{"id":6748,"date":"2020-10-08T17:13:07","date_gmt":"2020-10-08T17:13:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6748"},"modified":"2022-10-19T03:14:54","modified_gmt":"2022-10-19T03:14:54","slug":"4-4-applications-of-2x2-systems-of-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/4-4-applications-of-2x2-systems-of-equations\/","title":{"raw":"4.4: Applications of 2x2 Systems of Equations","rendered":"4.4: Applications of 2×2 Systems of Equations"},"content":{"raw":"
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section 4.4 Learning Objectives<\/h3>\r\n
\r\n
<\/div>\r\n
\r\n\r\n4.4:\u00a0 Applications of 2x2 Systems of Equations\u00a0<\/strong>\r\n\r\n<\/div>\r\n
\r\n
    \r\n \t
  • Set up and solve a linear system of equations by direct translation<\/li>\r\n \t
  • Use linear systems to solve value problems<\/li>\r\n \t
  • Use linear systems to solve mixture problems<\/li>\r\n \t
  • Use linear systems to solve motion problems<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n \r\n

    Write a system of linear equations from direct translations<\/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 of the methods learned in this module to solve the system of equations.\r\n\r\nTo introduce the topic, let us begin with direct translations.\r\n\r\nIn Section 1.7, we learned how to translate many key phrases in English into mathematical equations.\u00a0 Similarly, we can translate from English to form a system of equations.\r\n
    \r\n

    Example 1<\/h3>\r\nOne number is 3 less than 4 times another.\u00a0 If the sum of these numbers is 32, find the two numbers.\r\n\r\n[reveal-answer q=\"465773\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"465773\"]\r\n\r\nThere are two numbers in the problem, which we can represent by the variables x and y.\r\n\r\nExamining the first sentence, we recognize some familiar key phrases. Recall that \"is\" will correspond to\u00a0equals<\/em> while \"less than\" is\u00a0subtraction<\/em>, while being careful with the order as this implies 3 is being subtracted from 4 times the second number. So, the statement \"one number is 3 less than 4 times another\" results in the equation\r\n

    [latex]x=4y-3[\/latex]<\/p>\r\nThe second sentence discusses the \"sum,\" which indicates addition. Thus, \"the sum of these numbers is 32\" translates to the equation\r\n

    [latex]x+y=32[\/latex]<\/p>\r\nWe now have a system of equations to solve.\r\n

    [latex]\\begin{array}{l}x=4y-3 \\\\ x+y=32\\end{array}[\/latex]<\/p>\r\nWhile we can use any of the techniques learned in this module to achieve this, substitution will work particularly nicely since the first equation already has one of the variables isolated. Substituting [latex]x=4y-3[\/latex] into the second equation for [latex]x[\/latex] yields\r\n

    [latex]\\begin{array}{r}x+y=32 \\\\ 4y-3+y=32 \\\\ 5y-3=32 \\\\ 5y=35 \\\\ y=7\\hspace{.05in}\\end{array}[\/latex]<\/p>\r\nWe can now substitute [latex]y=7[\/latex] into either of the original equations to find [latex]x[\/latex].\r\n

    [latex]\\begin{array}{l}x=4y-3\\\\x=4(7)-3\\\\x=28-3\\\\x=25\\end{array}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\nThe two numbers are 25 and 7.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n

    Write a system of linear equations\u00a0representing\u00a0a value\u00a0problem<\/h2>\r\nOne application of systems of equations is 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\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nNumber (usually what you are trying to find)<\/th>\r\nValue<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Item 1<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    Item 2<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    Total<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/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
    \r\n

    Example 2<\/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\nRead 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\nDefine and Translate:\u00a0<\/strong>Let c<\/em> = the number of children and 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<\/em>. \u00a0$25.00 times c will give the revenue from the number of child tickets sold.\r\n\r\nWrite and Solve:\u00a0<\/strong>Although a table is not necessary, it can help you get started. \u00a0For this problem, we labeled columns as number, 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\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nNumber<\/th>\r\nValue<\/th>\r\nTotal Revenue<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Child Tickets<\/td>\r\n\u00a0c<\/td>\r\n\u00a0$25.00<\/td>\r\n25c<\/td>\r\n<\/tr>\r\n
    Adult Tickets<\/td>\r\n\u00a0a<\/em><\/td>\r\n\u00a0$50.00<\/td>\r\n50a<\/em><\/td>\r\n<\/tr>\r\n
    Total Tickets<\/td>\r\n2000<\/td>\r\n<\/td>\r\n$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.\u00a0 [latex]25c+50a=70,000[\/latex]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nNumber<\/th>\r\nValue<\/th>\r\nTotal Revenue<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Child Tickets<\/td>\r\n\u00a0c<\/td>\r\n\u00a0$25.00<\/td>\r\n25c<\/td>\r\n<\/tr>\r\n
    Adult Tickets<\/td>\r\n\u00a0a<\/em><\/td>\r\n\u00a0$50.00<\/td>\r\n\u00a050a<\/em><\/td>\r\n<\/tr>\r\n
    Total Tickets<\/td>\r\n2000<\/td>\r\n<\/td>\r\n[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\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nNumber<\/th>\r\nValue<\/th>\r\nTotal Revenue<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Child Tickets<\/td>\r\n\u00a0c<\/td>\r\n\u00a0$25.00<\/td>\r\n25c<\/td>\r\n<\/tr>\r\n
    Adult Tickets<\/td>\r\n\u00a0a<\/em><\/td>\r\n\u00a0$50.00<\/td>\r\n\u00a050a<\/em><\/td>\r\n<\/tr>\r\n
    Total Tickets<\/td>\r\n[latex]c+a=2,000[\/latex]<\/td>\r\n<\/td>\r\n[latex]25c+50a=70,000[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    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

    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

    [latex]\\begin{array}{c}c+a=2,000\\\\ a=2,000-c\\end{array}[\/latex]<\/p>\r\n

    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

    [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

    Substitute [latex]c=1,200[\/latex] into the first equation to solve for [latex]a[\/latex].<\/p>\r\n

    [latex]\\begin{array}{r}1,200+a=2,000 \\\\ a=800\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n

    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

    [\/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

    \r\n

    Example 3<\/h3>\r\nIn a change jar there\u00a0are 11 coins that have a value of $1.85. 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\nRead 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\nDefine 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\nWrite and Solve:<\/strong> We can label a new table with the information we are given.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nNumber<\/th>\r\nValue<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Quarters<\/td>\r\n\u00a0q<\/td>\r\n\u00a0$0.25<\/td>\r\n$0.25q<\/td>\r\n<\/tr>\r\n
    Dimes<\/td>\r\n\u00a0d<\/td>\r\n\u00a0$0.10<\/td>\r\n\u00a0$0.10d<\/td>\r\n<\/tr>\r\n
    Total number of coins<\/td>\r\nq+d=11<\/td>\r\n<\/td>\r\n$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

    \u00a0[latex]\\begin{array}{r}q+d=11\\,\\,\\,\\\\0.25q+0.10d=1.85\\end{array}[\/latex]<\/p>\r\n

    Substitution looks like the easiest path to a solution, solve for q.<\/p>\r\n

    [latex]\\begin{array}{c}q+d=11\\\\ q=11-d\\end{array}[\/latex]<\/p>\r\n

    Substitute this into the other equation, and solve for d.<\/p>\r\n

    [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

    Substitute [latex]d=6[\/latex] into the first equation to solve for [latex]q[\/latex].<\/p>\r\n

    [latex]\\begin{array}{r}q+6=11 \\\\q=5\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\nWe have 6 dimes and 5 quarters.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    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\r\nIn the next example we see a variation on the value problem, where it will make more sense to label the first column \"Amount\" rather than \"Number.\"\r\n

    \r\n

    Example 4<\/h3>\r\nIn a candy shop, chocolate which sells for $4.00 per pound is mixed with nuts which are sold for $2.50 per pound to produce a chocolate-nut candy which sells for $3.50 per pound. How much of each is used to make 30 pounds of the mixture?\r\n\r\n[reveal-answer q=\"966435\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"966435\"]\r\n\r\nLet\u00a0c<\/em> represent the pounds of chocolate and\u00a0n<\/em> the pounds of nuts.\r\n\r\nWe begin by making a table similar to those in the previous example. One important difference, however, is that this time, we have a \"value\" for the bottom row. For each row, multiply the amount by the value to obtain the total (in this case, signifying the total cost of the chocolate, nuts, and the mixture).\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nAmount<\/th>\r\nValue<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Chocolate<\/td>\r\n\u00a0[latex]c[\/latex]<\/td>\r\n\u00a0[latex]4.00[\/latex]<\/td>\r\n[latex]4c[\/latex]<\/td>\r\n<\/tr>\r\n
    Nuts<\/td>\r\n\u00a0[latex]n[\/latex]<\/td>\r\n\u00a0[latex]2.50[\/latex]<\/td>\r\n[latex]2.5n[\/latex]<\/td>\r\n<\/tr>\r\n
    Mixture<\/td>\r\n\u00a0[latex]30[\/latex]<\/td>\r\n\u00a0[latex]3.50[\/latex]<\/td>\r\n[latex]105[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAdding the amounts and total costs of the chocolate and nuts to those for the mixture produces the following system of equations:\r\n

    [latex]c+n=30[\/latex]<\/p>\r\n

    [latex]4c+2.5n=105[\/latex]<\/p>\r\nAs usual, we can use any method we have learned to solve this. Let us multiply the top equation by [latex]-2.5[\/latex] and add the equations to eliminate the n<\/em> terms.\r\n

    [latex]-2.5c-2.5n=-75[\/latex]<\/p>\r\n

    [latex]\\hspace{.25in}\\underline{4c+2.5n=105}[\/latex]<\/p>\r\n

    [latex]\\hspace{.25in}\\hspace{.3in}1.5c=30[\/latex]<\/p>\r\n

    [latex]\\hspace{.73in}c=20[\/latex]<\/p>\r\nWe now plug this into either equation to solve for\u00a0n<\/em>.\r\n

    [latex]20+n=30[\/latex]<\/p>\r\n

    [latex]\\hspace{.39in}n=10[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\nThe candy shop should use 20 pounds of chocolate and 10 pounds of nuts.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n

    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 describes the amount of a substance that is in a specific volume of liquid. \u00a0For example, if you have 30 grams of salt dissolved in water to create a 100mL saline solution, you have a concentration given by the following ratio:\r\n

    [latex]\\frac{30\\text{ grams }}{100\\text{ mL }}=0.30\\frac{\\text{ grams }}{\\text{ mL }}=30\\text{%}[\/latex]<\/p>\r\nConcentration is often given by a percentage.\u00a0 In the above example, this can be achieved since 1 mL of water has a mass of 1 gram.\u00a0 Often, we will mix two liquids directly.\u00a0 For example, if 20mL of ethanol is in a 50mL solution, the concentration of ethanol is given by\r\n

    [latex]\\frac{20\\text{ mL }}{50\\text{ mL }}=0.4=40\\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\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nAmount of Solution<\/th>\r\nConcentration<\/th>\r\nAmount of Solute<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Solution 1<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    Solution 2<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    Final Solution<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/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\u00a00.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex] solution, and the other is 75mL of a 0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex] 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\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nAmount of Solution<\/th>\r\nConcentration<\/th>\r\nAmount of Solute<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Solution 1<\/td>\r\n\u00a0120 mL<\/td>\r\n0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n\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
    Solution 2<\/td>\r\n\u00a075 mL<\/td>\r\n0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n\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
    Final Solution<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNext we add the new volumes and new masses.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nAmount of Solution<\/th>\r\nConcentration (%)<\/th>\r\nAmount of Solute<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Solution 1<\/td>\r\n\u00a0120 mL<\/td>\r\n0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n\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
    Solution 2<\/td>\r\n\u00a075 mL<\/td>\r\n0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n\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
    Final Solution<\/td>\r\n195 mL<\/td>\r\n[latex]\\frac{28.05\\text{ grams }}{ 195 \\text{ mL }}=0.14=14\\text{ % }[\/latex]<\/td>\r\n[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\nWe 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
    \r\n

    Example 5<\/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\nRead 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\nDefine 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\nWrite 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\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nAmount of Solution<\/th>\r\nConcentration<\/th>\r\nAmount of Methane<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Solution 1<\/td>\r\n\u00a070<\/td>\r\n\u00a00.5<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    Solution 2<\/td>\r\n\u00a0x<\/td>\r\n\u00a00.8<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    Final Solution<\/td>\r\n70+x<\/td>\r\n0.6<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nMultiply the amount of solution by the concentration\u00a0to get the amount of methane.\u00a0 Be sure to distribute on the last row: [latex]\\left(70 + x\\right)0.6[\/latex].\u00a0 Add the entries in the solution amount column to get the final amount. The concentration for this amount is 0.6 because we want the final solution to be 60% methane.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nAmount of Solution<\/th>\r\nConcentration<\/th>\r\nAmount of Methane<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Solution 1<\/td>\r\n\u00a070<\/td>\r\n\u00a00.5<\/td>\r\n\u00a035<\/td>\r\n<\/tr>\r\n
    Solution 2<\/td>\r\n\u00a0x<\/td>\r\n\u00a00.8<\/td>\r\n\u00a00.8x<\/em><\/td>\r\n<\/tr>\r\n
    Final Solution<\/td>\r\n\u00a070+x<\/td>\r\n0.6<\/td>\r\n\u00a0[latex]42+0.6x[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAdd the amounts of methane for solution 1 and solution 2 to get the total amount of methane 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

    [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

    Subtract 35 from both sides<\/p>\r\n

    [latex]\\begin{array}{c}35+0.2x=42\\\\\\underline{-35}\\,\\,\\,\\,\\,\\,\\,\\underline{-35}\\\\0.2x=7\\end{array}[\/latex]<\/p>\r\n

    Divide both sides by 0.2<\/p>\r\n

    [latex]\\begin{array}{c}0.2x=7\\\\\\frac{0.2x}{0.2}=\\frac{7}{0.2}\\end{array}[\/latex]\r\n[latex]x=35[\/latex]<\/p>\r\n\r\n

    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
    \r\n

    Example 6<\/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 that is 20% butterfat?\r\n[reveal-answer q=\"966963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"966963\"]\r\n\r\nRead 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\nDefine 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\nWrite and Solve:\u00a0<\/strong>Fill in the table with the information we know.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nAmount of Solution (Milk)<\/th>\r\nConcentration (%)<\/th>\r\nAmount of Butterfat<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Solution 1<\/td>\r\n\u00a0x<\/td>\r\n\u00a00.24<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    Solution 2<\/td>\r\n\u00a0y<\/td>\r\n\u00a00.18<\/td>\r\n<\/td>\r\n<\/tr>\r\n
    Final Solution<\/td>\r\n42<\/td>\r\n0.2<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFind the amount of butterfat by multiplying the amount of each solution by the concentration.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    <\/th>\r\nAmount<\/th>\r\nConcentration (%)<\/th>\r\nTotal Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Solution 1<\/td>\r\n\u00a0x<\/td>\r\n\u00a00.24<\/td>\r\n\u00a00.24x<\/td>\r\n<\/tr>\r\n
    Solution 2<\/td>\r\n\u00a0y<\/td>\r\n\u00a00.18<\/td>\r\n\u00a00.18y<\/td>\r\n<\/tr>\r\n
    Final Solution<\/td>\r\nx+y=42<\/td>\r\n0.2<\/td>\r\n0.24x+0.18y=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 = 0.84\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

    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 \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\r\n \r\n

    Write a system of linear equations representing a distance, rate, and time problem<\/h2>\r\nDistance, rate, and time applications are common in algebra. To solve this type of problem, we will use the following formula:\r\n

    [latex]\\mbox{distance}=\\mbox{rate}\\times\\mbox{time}[\/latex]<\/p>\r\n

    [latex]d=r\\cdot t[\/latex]<\/p>\r\nA problem that will typically result in a system of linear equations arises when there are multiple rates that contribute to the overall rate of an object. Before we jump into a full [latex]d=r\\cdot t[\/latex] problem, let us first explore an example that focuses solely on the rates.\r\n

    \r\n

    Example 7<\/h3>\r\nA boat traveling upstream on the Big River averages 5 mph, while traveling downstream with the same current, it averages 7 mph. Find the speed of the current and the speed of the boat in still water.\r\n\r\n[reveal-answer q=\"268762\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"268762\"]\r\n\r\nThere are two factors contributing to the overall rate of the boat. The boat is\u00a0trying<\/em> to travel at a certain speed, the \"speed of the boat in still water,\" but the current will either add to or subtract from this speed.\r\n\r\nLet\u00a0b<\/em> represent the speed of the boat in still water and\u00a0c<\/em> the speed of the current.\r\n\r\nWhen the boat travels upstream, it is fighting against the current, which slows the boat down. Given that the overall rate is 5 mph, this is given by the equation\r\n

    [latex]b-c=5[\/latex]<\/p>\r\nWhen the boat travels downstream, the current instead speeds the boat up. With he overall rate of 5 mph, this is given by\r\n

    [latex]b+c=7[\/latex]<\/p>\r\nThus, we now have a system of equations. This particular system can be quickly solved with the elimination method since adding the original equations together will immediately lead to the\u00a0c<\/em>\u00a0terms canceling each other out.\r\n

    [latex]b-c=5[\/latex]<\/p>\r\n

    [latex]\\underline{b+c=7}[\/latex]<\/p>\r\n

    [latex]\\hspace{.22in}2b=12[\/latex]<\/p>\r\n

    [latex]\\hspace{.23in}b=6[\/latex]<\/p>\r\n

    We can substitute this result into either equation to solve for c<\/em>.<\/p>\r\n

    [latex]6+c=7[\/latex]<\/p>\r\n

    [latex]\\hspace{.28in}c=1[\/latex]<\/p>\r\n

    .<\/p>\r\n\r\n

    Answer<\/h4>\r\nThe speed of the current is 1 mph and the speed of the boat in still water is 6 mph.\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can now utilize this idea to deal with a distance, rate, and time problem. As with the earlier types of problems, you may find it helpful to organize the information in a table.\r\n
    \r\n

    Example 8<\/h3>\r\nA boat can travel 20 miles downstream in 2 hours, while it takes 5 hours to make the same trip upstream.\u00a0 Find the speed of the boat in still water and the speed of the current.\r\n\r\n[reveal-answer q=\"5263\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"5263\"]\r\n\r\nFirst, note that since the boat makes the \"same trip\" upstream, this implies that it travels the same distance of 20 miles.\r\n\r\nLet\u00a0b<\/em> represent the speed of the boat in still water and\u00a0c<\/em> the speed of the current. Using what we learned about rate from the previous example helps us to fill in the following table:\r\n\r\n\r\n\r\n\r\n\r\n
    <\/td>\r\nRate<\/strong><\/td>\r\nTime<\/strong><\/td>\r\nDistance<\/strong><\/td>\r\n<\/tr>\r\n
    Downstream<\/td>\r\n[latex]b+c[\/latex]<\/td>\r\n[latex]2[\/latex]<\/td>\r\n[latex]20[\/latex]<\/td>\r\n<\/tr>\r\n
    Upstream<\/td>\r\n[latex]b-c[\/latex]<\/td>\r\n[latex]5[\/latex]<\/td>\r\n[latex]20[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow we can form our system of equations by applying the formula [latex]r\\cdot t=d[\/latex] to each row of the table. To indicate that we are multiplying the entire rate times the time, we use parentheses.\r\n

    [latex]\\begin{array}{r}(b+c)\\cdot 2=20 \\\\(b-c)\\cdot 5=20\\end{array}[\/latex]<\/p>\r\nWhile your first instinct might be to distribute (and this would work), we can form a much simpler system if we divide both sides by the time in each equation.\r\n

    [latex]\\begin{array}{r}\\displaystyle\\frac{(b+c)\\cdot 2}{2}=\\displaystyle\\frac{20}{2} \\\\ \\displaystyle\\frac{(b-c)\\cdot }{5}=\\displaystyle\\frac{20}{5}\\end{array}[\/latex]<\/p>\r\nThis results in the following system, which can be quickly solved using the elimination method.\r\n

    [latex]b+c=10[\/latex]<\/p>\r\n

    [latex]\\underline{b-c=4}[\/latex]<\/p>\r\n

    [latex]\\hspace{.23in} 2b=14[\/latex]<\/p>\r\n

    [latex]\\hspace{.25in}b=7[\/latex]<\/p>\r\nFor simplicity, we can substitue this result into either equation in our modified system to solve for\u00a0c<\/em>.\r\n

    [latex]7+c=10[\/latex]<\/p>\r\n

    [latex]\\hspace{.32in}c=3[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\nThe speed of the current is 3 mph and the speed of the boat in still water is 7 mph.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\nTo summarize the section, the following video provides several examples of applications of linear systems, including a ticket price problem, a mixture problem, and a motion problem.\r\n\r\n[embed]https:\/\/youtu.be\/FRaJv2Faass[\/embed]\r\n\r\n \r\n\r\n ","rendered":"
    \n

    section 4.4 Learning Objectives<\/h3>\n
    \n
    <\/div>\n
    \n

    4.4:\u00a0 Applications of 2×2 Systems of Equations\u00a0<\/strong><\/p>\n<\/div>\n

    \n
      \n
    • Set up and solve a linear system of equations by direct translation<\/li>\n
    • Use linear systems to solve value problems<\/li>\n
    • Use linear systems to solve mixture problems<\/li>\n
    • Use linear systems to solve motion problems<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n

       <\/p>\n

      Write a system of linear equations from direct translations<\/h2>\n

      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 of the methods learned in this module to solve the system of equations.<\/p>\n

      To introduce the topic, let us begin with direct translations.<\/p>\n

      In Section 1.7, we learned how to translate many key phrases in English into mathematical equations.\u00a0 Similarly, we can translate from English to form a system of equations.<\/p>\n

      \n

      Example 1<\/h3>\n

      One number is 3 less than 4 times another.\u00a0 If the sum of these numbers is 32, find the two numbers.<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      There are two numbers in the problem, which we can represent by the variables x and y.<\/p>\n

      Examining the first sentence, we recognize some familiar key phrases. Recall that “is” will correspond to\u00a0equals<\/em> while “less than” is\u00a0subtraction<\/em>, while being careful with the order as this implies 3 is being subtracted from 4 times the second number. So, the statement “one number is 3 less than 4 times another” results in the equation<\/p>\n

      [latex]x=4y-3[\/latex]<\/p>\n

      The second sentence discusses the “sum,” which indicates addition. Thus, “the sum of these numbers is 32” translates to the equation<\/p>\n

      [latex]x+y=32[\/latex]<\/p>\n

      We now have a system of equations to solve.<\/p>\n

      [latex]\\begin{array}{l}x=4y-3 \\\\ x+y=32\\end{array}[\/latex]<\/p>\n

      While we can use any of the techniques learned in this module to achieve this, substitution will work particularly nicely since the first equation already has one of the variables isolated. Substituting [latex]x=4y-3[\/latex] into the second equation for [latex]x[\/latex] yields<\/p>\n

      [latex]\\begin{array}{r}x+y=32 \\\\ 4y-3+y=32 \\\\ 5y-3=32 \\\\ 5y=35 \\\\ y=7\\hspace{.05in}\\end{array}[\/latex]<\/p>\n

      We can now substitute [latex]y=7[\/latex] into either of the original equations to find [latex]x[\/latex].<\/p>\n

      [latex]\\begin{array}{l}x=4y-3\\\\x=4(7)-3\\\\x=28-3\\\\x=25\\end{array}[\/latex]<\/p>\n

      Answer<\/h4>\n

      The two numbers are 25 and 7.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

       <\/p>\n

      Write a system of linear equations\u00a0representing\u00a0a value\u00a0problem<\/h2>\n

      One application of systems of equations is 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

      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\n\n\n\n\n\n\n
      <\/th>\nNumber (usually what you are trying to find)<\/th>\nValue<\/th>\nTotal<\/th>\n<\/tr>\n<\/thead>\n
      Item 1<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
      Item 2<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
      Total<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      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

      \n

      Example 2<\/h3>\n

      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

      Show Solution<\/span><\/p>\n
      \n

      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

      Define and Translate:\u00a0<\/strong>Let c<\/em> = the number of children and 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<\/em>. \u00a0$25.00 times c will give the revenue from the number of child tickets sold.<\/p>\n

      Write and Solve:\u00a0<\/strong>Although a table is not necessary, it can help you get started. \u00a0For this problem, we labeled columns as number, value, and total revenue because that is the information we are given.<\/p>\n

      The total number of people is [latex]2,000[\/latex].<\/p>\n\n\n\n\n\n\n\n
      <\/th>\nNumber<\/th>\nValue<\/th>\nTotal Revenue<\/th>\n<\/tr>\n<\/thead>\n
      Child Tickets<\/td>\n\u00a0c<\/td>\n\u00a0$25.00<\/td>\n25c<\/td>\n<\/tr>\n
      Adult Tickets<\/td>\n\u00a0a<\/em><\/td>\n\u00a0$50.00<\/td>\n50a<\/em><\/td>\n<\/tr>\n
      Total Tickets<\/td>\n2000<\/td>\n<\/td>\n$70,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      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.\u00a0 [latex]25c+50a=70,000[\/latex]<\/p>\n\n\n\n\n\n\n\n
      <\/th>\nNumber<\/th>\nValue<\/th>\nTotal Revenue<\/th>\n<\/tr>\n<\/thead>\n
      Child Tickets<\/td>\n\u00a0c<\/td>\n\u00a0$25.00<\/td>\n25c<\/td>\n<\/tr>\n
      Adult Tickets<\/td>\n\u00a0a<\/em><\/td>\n\u00a0$50.00<\/td>\n\u00a050a<\/em><\/td>\n<\/tr>\n
      Total Tickets<\/td>\n2000<\/td>\n<\/td>\n[latex]25c+50a=70,000[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      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\n\n\n\n\n\n\n
      <\/th>\nNumber<\/th>\nValue<\/th>\nTotal Revenue<\/th>\n<\/tr>\n<\/thead>\n
      Child Tickets<\/td>\n\u00a0c<\/td>\n\u00a0$25.00<\/td>\n25c<\/td>\n<\/tr>\n
      Adult Tickets<\/td>\n\u00a0a<\/em><\/td>\n\u00a0$50.00<\/td>\n\u00a050a<\/em><\/td>\n<\/tr>\n
      Total Tickets<\/td>\n[latex]c+a=2,000[\/latex]<\/td>\n<\/td>\n[latex]25c+50a=70,000[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      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

      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

      [latex]\\begin{array}{c}c+a=2,000\\\\ a=2,000-c\\end{array}[\/latex]<\/p>\n

      Substitute the expression [latex]2,000-c[\/latex] in the second equation for [latex]a[\/latex] and solve for [latex]c[\/latex].<\/p>\n

      [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

      Substitute [latex]c=1,200[\/latex] into the first equation to solve for [latex]a[\/latex].<\/p>\n

      [latex]\\begin{array}{r}1,200+a=2,000 \\\\ a=800\\,\\,\\,\\,\\,\\, \\end{array}[\/latex]<\/p>\n

      Answer<\/h4>\n

      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

      <\/div>\n<\/div>\n<\/div>\n

      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

      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