{"id":557,"date":"2024-05-23T22:49:39","date_gmt":"2024-05-23T22:49:39","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/?post_type=chapter&#038;p=557"},"modified":"2024-08-26T22:51:01","modified_gmt":"2024-08-26T22:51:01","slug":"2-1-applications-of-systems-of-two-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-1-applications-of-systems-of-two-linear-equations\/","title":{"raw":"2.1 Applications of Systems of Two Linear Equations","rendered":"2.1 Applications of Systems of Two Linear Equations"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Solve applications of systems of two linear equations with two variables using substitution or elimination.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn this section we present examples of applications which are solved using systems of linear equations.\r\n\r\nIf you need a refresher on how to solve systems of linear equations by substitution and elimination, first visit <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-r-solving-linear-systems-of-equations-by-substitution-and-elimination\/\">Section 2.R: Solving Linear Systems of Equations by Substitution and Elimination<\/a>.\r\n<h2 id=\"CRP\">Cost, Revenue, and Profit<\/h2>\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\r\nThe 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 per unit. 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 total cost of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The [latex]x[\/latex]-axis represents quantity in hundreds of units. The [latex]y[\/latex]-axis represents either cost or revenue in hundreds of dollars.\r\n\r\n<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 [latex]700[\/latex] units are produced, the cost is [latex]\\$3,300[\/latex] and the revenue is also [latex]\\$3,300[\/latex]. In other words, the company breaks even if they produce and sell [latex]700[\/latex] units. They neither make money nor lose money.\r\n\r\nThe shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The <strong>profit function<\/strong> is the revenue function minus the cost function, written as [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex]. Note that when [latex]R(x)=C(x),[\/latex] then [latex]P(x)=0.[\/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 [latex]x[\/latex] and [latex]y[\/latex] for the cost equation.<\/li>\r\n \t<li>Interpret [latex]x[\/latex] and [latex]y[\/latex] for the revenue equation.<\/li>\r\n \t<li>Find and interpret the break-even point for this business.<\/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<ol>\r\n \t<li>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, [latex]x[\/latex] would represent the number of frames produced (instead of skateboards) and [latex]y[\/latex] would represent the amount of money it would cost to produce them (the same as the skateboard problem).<\/li>\r\n \t<li>The revenue equation represents money coming into the company, so in this context [latex]x[\/latex] still represents the number of bike frames manufactured, and [latex]y[\/latex] now represents the amount of money made from selling them.<\/li>\r\n \t<li>We must solve the system of equations\r\n<p style=\"text-align: center;\">[latex]\\left\\{\\begin{array}{l}y=0.85x+35,000\\\\ y=1.55x \\end{array}\\right.[\/latex]<\/p>\r\nSubstitution is the best method since both equations are already solved for [latex]y[\/latex]. Substitute 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}{rl}0.85x+35,000&amp;=\\;1.55x\\\\ 35,000&amp;=\\;0.7x\\\\ 50,000&amp;=\\;x\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Then, we substitute [latex]x=50,000[\/latex] into either the cost function or the revenue function.<\/p>\r\n<p style=\"text-align: center;\">[latex]y=1.55\\left(50,000\\right)=77,500[\/latex]<\/p>\r\nThe break-even point is [latex]\\left(50,000,77,500\\right),[\/latex] which represents [latex]50,000[\/latex] bike frames sold and both cost and revenue of [latex]\\$77,500.[\/latex] If more than [latex]50,000[\/latex] bike frames are sold, a profit will result.<\/li>\r\n<\/ol>\r\n<h4>Analysis of the Solution<\/h4>\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]. The above problem could have also been solved by setting [latex]P\\left(x\\right)=0[\/latex] and solving. You may want to verify this yourself.\r\n\r\nThe cost to produce [latex]50,000[\/latex] units is [latex]\\$77,500[\/latex], and the revenue from the sales of [latex]50,000[\/latex] units is also [latex]\\$77,500.[\/latex] To make a profit, the business must produce and sell more than [latex]50,000[\/latex] units.\r\n\r\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 [latex]x[\/latex]-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 [latex]0[\/latex]. 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\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Total Value Problems<\/h2>\r\nIn the next set of examples, we are given two types of items and multiple descriptions of their \"value.\" For example, we may be given two types of tickets sold at a circus and the totals of number of tickets sold as well as the total sales (or revenue).\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe cost of a ticket to the circus is [latex]\\$25.00[\/latex] for children and [latex]\\$50.00[\/latex] for adults. On a certain day, attendance at the circus is [latex]2,000[\/latex] and the total gate revenue is [latex]\\$70,000[\/latex]. How many children and how many adults bought tickets?\r\n\r\n[reveal-answer q=\"455809\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"455809\"]\r\n\r\nLet [latex]c=[\/latex] the number of children and [latex]a=[\/latex] 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 [latex]\\$25.00[\/latex] by the number of children, [latex]25c[\/latex]. The revenue from all adults can be found by multiplying [latex]\\$50.00[\/latex] by the number of adults, [latex]50a[\/latex]. The total revenue is [latex]\\$70,000[\/latex]. We can use this to write an equation for the revenue.\r\n<p style=\"text-align: center;\">[latex]25c+50a=70,000[\/latex]<\/p>\r\nWe now have a system of linear equations in two variables.\r\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{llll}c&amp;+&amp;a&amp;=\\;2,000\\\\ 25c&amp;+&amp;50a&amp;=\\;70,000\\end{array} \\right.[\/latex]<\/p>\r\nIn the first equation the coefficient of both variables is [latex]1[\/latex] so we solve by substitution (elimination would also be possible.) We will solve the first equation for [latex]a[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}c+a&amp;=\\;2,000\\\\ a&amp;=\\;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}{rl} 25c+50\\left(2,000-c\\right)&amp;=\\;70,000\\hfill \\\\ 25c+100,000 - 50c&amp;=\\;70,000\\hfill \\\\ \\text{ }-25c&amp;=\\;-30,000\\hfill \\\\ \\text{ }c&amp;=\\;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}{rl}1,200+a&amp;=\\;2,000\\hfill \\\\ \\text{ }\\text{}a&amp;=\\;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 YOURSELF<\/h3>\r\n[ohm_question hide_question_numbers=1]202618[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 id=\"mixture\">Mixture Problems<\/h2>\r\nA <strong>solution<\/strong> is a mixture of two or more different substances like water and salt or vinegar and oil. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand.\r\n\r\nThe concentration or strength of a liquid solution is often described as a percentage. \u00a0This number comes from the ratio of how much mass is in a specific volume of liquid. \u00a0For example if you have [latex]50[\/latex] grams of salt in a [latex]100[\/latex] mL of water you have a [latex]50\\%[\/latex] 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{ % }\\frac{\\text{ grams }}{\\text{ mL }}[\/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. For example, you may have a [latex]9\\%[\/latex] solution and a [latex]20\\%[\/latex] solution on hand, and need to mix them to create a [latex]16\\%[\/latex] solution.\r\n\r\nWe will use the following table to help us solve mixture problems. The units on the last row will change from problem to problem based on what the concentration is describing. In the above example, the last row would represent grams of salt.\r\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.6061%;\"><strong>Value<\/strong><\/td>\r\n<td style=\"width: 12.2727%;\"><\/td>\r\n<td style=\"width: 14.2424%;\"><\/td>\r\n<td style=\"width: 23.5737%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nConsider mixing the following two solutions together:\r\n<ul>\r\n \t<li>One is [latex]120[\/latex] mL of a [latex]9\\%\\dfrac{\\text{grams}}{\\text{mL}}[\/latex] solution,<\/li>\r\n \t<li>The other is [latex]75[\/latex] mL of a [latex]23\\%\\dfrac{\\text{grams}}{\\text{mL}}[\/latex] solution.<\/li>\r\n<\/ul>\r\nLet's determine the concentration of the resulting solution using our table. First we place the values we know into the table. Note that while the concentrations are usually given to you as percentages, we will use their decimal form when performing computations.\r\n<table class=\"lines\" style=\"border-collapse: collapse; width: 85.9726%; height: 70px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 27.9324%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\r\n<td style=\"width: 12.3963%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\r\n<td style=\"width: 30.0367%; height: 14px;\"><strong>Totals<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\r\n<td style=\"width: 27.9324%; height: 14px;\">[latex]120[\/latex] mL<\/td>\r\n<td style=\"width: 12.3963%; height: 14px;\">[latex]75[\/latex] mL<\/td>\r\n<td style=\"width: 30.0367%; height: 14px;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 42px;\">\r\n<td style=\"width: 15.6061%; height: 42px;\"><strong>Concentration<\/strong><\/td>\r\n<td style=\"width: 27.9324%; height: 42px;\">[latex]0.09\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td style=\"width: 12.3963%; height: 42px;\">[latex]0.23\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td style=\"width: 30.0367%; height: 42px;\"><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.6061%;\"><strong>Value<\/strong><\/td>\r\n<td style=\"width: 27.9324%;\"><\/td>\r\n<td style=\"width: 12.3963%;\"><\/td>\r\n<td style=\"width: 30.0367%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow fill in the values in the bottom row by multiplying amount by concentration in each column.\r\n<table class=\"lines\" style=\"border-collapse: collapse; width: 85.9726%; height: 70px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 27.9324%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\r\n<td style=\"width: 12.3963%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\r\n<td style=\"width: 30.0367%; height: 14px;\"><strong>Totals<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\r\n<td style=\"width: 27.9324%; height: 14px;\">[latex]\\require{color}120[\/latex] mL<\/td>\r\n<td style=\"width: 12.3963%; height: 14px;\">[latex]75[\/latex] mL<\/td>\r\n<td style=\"width: 30.0367%; height: 14px;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 42px;\">\r\n<td style=\"width: 15.6061%; height: 42px;\"><strong>Concentration<\/strong><\/td>\r\n<td style=\"width: 27.9324%; height: 42px;\">[latex]0.09\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td style=\"width: 12.3963%; height: 42px;\">[latex]0.23\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td style=\"width: 30.0367%; height: 42px;\"><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.6061%;\"><strong>Value<\/strong><\/td>\r\n<td style=\"width: 27.9324%;\">[latex]\\left(120{\\color{red}\\cancel{\\color{black}{\\text{ mL}}}}\\right)\\left(0.09\\dfrac{\\text{ grams }}{{\\color{red}\\cancel{\\color{black}{\\text{ mL }}}}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\r\n<td style=\"width: 12.3963%;\">[latex]\\left(75\\color{red}\\cancel{\\color{black}{\\text{ mL}}}\\right)\\left(0.23\\dfrac{\\text{ grams }}{\\color{red}\\cancel{\\color{black}{\\text{ mL }}}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\r\n<td style=\"width: 30.0367%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFinally, complete the Totals column. Our final answer is the total of the Concentration row, found by dividing totals of Value and Amount.\r\n<table class=\"lines\" style=\"border-collapse: collapse; width: 86.5013%; height: 112px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 27.9324%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\r\n<td style=\"width: 12.3963%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\r\n<td style=\"width: 30.5658%; height: 14px; text-align: center;\"><strong>Totals<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\r\n<td style=\"width: 27.9324%; height: 14px;\">[latex]120[\/latex] mL<\/td>\r\n<td style=\"width: 12.3963%; height: 14px;\">[latex]75[\/latex] mL<\/td>\r\n<td style=\"width: 30.5658%; height: 14px; text-align: center;\">[latex]195[\/latex] mL<\/td>\r\n<\/tr>\r\n<tr style=\"height: 42px;\">\r\n<td style=\"width: 15.6061%; height: 42px;\"><strong>Concentration<\/strong><\/td>\r\n<td style=\"width: 27.9324%; height: 42px;\">[latex]0.09\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td style=\"width: 12.3963%; height: 42px;\">[latex]0.23\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td style=\"width: 30.5658%; height: 42px; text-align: center;\">[latex]\\frac{28.05\\text{ grams }}{ 195 \\text{ mL }}\\approx0.14=14\\text{ % }[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 42px;\">\r\n<td style=\"width: 15.6061%; height: 42px;\"><strong>Value<\/strong><\/td>\r\n<td style=\"width: 27.9324%; height: 42px;\">[latex]10.8 \\text{ grams }[\/latex]<\/td>\r\n<td style=\"width: 12.3963%; height: 42px;\">\u00a0[latex]17.25 \\text{ grams }[\/latex]<\/td>\r\n<td style=\"width: 30.5658%; height: 42px;\">[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\nThe final solution has a concentration of about [latex]14\\%\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]. If the context of this concentration is clear or the units are not clearly specified, we will often just say the concentration is\u00a0[latex]14\\%[\/latex] without specifying units.\r\n\r\nIn 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 [latex]70[\/latex] mL of a [latex]50\\%[\/latex] methane solution. How much of an [latex]80\\%[\/latex] solution must she add so the final solution is [latex]60\\%[\/latex] methane?\r\n[reveal-answer q=\"274848\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"274848\"]\r\n\r\nLet's make sure we understand the problem and the variables we will be using first before starting to write equations.\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 [latex]70[\/latex] mL of [latex]50\\%[\/latex] methane and solution 2 is the unknown amount with [latex]80\\%[\/latex] methane. \u00a0We can call our unknown amount [latex]x[\/latex]. We know the final mixture has a concentration of [latex]60\\%[\/latex].\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 style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]70[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]x[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]0.5[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]0.8[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]0.6[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.6061%;\"><strong>Value (Methane)<\/strong><\/td>\r\n<td style=\"width: 12.2727%;\"><\/td>\r\n<td style=\"width: 14.2424%;\"><\/td>\r\n<td style=\"width: 23.5737%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFirst complete the first row by adding [latex]70+x[\/latex]. Then complete the last row by multiplying Amount by Concentration\u00a0in each column.\r\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]70[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]x[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]70+x[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]0.5[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]0.8[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]0.6[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.6061%;\"><strong>Value (Methane)<\/strong><\/td>\r\n<td style=\"width: 12.2727%;\">[latex]70\\cdot 0.5[\/latex]<\/td>\r\n<td style=\"width: 14.2424%;\">[latex]x\\cdot 0.8[\/latex]<\/td>\r\n<td style=\"width: 23.5737%;\">[latex](70+x)\\cdot 0.6[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe last row now represents an equation we can solve since the parts must add to the total.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}70\\cdot 0.5+x\\cdot 0.8&amp;=(70+x)\\cdot 0.6\\\\\r\n35+0.8x&amp;=42+0.6x&amp;&amp;\\color{blue}{\\textsf{simplify}}\\\\\r\n0.8x&amp;=7+0.6x&amp;&amp;\\color{blue}{\\textsf{subtract }35\\textsf{ from both sides}}\\\\\r\n0.2x&amp;=7&amp;&amp;\\color{blue}{\\textsf{subtract }0.6x\\textsf{ from both sides}}\\\\\r\nx&amp;=35&amp;&amp;\\color{blue}{\\textsf{divide by }0.2\\textsf{ on both sides}}\\end{align}[\/latex]<\/p>\r\nWe must add [latex]35[\/latex] mL of the [latex]80\\%[\/latex] methane solution to the original\u00a0[latex]70[\/latex] mL of [latex]50\\%[\/latex] methane in order to get the desired mixture of [latex]60\\%[\/latex] methane.\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 [latex]24\\%[\/latex] butterfat and another that is [latex]18\\%[\/latex] butterfat. How much of each should he use to end up with [latex]42[\/latex] gallons of [latex]20\\%[\/latex] 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 [latex]42[\/latex] 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 [latex]24\\%[\/latex] solution [latex]x[\/latex], and the unknown volume of the [latex]18\\%[\/latex] solution [latex]y[\/latex].\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know.\r\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]x[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]y[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]42[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]0.24[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]0.18[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]0.2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.6061%;\"><strong>Value (Butterfat)<\/strong><\/td>\r\n<td style=\"width: 12.2727%;\"><\/td>\r\n<td style=\"width: 14.2424%;\"><\/td>\r\n<td style=\"width: 23.5737%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nComplete the last row by multiplying Amount by Concentration\u00a0in each column.\r\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]x[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]y[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]42[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]0.24[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]0.18[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]0.2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.6061%;\"><strong>Value (Butterfat)<\/strong><\/td>\r\n<td style=\"width: 12.2727%;\">[latex]0.24x[\/latex]<\/td>\r\n<td style=\"width: 14.2424%;\">[latex]0.18y[\/latex]<\/td>\r\n<td style=\"width: 23.5737%;\">[latex]0.2\\cdot42=8.4[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRemember the \"Amount\" row and the \"Value\" row represent the idea that two parts must add to a total. The first and third rows thus give us a system of equations:\r\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}\\phantom{0.24}x+\\phantom{0.18}y&amp;=\\text{ }42\\\\ 0.24x+0.18y&amp;=\\text{ }8.4\\end{array}\\right.[\/latex]<\/p>\r\nWe can use either substitution or elimination to solve. In this example we demonstrate the elimination method and begin by eliminating [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}x+y&amp;=42&amp;\\xrightarrow{\\cdot \\; -0.24} &amp; -0.24x-0.24y &amp;=-10.08\\\\\r\n0.24x+0.18y&amp;=8.4&amp;\\xrightarrow{\\phantom{\\cdot -0.24}} &amp; 0.24x+0.18y&amp;=8.4 \\\\\r\n&amp; &amp; &amp; \\text{__________}&amp;\\text{______} \\\\\r\n&amp; &amp; &amp; -0.06y&amp;=-1.68 \\\\\r\n&amp; &amp; &amp; y &amp;=28 \\\\ \\end{array}[\/latex]<\/p>\r\nWe can substitute [latex]y=28[\/latex] into either of the original equations to solve for [latex]x[\/latex]. We use the first equation because it has easier numbers to work with.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}x+(28)&amp;=42\\\\ x&amp;=14\\\\ \\end{array}[\/latex]<\/p>\r\nThis can be interpreted as [latex]14[\/latex] gallons of [latex]24\\%[\/latex] butterfat milk added to [latex]28[\/latex] gallons of [latex]18\\%[\/latex] butterfat milk will give the desired [latex]42[\/latex] gallons of [latex]20\\%[\/latex] butterfat milk.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\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\r\nIn the preceding mixture problems, our equations involved a concentration which had the form of a rate (which we wrote as a percent). Here is one more example that uses a different kind of rate.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nA store owner wants to develop a new snack mix by mixing cashews and raisins. The cashews cost [latex]\\$6.50[\/latex] per pound and the raisins cost [latex]\\$4.00[\/latex] per pound. How much of each should be used to obtain [latex]40[\/latex] pounds of snack mix worth [latex]\\$4.75[\/latex] per pound?\r\n\r\n[reveal-answer q=\"309899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"309899\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We are asked to find the starting quantities of cashews and raisins with known prices per pound. The total number of final pounds and final price per pound are known.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>We will set [latex]c[\/latex] = pounds of cashews and [latex]r[\/latex] = pounds of raisins.\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know.\r\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]c[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]r[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]40[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Price Per Pound<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]6.50[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]4.00[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]4.75[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.6061%;\"><strong>Value ($)<\/strong><\/td>\r\n<td style=\"width: 12.2727%;\"><\/td>\r\n<td style=\"width: 14.2424%;\"><\/td>\r\n<td style=\"width: 23.5737%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nComplete the last row by multiplying Amount by Price Per Pound in each column.\r\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]c[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]r[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]40[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]6.50[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]4.00[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]4.75[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.6061%;\"><strong>Value (Butterfat)<\/strong><\/td>\r\n<td style=\"width: 12.2727%;\">[latex]6.5c[\/latex]<\/td>\r\n<td style=\"width: 14.2424%;\">[latex]4r[\/latex]<\/td>\r\n<td style=\"width: 23.5737%;\">[latex]40\\cdot 4.75[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRemember the \"Amount\" row and the \"Value\" row represent the idea that two parts must add to a total. The first and third rows thus give us a system of equations:\r\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}\\phantom{6.5}c+\\phantom{4}r&amp;=\\text{ }40\\\\ 6.5c+4r&amp;=\\text{ }190\\end{array}\\right.[\/latex]<\/p>\r\nSubstitution is probably easier since the first equation can easily be solved for either variable. We solve it for [latex]c.[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}c+r&amp;=40\\\\\r\nc&amp;=40-r\\end{align}[\/latex]<\/p>\r\nNow substitute [latex]40-r[\/latex] for [latex]c[\/latex] in the second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}6.5c+4r&amp;=190\\\\\r\n6.5(40-r)+4r&amp;=190\\\\\r\n260-6.5r+4r&amp;=190\\\\\r\n260-2.5r&amp;=190\\\\\r\n-2.5r&amp;=-70\\\\\r\nr&amp;=28\\end{align}[\/latex]<\/p>\r\nSubstitute [latex]r=28[\/latex] into the original first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}c+(28)&amp;=40\\\\ c&amp;=12 \\end{align}[\/latex]<\/p>\r\nTo achieve the desired mixture, the store owner needs to use [latex]12[\/latex] pounds of cashews and [latex]28[\/latex] pounds of raisins.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 id=\"motion\">Motion Problems<\/h2>\r\nIn this section we will work with <strong>uniform motion<\/strong>, or motion with a constant speed. Uniform motion can be modeled by the following equation:\r\n<div class=\"textbox key-takeaways\">\r\n<h3>UNIFORM MOTION<\/h3>\r\nIf an object in uniform motion has speed [latex]r[\/latex] and travels for amount of time [latex]t[\/latex], the distance [latex]d[\/latex] that it travels is given by\r\n<p style=\"text-align: center;\">[latex]d = r \\cdot t.[\/latex]<\/p>\r\n\r\n<\/div>\r\nHere is an example of a motion problem with two unknowns to solve for. Each object in motion results in a separate instance of the uniform motion equation.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nA train leaves a station traveling north at [latex]56[\/latex] km\/hr. A second train leaves [latex]4[\/latex] hours later traveling north on a parallel track at [latex]84[\/latex] km\/hr. How far from the station will the second train pass the first train?\r\n\r\n[reveal-answer q=\"66394\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"66394\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>The trains' speeds are known. We do not know their times but we have a relationship between the two times. We do not know the distances for either train but the distance will be equal when the second train is passing the first train.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>Let [latex]d[\/latex] be the unknown distance from the station when the second train passes the first one. Let [latex]t[\/latex] be the amount of time the <strong>second<\/strong> train is traveling. Hence the <strong>first train<\/strong> is traveling 4 hours longer, or\u00a0[latex]t+4.[\/latex]\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know. The table is not necessary, but can help you organize.\r\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 54px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Distance<\/strong><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Rate (or speed)<\/strong><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Time<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Train 1<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]d[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]56[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]t+4[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Train 2<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]d[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]84[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]t[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nEach row represents a uniform motion, which can be modeled by [latex]d=rt.[\/latex] Thus we have the system of equations\r\n<p style=\"text-align: center;\">[latex]\\left\\{\\begin{array}{ll}d\\;=&amp;56(t+4)\\\\\r\nd\\;=&amp;84t\\end{array}\\right.[\/latex]<\/p>\r\nThis is best solved with substitution since [latex]d[\/latex] is already solved for in both equations. The substitution will effectively set the right sides equal to each other.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}56(t+4)&amp;=84t\\\\\r\n56t+224&amp;=84t\\\\\r\n224&amp;=28t\\\\\r\n8&amp;=t\\end{align}[\/latex]<\/p>\r\nThe second train will pass the first after [latex]8[\/latex] hours. The original question asked us for the distance from the station, which is found by using [latex]t=8[\/latex] in either equation, for example\r\n<p style=\"text-align: center;\">[latex]d=84t=84(8)=672.[\/latex]<\/p>\r\nThe second train will pass the first [latex]672[\/latex] km from the station.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\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]. In examples, sometimes one or both of [latex]b[\/latex] and [latex]c[\/latex] is given to you. It is important that [latex]b[\/latex] is always first when subtracting.\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 [latex]60[\/latex] miles downstream for [latex]4[\/latex] hours and then took [latex]5[\/latex] 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 and Understand:\u00a0<\/strong>A picture can help us visualize what is happening. Distance and time are known to us for both trips.\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>Define and Translate:\u00a0\u00a0<\/strong>The unknowns are the speed of the ship [latex]b[\/latex] in still water and the speed of the current [latex]c,[\/latex] which are the same for both trips.\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]b+c[\/latex].\u00a0Going upstream, the current slows the ship\u00a0and so the actual rate is [latex]b-c[\/latex].\r\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 54px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Distance<\/strong><\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Rate (or speed)<\/strong><\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Time<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Downstream<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]60[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]b+c[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Upstream<\/strong><\/td>\r\n<td style=\"width: 12.2727%; height: 14px;\">[latex]60[\/latex]<\/td>\r\n<td style=\"width: 14.2424%; height: 14px;\">[latex]b-c[\/latex]<\/td>\r\n<td style=\"width: 23.5737%; height: 14px;\">[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<strong>Write and Solve:<\/strong>\u00a0Since [latex]d=rt[\/latex], we can\u00a0write an equation for each direction.\r\n<p style=\"text-align: center;\">[latex]\\left\\{\\begin{array}{rl}60\\;&amp;=(b+c)4\\\\\r\n60\\;&amp;=(b-c)5\\end{array}\\right.[\/latex]<\/p>\r\nDistribute in each equation, then solve by elimination.\r\n<div>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}4b+4c&amp;=60&amp;\\xrightarrow{\\cdot \\; 5} &amp; 20b+20c &amp;=300\\\\\r\n5b-5c&amp;=60&amp;\\xrightarrow{\\cdot \\; 4} &amp; 20b-20c&amp;=240 \\\\\r\n&amp; &amp; &amp; \\text{_______}&amp;\\text{______} \\\\\r\n&amp; &amp; &amp; 40b&amp;=540 \\\\\r\n&amp; &amp; &amp; b&amp;=\\dfrac{540}{40}=13.5 \\\\ \\end{array}[\/latex]<\/p>\r\nSubstitute [latex]b=13.5[\/latex] into the first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}4(13.5)+4c&amp;=60\\\\ 54+4c&amp;=60\\\\4c&amp;=6\\\\c&amp;=1.5\\\\ \\end{array}[\/latex]<\/p>\r\nThe boat is traveling [latex]13.5[\/latex] mph in still water, and the current is [latex]1.5[\/latex] 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\r\n\r\nDepending on the information given in the problem, there are many different setups with which you can start a motion problem. Focus on using the table with Distance = Rate [latex]\\cdot[\/latex] Time, filling in any known quantities and representing unknown quantities by variables. Make sure if you know any relationships between unknown quantities, you use as few total variables as possible. Also make sure you represent Rate by two components if necessary (for example, boat speed and current with boat speed first).\r\n<h2>Summary<\/h2>\r\nSystems of equations in two variables are useful for solving a variety of application problems. They can find break-even points in business applications, solve total value problems and mixture problems, and solve for unknowns in problems of motion. Often a chart or table can be useful to organize information that is known or unknown, and also to set up the equations.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Solve applications of systems of two linear equations with two variables using substitution or elimination.<\/li>\n<\/ul>\n<\/div>\n<p>In this section we present examples of applications which are solved using systems of linear equations.<\/p>\n<p>If you need a refresher on how to solve systems of linear equations by substitution and elimination, first visit <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-r-solving-linear-systems-of-equations-by-substitution-and-elimination\/\">Section 2.R: Solving Linear Systems of Equations by Substitution and Elimination<\/a>.<\/p>\n<h2 id=\"CRP\">Cost, Revenue, and Profit<\/h2>\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<p>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 per unit. 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 total cost of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The [latex]x[\/latex]-axis represents quantity in hundreds of units. The [latex]y[\/latex]-axis represents either cost or revenue in hundreds of dollars.<\/p>\n<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 [latex]700[\/latex] units are produced, the cost is [latex]\\$3,300[\/latex] and the revenue is also [latex]\\$3,300[\/latex]. In other words, the company breaks even if they produce and sell [latex]700[\/latex] units. They neither make money nor lose money.<\/p>\n<p>The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The <strong>profit function<\/strong> is the revenue function minus the cost function, written as [latex]P\\left(x\\right)=R\\left(x\\right)-C\\left(x\\right)[\/latex]. Note that when [latex]R(x)=C(x),[\/latex] then [latex]P(x)=0.[\/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 [latex]x[\/latex] and [latex]y[\/latex] for the cost equation.<\/li>\n<li>Interpret [latex]x[\/latex] and [latex]y[\/latex] for the revenue equation.<\/li>\n<li>Find and interpret the break-even point for this business.<\/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<ol>\n<li>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, [latex]x[\/latex] would represent the number of frames produced (instead of skateboards) and [latex]y[\/latex] would represent the amount of money it would cost to produce them (the same as the skateboard problem).<\/li>\n<li>The revenue equation represents money coming into the company, so in this context [latex]x[\/latex] still represents the number of bike frames manufactured, and [latex]y[\/latex] now represents the amount of money made from selling them.<\/li>\n<li>We must solve the system of equations\n<p style=\"text-align: center;\">[latex]\\left\\{\\begin{array}{l}y=0.85x+35,000\\\\ y=1.55x \\end{array}\\right.[\/latex]<\/p>\n<p>Substitution is the best method since both equations are already solved for [latex]y[\/latex]. 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}{rl}0.85x+35,000&=\\;1.55x\\\\ 35,000&=\\;0.7x\\\\ 50,000&=\\;x\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">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]y=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] which represents [latex]50,000[\/latex] bike frames sold and both cost and revenue of [latex]\\$77,500.[\/latex] If more than [latex]50,000[\/latex] bike frames are sold, a profit will result.<\/li>\n<\/ol>\n<h4>Analysis of the Solution<\/h4>\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]. The above problem could have also been solved by setting [latex]P\\left(x\\right)=0[\/latex] and solving. You may want to verify this yourself.<\/p>\n<p>The cost to produce [latex]50,000[\/latex] units is [latex]\\$77,500[\/latex], and the revenue from the sales of [latex]50,000[\/latex] units is also [latex]\\$77,500.[\/latex] To make a profit, the business must produce and sell more than [latex]50,000[\/latex] units.<\/p>\n<p><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 [latex]x[\/latex]-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 [latex]0[\/latex]. 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\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Total Value Problems<\/h2>\n<p>In the next set of examples, we are given two types of items and multiple descriptions of their &#8220;value.&#8221; For example, we may be given two types of tickets sold at a circus and the totals of number of tickets sold as well as the total sales (or revenue).<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The cost of a ticket to the circus is [latex]\\$25.00[\/latex] for children and [latex]\\$50.00[\/latex] for adults. On a certain day, attendance at the circus is [latex]2,000[\/latex] and the total gate revenue is [latex]\\$70,000[\/latex]. How many children and how many adults bought tickets?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><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 [latex]c=[\/latex] the number of children and [latex]a=[\/latex] 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 [latex]\\$25.00[\/latex] by the number of children, [latex]25c[\/latex]. The revenue from all adults can be found by multiplying [latex]\\$50.00[\/latex] by the number of adults, [latex]50a[\/latex]. The total revenue is [latex]\\$70,000[\/latex]. We can use this to write an equation for the revenue.<\/p>\n<p style=\"text-align: center;\">[latex]25c+50a=70,000[\/latex]<\/p>\n<p>We now have a system of linear equations in two variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{llll}c&+&a&=\\;2,000\\\\ 25c&+&50a&=\\;70,000\\end{array} \\right.[\/latex]<\/p>\n<p>In the first equation the coefficient of both variables is [latex]1[\/latex] so we solve by substitution (elimination would also be possible.) We will solve the first equation for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}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}{rl} 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}{rl}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 YOURSELF<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm202618\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=202618&theme=oea&iframe_resize_id=ohm202618\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 id=\"mixture\">Mixture Problems<\/h2>\n<p>A <strong>solution<\/strong> is a mixture of two or more different substances like water and salt or vinegar and oil. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand.<\/p>\n<p>The concentration or strength of a liquid solution is often described as a percentage. \u00a0This number comes from the ratio of how much mass is in a specific volume of liquid. \u00a0For example if you have [latex]50[\/latex] grams of salt in a [latex]100[\/latex] mL of water you have a [latex]50\\%[\/latex] 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{ % }\\frac{\\text{ grams }}{\\text{ mL }}[\/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. For example, you may have a [latex]9\\%[\/latex] solution and a [latex]20\\%[\/latex] solution on hand, and need to mix them to create a [latex]16\\%[\/latex] solution.<\/p>\n<p>We will use the following table to help us solve mixture problems. The units on the last row will change from problem to problem based on what the concentration is describing. In the above example, the last row would represent grams of salt.<\/p>\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.6061%;\"><strong>Value<\/strong><\/td>\n<td style=\"width: 12.2727%;\"><\/td>\n<td style=\"width: 14.2424%;\"><\/td>\n<td style=\"width: 23.5737%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Consider mixing the following two solutions together:<\/p>\n<ul>\n<li>One is [latex]120[\/latex] mL of a [latex]9\\%\\dfrac{\\text{grams}}{\\text{mL}}[\/latex] solution,<\/li>\n<li>The other is [latex]75[\/latex] mL of a [latex]23\\%\\dfrac{\\text{grams}}{\\text{mL}}[\/latex] solution.<\/li>\n<\/ul>\n<p>Let&#8217;s determine the concentration of the resulting solution using our table. First we place the values we know into the table. Note that while the concentrations are usually given to you as percentages, we will use their decimal form when performing computations.<\/p>\n<table class=\"lines\" style=\"border-collapse: collapse; width: 85.9726%; height: 70px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 27.9324%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\n<td style=\"width: 12.3963%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\n<td style=\"width: 30.0367%; height: 14px;\"><strong>Totals<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\n<td style=\"width: 27.9324%; height: 14px;\">[latex]120[\/latex] mL<\/td>\n<td style=\"width: 12.3963%; height: 14px;\">[latex]75[\/latex] mL<\/td>\n<td style=\"width: 30.0367%; height: 14px;\"><\/td>\n<\/tr>\n<tr style=\"height: 42px;\">\n<td style=\"width: 15.6061%; height: 42px;\"><strong>Concentration<\/strong><\/td>\n<td style=\"width: 27.9324%; height: 42px;\">[latex]0.09\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td style=\"width: 12.3963%; height: 42px;\">[latex]0.23\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td style=\"width: 30.0367%; height: 42px;\"><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.6061%;\"><strong>Value<\/strong><\/td>\n<td style=\"width: 27.9324%;\"><\/td>\n<td style=\"width: 12.3963%;\"><\/td>\n<td style=\"width: 30.0367%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now fill in the values in the bottom row by multiplying amount by concentration in each column.<\/p>\n<table class=\"lines\" style=\"border-collapse: collapse; width: 85.9726%; height: 70px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 27.9324%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\n<td style=\"width: 12.3963%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\n<td style=\"width: 30.0367%; height: 14px;\"><strong>Totals<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\n<td style=\"width: 27.9324%; height: 14px;\">[latex]\\require{color}120[\/latex] mL<\/td>\n<td style=\"width: 12.3963%; height: 14px;\">[latex]75[\/latex] mL<\/td>\n<td style=\"width: 30.0367%; height: 14px;\"><\/td>\n<\/tr>\n<tr style=\"height: 42px;\">\n<td style=\"width: 15.6061%; height: 42px;\"><strong>Concentration<\/strong><\/td>\n<td style=\"width: 27.9324%; height: 42px;\">[latex]0.09\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td style=\"width: 12.3963%; height: 42px;\">[latex]0.23\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td style=\"width: 30.0367%; height: 42px;\"><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.6061%;\"><strong>Value<\/strong><\/td>\n<td style=\"width: 27.9324%;\">[latex]\\left(120{\\color{red}\\cancel{\\color{black}{\\text{ mL}}}}\\right)\\left(0.09\\dfrac{\\text{ grams }}{{\\color{red}\\cancel{\\color{black}{\\text{ mL }}}}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\n<td style=\"width: 12.3963%;\">[latex]\\left(75\\color{red}\\cancel{\\color{black}{\\text{ mL}}}\\right)\\left(0.23\\dfrac{\\text{ grams }}{\\color{red}\\cancel{\\color{black}{\\text{ mL }}}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\n<td style=\"width: 30.0367%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Finally, complete the Totals column. Our final answer is the total of the Concentration row, found by dividing totals of Value and Amount.<\/p>\n<table class=\"lines\" style=\"border-collapse: collapse; width: 86.5013%; height: 112px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 27.9324%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\n<td style=\"width: 12.3963%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\n<td style=\"width: 30.5658%; height: 14px; text-align: center;\"><strong>Totals<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\n<td style=\"width: 27.9324%; height: 14px;\">[latex]120[\/latex] mL<\/td>\n<td style=\"width: 12.3963%; height: 14px;\">[latex]75[\/latex] mL<\/td>\n<td style=\"width: 30.5658%; height: 14px; text-align: center;\">[latex]195[\/latex] mL<\/td>\n<\/tr>\n<tr style=\"height: 42px;\">\n<td style=\"width: 15.6061%; height: 42px;\"><strong>Concentration<\/strong><\/td>\n<td style=\"width: 27.9324%; height: 42px;\">[latex]0.09\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td style=\"width: 12.3963%; height: 42px;\">[latex]0.23\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td style=\"width: 30.5658%; height: 42px; text-align: center;\">[latex]\\frac{28.05\\text{ grams }}{ 195 \\text{ mL }}\\approx0.14=14\\text{ % }[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 42px;\">\n<td style=\"width: 15.6061%; height: 42px;\"><strong>Value<\/strong><\/td>\n<td style=\"width: 27.9324%; height: 42px;\">[latex]10.8 \\text{ grams }[\/latex]<\/td>\n<td style=\"width: 12.3963%; height: 42px;\">\u00a0[latex]17.25 \\text{ grams }[\/latex]<\/td>\n<td style=\"width: 30.5658%; height: 42px;\">[latex]10.8\\text{ grams }+17.25\\text{ grams }=28.05\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The final solution has a concentration of about [latex]14\\%\\dfrac{\\text{ grams }}{\\text{ mL }}[\/latex]. If the context of this concentration is clear or the units are not clearly specified, we will often just say the concentration is\u00a0[latex]14\\%[\/latex] without specifying units.<\/p>\n<p>In 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 [latex]70[\/latex] mL of a [latex]50\\%[\/latex] methane solution. How much of an [latex]80\\%[\/latex] solution must she add so the final solution is [latex]60\\%[\/latex] methane?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><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 make sure we understand the problem and the variables we will be using first before starting to write equations.<\/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 [latex]70[\/latex] mL of [latex]50\\%[\/latex] methane and solution 2 is the unknown amount with [latex]80\\%[\/latex] methane. \u00a0We can call our unknown amount [latex]x[\/latex]. We know the final mixture has a concentration of [latex]60\\%[\/latex].<\/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 style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]70[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]0.5[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]0.8[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]0.6[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.6061%;\"><strong>Value (Methane)<\/strong><\/td>\n<td style=\"width: 12.2727%;\"><\/td>\n<td style=\"width: 14.2424%;\"><\/td>\n<td style=\"width: 23.5737%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>First complete the first row by adding [latex]70+x[\/latex]. Then complete the last row by multiplying Amount by Concentration\u00a0in each column.<\/p>\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]70[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]70+x[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]0.5[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]0.8[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]0.6[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.6061%;\"><strong>Value (Methane)<\/strong><\/td>\n<td style=\"width: 12.2727%;\">[latex]70\\cdot 0.5[\/latex]<\/td>\n<td style=\"width: 14.2424%;\">[latex]x\\cdot 0.8[\/latex]<\/td>\n<td style=\"width: 23.5737%;\">[latex](70+x)\\cdot 0.6[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The last row now represents an equation we can solve since the parts must add to the total.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}70\\cdot 0.5+x\\cdot 0.8&=(70+x)\\cdot 0.6\\\\  35+0.8x&=42+0.6x&&\\color{blue}{\\textsf{simplify}}\\\\  0.8x&=7+0.6x&&\\color{blue}{\\textsf{subtract }35\\textsf{ from both sides}}\\\\  0.2x&=7&&\\color{blue}{\\textsf{subtract }0.6x\\textsf{ from both sides}}\\\\  x&=35&&\\color{blue}{\\textsf{divide by }0.2\\textsf{ on both sides}}\\end{align}[\/latex]<\/p>\n<p>We must add [latex]35[\/latex] mL of the [latex]80\\%[\/latex] methane solution to the original\u00a0[latex]70[\/latex] mL of [latex]50\\%[\/latex] methane in order to get the desired mixture of [latex]60\\%[\/latex] methane.<\/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 [latex]24\\%[\/latex] butterfat and another that is [latex]18\\%[\/latex] butterfat. How much of each should he use to end up with [latex]42[\/latex] gallons of [latex]20\\%[\/latex] 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 [latex]42[\/latex] 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 [latex]24\\%[\/latex] solution [latex]x[\/latex], and the unknown volume of the [latex]18\\%[\/latex] solution [latex]y[\/latex].<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know.<\/p>\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]y[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]42[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]0.24[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]0.18[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]0.2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.6061%;\"><strong>Value (Butterfat)<\/strong><\/td>\n<td style=\"width: 12.2727%;\"><\/td>\n<td style=\"width: 14.2424%;\"><\/td>\n<td style=\"width: 23.5737%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Complete the last row by multiplying Amount by Concentration\u00a0in each column.<\/p>\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]y[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]42[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]0.24[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]0.18[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]0.2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.6061%;\"><strong>Value (Butterfat)<\/strong><\/td>\n<td style=\"width: 12.2727%;\">[latex]0.24x[\/latex]<\/td>\n<td style=\"width: 14.2424%;\">[latex]0.18y[\/latex]<\/td>\n<td style=\"width: 23.5737%;\">[latex]0.2\\cdot42=8.4[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Remember the &#8220;Amount&#8221; row and the &#8220;Value&#8221; row represent the idea that two parts must add to a total. The first and third rows thus give us a system of equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}\\phantom{0.24}x+\\phantom{0.18}y&=\\text{ }42\\\\ 0.24x+0.18y&=\\text{ }8.4\\end{array}\\right.[\/latex]<\/p>\n<p>We can use either substitution or elimination to solve. In this example we demonstrate the elimination method and begin by eliminating [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}x+y&=42&\\xrightarrow{\\cdot \\; -0.24} & -0.24x-0.24y &=-10.08\\\\  0.24x+0.18y&=8.4&\\xrightarrow{\\phantom{\\cdot -0.24}} & 0.24x+0.18y&=8.4 \\\\  & & & \\text{__________}&\\text{______} \\\\  & & & -0.06y&=-1.68 \\\\  & & & y &=28 \\\\ \\end{array}[\/latex]<\/p>\n<p>We can substitute [latex]y=28[\/latex] into either of the original equations to solve for [latex]x[\/latex]. We use the first equation because it has easier numbers to work with.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}x+(28)&=42\\\\ x&=14\\\\ \\end{array}[\/latex]<\/p>\n<p>This can be interpreted as [latex]14[\/latex] gallons of [latex]24\\%[\/latex] butterfat milk added to [latex]28[\/latex] gallons of [latex]18\\%[\/latex] butterfat milk will give the desired [latex]42[\/latex] gallons of [latex]20\\%[\/latex] butterfat milk.<\/p>\n<\/div>\n<\/div>\n<\/div>\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<p>In the preceding mixture problems, our equations involved a concentration which had the form of a rate (which we wrote as a percent). Here is one more example that uses a different kind of rate.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>A store owner wants to develop a new snack mix by mixing cashews and raisins. The cashews cost [latex]\\$6.50[\/latex] per pound and the raisins cost [latex]\\$4.00[\/latex] per pound. How much of each should be used to obtain [latex]40[\/latex] pounds of snack mix worth [latex]\\$4.75[\/latex] per pound?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q309899\">Show Solution<\/span><\/p>\n<div id=\"q309899\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong>We are asked to find the starting quantities of cashews and raisins with known prices per pound. The total number of final pounds and final price per pound are known.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>We will set [latex]c[\/latex] = pounds of cashews and [latex]r[\/latex] = pounds of raisins.<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know.<\/p>\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]c[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]r[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]40[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Price Per Pound<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]6.50[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]4.00[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]4.75[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.6061%;\"><strong>Value ($)<\/strong><\/td>\n<td style=\"width: 12.2727%;\"><\/td>\n<td style=\"width: 14.2424%;\"><\/td>\n<td style=\"width: 23.5737%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Complete the last row by multiplying Amount by Price Per Pound in each column.<\/p>\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 42px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Solution 1<\/strong><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Solution 2<\/strong><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Totals<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Amount<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]c[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]r[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]40[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Concentration<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]6.50[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]4.00[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]4.75[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.6061%;\"><strong>Value (Butterfat)<\/strong><\/td>\n<td style=\"width: 12.2727%;\">[latex]6.5c[\/latex]<\/td>\n<td style=\"width: 14.2424%;\">[latex]4r[\/latex]<\/td>\n<td style=\"width: 23.5737%;\">[latex]40\\cdot 4.75[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Remember the &#8220;Amount&#8221; row and the &#8220;Value&#8221; row represent the idea that two parts must add to a total. The first and third rows thus give us a system of equations:<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}\\phantom{6.5}c+\\phantom{4}r&=\\text{ }40\\\\ 6.5c+4r&=\\text{ }190\\end{array}\\right.[\/latex]<\/p>\n<p>Substitution is probably easier since the first equation can easily be solved for either variable. We solve it for [latex]c.[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}c+r&=40\\\\  c&=40-r\\end{align}[\/latex]<\/p>\n<p>Now substitute [latex]40-r[\/latex] for [latex]c[\/latex] in the second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}6.5c+4r&=190\\\\  6.5(40-r)+4r&=190\\\\  260-6.5r+4r&=190\\\\  260-2.5r&=190\\\\  -2.5r&=-70\\\\  r&=28\\end{align}[\/latex]<\/p>\n<p>Substitute [latex]r=28[\/latex] into the original first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}c+(28)&=40\\\\ c&=12 \\end{align}[\/latex]<\/p>\n<p>To achieve the desired mixture, the store owner needs to use [latex]12[\/latex] pounds of cashews and [latex]28[\/latex] pounds of raisins.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 id=\"motion\">Motion Problems<\/h2>\n<p>In this section we will work with <strong>uniform motion<\/strong>, or motion with a constant speed. Uniform motion can be modeled by the following equation:<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>UNIFORM MOTION<\/h3>\n<p>If an object in uniform motion has speed [latex]r[\/latex] and travels for amount of time [latex]t[\/latex], the distance [latex]d[\/latex] that it travels is given by<\/p>\n<p style=\"text-align: center;\">[latex]d = r \\cdot t.[\/latex]<\/p>\n<\/div>\n<p>Here is an example of a motion problem with two unknowns to solve for. Each object in motion results in a separate instance of the uniform motion equation.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>A train leaves a station traveling north at [latex]56[\/latex] km\/hr. A second train leaves [latex]4[\/latex] hours later traveling north on a parallel track at [latex]84[\/latex] km\/hr. How far from the station will the second train pass the first train?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q66394\">Show Solution<\/span><\/p>\n<div id=\"q66394\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong>The trains&#8217; speeds are known. We do not know their times but we have a relationship between the two times. We do not know the distances for either train but the distance will be equal when the second train is passing the first train.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>Let [latex]d[\/latex] be the unknown distance from the station when the second train passes the first one. Let [latex]t[\/latex] be the amount of time the <strong>second<\/strong> train is traveling. Hence the <strong>first train<\/strong> is traveling 4 hours longer, or\u00a0[latex]t+4.[\/latex]<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know. The table is not necessary, but can help you organize.<\/p>\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 54px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Distance<\/strong><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Rate (or speed)<\/strong><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Time<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Train 1<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]d[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]56[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]t+4[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Train 2<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]d[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]84[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]t[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Each row represents a uniform motion, which can be modeled by [latex]d=rt.[\/latex] Thus we have the system of equations<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{\\begin{array}{ll}d\\;=&56(t+4)\\\\  d\\;=&84t\\end{array}\\right.[\/latex]<\/p>\n<p>This is best solved with substitution since [latex]d[\/latex] is already solved for in both equations. The substitution will effectively set the right sides equal to each other.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}56(t+4)&=84t\\\\  56t+224&=84t\\\\  224&=28t\\\\  8&=t\\end{align}[\/latex]<\/p>\n<p>The second train will pass the first after [latex]8[\/latex] hours. The original question asked us for the distance from the station, which is found by using [latex]t=8[\/latex] in either equation, for example<\/p>\n<p style=\"text-align: center;\">[latex]d=84t=84(8)=672.[\/latex]<\/p>\n<p>The second train will pass the first [latex]672[\/latex] km from the station.<\/p>\n<\/div>\n<\/div>\n<\/div>\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]. In examples, sometimes one or both of [latex]b[\/latex] and [latex]c[\/latex] is given to you. It is important that [latex]b[\/latex] is always first when subtracting.<\/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 [latex]60[\/latex] miles downstream for [latex]4[\/latex] hours and then took [latex]5[\/latex] 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 and Understand:\u00a0<\/strong>A picture can help us visualize what is happening. Distance and time are known to us for both trips.<\/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>Define and Translate:\u00a0\u00a0<\/strong>The unknowns are the speed of the ship [latex]b[\/latex] in still water and the speed of the current [latex]c,[\/latex] which are the same for both trips.<\/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]b+c[\/latex].\u00a0Going upstream, the current slows the ship\u00a0and so the actual rate is [latex]b-c[\/latex].<\/p>\n<table style=\"border-collapse: collapse; width: 65.6944%; height: 54px;\">\n<tbody>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><\/td>\n<td style=\"width: 12.2727%; height: 14px;\"><strong>Distance<\/strong><\/td>\n<td style=\"width: 14.2424%; height: 14px;\"><strong>Rate (or speed)<\/strong><\/td>\n<td style=\"width: 23.5737%; height: 14px;\"><strong>Time<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Downstream<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]60[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]b+c[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]4[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 15.6061%; height: 14px;\"><strong>Upstream<\/strong><\/td>\n<td style=\"width: 12.2727%; height: 14px;\">[latex]60[\/latex]<\/td>\n<td style=\"width: 14.2424%; height: 14px;\">[latex]b-c[\/latex]<\/td>\n<td style=\"width: 23.5737%; height: 14px;\">[latex]5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Write and Solve:<\/strong>\u00a0Since [latex]d=rt[\/latex], we can\u00a0write an equation for each direction.<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{\\begin{array}{rl}60\\;&=(b+c)4\\\\  60\\;&=(b-c)5\\end{array}\\right.[\/latex]<\/p>\n<p>Distribute in each equation, then solve by elimination.<\/p>\n<div>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}4b+4c&=60&\\xrightarrow{\\cdot \\; 5} & 20b+20c &=300\\\\  5b-5c&=60&\\xrightarrow{\\cdot \\; 4} & 20b-20c&=240 \\\\  & & & \\text{_______}&\\text{______} \\\\  & & & 40b&=540 \\\\  & & & b&=\\dfrac{540}{40}=13.5 \\\\ \\end{array}[\/latex]<\/p>\n<p>Substitute [latex]b=13.5[\/latex] into the first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}4(13.5)+4c&=60\\\\ 54+4c&=60\\\\4c&=6\\\\c&=1.5\\\\ \\end{array}[\/latex]<\/p>\n<p>The boat is traveling [latex]13.5[\/latex] mph in still water, and the current is [latex]1.5[\/latex] 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<p>Depending on the information given in the problem, there are many different setups with which you can start a motion problem. Focus on using the table with Distance = Rate [latex]\\cdot[\/latex] Time, filling in any known quantities and representing unknown quantities by variables. Make sure if you know any relationships between unknown quantities, you use as few total variables as possible. Also make sure you represent Rate by two components if necessary (for example, boat speed and current with boat speed first).<\/p>\n<h2>Summary<\/h2>\n<p>Systems of equations in two variables are useful for solving a variety of application problems. They can find break-even points in business applications, solve total value problems and mixture problems, and solve for unknowns in problems of motion. Often a chart or table can be useful to organize information that is known or unknown, and also to set up the equations.<\/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-557\">\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>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><li>Revision and Adaptation. <strong>Authored by<\/strong>: Lumen Learning. <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>: 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>: 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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":16925,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"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\":\"Lumen Learning\",\"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 Systems of Equations using Elimination\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ova8GSmPV4o\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 115164, 115120, 115110\",\"author\":\"Shabazian, Roy\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"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 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\":\"Greg Harbaugh\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 2239\",\"author\":\"Morales, Lawrence\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Ex: System of Equations Application - 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