p>With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let\u2019s focus on eliminating [latex]x[\/latex]. If we multiply both sides of the first equation by [latex]-3[\/latex], then we will be able to eliminate the [latex]x[\/latex] -variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+3y&=2 \\\\ \\left(-3\\right)\\left(x+3y\\right)&=\\left(-3\\right)\\left(2\\right) \\\\ -3x - 9y&=-6 \\end{align}[\/latex]<\/p>\n<p>Now add the equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \u22123x\u22129y&=\u22126 \\\\ +3x+9y&=6 \\\\ \\hline 0&=0 \\end{align}[\/latex]<\/p>\n<p>We can see that there will be an infinite number of solutions that satisfy both equations.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let\u2019s look at what happens when we convert the system to slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\begin{gathered}x+3y=2 \\\\ 3y=-x+2 \\\\ y=-\\frac{1}{3}x+\\frac{2}{3} \\end{gathered} \\hspace{2cm} \\begin{gathered} 3x+9y=6 \\\\9y=-3x+6 \\\\ y=-\\frac{3}{9}x+\\frac{6}{9} \\\\ y=-\\frac{1}{3}x+\\frac{2}{3} \\end{gathered}\\end{align}[\/latex]<\/p>\n<p>Look at the graph below. Notice the results are the same. The general solution to the system is [latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183615\/CNX_Precalc_Figure_09_01_0082.jpg\" alt=\"A graph of two lines that overlap each other. The first line's equation is x+3y=2. The second line's equation is 3x-9y=6.\" width=\"487\" height=\"366\" \/><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Writing the general solution<\/h3>\n<p>In the previous example, we presented an analysis of the solution to the following system of equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+3y=2\\\\ 3x+9y=6\\end{gathered}[\/latex]<\/p>\n<p>After a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as\u00a0[latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex]. Why would we write the solution this way? In some ways, this representation tells us a lot. \u00a0It tells us that <em>x<\/em> can be anything, <em>x<\/em> is <em>x<\/em>. \u00a0It also tells us that <em>y<\/em> is going to depend on <em>x<\/em>, just like when we write a function rule. \u00a0In this case, depending on what you put in for <em>x<\/em>, <em>y<\/em> will be defined in terms of <em>x<\/em> as [latex]-\\frac{1}{3}x+\\frac{2}{3}[\/latex].<\/p>\n<p>In other words, there are infinitely many (<em>x<\/em>,<em>y<\/em>) pairs that will satisfy this system of equations, and they all fall on the line\u00a0[latex]f(x)-\\frac{1}{3}x+\\frac{2}{3}[\/latex].<\/p>\n<\/div>\n<p>Watch the next video for an example of a dependent system.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Solve a System of Equations Using Substitution - Infinite Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Pcqb109yK5Q?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the following system of equations in two variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y - 2x=5 \\\\ -3y+6x=-15 \\end{gathered}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q218404\">Show Solution<\/span><\/p>\n<div id=\"q218404\" class=\"hidden-answer\" style=\"display: none\">\n<p>The system is dependent so there are infinitely many solutions of the form [latex]\\left(x,2x+5\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm15665\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15665&theme=oea&iframe_resize_id=ohm15665&show_question_numbers\" width=\"100%\" height=\"290\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use an online graphing calculator to write three different systems:<\/p>\n<ul>\n<li>A system of equations with one solution<\/li>\n<li>A system of equations with no solutions<\/li>\n<li>A system of equations with infinitely many solutions<\/li>\n<\/ul>\n<\/div>\n<h2><strong>Using Systems of Equations to Investigate Profits<\/strong><\/h2>\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 <em>y<\/em>-axis represents either cost or revenue in hundreds of dollars.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" 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\" \/><\/p>\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 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 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<div class=\"textbox exercises\">\n<h3>Example: Finding the Break-Even Point and the Profit Function Using Substitution<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q569292\">Show Solution<\/span><\/p>\n<div id=\"q569292\" class=\"hidden-answer\" style=\"display: none\">\n<p>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<h4>Analysis of the Solution<\/h4>\n<p>The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" 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=\"487\" height=\"390\" \/><\/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 <em>x<\/em>-axis. Then, the graph emerges into positive <em>y<\/em>-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 0. The area to the left of the break-even point represents operating at a loss.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" 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=\"731\" height=\"507\" \/>\u00a0<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will see an example of how to find the break-even point for a small sno-cone business.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-3\" title=\"System of Equations App:  Break-Even Point\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/qey3FmE8saQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Writing a System of Linear Equations Given a Situation<\/h2>\n<p>It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.<\/h3>\n<ol>\n<li>Identify the input and output of each linear model.<\/li>\n<li>Identify the slope and <em>y<\/em>-intercept of each linear model.<\/li>\n<li>Find the solution by setting the two linear functions equal to another and solving for <em>x<\/em>, or find the point of intersection on a graph.<\/li>\n<\/ol>\n<\/div>\n<p>Now let&#8217;s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Work through the following examples on paper perhaps more than once to be sure you understand how the situations can be translated into a mathematical model. They use techniques you used earlier in the course to write models for real-world situations.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing and Solving a System of Equations in Two Variables<\/h3>\n<p>The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q455809\">Show Solution<\/span><\/p>\n<div id=\"q455809\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let <em>c<\/em> = the number of children and <em>a<\/em> = the number of adults in attendance.<\/p>\n<p>The total number of people is 2,000. 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 $25.00 by the number of children, [latex]25c[\/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[\/latex]. The total revenue is $70,000. 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 1. 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>We find that 1,200 children and 800 adults bought tickets to the circus that day.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>See the video below for another example of translating a given situation into a system of two linear equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex: Solve an Application Problem Using a System of Linear Equations (09x-43)\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/euh9ksWrq0A?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q454145\">Show Solution<\/span><\/p>\n<div id=\"q454145\" class=\"hidden-answer\" style=\"display: none\">\n<p>700 children, 950 adults<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm23774\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23774&theme=oea&iframe_resize_id=ohm23774&show_question_numbers\" width=\"100%\" height=\"245\"><\/iframe><\/p>\n<\/div>\n<p>Sometimes, a system of equations can inform a decision. \u00a0In our next example, we help answer the question, &#8220;Which truck rental company will give the best value?&#8221;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Building a System of Linear Models to Choose a Truck Rental Company<\/h3>\n<p>Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 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-308-1\" href=\"#footnote-308-1\" 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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q869152\">Show Solution<\/span><\/p>\n<div id=\"q869152\" class=\"hidden-answer\" style=\"display: none\">\n<p>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><em>d<\/em>, distance driven in miles<\/td>\n<\/tr>\n<tr>\n<td>Outputs<\/td>\n<td><em>K<\/em>(<em>d<\/em>): cost, in dollars, for renting from Keep on Trucking<em>M<\/em>(<em>d<\/em>) 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: <em>K<\/em>(0) = 20 and <em>M<\/em>(0) = 16<\/td>\n<\/tr>\n<tr>\n<td>Rate of Change<\/td>\n<td><em>K<\/em>(<em>d<\/em>) = $0.59\/mile and <em>P<\/em>(<em>d<\/em>) = $0.63\/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<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" 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\" width=\"731\" height=\"340\" alt=\"image\" \/><\/p>\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 100 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 Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d>100[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The applications for systems seems almost endless, but we will just show one more. The next few problems involve mixtures.<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>These examples are challenging. Work through them on paper perhaps more than once or twice to get a feel for how the situations are being translated into mathematical models.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solve a Chemical Mixture Problem<\/h3>\n<p>A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q350379\">Show Solution<\/span><\/p>\n<div id=\"q350379\" class=\"hidden-answer\" style=\"display: none\">\n<p>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 70 mL of solution, and the unknown amount can be <em>x<\/em>. The part is the percentages, or concentration of solution 0.5 for start, 0.8 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>70mL<\/td>\n<td>0.5<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Add<\/td>\n<td>[latex]x[\/latex]<\/td>\n<td>0.8<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Final<\/td>\n<td>[latex]70+x[\/latex]<\/td>\n<td>0.6<\/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;\">70mL<\/td>\n<td style=\"height: 15px;\">0.5<\/td>\n<td style=\"height: 15px;\">35<\/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;\">0.8<\/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;\">0.6<\/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>35mL of 80% solution must be added to 70mL of 50% solution to get a 60% solution of Methane.<\/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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm8589\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=8589&theme=oea&iframe_resize_id=ohm8589&show_question_numbers\" width=\"100%\" height=\"260\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm2239\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2239&theme=oea&iframe_resize_id=ohm2239&show_question_numbers\" width=\"100%\" height=\"230\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-308\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>System of Equations App: Break-Even Point. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/qey3FmE8saQ\">https:\/\/youtu.be\/qey3FmE8saQ<\/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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Beginning and Intermediate Algebra. <strong>Authored by<\/strong>: Wallace, Tyler. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.wallace.ccfaculty.org\/book\/book.html\">http:\/\/www.wallace.ccfaculty.org\/book\/book.html<\/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>Question ID 29699. <strong>Authored by<\/strong>: McClure, Caren. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 15655. <strong>Authored by<\/strong>: Johns,Bryan, mb Lippman,David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 23774. <strong>Authored by<\/strong>: Shahbazian,Roy. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 8589. <strong>Authored by<\/strong>: Harbaugh, Greg. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 2239. <strong>Authored by<\/strong>: Morales,Lawrence, mb Sousa,James, mb Lippman,David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Ex: Solve an Application Problem Using a System of Linear Equations (09x-43).. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/euh9ksWrq0A\">https:\/\/youtu.be\/euh9ksWrq0A<\/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: Solve a System of Equations Using Substitution - No Solution.. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/kTtKfh5gFUc\">https:\/\/youtu.be\/kTtKfh5gFUc<\/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: Solve a System of Equations Using Substitution - Infinite Solutions.. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Pcqb109yK5Q\">https:\/\/youtu.be\/Pcqb109yK5Q<\/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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-308-1\">Rates retrieved Aug 2, 2010 from http:\/\/www.budgettruck.com and http:\/\/www.uhaul.com\/ <a href=\"#return-footnote-308-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":395986,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Beginning and Intermediate Algebra\",\"author\":\"Wallace, Tyler\",\"organization\":\"\",\"url\":\"http:\/\/www.wallace.ccfaculty.org\/book\/book.html\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 29699\",\"author\":\"McClure, Caren\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 15655\",\"author\":\"Johns,Bryan, mb Lippman,David\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 23774\",\"author\":\"Shahbazian,Roy\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 8589\",\"author\":\"Harbaugh, 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