{"id":4723,"date":"2017-12-26T16:56:14","date_gmt":"2017-12-26T16:56:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/chapter\/systems-of-linear-equations-two-variables\/"},"modified":"2023-11-08T13:19:28","modified_gmt":"2023-11-08T13:19:28","slug":"systems-of-linear-equations-two-variables","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/chapter\/systems-of-linear-equations-two-variables\/","title":{"raw":"5.2 - Applications of Systems of Linear Equations","rendered":"5.2 &#8211; Applications of Systems of Linear Equations"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>(5.2.1) - Solve cost and revenue problems\r\n<ul>\r\n \t<li>Specify what the variables in a cost\/ revenue system of linear equations represent<\/li>\r\n \t<li>Determine and apply an appropriate method for solving the system<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>(5.2.2) - Solve value problems with a system of linear equations<\/li>\r\n \t<li>(5.2.3) - Solve mixture problems with a system of linear equations<\/li>\r\n \t<li>(5.2.4) - Solve uniform motion problems with a system of linear equations<\/li>\r\n<\/ul>\r\n<\/div>\r\nA skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible?\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183558\/CNX_Precalc_Figure_09_01_0012.jpg\" alt=\"Skateboarders at a skating rink by the beach.\" width=\"487\" height=\"252\" \/> (credit: Thomas S\u00f8renes)[\/caption]\r\n<h1>(5.2.1) - Solve cost and revenue problems<\/h1>\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\/2862\/2017\/12\/26165607\/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<\/h3>\r\nA business wants to manufacture bike frames. Before they start production, they need to make sure they can make a profit with the materials and labor force they have. Their accountant has given them a cost equation of [latex]y=0.85x+35,000[\/latex] and a revenue equation of [latex]y=1.55x[\/latex]:\r\n<ol>\r\n \t<li>Interpret x and y for the cost equation<\/li>\r\n \t<li>Interpret x and y for the revenue equation<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"86281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"86281\"]\r\n\r\nCost: [latex]y=0.85x+35,000[\/latex]\r\n\r\nRevenue:[latex]y=1.55x[\/latex]\r\n\r\nThe cost equation represents money leaving the company, namely how much it costs to produce a given number of bike frames. If we use the skateboard example as a model, x would represent the number of frames produced (instead of skateboards) and y would represent the amount of money it would cost to produce them (the same as the skateboard problem).\r\n\r\nThe revenue equation represents money coming into the company, so in this context x still represents the number of bike frames manufactured, and y now represents the amount of money made from selling them. \u00a0Let's organize this information in a table:\r\n<table>\r\n<thead>\r\n<tr>\r\n<td>Equation Type<\/td>\r\n<td>x represents<\/td>\r\n<td>y represents<\/td>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Revenue Eqn.<\/td>\r\n<td>number of frames<\/td>\r\n<td>amount of money made selling frames<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Cost Eqn.<\/td>\r\n<td>number of frames<\/td>\r\n<td>cost for making frames<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\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\"]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{array}{l}\\begin{array}{l}\\\\ y=0.85x+35,000\\end{array}\\hfill \\\\ y=1.55x\\hfill \\end{array}[\/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{array}{c}0.85x+35,000=1.55x\\\\ 35,000=0.7x\\\\ 50,000=x\\end{array}[\/latex]<\/p>\r\nThen, we substitute [latex]x=50,000[\/latex] into either the cost function or the revenue function.\r\n[latex]1.55\\left(50,000\\right)=77,500[\/latex]\r\n\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{array}{l}P\\left(x\\right)=1.55x-\\left(0.85x+35,000\\right)\\hfill \\\\ \\text{ }=0.7x - 35,000\\hfill \\end{array}[\/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\/2862\/2017\/12\/26165609\/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\/2862\/2017\/12\/26165611\/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\n<h1>(5.2.2) - Solve value problems with a system of linear equations<\/h1>\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\n1) Identify unknown quantities in a problem represent them with variables.\r\n\r\n2) Write a system of equations which models the problem's conditions.\r\n\r\n3) Solve the system.\r\n\r\n4) Check proposed solution.\r\n\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 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\"]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 [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 $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{array}{c}c+a=2,000\\\\ 25c+50a=70,000\\end{array}[\/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{array}{c}c+a=2,000\\\\ a=2,000-c\\end{array}[\/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{array}{l} 25c+50\\left(2,000-c\\right)=70,000\\hfill \\\\ 25c+100,000 - 50c=70,000\\hfill \\\\ \\text{ }-25c=-30,000\\hfill \\\\ \\text{ }c=1,200\\hfill \\end{array}[\/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{array}{l}1,200+a=2,000\\hfill \\\\ \\text{ }\\text{}a=800\\hfill \\end{array}[\/latex]<\/p>\r\nWe find that [latex]1,200[\/latex] children and [latex]800[\/latex] adults bought tickets to the circus that day.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this video example we show how to set up a system of linear equations that represents the total cost for admission to a museum.\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\"]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<\/div>\r\nSometimes, a system 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\"]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{array}{l}K\\left(d\\right)=0.59d+20\\\\ M\\left(d\\right)=0.63d+16\\end{array}[\/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\/2862\/2017\/12\/26165613\/CNX_Precalc_Figure_02_03_0072.jpg\" alt=\"image\" 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{array}{l}K\\left(d\\right)=M\\left(d\\right)\\hfill \\\\ 0.59d+20=0.63d+16\\hfill \\\\ 4=0.04d\\hfill \\\\ 100=d\\hfill \\\\ d=100\\hfill \\end{array}[\/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\n<h1>(5.2.3) - Solve mixture problems with a system of linear equations<\/h1>\r\nOne application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. \u00a0A solution is a mixture of two or more different substances like water and salt or vinegar and oil. \u00a0Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. \u00a0There are many other disciplines that use solutions as well.\r\n\r\nThe concentration or strength of a liquid solution is often described\u00a0\u00a0as a percentage. \u00a0This number comes from the ratio of how much mass is in a specific volume of liquid. \u00a0For example if you have 50 grams of salt in a 100mL of water you have a 50% salt solution based on the following ratio:\r\n<p style=\"text-align: center\">[latex]\\frac{50\\text{ grams }}{100\\text{ mL }}=0.50\\frac{\\text{ grams }}{\\text{ mL }}=50\\text{ % }[\/latex]<\/p>\r\nSolutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. \u00a0In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.\r\n\r\nWe will use the following table to help us solve mixture problems:\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\"><\/td>\r\n<td class=\"border\"><\/td>\r\n<td class=\"border\"><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\"><\/td>\r\n<td class=\"border\"><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nTo demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is 120mL of a 9% solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.\r\n\r\nFirst, find the total mass of solids for each solution by multiplying the volume by the concentration.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0120 mL<\/td>\r\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a075 mL<\/td>\r\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNext we add the new volumes and new masses.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0120 mL<\/td>\r\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a075 mL<\/td>\r\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center\">195 mL<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\frac{28.05\\text{ grams }}{ 195 \\text{ mL }}=0.14=14\\text{ % }[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]10.8\\text{ grams }+17.25\\text{ grams }=28.05\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. \u00a0In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA chemist has 70 mL of a 50% methane solution. How much of an 80% solution must she add so the final solution is 60% methane?\r\n[reveal-answer q=\"274848\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"274848\"]\r\n\r\nLet's use the problem solving process outlined in Module 1 to help us work through a solution to the problem.\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We are looking for a new amount - in this case a volume - \u00a0based on the words \"how much\". \u00a0We know two starting \u00a0concentrations and the final concentration, as well as one volume.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>Solution 1 is the 70 mL of 50% methane and solution 2 is the unknown amount with 80% methane. \u00a0We can call our unknown amount x.\r\n\r\n<strong>Write and Solve: \u00a0<\/strong>Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a070<\/td>\r\n<td class=\"border\">\u00a00.5<\/td>\r\n<td class=\"border\"><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0x<\/td>\r\n<td class=\"border\">\u00a00.8<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<td class=\"border\" style=\"text-align: center\">0.6<\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nMultiply amount by concentration\u00a0to get total,\u00a0be sure to distribute on the last row: [latex]\\left(70 + x\\right)0.6[\/latex]Add the entries in the amount column to get final amount. The concentration for this amount is 0.6 because we want the final solution to be 60% methane.\r\n<table class=\" undefined alignleft\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a070<\/td>\r\n<td class=\"border\">\u00a00.5<\/td>\r\n<td class=\"border\">\u00a035<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0x<\/td>\r\n<td class=\"border\">\u00a00.8<\/td>\r\n<td>\u00a00.8<em>x<\/em><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: left\">\u00a070+x<\/td>\r\n<td class=\"border\" style=\"text-align: left\">0.6<\/td>\r\n<td class=\"border\" style=\"text-align: center\">\u00a0[latex]42+0.6x[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAdd the total mass for solution 1 and solution 2 to get the total mass for the 60% solution. This is our equation for finding the unknown volume.\r\n\r\n[latex]35+0.8x=42+0.6x[\/latex]\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}35+0.8x=42+0.6x\\\\\\underline{-0.6x}\\,\\,\\,\\,\\,\\,\\,\\underline{-0.6x}\\\\35+0.2x=42\\\\\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Subtract 35 from both sides<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}35+0.2x=42\\\\\\underline{-35}\\,\\,\\,\\,\\,\\,\\,\\underline{-35}\\\\0.2x=7\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Divide both sides by 0.2<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0.2x=7\\\\\\frac{0.2x}{0.2}=\\frac{7}{0.2}\\end{array}[\/latex]\r\n[latex]x=35[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n35mL must be added to the original 70 mL to gain a solution with a concentration of 60%\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe above problem illustrates how we can use\u00a0the mixture table\u00a0to define\u00a0an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons of 20% butterfat?\r\n[reveal-answer q=\"966963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"966963\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is 42 gallons. There are two unknowns in this problem.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>We will call the unknown volume of the \u00a024% solution x, and the unknown volume of the 18% solution y.\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0x<\/td>\r\n<td class=\"border\">\u00a00.24<\/td>\r\n<td class=\"border\"><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0y<\/td>\r\n<td class=\"border\">\u00a00.18<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: left\">42<\/td>\r\n<td class=\"border\" style=\"text-align: left\">0.2<\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFind the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0x<\/td>\r\n<td class=\"border\">\u00a00.24<\/td>\r\n<td class=\"border\">\u00a00.24x<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0y<\/td>\r\n<td class=\"border\">\u00a00.18<\/td>\r\n<td>\u00a00.18y<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: left\">x+y=42<\/td>\r\n<td class=\"border\" style=\"text-align: left\">0.2<\/td>\r\n<td class=\"border\" style=\"text-align: center\">8.4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen you sum the amount column you get\u00a0one equation: [latex]x+ y = 42[\/latex]\r\nWhen you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[\/latex]\r\n\r\nUse elimination to find a value for [latex]x[\/latex], and [latex]y[\/latex].\r\n\r\nMultiply the first equation by [latex]-0.18[\/latex]\r\n<p style=\"text-align: center\">[latex]\\begin{array}{cc}-0.18(x+y) &amp;= (42)(-0.18) \\\\ -0.18x-0.18y &amp;= -7.56 \\end{array}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nNow our system of equations looks like this:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{cc} -0.18x-0.18y &amp;= -7.56\\\\0.24x + 0.18y &amp;= 8.4 \\end{array}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nAdding the two equations together to eliminate the y terms gives this equation:\r\n<p style=\"text-align: center\">[latex]0.06x = 8.4[\/latex]<\/p>\r\nDivide by 0.06 on each side:\r\n<p style=\"text-align: center\">[latex]x = 14[\/latex]<\/p>\r\nNow substitute the value for x into one of the equations in order to solve for y.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{cc} (14) + y &amp;= 42\\\\ y &amp;= 28 \\end{array}[\/latex]<\/p>\r\n&nbsp;\r\n\r\n&nbsp;\r\n<h4>Answer<\/h4>\r\nThis can be interpreted as 14 gallons of 24% butterfat milk added to 28 gallons of 18% butterfat milk will give 42 gallons of 20% butterfat milk.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.\r\n\r\nhttps:\/\/youtu.be\/4s5MCqphpKo\r\n<h1>(5.2.4) - Solve uniform motion problems with a system of linear equations<\/h1>\r\n<p id=\"fs-id1167835306358\">Many real-world applications of uniform motion arise because of the effects of currents\u2014of water or air\u2014on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.<\/p>\r\n<p id=\"fs-id1167835191090\">Let\u2019s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.<\/p>\r\n<p id=\"fs-id1167834472600\">The images below show how a river current affects the speed at which a boat is actually travelling. We\u2019ll call the speed of the boat in still water [latex]b[\/latex]\u00a0and the speed of the river current [latex]c[\/latex].<\/p>\r\n<p id=\"fs-id1167834098208\">The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat\u2019s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is [latex]b+c[\/latex].<\/p>\r\n<img class=\"size-medium wp-image-5349 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/17204215\/Math101_5_2_4im1-300x261.jpg\" alt=\"\" width=\"300\" height=\"261\" \/>\r\n\r\nNow, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat\u2019s actual speed is slower than its speed in still water. The actual speed of the boat is [latex]b-c[\/latex].\r\n\r\n<img class=\"size-medium wp-image-5351 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/17204456\/Math101_5_2_4im2-300x261.jpg\" alt=\"\" width=\"300\" height=\"261\" \/>\r\n\r\nWe\u2019ll put some numbers to this situation in the next example.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\n<p id=\"fs-id1167835365522\">Translate to a system of equations and then solve.<\/p>\r\n<p id=\"fs-id1167835356872\">A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.<\/p>\r\n[reveal-answer q=\"12628\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"12628\"]\r\n\r\n<strong>Read<\/strong> the problem:\u00a0This is a uniform motion problem and a picture will help us visualize the situation.\r\n\r\n<img class=\"alignnone size-medium wp-image-5352\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/17205155\/Math101_5_2_4im3-300x113.jpg\" alt=\"\" width=\"300\" height=\"113\" \/>\r\n\r\n<strong>Identify<\/strong>\u00a0what we are looking for:\u00a0We are looking for the speed of the ship\u00a0in still water and the speed of the current.\r\n\r\n<strong>Name<\/strong>\u00a0what we are looking for:\r\n\r\nLet [latex]s=[\/latex]\u00a0the rate of the ship in still water.\r\n\r\nLet [latex]c=[\/latex] the rate of the current.\r\n\r\nA chart will help us organize the information.\u00a0The ship goes downstream and then upstream.\u00a0Going downstream, the current helps the\u00a0ship and so the ship's actual rate is [latex]s+c[\/latex].\u00a0Going upstream, the current slows the ship\u00a0and so the actual rate is [latex]s-c[\/latex].\u00a0Downstream it takes 4 hours.\u00a0Upstream it takes 5 hours.\u00a0Each way the distance is 60 miles.\r\n\r\n<img class=\"size-medium wp-image-5353 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/17210031\/Math101_5_2_4im4-300x74.jpg\" alt=\"\" width=\"300\" height=\"74\" \/>\r\n\r\n<strong>Translate<\/strong>\u00a0into a system of equations.\u00a0Since rate times time is distance, we can\u00a0write the system of equations.\r\n<p style=\"text-align: center\">\u00a0[latex]\\begin{array}{c}4(s+c)=60 \\\\ 5(s-c) = 60\\end{array}[\/latex]<\/p>\r\n<strong>Solve<\/strong>\u00a0the system of equations.\u00a0Distribute to put both equations in standard\u00a0form, then solve by elimination.\r\n<p style=\"text-align: center\">\u00a0[latex]\\begin{array}{c}4s+4c=60 \\\\ 5s-5c = 60\\end{array}[\/latex]<\/p>\r\n<strong>Multiply<\/strong> the top equation by 5 and the\u00a0bottom equation by 4.\u00a0Add the equations, then solve for [latex]s[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{cc}20s+20c &amp;= 300 \\\\ 20s-20c &amp;= 240 \\\\ \\hline \\\\ 40s &amp;= 540 \\\\ s &amp;= 13.5 \\end{array}[\/latex]<\/p>\r\n<strong>Substitute<\/strong> [latex]s=13.5[\/latex]\u00a0into one of the original\u00a0equations.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{cc}4(s+c) &amp;= 60 \\\\ 4(13.5+c) &amp;= 60 \\\\ 54 + 4c &amp;= 60 \\\\ 4c &amp;= 6 \\\\ c &amp;= 1.5 \\end{array}[\/latex]<\/p>\r\n<strong>Check<\/strong>\u00a0the answer in the problem.\u00a0The downstream rate would be:\r\n<p style=\"text-align: center\">[latex]13.5+1.5 = 15[\/latex] mph<\/p>\r\nIn 4 hours the ship would travel:\r\n<p style=\"text-align: center\">[latex]15 \\cdot 4 = 60[\/latex] miles.<\/p>\r\nThe upstream rate would be\r\n<p style=\"text-align: center\">[latex]13.5 - 1.5 = 12[\/latex] mph.<\/p>\r\nIn 5 hours the ship would travel\r\n<p style=\"text-align: center\">[latex]12\\cdot 5[\/latex] miles.<\/p>\r\n\r\n<div>\r\n\r\n<strong>Answer<\/strong>\u00a0the question.\u00a0The rate of the ship is 13.5 mph and\u00a0the rate of the current is 1.5 mph.\r\n<div><\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video, we present another example of a uniform motion problem which can be solved with a system of linear equations.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=OuxMYTqDhxw","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>(5.2.1) &#8211; Solve cost and revenue problems\n<ul>\n<li>Specify what the variables in a cost\/ revenue system of linear equations represent<\/li>\n<li>Determine and apply an appropriate method for solving the system<\/li>\n<\/ul>\n<\/li>\n<li>(5.2.2) &#8211; Solve value problems with a system of linear equations<\/li>\n<li>(5.2.3) &#8211; Solve mixture problems with a system of linear equations<\/li>\n<li>(5.2.4) &#8211; Solve uniform motion problems with a system of linear equations<\/li>\n<\/ul>\n<\/div>\n<p>A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible?<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183558\/CNX_Precalc_Figure_09_01_0012.jpg\" alt=\"Skateboarders at a skating rink by the beach.\" width=\"487\" height=\"252\" \/><\/p>\n<p class=\"wp-caption-text\">(credit: Thomas S\u00f8renes)<\/p>\n<\/div>\n<h1>(5.2.1) &#8211; Solve cost and revenue problems<\/h1>\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\/2862\/2017\/12\/26165607\/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<\/h3>\n<p>A business wants to manufacture bike frames. Before they start production, they need to make sure they can make a profit with the materials and labor force they have. Their accountant has given them a cost equation of [latex]y=0.85x+35,000[\/latex] and a revenue equation of [latex]y=1.55x[\/latex]:<\/p>\n<ol>\n<li>Interpret x and y for the cost equation<\/li>\n<li>Interpret x and y for the revenue equation<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q86281\">Show Solution<\/span><\/p>\n<div id=\"q86281\" class=\"hidden-answer\" style=\"display: none\">\n<p>Cost: [latex]y=0.85x+35,000[\/latex]<\/p>\n<p>Revenue:[latex]y=1.55x[\/latex]<\/p>\n<p>The cost equation represents money leaving the company, namely how much it costs to produce a given number of bike frames. If we use the skateboard example as a model, x would represent the number of frames produced (instead of skateboards) and y would represent the amount of money it would cost to produce them (the same as the skateboard problem).<\/p>\n<p>The revenue equation represents money coming into the company, so in this context x still represents the number of bike frames manufactured, and y now represents the amount of money made from selling them. \u00a0Let&#8217;s organize this information in a table:<\/p>\n<table>\n<thead>\n<tr>\n<td>Equation Type<\/td>\n<td>x represents<\/td>\n<td>y represents<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Revenue Eqn.<\/td>\n<td>number of frames<\/td>\n<td>amount of money made selling frames<\/td>\n<\/tr>\n<tr>\n<td>Cost Eqn.<\/td>\n<td>number of frames<\/td>\n<td>cost for making frames<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\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\">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{array}{l}\\begin{array}{l}\\\\ y=0.85x+35,000\\end{array}\\hfill \\\\ y=1.55x\\hfill \\end{array}[\/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{array}{c}0.85x+35,000=1.55x\\\\ 35,000=0.7x\\\\ 50,000=x\\end{array}[\/latex]<\/p>\n<p>Then, we substitute [latex]x=50,000[\/latex] into either the cost function or the revenue function.<br \/>\n[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{array}{l}P\\left(x\\right)=1.55x-\\left(0.85x+35,000\\right)\\hfill \\\\ \\text{ }=0.7x - 35,000\\hfill \\end{array}[\/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\/2862\/2017\/12\/26165609\/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\/2862\/2017\/12\/26165611\/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<h1>(5.2.2) &#8211; Solve value problems with a system of linear equations<\/h1>\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<p>1) Identify unknown quantities in a problem represent them with variables.<\/p>\n<p>2) Write a system of equations which models the problem&#8217;s conditions.<\/p>\n<p>3) Solve the system.<\/p>\n<p>4) Check proposed solution.<\/p>\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 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\">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 [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 $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{array}{c}c+a=2,000\\\\ 25c+50a=70,000\\end{array}[\/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{array}{c}c+a=2,000\\\\ a=2,000-c\\end{array}[\/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{array}{l} 25c+50\\left(2,000-c\\right)=70,000\\hfill \\\\ 25c+100,000 - 50c=70,000\\hfill \\\\ \\text{ }-25c=-30,000\\hfill \\\\ \\text{ }c=1,200\\hfill \\end{array}[\/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{array}{l}1,200+a=2,000\\hfill \\\\ \\text{ }\\text{}a=800\\hfill \\end{array}[\/latex]<\/p>\n<p>We find that [latex]1,200[\/latex] children and [latex]800[\/latex] adults bought tickets to the circus that day.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In this video example we show how to set up a system of linear equations that represents the total cost for admission to a museum.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" 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\">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<\/div>\n<p>Sometimes, a system 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-4723-1\" href=\"#footnote-4723-1\" 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\">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{array}{l}K\\left(d\\right)=0.59d+20\\\\ M\\left(d\\right)=0.63d+16\\end{array}[\/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\/2862\/2017\/12\/26165613\/CNX_Precalc_Figure_02_03_0072.jpg\" alt=\"image\" width=\"731\" height=\"340\" \/><\/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{array}{l}K\\left(d\\right)=M\\left(d\\right)\\hfill \\\\ 0.59d+20=0.63d+16\\hfill \\\\ 4=0.04d\\hfill \\\\ 100=d\\hfill \\\\ d=100\\hfill \\end{array}[\/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<h1>(5.2.3) &#8211; Solve mixture problems with a system of linear equations<\/h1>\n<p>One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. \u00a0A solution is a mixture of two or more different substances like water and salt or vinegar and oil. \u00a0Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. \u00a0There are many other disciplines that use solutions as well.<\/p>\n<p>The concentration or strength of a liquid solution is often described\u00a0\u00a0as a percentage. \u00a0This number comes from the ratio of how much mass is in a specific volume of liquid. \u00a0For example if you have 50 grams of salt in a 100mL of water you have a 50% salt solution based on the following ratio:<\/p>\n<p style=\"text-align: center\">[latex]\\frac{50\\text{ grams }}{100\\text{ mL }}=0.50\\frac{\\text{ grams }}{\\text{ mL }}=50\\text{ % }[\/latex]<\/p>\n<p>Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. \u00a0In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.<\/p>\n<p>We will use the following table to help us solve mixture problems:<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\"><\/td>\n<td class=\"border\"><\/td>\n<td class=\"border\"><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\"><\/td>\n<td class=\"border\"><\/td>\n<td><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is 120mL of a 9% solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.<\/p>\n<p>First, find the total mass of solids for each solution by multiplying the volume by the concentration.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0120 mL<\/td>\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a075 mL<\/td>\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Next we add the new volumes and new masses.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0120 mL<\/td>\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a075 mL<\/td>\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center\">195 mL<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]\\frac{28.05\\text{ grams }}{ 195 \\text{ mL }}=0.14=14\\text{ % }[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]10.8\\text{ grams }+17.25\\text{ grams }=28.05\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. \u00a0In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A chemist has 70 mL of a 50% methane solution. How much of an 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=\"q274848\">Show Solution<\/span><\/p>\n<div id=\"q274848\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let&#8217;s use the problem solving process outlined in Module 1 to help us work through a solution to the problem.<\/p>\n<p><strong>Read and Understand:\u00a0<\/strong>We are looking for a new amount &#8211; in this case a volume &#8211; \u00a0based on the words &#8220;how much&#8221;. \u00a0We know two starting \u00a0concentrations and the final concentration, as well as one volume.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>Solution 1 is the 70 mL of 50% methane and solution 2 is the unknown amount with 80% methane. \u00a0We can call our unknown amount x.<\/p>\n<p><strong>Write and Solve: \u00a0<\/strong>Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a070<\/td>\n<td class=\"border\">\u00a00.5<\/td>\n<td class=\"border\"><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0x<\/td>\n<td class=\"border\">\u00a00.8<\/td>\n<td><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<td class=\"border\" style=\"text-align: center\">0.6<\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Multiply amount by concentration\u00a0to get total,\u00a0be sure to distribute on the last row: [latex]\\left(70 + x\\right)0.6[\/latex]Add the entries in the amount column to get final amount. The concentration for this amount is 0.6 because we want the final solution to be 60% methane.<\/p>\n<table class=\"undefined alignleft\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a070<\/td>\n<td class=\"border\">\u00a00.5<\/td>\n<td class=\"border\">\u00a035<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0x<\/td>\n<td class=\"border\">\u00a00.8<\/td>\n<td>\u00a00.8<em>x<\/em><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: left\">\u00a070+x<\/td>\n<td class=\"border\" style=\"text-align: left\">0.6<\/td>\n<td class=\"border\" style=\"text-align: center\">\u00a0[latex]42+0.6x[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Add the total mass for solution 1 and solution 2 to get the total mass for the 60% solution. This is our equation for finding the unknown volume.<\/p>\n<p>[latex]35+0.8x=42+0.6x[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}35+0.8x=42+0.6x\\\\\\underline{-0.6x}\\,\\,\\,\\,\\,\\,\\,\\underline{-0.6x}\\\\35+0.2x=42\\\\\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Subtract 35 from both sides<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}35+0.2x=42\\\\\\underline{-35}\\,\\,\\,\\,\\,\\,\\,\\underline{-35}\\\\0.2x=7\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Divide both sides by 0.2<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0.2x=7\\\\\\frac{0.2x}{0.2}=\\frac{7}{0.2}\\end{array}[\/latex]<br \/>\n[latex]x=35[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>35mL must be added to the original 70 mL to gain a solution with a concentration of 60%<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The above problem illustrates how we can use\u00a0the mixture table\u00a0to define\u00a0an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons of 20% butterfat?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q966963\">Show Solution<\/span><\/p>\n<div id=\"q966963\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong>We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is 42 gallons. There are two unknowns in this problem.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>We will call the unknown volume of the \u00a024% solution x, and the unknown volume of the 18% solution y.<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0x<\/td>\n<td class=\"border\">\u00a00.24<\/td>\n<td class=\"border\"><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0y<\/td>\n<td class=\"border\">\u00a00.18<\/td>\n<td><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: left\">42<\/td>\n<td class=\"border\" style=\"text-align: left\">0.2<\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0x<\/td>\n<td class=\"border\">\u00a00.24<\/td>\n<td class=\"border\">\u00a00.24x<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0y<\/td>\n<td class=\"border\">\u00a00.18<\/td>\n<td>\u00a00.18y<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: left\">x+y=42<\/td>\n<td class=\"border\" style=\"text-align: left\">0.2<\/td>\n<td class=\"border\" style=\"text-align: center\">8.4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When you sum the amount column you get\u00a0one equation: [latex]x+ y = 42[\/latex]<br \/>\nWhen you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[\/latex]<\/p>\n<p>Use elimination to find a value for [latex]x[\/latex], and [latex]y[\/latex].<\/p>\n<p>Multiply the first equation by [latex]-0.18[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{cc}-0.18(x+y) &= (42)(-0.18) \\\\ -0.18x-0.18y &= -7.56 \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Now our system of equations looks like this:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{cc} -0.18x-0.18y &= -7.56\\\\0.24x + 0.18y &= 8.4 \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Adding the two equations together to eliminate the y terms gives this equation:<\/p>\n<p style=\"text-align: center\">[latex]0.06x = 8.4[\/latex]<\/p>\n<p>Divide by 0.06 on each side:<\/p>\n<p style=\"text-align: center\">[latex]x = 14[\/latex]<\/p>\n<p>Now substitute the value for x into one of the equations in order to solve for y.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{cc} (14) + y &= 42\\\\ y &= 28 \\end{array}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<h4>Answer<\/h4>\n<p>This can be interpreted as 14 gallons of 24% butterfat milk added to 28 gallons of 18% butterfat milk will give 42 gallons of 20% butterfat milk.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  System of Equations Application - Mixture Problem\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/4s5MCqphpKo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(5.2.4) &#8211; Solve uniform motion problems with a system of linear equations<\/h1>\n<p id=\"fs-id1167835306358\">Many real-world applications of uniform motion arise because of the effects of currents\u2014of water or air\u2014on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.<\/p>\n<p id=\"fs-id1167835191090\">Let\u2019s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.<\/p>\n<p id=\"fs-id1167834472600\">The images below show how a river current affects the speed at which a boat is actually travelling. We\u2019ll call the speed of the boat in still water [latex]b[\/latex]\u00a0and the speed of the river current [latex]c[\/latex].<\/p>\n<p id=\"fs-id1167834098208\">The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat\u2019s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is [latex]b+c[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-5349 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/17204215\/Math101_5_2_4im1-300x261.jpg\" alt=\"\" width=\"300\" height=\"261\" \/><\/p>\n<p>Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat\u2019s actual speed is slower than its speed in still water. The actual speed of the boat is [latex]b-c[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-5351 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/17204456\/Math101_5_2_4im2-300x261.jpg\" alt=\"\" width=\"300\" height=\"261\" \/><\/p>\n<p>We\u2019ll put some numbers to this situation in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p id=\"fs-id1167835365522\">Translate to a system of equations and then solve.<\/p>\n<p id=\"fs-id1167835356872\">A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q12628\">Show Answer<\/span><\/p>\n<div id=\"q12628\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read<\/strong> the problem:\u00a0This is a uniform motion problem and a picture will help us visualize the situation.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-5352\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/17205155\/Math101_5_2_4im3-300x113.jpg\" alt=\"\" width=\"300\" height=\"113\" \/><\/p>\n<p><strong>Identify<\/strong>\u00a0what we are looking for:\u00a0We are looking for the speed of the ship\u00a0in still water and the speed of the current.<\/p>\n<p><strong>Name<\/strong>\u00a0what we are looking for:<\/p>\n<p>Let [latex]s=[\/latex]\u00a0the rate of the ship in still water.<\/p>\n<p>Let [latex]c=[\/latex] the rate of the current.<\/p>\n<p>A chart will help us organize the information.\u00a0The ship goes downstream and then upstream.\u00a0Going downstream, the current helps the\u00a0ship and so the ship&#8217;s actual rate is [latex]s+c[\/latex].\u00a0Going upstream, the current slows the ship\u00a0and so the actual rate is [latex]s-c[\/latex].\u00a0Downstream it takes 4 hours.\u00a0Upstream it takes 5 hours.\u00a0Each way the distance is 60 miles.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-5353 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/17210031\/Math101_5_2_4im4-300x74.jpg\" alt=\"\" width=\"300\" height=\"74\" \/><\/p>\n<p><strong>Translate<\/strong>\u00a0into a system of equations.\u00a0Since rate times time is distance, we can\u00a0write the system of equations.<\/p>\n<p style=\"text-align: center\">\u00a0[latex]\\begin{array}{c}4(s+c)=60 \\\\ 5(s-c) = 60\\end{array}[\/latex]<\/p>\n<p><strong>Solve<\/strong>\u00a0the system of equations.\u00a0Distribute to put both equations in standard\u00a0form, then solve by elimination.<\/p>\n<p style=\"text-align: center\">\u00a0[latex]\\begin{array}{c}4s+4c=60 \\\\ 5s-5c = 60\\end{array}[\/latex]<\/p>\n<p><strong>Multiply<\/strong> the top equation by 5 and the\u00a0bottom equation by 4.\u00a0Add the equations, then solve for [latex]s[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{cc}20s+20c &= 300 \\\\ 20s-20c &= 240 \\\\ \\hline \\\\ 40s &= 540 \\\\ s &= 13.5 \\end{array}[\/latex]<\/p>\n<p><strong>Substitute<\/strong> [latex]s=13.5[\/latex]\u00a0into one of the original\u00a0equations.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{cc}4(s+c) &= 60 \\\\ 4(13.5+c) &= 60 \\\\ 54 + 4c &= 60 \\\\ 4c &= 6 \\\\ c &= 1.5 \\end{array}[\/latex]<\/p>\n<p><strong>Check<\/strong>\u00a0the answer in the problem.\u00a0The downstream rate would be:<\/p>\n<p style=\"text-align: center\">[latex]13.5+1.5 = 15[\/latex] mph<\/p>\n<p>In 4 hours the ship would travel:<\/p>\n<p style=\"text-align: center\">[latex]15 \\cdot 4 = 60[\/latex] miles.<\/p>\n<p>The upstream rate would be<\/p>\n<p style=\"text-align: center\">[latex]13.5 - 1.5 = 12[\/latex] mph.<\/p>\n<p>In 5 hours the ship would travel<\/p>\n<p style=\"text-align: center\">[latex]12\\cdot 5[\/latex] miles.<\/p>\n<div>\n<p><strong>Answer<\/strong>\u00a0the question.\u00a0The rate of the ship is 13.5 mph and\u00a0the rate of the current is 1.5 mph.<\/p>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next video, we present another example of a uniform motion problem which can be solved with a system of linear equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  System of Equations Application - Plane and Wind Problem\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/OuxMYTqDhxw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\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-4723\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Solving Systems of Equations using Elimination. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ova8GSmPV4o\">https:\/\/youtu.be\/ova8GSmPV4o<\/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 115164, 115120, 115110. <strong>Authored by<\/strong>: Shabazian, 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>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 23774. <strong>Authored by<\/strong>: Roy Shahbazian. <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>: Greg Harbaugh. <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. <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: System of Equations Application - Mixture Problem.. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning.. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/4s5MCqphpKo.\">https:\/\/youtu.be\/4s5MCqphpKo.<\/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>Beginning and Intermediate Algebra Textbook. . <strong>Authored by<\/strong>: Tyler Wallace. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.wallace.ccfaculty.org\/book\/book.html.%20\">http:\/\/www.wallace.ccfaculty.org\/book\/book.html.%20<\/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: System of Equations Application - Plane and Wind problem. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=OuxMYTqDhxw\">https:\/\/www.youtube.com\/watch?v=OuxMYTqDhxw<\/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>Intermediate Algebra . <strong>Authored by<\/strong>: Lynn Marecek et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/02776133-d49d-49cb-bfaa-67c7f61b25a1@4.13\">http:\/\/cnx.org\/contents\/02776133-d49d-49cb-bfaa-67c7f61b25a1@4.13<\/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\/02776133-d49d-49cb-bfaa-67c7f61b25a1@4.13<\/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-4723-1\">Rates retrieved Aug 2, 2010 from http:\/\/www.budgettruck.com and http:\/\/www.uhaul.com\/ <a href=\"#return-footnote-4723-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":23485,"menu_order":3,"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\":\"Solving 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