{"id":2015,"date":"2016-03-18T18:56:32","date_gmt":"2016-03-18T18:56:32","guid":{"rendered":"https:\/\/courses.candelalearning.com\/nrocarithmetic\/?post_type=chapter&#038;p=2015"},"modified":"2018-01-03T23:56:15","modified_gmt":"2018-01-03T23:56:15","slug":"applications-of-systems-of-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-beginalgebra\/chapter\/applications-of-systems-of-equations\/","title":{"raw":"Applications of Systems","rendered":"Applications of Systems"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Solve mixture problems\r\n<ul>\r\n \t<li>Write a system of linear equations representing a mixture problem, solve the system and interpret the results<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Solve value problems\r\n<ul>\r\n \t<li>Write a system of linear equations representing a number problem<\/li>\r\n \t<li>Determine and apply an appropriate method for solving the system<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Solve cost and revenue problems\r\n<ul>\r\n \t<li>Specify what the variables in a cost\/ revenue system of linear equations represent<\/li>\r\n \t<li>Determine and apply an appropriate method for solving the system<\/li>\r\n \t<li>Write a system of inequalites that represents the profit region<\/li>\r\n \t<li>Interpret the solutions to a system of cost\/ revenue equations and inequalities<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 id=\"title1\">Write a system of linear equations representing a mixture problem, solve the system and interpret the results<\/h2>\r\nOne application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. \u00a0A solution is a mixture of two or more different substances like water and salt or vinegar and oil. \u00a0Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. \u00a0There are many other disciplines that use solutions as well.\r\n\r\nThe concentration or strength of a liquid solution is often described\u00a0\u00a0as a percentage. \u00a0This number comes from the ratio of how much mass is in a specific volume of liquid. \u00a0For example if you have 50 grams of salt in a 100mL of water you have a 50% salt solution based on the following ratio:\r\n<p style=\"text-align: center;\">[latex]\\frac{50\\text{ grams }}{100\\text{ mL }}=0.50\\frac{\\text{ grams }}{\\text{ mL }}=50\\text{ % }[\/latex]<\/p>\r\nSolutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. \u00a0In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.\r\n\r\nWe will use the following table to help us solve mixture problems:\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\"><\/td>\r\n<td class=\"border\"><\/td>\r\n<td class=\"border\"><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\"><\/td>\r\n<td class=\"border\"><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center;\"><\/td>\r\n<td class=\"border\" style=\"text-align: center;\"><\/td>\r\n<td class=\"border\" style=\"text-align: center;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nTo demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is 120mL of a 9% solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.\r\n\r\nFirst, find the total mass of solids for each solution by multiplying the volume by the concentration.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0120 mL<\/td>\r\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a075 mL<\/td>\r\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center;\"><\/td>\r\n<td class=\"border\" style=\"text-align: center;\"><\/td>\r\n<td class=\"border\" style=\"text-align: center;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNext we add the new volumes and new masses.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0120 mL<\/td>\r\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a075 mL<\/td>\r\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">195 mL<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]\\frac{28.05\\text{ grams }}{ 195 \\text{ mL }}=0.14=14\\text{ % }[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]10.8\\text{ grams }+17.25\\text{ grams }=28.05\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. \u00a0In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA chemist has 70 mL of a 50% methane solution. How much of an 80% solution must she add so the final solution is 60% methane?\r\n[reveal-answer q=\"274848\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"274848\"]\r\n\r\nLet's use the problem solving process outlined in Module 1 to help us work through a solution to the problem.\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We are looking for a new amount - in this case a volume - \u00a0based on the words \"how much\". \u00a0We know two starting \u00a0concentrations and the final concentration, as well as one volume.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>Solution 1 is the 70 mL of 50% methane and solution 2 is the unknown amount with 80% methane. \u00a0We can call our unknown amount x.\r\n\r\n<strong>Write and Solve: \u00a0<\/strong>Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a070<\/td>\r\n<td class=\"border\">\u00a00.5<\/td>\r\n<td class=\"border\"><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0x<\/td>\r\n<td class=\"border\">\u00a00.8<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center;\"><\/td>\r\n<td class=\"border\" style=\"text-align: center;\">0.6<\/td>\r\n<td class=\"border\" style=\"text-align: center;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nMultiply amount by concentration\u00a0to get total,\u00a0be sure to distribute on the last row: [latex]\\left(70 + x\\right)0.6[\/latex]Add the entries in the amount column to get final amount. The concentration for this amount is 0.6 because we want the final solution to be 60% methane.\r\n<table class=\" undefined alignleft\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a070<\/td>\r\n<td class=\"border\">\u00a00.5<\/td>\r\n<td class=\"border\">\u00a035<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0x<\/td>\r\n<td class=\"border\">\u00a00.8<\/td>\r\n<td>\u00a00.8<em>x<\/em><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: left;\">\u00a070+x<\/td>\r\n<td class=\"border\" style=\"text-align: left;\">0.6<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">\u00a0[latex]42+0.6x[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAdd the total mass for solution 1 and solution 2 to get the total mass for the 60% solution. This is our equation for finding the unknown volume.\r\n\r\n[latex]35+0.8x=42+0.6x[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}35+0.8x=42+0.6x\\\\\\underline{-0.6x}\\,\\,\\,\\,\\,\\,\\,\\underline{-0.6x}\\\\35+0.2x=42\\\\\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Subtract 35 from both sides<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}35+0.2x=42\\\\\\underline{-35}\\,\\,\\,\\,\\,\\,\\,\\underline{-35}\\\\0.2x=7\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Divide both sides by 0.2<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}0.2x=7\\\\\\frac{0.2x}{0.2}=\\frac{7}{0.2}\\end{array}[\/latex]\r\nx=35<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n35mL must be added to the original 70 mL to gain a solution with a concentration of 60%\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe above problem illustrates how we can use\u00a0the mixture table\u00a0to define\u00a0an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons of 20% butterfat?\r\n[reveal-answer q=\"966963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"966963\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is 42 gallons. There are two unknowns in this problem.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>We will call the unknown volume of the \u00a024% solution x, and the unknown volume of the 18% solution y.\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0x<\/td>\r\n<td class=\"border\">\u00a00.24<\/td>\r\n<td class=\"border\"><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0y<\/td>\r\n<td class=\"border\">\u00a00.18<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: left;\">42<\/td>\r\n<td class=\"border\" style=\"text-align: left;\">0.2<\/td>\r\n<td class=\"border\" style=\"text-align: center;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFind the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0x<\/td>\r\n<td class=\"border\">\u00a00.24<\/td>\r\n<td class=\"border\">\u00a00.24x<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0y<\/td>\r\n<td class=\"border\">\u00a00.18<\/td>\r\n<td>\u00a00.18y<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: left;\">x+y=42<\/td>\r\n<td class=\"border\" style=\"text-align: left;\">0.2<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">8.4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen you sum the amount column you get\u00a0one equation: x+ y = 42\r\nWhen you sum the total column you get a second equation: 0.24x + 0.18y = 8.4\r\n\r\nUse elimination to find a value for x, and y.\r\n\r\nMultiply the first equation by -0.18\r\n\r\n-0.18(x+y) = (42)(-0.18)\r\n\r\n-0.18x - -0.18y = -7.56\r\n\r\nNow our system of equations looks like this:\r\n\r\n-0.18x - -0.18y = -7.56\r\n\r\n0.24x + 0.18y = 8.4\r\n\r\nAdding the two equations together to eliminate the y terms gives this equation:\r\n\r\n0.06x = 8.4\r\n\r\nDivide by 0.06 on each side:\r\n\r\nx = 14\r\n\r\nNow substitute the value for x into one of the equations in order to solve for y.\r\n\r\n(14) + y = 42\r\n\r\ny = 28\r\n<h4>Answer<\/h4>\r\nThis can be interpreted as 14 gallons of 24% butterfat milk added to 28 gallons of 18% butterfat milk will give 42 gallons of 20% butterfat milk.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.\r\n\r\nhttps:\/\/youtu.be\/4s5MCqphpKo\r\n<h2>Write a system of linear equations\u00a0representing\u00a0a value\u00a0problem<\/h2>\r\nSystems of equations are a very useful tool for modeling real-life situations and answering questions about them. If you can translate the application into two linear equations with two variables, then you have a system of equations that you can solve to find the solution. You can use any method to solve the system of equations.\r\n\r\nOne application of system of equations are known as value problems. Value problems are ones where each variable has a value attached to it. For example, the marketing team for an event venue wants to know how to focus their advertising based on who is attending specific events\u2014children, or adults? \u00a0They know the cost of a ticket to a basketball game is $25.00 for children and $50.00 for adults. Additionally, on a certain day, attendance at the game is 2,000 and the total gate revenue is $70,000. \u00a0How can the marketing team use this information to find out whether to spend more money on advertising directed toward children or adults?\r\n\r\nWe will use a table to help us set up and solve this value problem. The basic structure of the table is shown below:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Number (usually what you are trying to find)<\/th>\r\n<th>Value<\/th>\r\n<th>Total<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Item 1<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Item 2<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Total<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe first column in the table is used for the number of things we have. Quite often, this will be our variables. The second column is used for the value each item has. The third column is used for the total value which we calculate by multiplying the number by the value.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the total number of child and adult tickets sold given that the cost of a ticket to a basketball game is $25.00 for children and $50.00 for adults. Additionally, on a certain day, attendance at the game is 2,000 and the total gate revenue is $70,000.\r\n[reveal-answer q=\"181202\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"181202\"]\r\n\r\n<strong>Read and Understand:<\/strong> We want to find the number of child and adult tickets, we know the\u00a0total number of tickets sold, the total revenue and the cost of a child and adult ticket.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>Let <em>c<\/em> = the number of children and <em>a<\/em> = the number of adults in attendance. \u00a0Revenue comes from number of tickets sold multiplied by the price of the ticket. \u00a0We will get revenue for adults by multiplying $50.00 times a. \u00a0$25.00 times c will give the revenue from the number of child tickets sold.\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>We can use a table as we did in the mixture problems section to organize the information we have. \u00a0Although a table is not necessary, it can help you get started. \u00a0For this problem, we labeled columns as amount, value, and total revenue because that is the information we are given.\r\n\r\nThe total number of people is [latex]2,000[\/latex].\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Value<\/th>\r\n<th>Total Revenue<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Child Tickets<\/td>\r\n<td class=\"border\">\u00a0c<\/td>\r\n<td class=\"border\">\u00a0$25.00<\/td>\r\n<td class=\"border\">25c<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Adult Tickets<\/td>\r\n<td class=\"border\">\u00a0a<\/td>\r\n<td class=\"border\">\u00a0$50.00<\/td>\r\n<td>\u00a050a<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">Total Tickets<\/td>\r\n<td class=\"border\" style=\"text-align: left;\">2000<\/td>\r\n<td class=\"border\" style=\"text-align: left;\"><\/td>\r\n<td class=\"border\" style=\"text-align: center;\">$70,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe total revenue is $70,000. We can use this and the revenue from child and adult tickets to write an equation for the revenue.[latex]25c+50a=70,000[\/latex]\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Value<\/th>\r\n<th>Total Revenue<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Child Tickets<\/td>\r\n<td class=\"border\">\u00a0c<\/td>\r\n<td class=\"border\">\u00a0$25.00<\/td>\r\n<td class=\"border\">25c<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Adult Tickets<\/td>\r\n<td class=\"border\">\u00a0a<\/td>\r\n<td class=\"border\">\u00a0$50.00<\/td>\r\n<td>\u00a050a<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">Total Tickets<\/td>\r\n<td class=\"border\" style=\"text-align: left;\">2000<\/td>\r\n<td class=\"border\" style=\"text-align: left;\"><\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]25c+50a=70,000[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe number of people at the game\u00a0that day is the total number of child tickets sold plus the total number of adult tickets, [latex]c+a=2,000[\/latex]\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Value<\/th>\r\n<th>Total Revenue<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Child Tickets<\/td>\r\n<td class=\"border\">\u00a0c<\/td>\r\n<td class=\"border\">\u00a0$25.00<\/td>\r\n<td class=\"border\">25c<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Adult Tickets<\/td>\r\n<td class=\"border\">\u00a0a<\/td>\r\n<td class=\"border\">\u00a0$50.00<\/td>\r\n<td>\u00a050a<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">Total Tickets<\/td>\r\n<td class=\"border\" style=\"text-align: left;\">[latex]c+a=2,000[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: left;\"><\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]25c+50a=70,000[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: center;\">We now have a system of linear equations in two variables.[latex]\\begin{array}{r}c+a=2,000\\,\\,\\,\\\\ 25c+50a=70,000\\end{array}[\/latex].<\/p>\r\n<p style=\"text-align: left;\">We can use any method of solving systems of equations to solve this system for a and c. \u00a0Substitution looks easiest because we can \u00a0solve the first equation for either [latex]c[\/latex] or [latex]a[\/latex]. We will solve for [latex]a[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}c+a=2,000\\\\ a=2,000-c\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Substitute the expression [latex]2,000-c[\/latex] in the second equation for [latex]a[\/latex] and solve for [latex]c[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r} 25c+50\\left(2,000-c\\right)=70,000\\,\\,\\,\\, \\\\ 25c+100,000 - 50c=70,000\\,\\,\\,\\, \\\\ -25c=-30,000 \\\\ c=1,200\\,\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Substitute [latex]c=1,200[\/latex] into the first equation to solve for [latex]a[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}1,200+a=2,000 \\\\ a=800\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n<p style=\"text-align: left;\">We find that [latex]1,200[\/latex] children and [latex]800[\/latex] adults bought tickets to the game\u00a0that day. The marketing group may want to focus their advertising toward attracting young people.<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nThis example showed you how to find two unknown values given information that connected the two unknowns. With two equations, you are able to find a solution for two unknowns. \u00a0If you were to have three unknowns, you would need three equations to find them, and so on.\r\n\r\nIn the following video, you are given an example of how to use a system of equations to find the number of children and adults admitted to an amusement park based on entrance fees and total revenue. This example shows how to write equations and solve the system without a table.\r\n\r\nhttps:\/\/youtu.be\/uH4CgUhuDv0\r\n\r\nIn the next example, we will find the number of coins in a change jar given the total amount of money in the jar and the fact that the coins are either quarters or dimes.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn a change jar there\u00a0are 11 coins that have a value of S1.85. The coins are either quarters or dimes. How many of each kind of coin is in the jar?\r\n[reveal-answer q=\"698872\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"698872\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We want to find the number of quarters and the number of dimes in the jar. \u00a0We know that dimes are $0.10 and quarters are $0.25, and the total number of coins is 11.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>We will call the number of quarters q and the number of dimes d. The part of the total $1.85 that comes from quarters will be determined by how many quarters and the fact that each one is worth $0.25, so $0.25q represents the amount of $1.85 that is quarters. \u00a0The same idea can be used for dimes, so $0.10d represents the amount of $1.85 that is dimes.\r\n\r\n<strong>Write and Solve:<\/strong> We can label a new table with the information we are given.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">number<\/th>\r\n<th class=\"border\">value<\/th>\r\n<th>total<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">quarters<\/td>\r\n<td class=\"border\">\u00a0q<\/td>\r\n<td class=\"border\">\u00a0$0.25<\/td>\r\n<td class=\"border\">$0.25q<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">dimes<\/td>\r\n<td class=\"border\">\u00a0d<\/td>\r\n<td class=\"border\">\u00a0$0.10<\/td>\r\n<td>\u00a0$0.10d<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left;\">total number of coins<\/td>\r\n<td class=\"border\" style=\"text-align: left;\">q+d=11<\/td>\r\n<td class=\"border\" style=\"text-align: left;\"><\/td>\r\n<td class=\"border\" style=\"text-align: center;\">$0.25q+$0.10d=$1.85<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe can write our two equations, remember that we need two to solve for two unknowns.\r\n<p style=\"text-align: center;\">\u00a0[latex]\\begin{array}{r}q+d=11\\,\\,\\,\\\\0.25q+0.10d=1.85\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Substitution looks like the easiest path to a solution, solve for q.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}q+d=11\\\\ q=11-d\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Substitute this into the other equation, and solve for d.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}0.25\\left(11-d\\right)+0.10d=1.85\\,\\,\\,\\, \\\\2.75-0.25d+0.10d=1.85\\,\\,\\,\\,\\\\ 2.75-0.15d=1.85\\\\-0.15d=-0.9\\\\\\,\\,\\,\\,\\,\\,\\, d=6\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Substitute [latex]d=6[\/latex] into the first equation to solve for [latex]q[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}q+6=11 \\\\q=5\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\nWe have 6 dimes and 5 quarters.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p class=\"no-indent\">In the following video, you will see an example similar to the previous one, except that the equations are written and solved without the use of a table.<\/p>\r\nhttps:\/\/youtu.be\/GZYtSP-X_is\r\n<h2 class=\"no-indent\">Cost and Revenue Problems<\/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<img class=\"size-full wp-image-5880 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/08\/01202934\/CNX_Precalc_Figure_09_01_0012.jpg\" alt=\"Skateboarders at a skating rink by the beach.\" width=\"487\" height=\"252\" \/>\r\n\r\n<b><\/b>Using what we have learned about systems of equations, we can answer these questions.\u00a0The skateboard manufacturer\u2019s revenue equation is the equation used to calculate the amount of money that comes into the business. It can be represented as [latex]y=xp[\/latex], where [latex]x=[\/latex] quantity and [latex]p=[\/latex] price. The revenue equation is shown in orange in the graph below.\r\n\r\nThe cost equation is the equation used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost equation is shown in blue in the graph below. The [latex]x[\/latex] -axis represents quantity in hundreds of units. The <em>y<\/em>-axis represents both\u00a0cost and revenue in hundreds of dollars.\u00a0We won't learn how to write a cost equation in this example, they will be given to you. If you take any business or economics courses, you will learn more about how to write a cost equation.\r\n\r\n<img class=\"aligncenter wp-image-5882 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/08\/01203010\/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 break-even point, we learned that this is the solution to the system of linear equations that in this case comprise the cost and revenue equations.\r\n\r\nRead the axes of the graph carefully, note that quantity is in hundreds, and money is in thousands. The solution to the graphed system is (7, 33). This means\u00a0that if 700 units are produced, the cost to make them is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.\r\n\r\nThe shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA business wants to manufacture bike frames. Before they start production, they need to make sure they can make a profit with the materials and labor force they have. Their accountant has given them a cost equation of [latex]y=0.85x+35,000[\/latex] and a revenue equation of [latex]y=1.55x[\/latex]:\r\n<ol>\r\n \t<li>Interpret x and y for the cost equation<\/li>\r\n \t<li>Interpret x and y for the revenue equation<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"86281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"86281\"]\r\n\r\nCost: [latex]y=0.85x+35,000[\/latex]\r\n\r\nRevenue:[latex]y=1.55x[\/latex]\r\n\r\nThe cost equation represents money leaving the company, namely how much it costs to produce a given number of bike frames. If we use the skateboard example as a model, x would represent the number of frames produced (instead of skateboards) and y would represent the amount of money it would cost to produce them (the same as the skateboard problem).\r\n\r\nThe revenue equation represents money coming into the company, so in this context x still represents the number of bike frames manufactured, and y now represents the amount of money made from selling them. \u00a0Let's organize this information in a table:\r\n<table>\r\n<thead>\r\n<tr>\r\n<td>Equation Type<\/td>\r\n<td>x represents<\/td>\r\n<td>y represents<\/td>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Revenue Eqn.<\/td>\r\n<td>number of frames<\/td>\r\n<td>amount of money made selling frames<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Cost Eqn.<\/td>\r\n<td>number of frames<\/td>\r\n<td>cost for making frames<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGiven the same cost and revenue equations from the previous example, find the break-even point for the bike manufacturer. \u00a0Interpret the solution with words.\r\n\r\nCost: [latex]y=0.85x+35,000[\/latex]\r\n\r\nRevenue: [latex]y=1.55x[\/latex]\r\n[reveal-answer q=\"145700\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"145700\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We want the break even point for this system that represents cost and revenue. \u00a0This means we want to find where the two lines cross, and we have learned a few different methods for doing this because this is the solution to the system of equations! \u00a0Substitution looks like the easiest method since the revenue equation is already solved for y,\u00a0[latex]y=1.55x[\/latex].\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>Write the system of equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ y=0.85x+35,000\\hfill \\\\ y=1.55x\\hfill \\end{array}\\\\[\/latex]<\/p>\r\n<strong>Write and Solve:\u00a0<\/strong>The equations are already written for us, so we just need to solve the system using substitution.\r\n\r\nSubstitute the expression [latex]0.85x+35,000[\/latex] from the first equation into the second equation and solve for [latex]x[\/latex].\r\n[latex]\\begin{array}{r}0.85x+35,000=1.55x\\\\ 35,000=0.7x\\,\\,\\,\\\\ 50,000=x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}\\\\[\/latex]\r\nThen, we substitute [latex]x=50,000[\/latex] into either the cost equation or the revenue equation.\r\n[latex]1.55\\left(50,000\\right)=77,500\\\\[\/latex]\r\nThe break-even point is [latex]\\left(50,000,77,500\\right)[\/latex].\r\n<h4><strong>Check and Interpret:<\/strong><\/h4>\r\nThe solution to this system is[latex]\\left(50,000,77,500\\right)[\/latex], but what does that mean? Think of a point as \u00a0[latex]\\left(x,y)[\/latex], where in this case x is the quantity of bikes manufactured and y is an amount of money. For our system y represents two different things and x represents one thing. \u00a0Refer to tehe table we made in the first example, shown below:\r\n<table>\r\n<thead>\r\n<tr>\r\n<td>Equation Type<\/td>\r\n<td>x represents<\/td>\r\n<td>y represents<\/td>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Revenue Eqn.<\/td>\r\n<td>number of frames<\/td>\r\n<td>amount of money made selling frames<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Cost Eqn.<\/td>\r\n<td>number of frames<\/td>\r\n<td>cost for making frames<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nLet's interpret the solution with respect to the Cost equation first. x = 50,000 and y = 77,500. \u00a0Using our table, we can translate this as \"the cost for producing 50,000 bike frames is $75,500\".\r\n\r\nIn the same way, the Revenue equation can be interpreted as \"the amount of money the company makes from selling 50,000 bike frames is $77,500\".\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, you will see how the information you learned about systems of linear inequalities can be applied to answering questions about cost and revenue. \u00a0Below is a graph\u00a0of the Cost\/ Revenue system in the previous system:\r\n\r\n<img class=\"size-full wp-image-5883 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/08\/01203134\/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\nNote how the blue shaded region between the Cost and Revenue equations is labeled Profit. This is the \"sweet spot\" that the company wants to achieve\u00a0where they\u00a0produce enough bike frames at a minimal enough cost to\u00a0make money. They don't want more money going out than coming in!\r\n\r\nThe following example shows how to write the system of linear equations\u00a0as a system of linear inequalities whose solution set is the profit region for the system.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nDefine the profit region for the bike manufacturing business using inequalities, given the system of linear equations:\r\n\r\nCost: [latex]y=0.85x+35,000[\/latex]\r\n\r\nRevenue: [latex]y=1.55x[\/latex]\r\n[reveal-answer q=\"563864\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"563864\"]\r\n\r\n<strong>Read and Understand:<\/strong> We know that graphically, \u00a0solutions to\u00a0linear inequalities are entire regions, and we learned how to graph systems of linear inequalities earlier in this module. Based on the graph below and the equations that define cost and revenue, we can use inequalities to define the region for which the bike manufacturer will make a profit.\r\n\r\n<img class=\" wp-image-4210 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/18223321\/Screen-Shot-2016-05-18-at-3.33.11-PM-300x280.png\" alt=\"Cost\/ Revenue with Profit\" width=\"358\" height=\"334\" \/>\r\n\r\nLet's start with the revenue equation. \u00a0We know that the break even point is at (50,000, 77,500) and the profit region is\u00a0the blue area. \u00a0If we choose a point in the region and test it like we did for finding solution regions to inequalities, we will know which kind of inequality sign to use.\r\n\r\nLet's test the point [latex]\\left(65,00,100,000\\right)[\/latex] in both equations to determine which inequality sign to use.\r\n\r\nCost:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=0.85x+{35,000}\\\\{100,000}\\text{ ? }0.85\\left(65,000\\right)+35,000\\\\100,000\\text{ ? }90,250\\end{array}[\/latex]<\/p>\r\nWe need to use &gt; because 100,000 is greater than 90,250\r\n\r\nThe cost inequality that will ensure the company makes profit - not just break even - is\u00a0[latex]y&gt;0.85x+35,000[\/latex]\r\n\r\nNow test the point in the revenue equation:\r\n\r\nRevenue:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=1.55x\\\\100,000\\text{ ? }1.55\\left(65,000\\right)\\\\100,000\\text{ ? }100,750\\end{array}[\/latex]<\/p>\r\nWe need to use &lt; because 100,000 is less than 100,750\r\n\r\nThe revenue inequality that will ensure the company makes profit - not just break even - is\u00a0[latex]y&lt;1.55x[\/latex]\r\n\r\nThe systems of inequalities that defines the profit region for the bike manufacturer:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y&gt;0.85x+35,000\\\\y&lt;1.55x\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. The system of linear inequalities that represents the number of units that the company must produce in order to earn a profit is:\r\n\r\n[latex]\\begin{array}{l}y&gt;0.85x+35,000\\\\y&lt;1.55x\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video you will see an example of how to find the break even point for a small sno-cone business.\r\n\r\nhttps:\/\/youtu.be\/qey3FmE8saQ\r\n<h2 class=\"no-indent\">Summary<\/h2>\r\n<p id=\"video1\" class=\"no-indent\"><span style=\"color: #ff0000;\"><span style=\"color: #000000;\">In this section, we saw two examples of writing a system of two linear equations to find two unknowns that were related to each other. \u00a0In the first, the equations were related by the sum of the number of tickets bought and the sum of the total revenue brought in by the tickets sold. \u00a0In the second problem, the relationships were similar. \u00a0The two variables were related by the sum of the number of coins, and the total value of the coins.<\/span><\/span><\/p>\r\n<p class=\"no-indent\">We have seen that systems of linear equations and inequalities can help to define market behaviors that are very helpful to businesses. \u00a0The intersection of cost and revenue equations gives the break even point, and also helps define the region for which a company will make a profit.<\/p>\r\n&nbsp;","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Solve mixture problems\n<ul>\n<li>Write a system of linear equations representing a mixture problem, solve the system and interpret the results<\/li>\n<\/ul>\n<\/li>\n<li>Solve value problems\n<ul>\n<li>Write a system of linear equations representing a number problem<\/li>\n<li>Determine and apply an appropriate method for solving the system<\/li>\n<\/ul>\n<\/li>\n<li>Solve cost and revenue problems\n<ul>\n<li>Specify what the variables in a cost\/ revenue system of linear equations represent<\/li>\n<li>Determine and apply an appropriate method for solving the system<\/li>\n<li>Write a system of inequalites that represents the profit region<\/li>\n<li>Interpret the solutions to a system of cost\/ revenue equations and inequalities<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h2 id=\"title1\">Write a system of linear equations representing a mixture problem, solve the system and interpret the results<\/h2>\n<p>One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. \u00a0A solution is a mixture of two or more different substances like water and salt or vinegar and oil. \u00a0Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. \u00a0There are many other disciplines that use solutions as well.<\/p>\n<p>The concentration or strength of a liquid solution is often described\u00a0\u00a0as a percentage. \u00a0This number comes from the ratio of how much mass is in a specific volume of liquid. \u00a0For example if you have 50 grams of salt in a 100mL of water you have a 50% salt solution based on the following ratio:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{50\\text{ grams }}{100\\text{ mL }}=0.50\\frac{\\text{ grams }}{\\text{ mL }}=50\\text{ % }[\/latex]<\/p>\n<p>Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. \u00a0In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.<\/p>\n<p>We will use the following table to help us solve mixture problems:<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\"><\/td>\n<td class=\"border\"><\/td>\n<td class=\"border\"><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\"><\/td>\n<td class=\"border\"><\/td>\n<td><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center;\"><\/td>\n<td class=\"border\" style=\"text-align: center;\"><\/td>\n<td class=\"border\" style=\"text-align: center;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is 120mL of a 9% solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.<\/p>\n<p>First, find the total mass of solids for each solution by multiplying the volume by the concentration.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0120 mL<\/td>\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a075 mL<\/td>\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center;\"><\/td>\n<td class=\"border\" style=\"text-align: center;\"><\/td>\n<td class=\"border\" style=\"text-align: center;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Next we add the new volumes and new masses.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0120 mL<\/td>\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a075 mL<\/td>\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center;\">195 mL<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]\\frac{28.05\\text{ grams }}{ 195 \\text{ mL }}=0.14=14\\text{ % }[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]10.8\\text{ grams }+17.25\\text{ grams }=28.05\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. \u00a0In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A chemist has 70 mL of a 50% methane solution. How much of an 80% solution must she add so the final solution is 60% methane?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q274848\">Show Solution<\/span><\/p>\n<div id=\"q274848\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let&#8217;s use the problem solving process outlined in Module 1 to help us work through a solution to the problem.<\/p>\n<p><strong>Read and Understand:\u00a0<\/strong>We are looking for a new amount &#8211; in this case a volume &#8211; \u00a0based on the words &#8220;how much&#8221;. \u00a0We know two starting \u00a0concentrations and the final concentration, as well as one volume.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>Solution 1 is the 70 mL of 50% methane and solution 2 is the unknown amount with 80% methane. \u00a0We can call our unknown amount x.<\/p>\n<p><strong>Write and Solve: \u00a0<\/strong>Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a070<\/td>\n<td class=\"border\">\u00a00.5<\/td>\n<td class=\"border\"><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0x<\/td>\n<td class=\"border\">\u00a00.8<\/td>\n<td><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center;\"><\/td>\n<td class=\"border\" style=\"text-align: center;\">0.6<\/td>\n<td class=\"border\" style=\"text-align: center;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Multiply amount by concentration\u00a0to get total,\u00a0be sure to distribute on the last row: [latex]\\left(70 + x\\right)0.6[\/latex]Add the entries in the amount column to get final amount. The concentration for this amount is 0.6 because we want the final solution to be 60% methane.<\/p>\n<table class=\"undefined alignleft\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a070<\/td>\n<td class=\"border\">\u00a00.5<\/td>\n<td class=\"border\">\u00a035<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0x<\/td>\n<td class=\"border\">\u00a00.8<\/td>\n<td>\u00a00.8<em>x<\/em><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: left;\">\u00a070+x<\/td>\n<td class=\"border\" style=\"text-align: left;\">0.6<\/td>\n<td class=\"border\" style=\"text-align: center;\">\u00a0[latex]42+0.6x[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Add the total mass for solution 1 and solution 2 to get the total mass for the 60% solution. This is our equation for finding the unknown volume.<\/p>\n<p>[latex]35+0.8x=42+0.6x[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}35+0.8x=42+0.6x\\\\\\underline{-0.6x}\\,\\,\\,\\,\\,\\,\\,\\underline{-0.6x}\\\\35+0.2x=42\\\\\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Subtract 35 from both sides<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}35+0.2x=42\\\\\\underline{-35}\\,\\,\\,\\,\\,\\,\\,\\underline{-35}\\\\0.2x=7\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Divide both sides by 0.2<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}0.2x=7\\\\\\frac{0.2x}{0.2}=\\frac{7}{0.2}\\end{array}[\/latex]<br \/>\nx=35<\/p>\n<h4>Answer<\/h4>\n<p>35mL must be added to the original 70 mL to gain a solution with a concentration of 60%<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The above problem illustrates how we can use\u00a0the mixture table\u00a0to define\u00a0an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons of 20% butterfat?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q966963\">Show Solution<\/span><\/p>\n<div id=\"q966963\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong>We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is 42 gallons. There are two unknowns in this problem.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>We will call the unknown volume of the \u00a024% solution x, and the unknown volume of the 18% solution y.<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0x<\/td>\n<td class=\"border\">\u00a00.24<\/td>\n<td class=\"border\"><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0y<\/td>\n<td class=\"border\">\u00a00.18<\/td>\n<td><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: left;\">42<\/td>\n<td class=\"border\" style=\"text-align: left;\">0.2<\/td>\n<td class=\"border\" style=\"text-align: center;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0x<\/td>\n<td class=\"border\">\u00a00.24<\/td>\n<td class=\"border\">\u00a00.24x<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0y<\/td>\n<td class=\"border\">\u00a00.18<\/td>\n<td>\u00a00.18y<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: left;\">x+y=42<\/td>\n<td class=\"border\" style=\"text-align: left;\">0.2<\/td>\n<td class=\"border\" style=\"text-align: center;\">8.4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When you sum the amount column you get\u00a0one equation: x+ y = 42<br \/>\nWhen you sum the total column you get a second equation: 0.24x + 0.18y = 8.4<\/p>\n<p>Use elimination to find a value for x, and y.<\/p>\n<p>Multiply the first equation by -0.18<\/p>\n<p>-0.18(x+y) = (42)(-0.18)<\/p>\n<p>-0.18x &#8211; -0.18y = -7.56<\/p>\n<p>Now our system of equations looks like this:<\/p>\n<p>-0.18x &#8211; -0.18y = -7.56<\/p>\n<p>0.24x + 0.18y = 8.4<\/p>\n<p>Adding the two equations together to eliminate the y terms gives this equation:<\/p>\n<p>0.06x = 8.4<\/p>\n<p>Divide by 0.06 on each side:<\/p>\n<p>x = 14<\/p>\n<p>Now substitute the value for x into one of the equations in order to solve for y.<\/p>\n<p>(14) + y = 42<\/p>\n<p>y = 28<\/p>\n<h4>Answer<\/h4>\n<p>This can be interpreted as 14 gallons of 24% butterfat milk added to 28 gallons of 18% butterfat milk will give 42 gallons of 20% butterfat milk.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" 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<h2>Write a system of linear equations\u00a0representing\u00a0a value\u00a0problem<\/h2>\n<p>Systems of equations are a very useful tool for modeling real-life situations and answering questions about them. If you can translate the application into two linear equations with two variables, then you have a system of equations that you can solve to find the solution. You can use any method to solve the system of equations.<\/p>\n<p>One application of system of equations are known as value problems. Value problems are ones where each variable has a value attached to it. For example, the marketing team for an event venue wants to know how to focus their advertising based on who is attending specific events\u2014children, or adults? \u00a0They know the cost of a ticket to a basketball game is $25.00 for children and $50.00 for adults. Additionally, on a certain day, attendance at the game is 2,000 and the total gate revenue is $70,000. \u00a0How can the marketing team use this information to find out whether to spend more money on advertising directed toward children or adults?<\/p>\n<p>We will use a table to help us set up and solve this value problem. The basic structure of the table is shown below:<\/p>\n<table>\n<thead>\n<tr>\n<th>Number (usually what you are trying to find)<\/th>\n<th>Value<\/th>\n<th>Total<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Item 1<\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Item 2<\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Total<\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The first column in the table is used for the number of things we have. Quite often, this will be our variables. The second column is used for the value each item has. The third column is used for the total value which we calculate by multiplying the number by the value.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the total number of child and adult tickets sold given that the cost of a ticket to a basketball game is $25.00 for children and $50.00 for adults. Additionally, on a certain day, attendance at the game is 2,000 and the total gate revenue is $70,000.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q181202\">Show Solution<\/span><\/p>\n<div id=\"q181202\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:<\/strong> We want to find the number of child and adult tickets, we know the\u00a0total number of tickets sold, the total revenue and the cost of a child and adult ticket.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>Let <em>c<\/em> = the number of children and <em>a<\/em> = the number of adults in attendance. \u00a0Revenue comes from number of tickets sold multiplied by the price of the ticket. \u00a0We will get revenue for adults by multiplying $50.00 times a. \u00a0$25.00 times c will give the revenue from the number of child tickets sold.<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>We can use a table as we did in the mixture problems section to organize the information we have. \u00a0Although a table is not necessary, it can help you get started. \u00a0For this problem, we labeled columns as amount, value, and total revenue because that is the information we are given.<\/p>\n<p>The total number of people is [latex]2,000[\/latex].<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Value<\/th>\n<th>Total Revenue<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Child Tickets<\/td>\n<td class=\"border\">\u00a0c<\/td>\n<td class=\"border\">\u00a0$25.00<\/td>\n<td class=\"border\">25c<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Adult Tickets<\/td>\n<td class=\"border\">\u00a0a<\/td>\n<td class=\"border\">\u00a0$50.00<\/td>\n<td>\u00a050a<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">Total Tickets<\/td>\n<td class=\"border\" style=\"text-align: left;\">2000<\/td>\n<td class=\"border\" style=\"text-align: left;\"><\/td>\n<td class=\"border\" style=\"text-align: center;\">$70,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The total revenue is $70,000. We can use this and the revenue from child and adult tickets to write an equation for the revenue.[latex]25c+50a=70,000[\/latex]<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Value<\/th>\n<th>Total Revenue<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Child Tickets<\/td>\n<td class=\"border\">\u00a0c<\/td>\n<td class=\"border\">\u00a0$25.00<\/td>\n<td class=\"border\">25c<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Adult Tickets<\/td>\n<td class=\"border\">\u00a0a<\/td>\n<td class=\"border\">\u00a0$50.00<\/td>\n<td>\u00a050a<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">Total Tickets<\/td>\n<td class=\"border\" style=\"text-align: left;\">2000<\/td>\n<td class=\"border\" style=\"text-align: left;\"><\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]25c+50a=70,000[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The number of people at the game\u00a0that day is the total number of child tickets sold plus the total number of adult tickets, [latex]c+a=2,000[\/latex]<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Value<\/th>\n<th>Total Revenue<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Child Tickets<\/td>\n<td class=\"border\">\u00a0c<\/td>\n<td class=\"border\">\u00a0$25.00<\/td>\n<td class=\"border\">25c<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Adult Tickets<\/td>\n<td class=\"border\">\u00a0a<\/td>\n<td class=\"border\">\u00a0$50.00<\/td>\n<td>\u00a050a<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">Total Tickets<\/td>\n<td class=\"border\" style=\"text-align: left;\">[latex]c+a=2,000[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: left;\"><\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]25c+50a=70,000[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">We now have a system of linear equations in two variables.[latex]\\begin{array}{r}c+a=2,000\\,\\,\\,\\\\ 25c+50a=70,000\\end{array}[\/latex].<\/p>\n<p style=\"text-align: left;\">We can use any method of solving systems of equations to solve this system for a and c. \u00a0Substitution looks easiest because we can \u00a0solve the first equation for either [latex]c[\/latex] or [latex]a[\/latex]. We will solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}c+a=2,000\\\\ a=2,000-c\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">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}{r} 25c+50\\left(2,000-c\\right)=70,000\\,\\,\\,\\, \\\\ 25c+100,000 - 50c=70,000\\,\\,\\,\\, \\\\ -25c=-30,000 \\\\ c=1,200\\,\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">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}{r}1,200+a=2,000 \\\\ a=800\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p style=\"text-align: left;\">We find that [latex]1,200[\/latex] children and [latex]800[\/latex] adults bought tickets to the game\u00a0that day. The marketing group may want to focus their advertising toward attracting young people.<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<p>This example showed you how to find two unknown values given information that connected the two unknowns. With two equations, you are able to find a solution for two unknowns. \u00a0If you were to have three unknowns, you would need three equations to find them, and so on.<\/p>\n<p>In the following video, you are given an example of how to use a system of equations to find the number of children and adults admitted to an amusement park based on entrance fees and total revenue. This example shows how to write equations and solve the system without a table.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  System of Equations Application - Entrance Fees\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/uH4CgUhuDv0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next example, we will find the number of coins in a change jar given the total amount of money in the jar and the fact that the coins are either quarters or dimes.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In a change jar there\u00a0are 11 coins that have a value of S1.85. The coins are either quarters or dimes. How many of each kind of coin is in the jar?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q698872\">Show Solution<\/span><\/p>\n<div id=\"q698872\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong>We want to find the number of quarters and the number of dimes in the jar. \u00a0We know that dimes are $0.10 and quarters are $0.25, and the total number of coins is 11.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>We will call the number of quarters q and the number of dimes d. The part of the total $1.85 that comes from quarters will be determined by how many quarters and the fact that each one is worth $0.25, so $0.25q represents the amount of $1.85 that is quarters. \u00a0The same idea can be used for dimes, so $0.10d represents the amount of $1.85 that is dimes.<\/p>\n<p><strong>Write and Solve:<\/strong> We can label a new table with the information we are given.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">number<\/th>\n<th class=\"border\">value<\/th>\n<th>total<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">quarters<\/td>\n<td class=\"border\">\u00a0q<\/td>\n<td class=\"border\">\u00a0$0.25<\/td>\n<td class=\"border\">$0.25q<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">dimes<\/td>\n<td class=\"border\">\u00a0d<\/td>\n<td class=\"border\">\u00a0$0.10<\/td>\n<td>\u00a0$0.10d<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left;\">total number of coins<\/td>\n<td class=\"border\" style=\"text-align: left;\">q+d=11<\/td>\n<td class=\"border\" style=\"text-align: left;\"><\/td>\n<td class=\"border\" style=\"text-align: center;\">$0.25q+$0.10d=$1.85<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We can write our two equations, remember that we need two to solve for two unknowns.<\/p>\n<p style=\"text-align: center;\">\u00a0[latex]\\begin{array}{r}q+d=11\\,\\,\\,\\\\0.25q+0.10d=1.85\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Substitution looks like the easiest path to a solution, solve for q.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}q+d=11\\\\ q=11-d\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Substitute this into the other equation, and solve for d.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}0.25\\left(11-d\\right)+0.10d=1.85\\,\\,\\,\\, \\\\2.75-0.25d+0.10d=1.85\\,\\,\\,\\,\\\\ 2.75-0.15d=1.85\\\\-0.15d=-0.9\\\\\\,\\,\\,\\,\\,\\,\\, d=6\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Substitute [latex]d=6[\/latex] into the first equation to solve for [latex]q[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}q+6=11 \\\\q=5\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p>We have 6 dimes and 5 quarters.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p class=\"no-indent\">In the following video, you will see an example similar to the previous one, except that the equations are written and solved without the use of a table.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  System of Equations Application - Coin Problem\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/GZYtSP-X_is?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 class=\"no-indent\">Cost and Revenue Problems<\/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<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-5880 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/08\/01202934\/CNX_Precalc_Figure_09_01_0012.jpg\" alt=\"Skateboarders at a skating rink by the beach.\" width=\"487\" height=\"252\" \/><\/p>\n<p><b><\/b>Using what we have learned about systems of equations, we can answer these questions.\u00a0The skateboard manufacturer\u2019s revenue equation is the equation used to calculate the amount of money that comes into the business. It can be represented as [latex]y=xp[\/latex], where [latex]x=[\/latex] quantity and [latex]p=[\/latex] price. The revenue equation is shown in orange in the graph below.<\/p>\n<p>The cost equation is the equation used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost equation is shown in blue in the graph below. The [latex]x[\/latex] -axis represents quantity in hundreds of units. The <em>y<\/em>-axis represents both\u00a0cost and revenue in hundreds of dollars.\u00a0We won&#8217;t learn how to write a cost equation in this example, they will be given to you. If you take any business or economics courses, you will learn more about how to write a cost equation.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-5882 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/08\/01203010\/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 break-even point, we learned that this is the solution to the system of linear equations that in this case comprise the cost and revenue equations.<\/p>\n<p>Read the axes of the graph carefully, note that quantity is in hundreds, and money is in thousands. The solution to the graphed system is (7, 33). This means\u00a0that if 700 units are produced, the cost to make them is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.<\/p>\n<p>The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A business wants to manufacture bike frames. Before they start production, they need to make sure they can make a profit with the materials and labor force they have. Their accountant has given them a cost equation of [latex]y=0.85x+35,000[\/latex] and a revenue equation of [latex]y=1.55x[\/latex]:<\/p>\n<ol>\n<li>Interpret x and y for the cost equation<\/li>\n<li>Interpret x and y for the revenue equation<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q86281\">Show Solution<\/span><\/p>\n<div id=\"q86281\" class=\"hidden-answer\" style=\"display: none\">\n<p>Cost: [latex]y=0.85x+35,000[\/latex]<\/p>\n<p>Revenue:[latex]y=1.55x[\/latex]<\/p>\n<p>The cost equation represents money leaving the company, namely how much it costs to produce a given number of bike frames. If we use the skateboard example as a model, x would represent the number of frames produced (instead of skateboards) and y would represent the amount of money it would cost to produce them (the same as the skateboard problem).<\/p>\n<p>The revenue equation represents money coming into the company, so in this context x still represents the number of bike frames manufactured, and y now represents the amount of money made from selling them. \u00a0Let&#8217;s organize this information in a table:<\/p>\n<table>\n<thead>\n<tr>\n<td>Equation Type<\/td>\n<td>x represents<\/td>\n<td>y represents<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Revenue Eqn.<\/td>\n<td>number of frames<\/td>\n<td>amount of money made selling frames<\/td>\n<\/tr>\n<tr>\n<td>Cost Eqn.<\/td>\n<td>number of frames<\/td>\n<td>cost for making frames<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Given the same cost and revenue equations from the previous example, find the break-even point for the bike manufacturer. \u00a0Interpret the solution with words.<\/p>\n<p>Cost: [latex]y=0.85x+35,000[\/latex]<\/p>\n<p>Revenue: [latex]y=1.55x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q145700\">Show Solution<\/span><\/p>\n<div id=\"q145700\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong>We want the break even point for this system that represents cost and revenue. \u00a0This means we want to find where the two lines cross, and we have learned a few different methods for doing this because this is the solution to the system of equations! \u00a0Substitution looks like the easiest method since the revenue equation is already solved for y,\u00a0[latex]y=1.55x[\/latex].<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>Write the system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ y=0.85x+35,000\\hfill \\\\ y=1.55x\\hfill \\end{array}\\\\[\/latex]<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>The equations are already written for us, so we just need to solve the system using substitution.<\/p>\n<p>Substitute the expression [latex]0.85x+35,000[\/latex] from the first equation into the second equation and solve for [latex]x[\/latex].<br \/>\n[latex]\\begin{array}{r}0.85x+35,000=1.55x\\\\ 35,000=0.7x\\,\\,\\,\\\\ 50,000=x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}\\\\[\/latex]<br \/>\nThen, we substitute [latex]x=50,000[\/latex] into either the cost equation or the revenue equation.<br \/>\n[latex]1.55\\left(50,000\\right)=77,500\\\\[\/latex]<br \/>\nThe break-even point is [latex]\\left(50,000,77,500\\right)[\/latex].<\/p>\n<h4><strong>Check and Interpret:<\/strong><\/h4>\n<p>The solution to this system is[latex]\\left(50,000,77,500\\right)[\/latex], but what does that mean? Think of a point as \u00a0[latex]\\left(x,y)[\/latex], where in this case x is the quantity of bikes manufactured and y is an amount of money. For our system y represents two different things and x represents one thing. \u00a0Refer to tehe table we made in the first example, shown below:<\/p>\n<table>\n<thead>\n<tr>\n<td>Equation Type<\/td>\n<td>x represents<\/td>\n<td>y represents<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Revenue Eqn.<\/td>\n<td>number of frames<\/td>\n<td>amount of money made selling frames<\/td>\n<\/tr>\n<tr>\n<td>Cost Eqn.<\/td>\n<td>number of frames<\/td>\n<td>cost for making frames<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Let&#8217;s interpret the solution with respect to the Cost equation first. x = 50,000 and y = 77,500. \u00a0Using our table, we can translate this as &#8220;the cost for producing 50,000 bike frames is $75,500&#8221;.<\/p>\n<p>In the same way, the Revenue equation can be interpreted as &#8220;the amount of money the company makes from selling 50,000 bike frames is $77,500&#8221;.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, you will see how the information you learned about systems of linear inequalities can be applied to answering questions about cost and revenue. \u00a0Below is a graph\u00a0of the Cost\/ Revenue system in the previous system:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-5883 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/08\/01203134\/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>Note how the blue shaded region between the Cost and Revenue equations is labeled Profit. This is the &#8220;sweet spot&#8221; that the company wants to achieve\u00a0where they\u00a0produce enough bike frames at a minimal enough cost to\u00a0make money. They don&#8217;t want more money going out than coming in!<\/p>\n<p>The following example shows how to write the system of linear equations\u00a0as a system of linear inequalities whose solution set is the profit region for the system.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Define the profit region for the bike manufacturing business using inequalities, given the system of linear equations:<\/p>\n<p>Cost: [latex]y=0.85x+35,000[\/latex]<\/p>\n<p>Revenue: [latex]y=1.55x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q563864\">Show Solution<\/span><\/p>\n<div id=\"q563864\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:<\/strong> We know that graphically, \u00a0solutions to\u00a0linear inequalities are entire regions, and we learned how to graph systems of linear inequalities earlier in this module. Based on the graph below and the equations that define cost and revenue, we can use inequalities to define the region for which the bike manufacturer will make a profit.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4210 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/05\/18223321\/Screen-Shot-2016-05-18-at-3.33.11-PM-300x280.png\" alt=\"Cost\/ Revenue with Profit\" width=\"358\" height=\"334\" \/><\/p>\n<p>Let&#8217;s start with the revenue equation. \u00a0We know that the break even point is at (50,000, 77,500) and the profit region is\u00a0the blue area. \u00a0If we choose a point in the region and test it like we did for finding solution regions to inequalities, we will know which kind of inequality sign to use.<\/p>\n<p>Let&#8217;s test the point [latex]\\left(65,00,100,000\\right)[\/latex] in both equations to determine which inequality sign to use.<\/p>\n<p>Cost:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=0.85x+{35,000}\\\\{100,000}\\text{ ? }0.85\\left(65,000\\right)+35,000\\\\100,000\\text{ ? }90,250\\end{array}[\/latex]<\/p>\n<p>We need to use &gt; because 100,000 is greater than 90,250<\/p>\n<p>The cost inequality that will ensure the company makes profit &#8211; not just break even &#8211; is\u00a0[latex]y>0.85x+35,000[\/latex]<\/p>\n<p>Now test the point in the revenue equation:<\/p>\n<p>Revenue:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=1.55x\\\\100,000\\text{ ? }1.55\\left(65,000\\right)\\\\100,000\\text{ ? }100,750\\end{array}[\/latex]<\/p>\n<p>We need to use &lt; because 100,000 is less than 100,750<\/p>\n<p>The revenue inequality that will ensure the company makes profit &#8211; not just break even &#8211; is\u00a0[latex]y<1.55x[\/latex]\n\nThe systems of inequalities that defines the profit region for the bike manufacturer:\n\n\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y>0.85x+35,000\\\\y<1.55x\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. The system of linear inequalities that represents the number of units that the company must produce in order to earn a profit is:<\/p>\n<p>[latex]\\begin{array}{l}y>0.85x+35,000\\\\y<1.55x\\end{array}[\/latex]\n\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video you will see an example of how to find the break even point for a small sno-cone business.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"System of Equations App:  Break-Even Point\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/qey3FmE8saQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 class=\"no-indent\">Summary<\/h2>\n<p id=\"video1\" class=\"no-indent\"><span style=\"color: #ff0000;\"><span style=\"color: #000000;\">In this section, we saw two examples of writing a system of two linear equations to find two unknowns that were related to each other. \u00a0In the first, the equations were related by the sum of the number of tickets bought and the sum of the total revenue brought in by the tickets sold. \u00a0In the second problem, the relationships were similar. \u00a0The two variables were related by the sum of the number of coins, and the total value of the coins.<\/span><\/span><\/p>\n<p class=\"no-indent\">We have seen that systems of linear equations and inequalities can help to define market behaviors that are very helpful to businesses. \u00a0The intersection of cost and revenue equations gives the break even point, and also helps define the region for which a company will make a profit.<\/p>\n<p>&nbsp;<\/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-2015\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>System of Equations App: Break-Even Point. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/qey3FmE8saQ\">https:\/\/youtu.be\/qey3FmE8saQ<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>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>Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/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>College Algebra. <strong>Authored by<\/strong>: Jay Abrams, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstaxcollege.org\/textbooks\/college-algebra.\">https:\/\/openstaxcollege.org\/textbooks\/college-algebra.<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>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\">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>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstaxcollege.org\/textbooks\/college-algebra\">https:\/\/openstaxcollege.org\/textbooks\/college-algebra<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":20,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Beginning and Intermediate Algebra Textbook\",\"author\":\"Tyler Wallace\",\"organization\":\"\",\"url\":\"www.wallace.ccfaculty.org\/book\/book.html\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstaxcollege.org\/textbooks\/college-algebra\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: System of Equations Application - 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