{"id":2934,"date":"2016-07-22T16:55:46","date_gmt":"2016-07-22T16:55:46","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2934"},"modified":"2023-11-08T13:20:13","modified_gmt":"2023-11-08T13:20:13","slug":"applications_of_rational_equations","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/chapter\/applications_of_rational_equations\/","title":{"raw":"8.5 - Applications of Rational Equations","rendered":"8.5 &#8211; Applications of Rational Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>(8.5.1) - Solve a rational formula for a specified variable<\/li>\r\n \t<li>(8.5.2) - Solving applications of rational equations\r\n<ul>\r\n \t<li>Average cost problems<\/li>\r\n \t<li>Work problems<\/li>\r\n \t<li>Motion problems<\/li>\r\n \t<li>Concentration of a mixture problems<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1>(8.5.1) - Solve a rational formula for a specified variable<\/h1>\r\n<strong>Rational formulas<\/strong> can be useful tools for representing real-life situations and for finding answers to real problems. Equations representing direct, inverse, and joint variation are examples of rational formulas that can model many real-life situations. As you will see, if you can find a formula, you can usually make sense of a situation.\r\n\r\nWhen solving problems using rational formulas, it is often helpful to first solve the formula for the specified variable. For example, work problems ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, [latex]W=rt[\/latex]. The amount of work done [latex]W[\/latex]\u00a0is the product of the rate of work [latex]r[\/latex]\u00a0and the time spent working [latex]t[\/latex]. Using algebra, you can write the work formula 3 ways:\r\n<p style=\"text-align: center\">[latex]W=rt[\/latex]<\/p>\r\nFind the time [latex]t[\/latex]:<i> <\/i>\r\n<p style=\"text-align: center\">[latex]\\displaystyle t=\\frac{W}{r}[\/latex]<i> (divide both sides by <\/i>[latex]r[\/latex]<i>)<\/i><\/p>\r\nFind the rate [latex]r[\/latex]:<i> <\/i>\r\n<p style=\"text-align: center\">[latex]\\displaystyle r=\\frac{W}{t}[\/latex]<i>(divide both sides by <\/i>[latex]t[\/latex]<i>)<\/i><\/p>\r\n\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nThe formula for finding the density of an object is [latex]\\displaystyle D=\\frac{m}{v}[\/latex], where [latex]D[\/latex] is the density, [latex]m[\/latex] is the mass of the object and [latex]v[\/latex] is the volume of the object. Rearrange the formula to solve for the mass [latex]m[\/latex]\u00a0and then for the volume [latex]v[\/latex].\r\n\r\n[reveal-answer q=\"537110\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"537110\"]Start with the formula for density.\r\n<p style=\"text-align: center\">[latex]\\displaystyle D=\\frac{m}{v}[\/latex]<\/p>\r\nMultiply both side of the equation by [latex]v[\/latex] to isolate [latex]m[latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle v\\cdot D=\\frac{m}{v}\\cdot v[\/latex]<\/p>\r\nSimplify and rewrite the equation, solving for [latex]m[\/latex].\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}v\\cdot D=m\\cdot \\frac{v}{v}\\\\v\\cdot D=m\\cdot 1\\\\v\\cdot D=m\\end{array}[\/latex]<\/p>\r\nTo solve the equation [latex]\\displaystyle D=\\frac{m}{v}[\/latex] in terms of [latex]v[\/latex], you will need do the same steps to this point, and then divide both sides by [latex]D[\/latex].\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{r}\\frac{v\\cdot D}{D}=\\frac{m}{D}\\\\\\\\\\frac{D}{D}\\cdot v=\\frac{m}{D}\\\\\\\\1\\cdot v=\\frac{m}{D}\\\\\\\\v=\\frac{m}{D}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex] m=D\\cdot v[\/latex] and [latex]\\displaystyle v=\\frac{m}{D}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow let\u2019s look at an example using the formula for the volume of a cylinder.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nThe formula for finding the volume of a cylinder is [latex]V=\\pi{r^{2}}h[\/latex], where [latex]V[\/latex] is the volume, [latex]r[\/latex] is the radius and [latex]h[\/latex] is the height of the cylinder. Rearrange the formula to solve for the height [latex]h[\/latex].\r\n\r\n[reveal-answer q=\"644317\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"644317\"]Start with the formula for the volume of a cylinder.\r\n<p style=\"text-align: center\">[latex] V=\\pi{{r}^{2}}h[\/latex]<\/p>\r\nDivide both sides by [latex] \\pi {{r}^{2}}[\/latex] to isolate [latex]h[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{V}{\\pi {{r}^{2}}}=\\frac{\\pi {{r}^{2}}h}{\\pi {{r}^{2}}}[\/latex]<\/p>\r\nSimplify. You find the height, [latex]h[\/latex], is equal to [latex] \\frac{V}{\\pi {{r}^{2}}}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{V}{\\pi {{r}^{2}}}=h[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\displaystyle h=\\frac{V}{\\pi {{r}^{2}}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we give another example of solving for a variable in a formula, or as they are also called, a literal equation.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=ecEUUbRLDQs&amp;feature=youtu.be\r\n<h1>(8.5.2) - Solving applications of rational equations<\/h1>\r\n<h3>Average cost problems<\/h3>\r\nSuppose we know that the cost of making a product is dependent on the number of items, [latex]x[\/latex], produced. This is given by the equation [latex]C\\left(x\\right)=15,000+500x[\/latex]. Note, the cost function consists of a <em>fixed cost<\/em> ([latex]\\$15,000[\/latex]) and a <em>variable cost<\/em> [latex](500x)[\/latex]. The <em>fixed cost<\/em> doesn't change when more items are produced, whereas the <em>variable cost<\/em> increases as more cars are produced.\u00a0 If we want to know the average cost for producing [latex]x[\/latex] items, we would divide the cost function by the number of items, [latex]x[\/latex].\r\n\r\nThe average cost function, which yields the average cost per item for [latex]x[\/latex]\u00a0items produced, is\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\overline{C}\\left(x\\right)=\\frac{15,000 +500x}{x}[\/latex]<\/p>\r\nMany other application problems require finding an average value in a similar way, giving us variables in the denominator.\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nLet [latex]x[\/latex] represent the number of cars produced in a factory in a typical week. Suppose that the the cost in dollars to produce [latex]x[\/latex] cars is given by [latex]C(x) = 15,000+500x[\/latex]. How many cars would the factory have to produce in one week, in order for the average production cost per car to be [latex]\\$1,000[\/latex]?\r\n\r\n[reveal-answer q=\"752418\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"752418\"]\r\n\r\nThe average cost of producing [latex]x[\/latex]:\u00a0cars per week is:\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\overline{C}\\left(x\\right)=\\frac{15,000 +500x}{x}[\/latex]<\/p>\r\nIf we want for the average production cost per car to be [latex]\\$1,000[\/latex], we set up the following rational equation:\r\n<p style=\"text-align: center\">[latex]\\displaystyle 1,000=\\frac{15,000 +500x}{x}[\/latex]<\/p>\r\nWe multiply both sides [latex]x[\/latex]:\r\n<p style=\"text-align: center\">[latex]1,000x = 15,000+500x[\/latex]<\/p>\r\nSubtract [latex]500x[\/latex] from both sides:\r\n<p style=\"text-align: center\">[latex]500x = 15,000[\/latex]<\/p>\r\nDivide both sides by [latex]500[\/latex]:\r\n<p style=\"text-align: center\">[latex]x=30[\/latex]<\/p>\r\n<strong>Answer<\/strong>\r\n\r\nThe factory must produce 30 cars per week.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Work problems<\/h3>\r\nRational equations can be used to solve a variety of problems that involve rates, times and work. Using rational expressions and equations can help you answer questions about how to combine workers or machines to complete a job on schedule.\r\n\r\n[caption id=\"attachment_5024\" align=\"aligncenter\" width=\"326\"]<img class=\"wp-image-5024\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/22162804\/Screen-Shot-2016-06-19-at-9.54.28-PM-233x300.png\" alt=\"Man with a lunch box walking. THere is a caption above him that says &quot;Boy! I sure did a good day's work today&quot;\" width=\"326\" height=\"420\" \/> A Good Day's Work[\/caption]\r\n\r\nA \u201cwork problem\u201d is an example of a real life situation that can be modeled and solved using a rational equation. Work problems often ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, [latex]W=rt[\/latex].\u00a0(Notice that the work formula is very similar to the relationship between distance, rate, and time, or [latex]d=rt[\/latex].) The amount of work done [latex]W[\/latex]\u00a0is the product of the rate of work [latex]r[\/latex]\u00a0and the time spent working [latex]t[\/latex]. The work formula has 3 versions.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}W=rt\\\\\\\\\\,\\,\\,\\,\\,t=\\frac{W}{r}\\\\\\\\\\,\\,\\,\\,\\,r=\\frac{W}{t}\\end{array}[\/latex]<\/p>\r\nSome work problems include multiple machines or people working on a project together for the same amount of time but at different rates. In that case, you can add their individual work rates together to get a total work rate. Let\u2019s look at an example.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nMyra takes 2 hours to plant 50 flower bulbs. Francis takes 3 hours to plant 45 flower bulbs. Working together, how long should it take them to plant 150 bulbs?\r\n\r\n[reveal-answer q=\"550322\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"550322\"]Think about how many bulbs each person can plant in one hour. This is their planting rate.\r\n\r\nMyra: [latex]\\displaystyle \\frac{50\\,\\,\\text{bulbs}}{2\\,\\,\\text{hours}}[\/latex], or [latex]\\displaystyle \\frac{25\\,\\,\\text{bulbs}}{1\\,\\,\\text{hour}}[\/latex]\r\n\r\nFrancis: [latex]\\displaystyle \\frac{45\\,\\,\\text{bulbs}}{3\\,\\,\\text{hours}}[\/latex], or [latex]\\displaystyle \\frac{15\\,\\,\\text{bulbs}}{1\\,\\,\\text{hour}}[\/latex]\r\n\r\nCombine their hourly rates to determine the rate they work together.\r\n\r\nMyra and Francis together:\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{25\\,\\,\\text{bulbs}}{1\\,\\,\\text{hour}}+\\frac{15\\,\\,\\text{bulbs}}{1\\,\\,\\text{hour}}=\\frac{40\\,\\,\\text{bulbs}}{1\\,\\,\\text{hour}}[\/latex]<\/p>\r\nUse one of the work formulas to write a rational equation, for example [latex]\\displaystyle r=\\frac{W}{t}[\/latex]. You know [latex]r[\/latex], the combined work rate, and you know [latex]W[\/latex], the amount of work that must be done. What you don't know is how much time it will take to do the required work at the designated rate.\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{40}{1}=\\frac{150}{t}[\/latex]<\/p>\r\nSolve the equation by multiplying both sides by the common denominator, then isolating [latex]t[\/latex].\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{c}\\frac{40}{1}\\cdot t=\\frac{150}{t}\\cdot t\\\\\\\\40t=150\\\\\\\\t=\\frac{150}{40}=\\frac{15}{4}\\\\\\\\t=3\\frac{3}{4}\\text{hours}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nIt should take 3 hours 45 minutes for Myra and Francis to plant 150 bulbs together.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/www.youtube.com\/watch?v=SzSasnDF7Ms&amp;feature=youtu.be\r\n\r\nOther work problems go the other way. You can calculate how long it will take one person to do a job alone when you know how long it takes people working together to complete the job.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nJoe and John are planning to paint a house together. John thinks that if he worked alone, it would take him 3 times as long as it would take Joe to paint the entire house. Working together, they can complete the job in 24 hours. How long would it take each of them, working alone, to complete the job?\r\n\r\n[reveal-answer q=\"593775\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"593775\"]Choose variables to represent the unknowns. Since it takes John 3 times as long as Joe to paint the house, his time is represented as [latex]3x[\/latex].\r\n\r\nLet [latex]x[\/latex]\u00a0= time it takes Joe\u00a0to complete the job\r\n\r\n[latex]3x[\/latex]<i>\u00a0<\/i>= time it takes John\u00a0to complete the job\r\n\r\nThe work is painting 1 house or 1. Write an expression to represent each person\u2019s rate using the formula\u00a0[latex] \\displaystyle r=\\frac{W}{t}[\/latex]<sub>.<\/sub>\r\n\r\nJoe\u2019s rate: [latex]\\displaystyle \\frac{1}{x}[\/latex]\r\n\r\nJohn\u2019s rate: [latex]\\displaystyle \\frac{1}{3x}[\/latex]\r\n\r\nTheir combined rate is the sum of their individual rates. Use this rate to write a new equation using the formula [latex]W=rt[\/latex].\r\n\r\ncombined rate: [latex]\\displaystyle \\frac{1}{x}+\\frac{1}{3x}[\/latex]\r\n\r\nThe problem states that it takes them 24 hours together to paint a house, so if you multiply their combined hourly rate [latex]\\displaystyle \\left( \\frac{1}{x}+\\frac{1}{3x} \\right)[\/latex] by 24, you will get 1, which is the number of houses they can paint in 24 hours.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}1=\\left( \\frac{1}{x}+\\frac{1}{3x} \\right)24\\\\\\\\1=\\frac{24}{x}+\\frac{24}{3x}\\end{array}[\/latex]<\/p>\r\nNow solve the equation for [latex]x[\/latex]. (Remember that [latex]x[\/latex] represents the number of hours it will take Joe to finish the job.)\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}\\,\\,\\,1=\\frac{3}{3}\\cdot \\frac{24}{x}+\\frac{24}{3x}\\\\\\\\\\,\\,\\,1=\\frac{3\\cdot 24}{3x}+\\frac{24}{3x}\\\\\\\\\\,\\,\\,1=\\frac{72}{3x}+\\frac{24}{3x}\\\\\\\\\\,\\,\\,1=\\frac{72+24}{3x}\\\\\\\\\\,\\,\\,1=\\frac{96}{3x}\\\\\\\\3x=96\\\\\\\\\\,\\,\\,x=32\\end{array}[\/latex]<\/p>\r\nCheck the solutions in the original equation.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}1=\\left( \\frac{1}{x}+\\frac{1}{3x} \\right)24\\\\\\\\1=\\left[ \\frac{\\text{1}}{\\text{32}}+\\frac{1}{3\\text{(32})} \\right]24\\\\\\\\1=\\frac{24}{\\text{32}}+\\frac{24}{3\\text{(32})}\\\\\\\\1=\\frac{24}{\\text{32}}+\\frac{24}{96}\\\\\\\\1=\\frac{3}{3}\\cdot \\frac{24}{\\text{32}}+\\frac{24}{96}\\\\\\\\1=\\frac{72}{96}+\\frac{24}{96}\\end{array}[\/latex]<\/p>\r\nThe solution checks. Since [latex]x=32[\/latex], it takes Joe 32 hours to paint the house by himself. John\u2019s time is 3<i>x<\/i>, so it would take him 96 hours to do the same amount of work.\r\n<h4>Answer<\/h4>\r\nIt takes 32 hours for Joe to paint the house by himself and 96 hours for John the paint the house himself.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video that follows, we show another example of finding one person's work rate given a combined work rate.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=kbRSYb8UYqU&amp;feature=youtu.be\r\n\r\nAs shown above, many work problems can be represented by the equation [latex]\\displaystyle \\frac{t}{a}+\\frac{t}{b}=1[\/latex], where [latex]t[\/latex] is the time to do the job together, [latex]a[\/latex] is the time it takes person A to do the job, and [latex]b[\/latex] is the time it takes person B to do the job. The 1 refers to the total work done\u2014in this case, the work was to paint 1 house.\r\n\r\nThe key idea here is to figure out each worker\u2019s individual rate of work. Then, once those rates are identified, add them together, multiply by the time [latex]t[\/latex], set it equal to the amount of work done, and solve the rational equation. Here is a summary:\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Formula for work problems<\/h3>\r\n[latex]a=[\/latex] the time needed for A to complete 1 job alone\r\n\r\n[latex]b=[\/latex] the time needed for B to complete 1 job alone\r\n\r\n[latex]t=[\/latex] the time needed for A and B to complete the 1 job together:\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{t}{a}+\\frac{t}{b} = 1[\/latex]<\/p>\r\n\r\n<\/div>\r\nWe present another example of two people painting at different rates in the following video.\r\n\r\nhttps:\/\/youtu.be\/SzSasnDF7Ms\r\n<h3>Motion problems<\/h3>\r\nWe have solved uniform motion problems using the formula [latex]D = rt[\/latex] in previous chapters. We used a table like the one below to organize the information and lead us to the equation.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{|c|c|c|c|}\r\n\\hline\r\n&amp; \\,\\text{rate}\\,&amp;\\,\\text{time}\\,&amp;\\,\\text{distance}\\\\\r\n\\hline\r\n\\,\\text{First}\\,&amp;\\,&amp;\\,&amp;\\\\\r\n\\hline\r\n\\,\\text{Second}\\,&amp;\\,&amp;\\,&amp;\\\\\r\n\\hline\r\n\\end{array} [\/latex]<\/p>\r\nThe formula [latex]D=rt[\/latex] assumes we know [latex]r[\/latex] and\u00a0 [latex]t[\/latex] and use them to find\u00a0 [latex]D[\/latex]. If we know\u00a0 [latex]D[\/latex] and [latex]r[\/latex] and need to find\u00a0 [latex]t[\/latex], we would solve the equation for\u00a0 [latex]t[\/latex] and get the formula [latex]\\displaystyle t=\\frac{D}{r}[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nGreg went to a conference in a city 120 miles away. On the way back, due to\u00a0road construction he had to drive 10 mph slower which resulted in the return\u00a0trip taking 2 hours longer. How fast did he drive on the way to the\u00a0conference?\r\n\r\n[reveal-answer q=\"519818\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"519818\"]\r\n<p style=\"text-align: center\">[latex]\\begin{eqnarray*}\r\n\\begin{array}{|c|c|c|c|}\r\n\\hline\r\n&amp; \\,\\text{rate}\\,&amp; \\,\\text{time}\\,&amp; \\,\\text{distance}\\\\\r\n\\hline\r\n\\,\\text{There}\\,&amp; r &amp; t &amp; 120\\\\\r\n\\hline\r\n\\,\\text{Back}\\,&amp; &amp; &amp; \\\\\r\n\\hline\r\n\\end{array}\\,&amp; &amp; \\begin{array}{l}\r\n\\,\\text{We}\\,\\,\\text{do}\\,\\,\\text{not}\\,\\,\\text{know}\\,\\,\\text{rate}, r, \\,\\text{or}\r\n\\,\\text{time}, t \\,\\text{he}\\,\\,\\text{traveled}\\\\\r\n\\,\\text{on}\\,\\,\\text{the}\\,\\,\\text{way}\\,\\,\\text{to}\\,\\,\\text{the}\\,\\,\\text{conference}\\,.\r\n\\,\\text{But}\\,\\,\\text{we}\\,\\,\\text{do}\\,\\,\\text{know}\\\\\r\n\\,\\text{the}\\,\\,\\text{distance}\\,\\,\\text{was}\\,120 \\,\\text{miles}\\,.\r\n\\end{array}\\\\\r\n&amp; &amp; \\\\\r\n\\begin{array}{|c|c|c|c|}\r\n\\hline\r\n&amp; \\,\\text{rate}\\,&amp; \\,\\text{time}\\,&amp; \\,\\text{distance}\\\\\r\n\\hline\r\n\\,\\text{There}\\,&amp; r &amp; t &amp; 120\\\\\r\n\\hline\r\n\\,\\text{Back}\\,&amp; r - 10 &amp; t + 2 &amp; 120\\\\\r\n\\hline\r\n\\end{array}&amp; &amp; \\begin{array}{l}\r\n\\,\\text{Coming}\\,\\,\\text{back}\\,\\,\\text{he}\\,\\,\\text{drove}\\,10 \\,\\text{mph}\r\n\\,\\text{slower}\\,(r - 10)\\\\\r\n\\,\\text{and}\\,\\,\\text{took}\\,2 \\,\\text{hours}\\,\\,\\text{longer}\\,(t + 2) . \\,\\text{The}\r\n\\,\\text{distance}\\\\\r\n\\,\\text{was}\\,\\,\\text{still}\\,120 \\,\\text{miles}\\,.\r\n\\end{array}\\\\\r\n&amp; &amp; \\\\\r\nr t = 120 &amp; &amp; \\,\\text{Equations}\\,\\,\\text{are}\\,\\,\\text{product}\\,\\,\\text{of}\r\n\\,\\text{rate}\\,\\,\\text{and}\\,\\,\\text{time}\\\\\r\n(r - 10) (t + 2) = 120 &amp; &amp; \\,\\text{We}\\,\\,\\text{have}\\,\\,\\text{simultaneous}\r\n\\,\\text{product}\\,\\,\\text{equations}\\\\\r\n&amp; &amp; \\\\\r\nt = \\frac{120}{r}\\,\\,\\text{and}\\,t + 2 = \\frac{120}{r - 10} &amp; &amp;\r\n\\,\\text{Solving}\\,\\,\\text{for}\\,\\,\\text{rate}, \\,\\text{divide}\\,\\,\\text{by}\\,r\r\n\\,\\text{and}\\,r - 10\\\\\r\n&amp; &amp; \\\\\r\n\\frac{120}{r}\\,+ 2 = \\frac{120}{r - 10}&amp; &amp; \\,\\text{Substitute}\r\n\\frac{120}{r} \\,\\text{for}\\,t \\,\\text{in}\\,\\,\\text{the}\\,\\,\\text{second}\r\n\\,\\text{equation}\\\\\r\n&amp; &amp; \\\\\r\n\\frac{120 r (r - 10)}{r} + 2 r (r - 10) = \\frac{120 r (r - 10)}{r - 10} &amp;\r\n&amp; \\,\\text{Multiply}\\,\\,\\text{each}\\,\\,\\text{term}\\,\\,\\text{by}\\,\\,\\text{LCD}\\,: r (r -\r\n10)\\\\\r\n&amp; &amp; \\\\\r\n120 (r - 10) + 2 r^2 - 20 r = 120 r &amp; &amp; \\,\\text{Reduce}\\,\\,\\text{each}\r\n\\,\\text{fraction}\\\\\r\n120 r - 1200 + 2 r^2 - 20 r = 120 r &amp; &amp; \\,\\text{Distribute}\\\\\r\n2 r^2 + 100 r - 1200 = 120 r &amp; &amp; \\,\\text{Combine}\\,\\,\\text{like}\r\n\\,\\text{terms}\\\\\r\n\\underline{- 120 r - 120 r} &amp; &amp; \\,\\text{Make}\\,\\,\\text{equation}\\,\\,\\text{equal}\r\n\\,\\text{to}\\,\\,\\text{zero}\\\\\r\n2 r^2 - 20 r - 1200 = 0 &amp; &amp; \\,\\text{Divide}\\,\\,\\text{each}\\,\\,\\text{term}\r\n\\,\\text{by}\\,2\\\\\r\nr^2 - 10 r - 600 = 0 &amp; &amp; \\,\\text{Factor}\\\\\r\n(r - 30) (r + 20) = 0 &amp; &amp; \\,\\text{Set}\\,\\,\\text{each}\\,\\,\\text{factor}\r\n\\,\\text{equal}\\,\\,\\text{to}\\,\\,\\text{zero}\\\\\r\nr - 30 = 0 \\,\\text{and}\\,r + 20 = 0 &amp; &amp; \\,\\text{Solve}\\,\\,\\text{each}\r\n\\,\\text{equation}\\\\\r\n\\underline{+ 30 + 30} \\underline{- 20 - 20} &amp; &amp; \\\\\r\nr = 30 \\,\\text{and}\\,r = - 20 &amp; &amp; \\,\\text{Can}' t \\,\\text{have}\\,a\r\n\\,\\text{negative}\\,\\,\\text{rate}\\\\\r\n30 \\,\\text{mph}\\,&amp; &amp; \\,\\text{Our}\\,\\,\\text{Solution}\r\n\\end{eqnarray*}[\/latex]<\/p>\r\n<strong>Answer:<\/strong>\r\n\r\n30mph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nA man rows downstream for 30 miles then turns around and returns to his\u00a0original location, the total trip took 8 hours. If the current flows at 2\u00a0miles per hour, how fast would the man row in still water?\r\n<p style=\"text-align: center\">[reveal-answer q=\"2397\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"2397\"]\\begin{eqnarray*}\r\n\\begin{array}{|c|c|c|c|}\r\n\\hline\r\n&amp; \\,\\text{rate}\\,&amp; \\,\\text{time}\\,&amp; \\,\\text{distance}\\\\\r\n\\hline\r\n\\,\\text{down}\\,&amp; &amp; t &amp; 30\\\\\r\n\\hline\r\n\\,\\text{up}\\,&amp; &amp; 8 - t &amp; 30\\\\\r\n\\hline\r\n\\end{array} &amp; &amp; \\begin{array}{l}\r\n\\,\\text{We}\\,\\,\\text{know}\\,\\,\\text{the}\\,\\,\\text{distance}\\,\\,\\text{up}\\,\\,\\text{and}\\,\r\n\\,\\text{down}\\,\\,\\text{is}\\,30.\\\\\r\n\\,\\text{Put}\\,t \\,\\text{for}\\,\\,\\text{time}\\,\\,\\text{downstream}\\,.\r\n\\,\\text{Subtracting}\\\\\r\n8 - t \\,\\text{becomes}\\,\\,\\text{time}\\,\\,\\text{upstream}\\,\r\n\\end{array}\\\\\r\n&amp; &amp; \\\\\r\n\\begin{array}{|c|c|c|c|}\r\n\\hline\r\n&amp; \\,\\text{rate}\\,&amp; \\,\\text{time}\\,&amp; \\,\\text{distance}\\\\\r\n\\hline\r\n\\,\\text{down}\\,&amp; r + 2 &amp; t &amp; 30\\\\\r\n\\hline\r\n\\,\\text{up}\\,&amp; r - 2 &amp; 8 - t &amp; 30\\\\\r\n\\hline\r\n\\end{array}&amp; &amp; \\begin{array}{l}\r\n\\,\\text{Downstream}\\,\\,\\text{the}\\,\\,\\text{current}\\,\\,\\text{of}\\,2 \\,\\text{mph}\r\n\\,\\text{pushes}\\,\\\\\r\n\\,\\text{the}\\,\\,\\text{boat} (r + 2) \\,\\text{and}\\,\\,\\text{upstream}\\,\\,\\text{the}\\,\r\n\\,\\text{current}\\,\\\\\r\n\\,\\text{pulls}\\,\\,\\text{the}\\,\\,\\text{boat}\\,(r - 2)\r\n\\end{array}\\\\\r\n&amp; &amp; \\\\\r\n(r + 2) t = 30 &amp; &amp; \\,\\text{Multiply}\\,\\,\\text{rate}\\,\\,\\text{by}\\,\\,\\text{time}\r\n\\,\\text{to}\\,\\,\\text{get}\\,\\,\\text{equations}\\,\\\\\r\n(r - 2) (8 - t) = 30 &amp; &amp; \\,\\text{We}\\,\\,\\text{have}\\,a \\,\\text{simultaneous}\r\n\\,\\text{product}\\\\\r\n&amp; &amp; \\\\\r\nt = \\frac{30}{r + 2} \\,\\text{and}\\,8 - t = \\frac{30}{r - 2} &amp; &amp;\r\n\\,\\text{Solving}\\,\\,\\text{for}\\,\\,\\text{rate}, \\,\\text{divide}\\,\\,\\text{by}\\,r + 2\r\n\\,\\text{or}\\,r - 2\\\\\r\n&amp; &amp; \\\\\r\n8 - \\frac{30}{r + 2}\\,= \\frac{30}{r - 2} &amp; &amp; \\,\\text{Substitute}\\,\\frac{30}{r\r\n+ 2} \\,\\text{for}\\,t \\,\\text{in}\\,\\,\\text{second}\\,\\,\\text{equation}\\,\\\\\r\n&amp; &amp; \\\\\r\n8 (r + 2) (r - 2) - \\frac{30 (r + 2) (r - 2)}{r + 2} = \\frac{30 (r + 2) (r\r\n- 2)}{r - 2} &amp; &amp; \\,\\text{Multiply}\\,\\,\\text{each}\\,\\,\\text{term}\\,\\,\\text{by}\r\n\\,\\text{LCD}\\,: (r + 2) (r - 2)\\\\\r\n&amp; &amp; \\\\\r\n8 (r + 2) (r - 2) - 30 (r - 2) = 30 (r + 2) &amp; &amp; \\,\\text{Reduce}\r\n\\,\\text{fractions}\\,\\\\\r\n8 r^2 - 32 - 30 r + 60 = 30 r + 60 &amp; &amp; \\,\\text{Multiply} \\,\\text{and}\r\n\\,\\text{distribute}\\\\\r\n8 r^2 - 30 r + 28 = 30 r + 60 &amp; &amp; \\,\\text{Make} \\,\\text{equation}\r\n\\,\\text{equal}\\,\\,\\text{zero}\\,\\\\\r\n\\underline{- 30 r - 60 - 30 r - 60} &amp; &amp; \\\\\r\n8 r^2 - 60 r - 32 = 0 &amp; &amp; \\,\\text{Divide}\\,\\,\\text{each}\\,\\,\\text{term}\\,\\,\\text{by}\r\n4\\\\\r\n2 r^2 - 15 r - 8 = 0 &amp; &amp; \\,\\text{Factor}\\\\\r\n(2 r + 1) (r - 8) = 0 &amp; &amp; \\,\\text{Set}\\,\\,\\text{each}\\,\\,\\text{factor}\r\n\\,\\text{equal}\\,\\,\\text{to}\\,\\,\\text{zero}\\,\\\\\r\n2 r + 1 = 0 \\,\\text{or}\\,r - 8 = 0 &amp; &amp; \\,\\text{Solve}\\,\\,\\text{each}\\,\r\n\\,\\text{equation}\\\\\r\n\\underline{- 1 - 1} \\underline{+ 8 + 8} &amp; &amp; \\\\\r\n2 r = - 1 \\,\\text{or}\\,r = 8 &amp; &amp; \\\\\r\n\\overline{2} \\overline{2} &amp; &amp; \\\\\r\nr = - \\frac{1}{2} \\,\\text{or}\\, r = 8 &amp; &amp; \\,\\text{Can}' t \\,\\text{have}\\,a\r\n\\,\\text{negative}\\,\\,\\text{rate}\\,\\\\\r\n8 \\,\\text{mph}\\,&amp; &amp; \\,\\text{Our}\\,\\,\\text{Solution}\r\n\\end{eqnarray*}<\/p>\r\n<strong>Answer:<\/strong>\r\n\r\n8 mph\r\n<p style=\"text-align: left\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nhttps:\/\/www.youtube.com\/watch?v=Zjx2hCjKtiQ\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=BpFVUXrChLM\r\n<h3>Concentration of a mixture problems<\/h3>\r\nMixtures are made of\u00a0ratios of different substances that may include chemicals, foods, water, or gases. There are many different situations where mixtures may occur both in nature and as a means to produce a desired product or outcome. \u00a0For example, chemical spills, manufacturing and even biochemical reactions involve mixtures. \u00a0The thing that can make mixtures interesting mathematically is when components of the mixture are\u00a0added at different rates and concentrations. In our last example we will define an equation that models the\u00a0concentration \u00a0- or ratio of sugar to water - in a large mixing tank over time. You are asked whether the final concentration of sugar is greater than the concentration at the beginning.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?\r\n[reveal-answer q=\"332373\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"332373\"]\r\n\r\nLet [latex]t[\/latex]\u00a0be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is a linear equation, as is the amount of sugar in the tank. We can write an equation independently for each:\r\n<p style=\"text-align: center\">[latex]\\begin{cases}\\text{water: }W\\left(t\\right)=100+10t\\text{ in gallons}\\\\ \\text{sugar: }S\\left(t\\right)=5+1t\\text{ in pounds}\\end{cases}[\/latex]<\/p>\r\nThe concentration, [latex]C[\/latex], will be the ratio of pounds of sugar to gallons of water\r\n<p style=\"text-align: center\">[latex]\\displaystyle C\\left(t\\right)=\\frac{5+t}{100+10t}[\/latex]<\/p>\r\nThe concentration after 12 minutes is given by evaluating [latex]C\\left(t\\right)[\/latex] at [latex]t=\\text{ }12[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle C\\left(12\\right)=\\frac{5+12}{100+10\\left(12\\right)} =\\frac{17}{220}[\/latex]<\/p>\r\nThis means the concentration is 17 pounds of sugar to 220 gallons of water.\r\n\r\nAt the beginning, the concentration is\r\n<p style=\"text-align: center\">[latex]\\displaystyle C\\left(0\\right)=\\frac{5+0}{100+10\\left(0\\right)}=\\frac{1}{20}[\/latex]<\/p>\r\nSince [latex]\\displaystyle \\frac{17}{220}\\approx 0.08&gt;\\frac{1}{20}=0.05[\/latex], the concentration is greater after 12 minutes than at the beginning.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of how to use rational functions\u00a0to model mixing.\r\n\r\nhttps:\/\/youtu.be\/GD6H7BE_0EI\r\n<h2><\/h2>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>(8.5.1) &#8211; Solve a rational formula for a specified variable<\/li>\n<li>(8.5.2) &#8211; Solving applications of rational equations\n<ul>\n<li>Average cost problems<\/li>\n<li>Work problems<\/li>\n<li>Motion problems<\/li>\n<li>Concentration of a mixture problems<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h1>(8.5.1) &#8211; Solve a rational formula for a specified variable<\/h1>\n<p><strong>Rational formulas<\/strong> can be useful tools for representing real-life situations and for finding answers to real problems. Equations representing direct, inverse, and joint variation are examples of rational formulas that can model many real-life situations. As you will see, if you can find a formula, you can usually make sense of a situation.<\/p>\n<p>When solving problems using rational formulas, it is often helpful to first solve the formula for the specified variable. For example, work problems ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, [latex]W=rt[\/latex]. The amount of work done [latex]W[\/latex]\u00a0is the product of the rate of work [latex]r[\/latex]\u00a0and the time spent working [latex]t[\/latex]. Using algebra, you can write the work formula 3 ways:<\/p>\n<p style=\"text-align: center\">[latex]W=rt[\/latex]<\/p>\n<p>Find the time [latex]t[\/latex]:<i> <\/i><\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle t=\\frac{W}{r}[\/latex]<i> (divide both sides by <\/i>[latex]r[\/latex]<i>)<\/i><\/p>\n<p>Find the rate [latex]r[\/latex]:<i> <\/i><\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle r=\\frac{W}{t}[\/latex]<i>(divide both sides by <\/i>[latex]t[\/latex]<i>)<\/i><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>The formula for finding the density of an object is [latex]\\displaystyle D=\\frac{m}{v}[\/latex], where [latex]D[\/latex] is the density, [latex]m[\/latex] is the mass of the object and [latex]v[\/latex] is the volume of the object. Rearrange the formula to solve for the mass [latex]m[\/latex]\u00a0and then for the volume [latex]v[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q537110\">Show Solution<\/span><\/p>\n<div id=\"q537110\" class=\"hidden-answer\" style=\"display: none\">Start with the formula for density.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle D=\\frac{m}{v}[\/latex]<\/p>\n<p>Multiply both side of the equation by [latex]v[\/latex] to isolate [latex]m[latex].  <\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle v\\cdot D=\\frac{m}{v}\\cdot v[\/latex]<\/p>\n<p>Simplify and rewrite the equation, solving for [latex]m[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}v\\cdot D=m\\cdot \\frac{v}{v}\\\\v\\cdot D=m\\cdot 1\\\\v\\cdot D=m\\end{array}[\/latex]<\/p>\n<p>To solve the equation [latex]\\displaystyle D=\\frac{m}{v}[\/latex] in terms of [latex]v[\/latex], you will need do the same steps to this point, and then divide both sides by [latex]D[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{r}\\frac{v\\cdot D}{D}=\\frac{m}{D}\\\\\\\\\\frac{D}{D}\\cdot v=\\frac{m}{D}\\\\\\\\1\\cdot v=\\frac{m}{D}\\\\\\\\v=\\frac{m}{D}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]m=D\\cdot v[\/latex] and [latex]\\displaystyle v=\\frac{m}{D}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Now let\u2019s look at an example using the formula for the volume of a cylinder.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>The formula for finding the volume of a cylinder is [latex]V=\\pi{r^{2}}h[\/latex], where [latex]V[\/latex] is the volume, [latex]r[\/latex] is the radius and [latex]h[\/latex] is the height of the cylinder. Rearrange the formula to solve for the height [latex]h[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q644317\">Show Solution<\/span><\/p>\n<div id=\"q644317\" class=\"hidden-answer\" style=\"display: none\">Start with the formula for the volume of a cylinder.<\/p>\n<p style=\"text-align: center\">[latex]V=\\pi{{r}^{2}}h[\/latex]<\/p>\n<p>Divide both sides by [latex]\\pi {{r}^{2}}[\/latex] to isolate [latex]h[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{V}{\\pi {{r}^{2}}}=\\frac{\\pi {{r}^{2}}h}{\\pi {{r}^{2}}}[\/latex]<\/p>\n<p>Simplify. You find the height, [latex]h[\/latex], is equal to [latex]\\frac{V}{\\pi {{r}^{2}}}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{V}{\\pi {{r}^{2}}}=h[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\displaystyle h=\\frac{V}{\\pi {{r}^{2}}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we give another example of solving for a variable in a formula, or as they are also called, a literal equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 2: Solve a Literal Equation for a Variable\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/ecEUUbRLDQs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>(8.5.2) - Solving applications of rational equations<\/h1>\n<h3>Average cost problems<\/h3>\n<p>Suppose we know that the cost of making a product is dependent on the number of items, [latex]x[\/latex], produced. This is given by the equation [latex]C\\left(x\\right)=15,000+500x[\/latex]. Note, the cost function consists of a <em>fixed cost<\/em> ([latex]\\$15,000[\/latex]) and a <em>variable cost<\/em> [latex](500x)[\/latex]. The <em>fixed cost<\/em> doesn't change when more items are produced, whereas the <em>variable cost<\/em> increases as more cars are produced.\u00a0 If we want to know the average cost for producing [latex]x[\/latex] items, we would divide the cost function by the number of items, [latex]x[\/latex].<\/p>\n<p>The average cost function, which yields the average cost per item for [latex]x[\/latex]\u00a0items produced, is<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\overline{C}\\left(x\\right)=\\frac{15,000 +500x}{x}[\/latex]<\/p>\n<p>Many other application problems require finding an average value in a similar way, giving us variables in the denominator.<\/p>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Let [latex]x[\/latex] represent the number of cars produced in a factory in a typical week. Suppose that the the cost in dollars to produce [latex]x[\/latex] cars is given by [latex]C(x) = 15,000+500x[\/latex]. How many cars would the factory have to produce in one week, in order for the average production cost per car to be [latex]\\$1,000[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q752418\">Show Answer<\/span><\/p>\n<div id=\"q752418\" class=\"hidden-answer\" style=\"display: none\">\n<p>The average cost of producing [latex]x[\/latex]:\u00a0cars per week is:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\overline{C}\\left(x\\right)=\\frac{15,000 +500x}{x}[\/latex]<\/p>\n<p>If we want for the average production cost per car to be [latex]\\$1,000[\/latex], we set up the following rational equation:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle 1,000=\\frac{15,000 +500x}{x}[\/latex]<\/p>\n<p>We multiply both sides [latex]x[\/latex]:<\/p>\n<p style=\"text-align: center\">[latex]1,000x = 15,000+500x[\/latex]<\/p>\n<p>Subtract [latex]500x[\/latex] from both sides:<\/p>\n<p style=\"text-align: center\">[latex]500x = 15,000[\/latex]<\/p>\n<p>Divide both sides by [latex]500[\/latex]:<\/p>\n<p style=\"text-align: center\">[latex]x=30[\/latex]<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p>The factory must produce 30 cars per week.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Work problems<\/h3>\n<p>Rational equations can be used to solve a variety of problems that involve rates, times and work. Using rational expressions and equations can help you answer questions about how to combine workers or machines to complete a job on schedule.<\/p>\n<div id=\"attachment_5024\" style=\"width: 336px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5024\" class=\"wp-image-5024\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/22162804\/Screen-Shot-2016-06-19-at-9.54.28-PM-233x300.png\" alt=\"Man with a lunch box walking. THere is a caption above him that says &quot;Boy! I sure did a good day's work today&quot;\" width=\"326\" height=\"420\" \/><\/p>\n<p id=\"caption-attachment-5024\" class=\"wp-caption-text\">A Good Day's Work<\/p>\n<\/div>\n<p>A \u201cwork problem\u201d is an example of a real life situation that can be modeled and solved using a rational equation. Work problems often ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, [latex]W=rt[\/latex].\u00a0(Notice that the work formula is very similar to the relationship between distance, rate, and time, or [latex]d=rt[\/latex].) The amount of work done [latex]W[\/latex]\u00a0is the product of the rate of work [latex]r[\/latex]\u00a0and the time spent working [latex]t[\/latex]. The work formula has 3 versions.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}W=rt\\\\\\\\\\,\\,\\,\\,\\,t=\\frac{W}{r}\\\\\\\\\\,\\,\\,\\,\\,r=\\frac{W}{t}\\end{array}[\/latex]<\/p>\n<p>Some work problems include multiple machines or people working on a project together for the same amount of time but at different rates. In that case, you can add their individual work rates together to get a total work rate. Let\u2019s look at an example.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Myra takes 2 hours to plant 50 flower bulbs. Francis takes 3 hours to plant 45 flower bulbs. Working together, how long should it take them to plant 150 bulbs?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q550322\">Show Solution<\/span><\/p>\n<div id=\"q550322\" class=\"hidden-answer\" style=\"display: none\">Think about how many bulbs each person can plant in one hour. This is their planting rate.<\/p>\n<p>Myra: [latex]\\displaystyle \\frac{50\\,\\,\\text{bulbs}}{2\\,\\,\\text{hours}}[\/latex], or [latex]\\displaystyle \\frac{25\\,\\,\\text{bulbs}}{1\\,\\,\\text{hour}}[\/latex]<\/p>\n<p>Francis: [latex]\\displaystyle \\frac{45\\,\\,\\text{bulbs}}{3\\,\\,\\text{hours}}[\/latex], or [latex]\\displaystyle \\frac{15\\,\\,\\text{bulbs}}{1\\,\\,\\text{hour}}[\/latex]<\/p>\n<p>Combine their hourly rates to determine the rate they work together.<\/p>\n<p>Myra and Francis together:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{25\\,\\,\\text{bulbs}}{1\\,\\,\\text{hour}}+\\frac{15\\,\\,\\text{bulbs}}{1\\,\\,\\text{hour}}=\\frac{40\\,\\,\\text{bulbs}}{1\\,\\,\\text{hour}}[\/latex]<\/p>\n<p>Use one of the work formulas to write a rational equation, for example [latex]\\displaystyle r=\\frac{W}{t}[\/latex]. You know [latex]r[\/latex], the combined work rate, and you know [latex]W[\/latex], the amount of work that must be done. What you don't know is how much time it will take to do the required work at the designated rate.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{40}{1}=\\frac{150}{t}[\/latex]<\/p>\n<p>Solve the equation by multiplying both sides by the common denominator, then isolating [latex]t[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{c}\\frac{40}{1}\\cdot t=\\frac{150}{t}\\cdot t\\\\\\\\40t=150\\\\\\\\t=\\frac{150}{40}=\\frac{15}{4}\\\\\\\\t=3\\frac{3}{4}\\text{hours}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>It should take 3 hours 45 minutes for Myra and Francis to plant 150 bulbs together.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 1:  Rational Equation Application - Painting Together\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/SzSasnDF7Ms?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Other work problems go the other way. You can calculate how long it will take one person to do a job alone when you know how long it takes people working together to complete the job.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Joe and John are planning to paint a house together. John thinks that if he worked alone, it would take him 3 times as long as it would take Joe to paint the entire house. Working together, they can complete the job in 24 hours. How long would it take each of them, working alone, to complete the job?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q593775\">Show Solution<\/span><\/p>\n<div id=\"q593775\" class=\"hidden-answer\" style=\"display: none\">Choose variables to represent the unknowns. Since it takes John 3 times as long as Joe to paint the house, his time is represented as [latex]3x[\/latex].<\/p>\n<p>Let [latex]x[\/latex]\u00a0= time it takes Joe\u00a0to complete the job<\/p>\n<p>[latex]3x[\/latex]<i>\u00a0<\/i>= time it takes John\u00a0to complete the job<\/p>\n<p>The work is painting 1 house or 1. Write an expression to represent each person\u2019s rate using the formula\u00a0[latex]\\displaystyle r=\\frac{W}{t}[\/latex]<sub>.<\/sub><\/p>\n<p>Joe\u2019s rate: [latex]\\displaystyle \\frac{1}{x}[\/latex]<\/p>\n<p>John\u2019s rate: [latex]\\displaystyle \\frac{1}{3x}[\/latex]<\/p>\n<p>Their combined rate is the sum of their individual rates. Use this rate to write a new equation using the formula [latex]W=rt[\/latex].<\/p>\n<p>combined rate: [latex]\\displaystyle \\frac{1}{x}+\\frac{1}{3x}[\/latex]<\/p>\n<p>The problem states that it takes them 24 hours together to paint a house, so if you multiply their combined hourly rate [latex]\\displaystyle \\left( \\frac{1}{x}+\\frac{1}{3x} \\right)[\/latex] by 24, you will get 1, which is the number of houses they can paint in 24 hours.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}1=\\left( \\frac{1}{x}+\\frac{1}{3x} \\right)24\\\\\\\\1=\\frac{24}{x}+\\frac{24}{3x}\\end{array}[\/latex]<\/p>\n<p>Now solve the equation for [latex]x[\/latex]. (Remember that [latex]x[\/latex] represents the number of hours it will take Joe to finish the job.)<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}\\,\\,\\,1=\\frac{3}{3}\\cdot \\frac{24}{x}+\\frac{24}{3x}\\\\\\\\\\,\\,\\,1=\\frac{3\\cdot 24}{3x}+\\frac{24}{3x}\\\\\\\\\\,\\,\\,1=\\frac{72}{3x}+\\frac{24}{3x}\\\\\\\\\\,\\,\\,1=\\frac{72+24}{3x}\\\\\\\\\\,\\,\\,1=\\frac{96}{3x}\\\\\\\\3x=96\\\\\\\\\\,\\,\\,x=32\\end{array}[\/latex]<\/p>\n<p>Check the solutions in the original equation.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}1=\\left( \\frac{1}{x}+\\frac{1}{3x} \\right)24\\\\\\\\1=\\left[ \\frac{\\text{1}}{\\text{32}}+\\frac{1}{3\\text{(32})} \\right]24\\\\\\\\1=\\frac{24}{\\text{32}}+\\frac{24}{3\\text{(32})}\\\\\\\\1=\\frac{24}{\\text{32}}+\\frac{24}{96}\\\\\\\\1=\\frac{3}{3}\\cdot \\frac{24}{\\text{32}}+\\frac{24}{96}\\\\\\\\1=\\frac{72}{96}+\\frac{24}{96}\\end{array}[\/latex]<\/p>\n<p>The solution checks. Since [latex]x=32[\/latex], it takes Joe 32 hours to paint the house by himself. John\u2019s time is 3<i>x<\/i>, so it would take him 96 hours to do the same amount of work.<\/p>\n<h4>Answer<\/h4>\n<p>It takes 32 hours for Joe to paint the house by himself and 96 hours for John the paint the house himself.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video that follows, we show another example of finding one person's work rate given a combined work rate.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex: Rational Equation App - Find Individual Working Time Given Time Working Together\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/kbRSYb8UYqU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>As shown above, many work problems can be represented by the equation [latex]\\displaystyle \\frac{t}{a}+\\frac{t}{b}=1[\/latex], where [latex]t[\/latex] is the time to do the job together, [latex]a[\/latex] is the time it takes person A to do the job, and [latex]b[\/latex] is the time it takes person B to do the job. The 1 refers to the total work done\u2014in this case, the work was to paint 1 house.<\/p>\n<p>The key idea here is to figure out each worker\u2019s individual rate of work. Then, once those rates are identified, add them together, multiply by the time [latex]t[\/latex], set it equal to the amount of work done, and solve the rational equation. Here is a summary:<\/p>\n<div class=\"textbox learning-objectives\">\n<h3>Formula for work problems<\/h3>\n<p>[latex]a=[\/latex] the time needed for A to complete 1 job alone<\/p>\n<p>[latex]b=[\/latex] the time needed for B to complete 1 job alone<\/p>\n<p>[latex]t=[\/latex] the time needed for A and B to complete the 1 job together:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\frac{t}{a}+\\frac{t}{b} = 1[\/latex]<\/p>\n<\/div>\n<p>We present another example of two people painting at different rates in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex 1:  Rational Equation Application - Painting Together\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/SzSasnDF7Ms?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Motion problems<\/h3>\n<p>We have solved uniform motion problems using the formula [latex]D = rt[\/latex] in previous chapters. We used a table like the one below to organize the information and lead us to the equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{|c|c|c|c|}  \\hline  & \\,\\text{rate}\\,&\\,\\text{time}\\,&\\,\\text{distance}\\\\  \\hline  \\,\\text{First}\\,&\\,&\\,&\\\\  \\hline  \\,\\text{Second}\\,&\\,&\\,&\\\\  \\hline  \\end{array}[\/latex]<\/p>\n<p>The formula [latex]D=rt[\/latex] assumes we know [latex]r[\/latex] and\u00a0 [latex]t[\/latex] and use them to find\u00a0 [latex]D[\/latex]. If we know\u00a0 [latex]D[\/latex] and [latex]r[\/latex] and need to find\u00a0 [latex]t[\/latex], we would solve the equation for\u00a0 [latex]t[\/latex] and get the formula [latex]\\displaystyle t=\\frac{D}{r}[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Greg went to a conference in a city 120 miles away. On the way back, due to\u00a0road construction he had to drive 10 mph slower which resulted in the return\u00a0trip taking 2 hours longer. How fast did he drive on the way to the\u00a0conference?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q519818\">Show Solution<\/span><\/p>\n<div id=\"q519818\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{eqnarray*}  \\begin{array}{|c|c|c|c|}  \\hline  & \\,\\text{rate}\\,& \\,\\text{time}\\,& \\,\\text{distance}\\\\  \\hline  \\,\\text{There}\\,& r & t & 120\\\\  \\hline  \\,\\text{Back}\\,& & & \\\\  \\hline  \\end{array}\\,& & \\begin{array}{l}  \\,\\text{We}\\,\\,\\text{do}\\,\\,\\text{not}\\,\\,\\text{know}\\,\\,\\text{rate}, r, \\,\\text{or}  \\,\\text{time}, t \\,\\text{he}\\,\\,\\text{traveled}\\\\  \\,\\text{on}\\,\\,\\text{the}\\,\\,\\text{way}\\,\\,\\text{to}\\,\\,\\text{the}\\,\\,\\text{conference}\\,.  \\,\\text{But}\\,\\,\\text{we}\\,\\,\\text{do}\\,\\,\\text{know}\\\\  \\,\\text{the}\\,\\,\\text{distance}\\,\\,\\text{was}\\,120 \\,\\text{miles}\\,.  \\end{array}\\\\  & & \\\\  \\begin{array}{|c|c|c|c|}  \\hline  & \\,\\text{rate}\\,& \\,\\text{time}\\,& \\,\\text{distance}\\\\  \\hline  \\,\\text{There}\\,& r & t & 120\\\\  \\hline  \\,\\text{Back}\\,& r - 10 & t + 2 & 120\\\\  \\hline  \\end{array}& & \\begin{array}{l}  \\,\\text{Coming}\\,\\,\\text{back}\\,\\,\\text{he}\\,\\,\\text{drove}\\,10 \\,\\text{mph}  \\,\\text{slower}\\,(r - 10)\\\\  \\,\\text{and}\\,\\,\\text{took}\\,2 \\,\\text{hours}\\,\\,\\text{longer}\\,(t + 2) . \\,\\text{The}  \\,\\text{distance}\\\\  \\,\\text{was}\\,\\,\\text{still}\\,120 \\,\\text{miles}\\,.  \\end{array}\\\\  & & \\\\  r t = 120 & & \\,\\text{Equations}\\,\\,\\text{are}\\,\\,\\text{product}\\,\\,\\text{of}  \\,\\text{rate}\\,\\,\\text{and}\\,\\,\\text{time}\\\\  (r - 10) (t + 2) = 120 & & \\,\\text{We}\\,\\,\\text{have}\\,\\,\\text{simultaneous}  \\,\\text{product}\\,\\,\\text{equations}\\\\  & & \\\\  t = \\frac{120}{r}\\,\\,\\text{and}\\,t + 2 = \\frac{120}{r - 10} & &  \\,\\text{Solving}\\,\\,\\text{for}\\,\\,\\text{rate}, \\,\\text{divide}\\,\\,\\text{by}\\,r  \\,\\text{and}\\,r - 10\\\\  & & \\\\  \\frac{120}{r}\\,+ 2 = \\frac{120}{r - 10}& & \\,\\text{Substitute}  \\frac{120}{r} \\,\\text{for}\\,t \\,\\text{in}\\,\\,\\text{the}\\,\\,\\text{second}  \\,\\text{equation}\\\\  & & \\\\  \\frac{120 r (r - 10)}{r} + 2 r (r - 10) = \\frac{120 r (r - 10)}{r - 10} &  & \\,\\text{Multiply}\\,\\,\\text{each}\\,\\,\\text{term}\\,\\,\\text{by}\\,\\,\\text{LCD}\\,: r (r -  10)\\\\  & & \\\\  120 (r - 10) + 2 r^2 - 20 r = 120 r & & \\,\\text{Reduce}\\,\\,\\text{each}  \\,\\text{fraction}\\\\  120 r - 1200 + 2 r^2 - 20 r = 120 r & & \\,\\text{Distribute}\\\\  2 r^2 + 100 r - 1200 = 120 r & & \\,\\text{Combine}\\,\\,\\text{like}  \\,\\text{terms}\\\\  \\underline{- 120 r - 120 r} & & \\,\\text{Make}\\,\\,\\text{equation}\\,\\,\\text{equal}  \\,\\text{to}\\,\\,\\text{zero}\\\\  2 r^2 - 20 r - 1200 = 0 & & \\,\\text{Divide}\\,\\,\\text{each}\\,\\,\\text{term}  \\,\\text{by}\\,2\\\\  r^2 - 10 r - 600 = 0 & & \\,\\text{Factor}\\\\  (r - 30) (r + 20) = 0 & & \\,\\text{Set}\\,\\,\\text{each}\\,\\,\\text{factor}  \\,\\text{equal}\\,\\,\\text{to}\\,\\,\\text{zero}\\\\  r - 30 = 0 \\,\\text{and}\\,r + 20 = 0 & & \\,\\text{Solve}\\,\\,\\text{each}  \\,\\text{equation}\\\\  \\underline{+ 30 + 30} \\underline{- 20 - 20} & & \\\\  r = 30 \\,\\text{and}\\,r = - 20 & & \\,\\text{Can}' t \\,\\text{have}\\,a  \\,\\text{negative}\\,\\,\\text{rate}\\\\  30 \\,\\text{mph}\\,& & \\,\\text{Our}\\,\\,\\text{Solution}  \\end{eqnarray*}[\/latex]<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>30mph<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>A man rows downstream for 30 miles then turns around and returns to his\u00a0original location, the total trip took 8 hours. If the current flows at 2\u00a0miles per hour, how fast would the man row in still water?<\/p>\n<p style=\"text-align: center\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q2397\">Show Answer<\/span><\/p>\n<div id=\"q2397\" class=\"hidden-answer\" style=\"display: none\">\\begin{eqnarray*}<br \/>\n\\begin{array}{|c|c|c|c|}<br \/>\n\\hline<br \/>\n&amp; \\,\\text{rate}\\,&amp; \\,\\text{time}\\,&amp; \\,\\text{distance}\\\\<br \/>\n\\hline<br \/>\n\\,\\text{down}\\,&amp; &amp; t &amp; 30\\\\<br \/>\n\\hline<br \/>\n\\,\\text{up}\\,&amp; &amp; 8 - t &amp; 30\\\\<br \/>\n\\hline<br \/>\n\\end{array} &amp; &amp; \\begin{array}{l}<br \/>\n\\,\\text{We}\\,\\,\\text{know}\\,\\,\\text{the}\\,\\,\\text{distance}\\,\\,\\text{up}\\,\\,\\text{and}\\,<br \/>\n\\,\\text{down}\\,\\,\\text{is}\\,30.\\\\<br \/>\n\\,\\text{Put}\\,t \\,\\text{for}\\,\\,\\text{time}\\,\\,\\text{downstream}\\,.<br \/>\n\\,\\text{Subtracting}\\\\<br \/>\n8 - t \\,\\text{becomes}\\,\\,\\text{time}\\,\\,\\text{upstream}\\,<br \/>\n\\end{array}\\\\<br \/>\n&amp; &amp; \\\\<br \/>\n\\begin{array}{|c|c|c|c|}<br \/>\n\\hline<br \/>\n&amp; \\,\\text{rate}\\,&amp; \\,\\text{time}\\,&amp; \\,\\text{distance}\\\\<br \/>\n\\hline<br \/>\n\\,\\text{down}\\,&amp; r + 2 &amp; t &amp; 30\\\\<br \/>\n\\hline<br \/>\n\\,\\text{up}\\,&amp; r - 2 &amp; 8 - t &amp; 30\\\\<br \/>\n\\hline<br \/>\n\\end{array}&amp; &amp; \\begin{array}{l}<br \/>\n\\,\\text{Downstream}\\,\\,\\text{the}\\,\\,\\text{current}\\,\\,\\text{of}\\,2 \\,\\text{mph}<br \/>\n\\,\\text{pushes}\\,\\\\<br \/>\n\\,\\text{the}\\,\\,\\text{boat} (r + 2) \\,\\text{and}\\,\\,\\text{upstream}\\,\\,\\text{the}\\,<br \/>\n\\,\\text{current}\\,\\\\<br \/>\n\\,\\text{pulls}\\,\\,\\text{the}\\,\\,\\text{boat}\\,(r - 2)<br \/>\n\\end{array}\\\\<br \/>\n&amp; &amp; \\\\<br \/>\n(r + 2) t = 30 &amp; &amp; \\,\\text{Multiply}\\,\\,\\text{rate}\\,\\,\\text{by}\\,\\,\\text{time}<br \/>\n\\,\\text{to}\\,\\,\\text{get}\\,\\,\\text{equations}\\,\\\\<br \/>\n(r - 2) (8 - t) = 30 &amp; &amp; \\,\\text{We}\\,\\,\\text{have}\\,a \\,\\text{simultaneous}<br \/>\n\\,\\text{product}\\\\<br \/>\n&amp; &amp; \\\\<br \/>\nt = \\frac{30}{r + 2} \\,\\text{and}\\,8 - t = \\frac{30}{r - 2} &amp; &amp;<br \/>\n\\,\\text{Solving}\\,\\,\\text{for}\\,\\,\\text{rate}, \\,\\text{divide}\\,\\,\\text{by}\\,r + 2<br \/>\n\\,\\text{or}\\,r - 2\\\\<br \/>\n&amp; &amp; \\\\<br \/>\n8 - \\frac{30}{r + 2}\\,= \\frac{30}{r - 2} &amp; &amp; \\,\\text{Substitute}\\,\\frac{30}{r<br \/>\n+ 2} \\,\\text{for}\\,t \\,\\text{in}\\,\\,\\text{second}\\,\\,\\text{equation}\\,\\\\<br \/>\n&amp; &amp; \\\\<br \/>\n8 (r + 2) (r - 2) - \\frac{30 (r + 2) (r - 2)}{r + 2} = \\frac{30 (r + 2) (r<br \/>\n- 2)}{r - 2} &amp; &amp; \\,\\text{Multiply}\\,\\,\\text{each}\\,\\,\\text{term}\\,\\,\\text{by}<br \/>\n\\,\\text{LCD}\\,: (r + 2) (r - 2)\\\\<br \/>\n&amp; &amp; \\\\<br \/>\n8 (r + 2) (r - 2) - 30 (r - 2) = 30 (r + 2) &amp; &amp; \\,\\text{Reduce}<br \/>\n\\,\\text{fractions}\\,\\\\<br \/>\n8 r^2 - 32 - 30 r + 60 = 30 r + 60 &amp; &amp; \\,\\text{Multiply} \\,\\text{and}<br \/>\n\\,\\text{distribute}\\\\<br \/>\n8 r^2 - 30 r + 28 = 30 r + 60 &amp; &amp; \\,\\text{Make} \\,\\text{equation}<br \/>\n\\,\\text{equal}\\,\\,\\text{zero}\\,\\\\<br \/>\n\\underline{- 30 r - 60 - 30 r - 60} &amp; &amp; \\\\<br \/>\n8 r^2 - 60 r - 32 = 0 &amp; &amp; \\,\\text{Divide}\\,\\,\\text{each}\\,\\,\\text{term}\\,\\,\\text{by}<br \/>\n4\\\\<br \/>\n2 r^2 - 15 r - 8 = 0 &amp; &amp; \\,\\text{Factor}\\\\<br \/>\n(2 r + 1) (r - 8) = 0 &amp; &amp; \\,\\text{Set}\\,\\,\\text{each}\\,\\,\\text{factor}<br \/>\n\\,\\text{equal}\\,\\,\\text{to}\\,\\,\\text{zero}\\,\\\\<br \/>\n2 r + 1 = 0 \\,\\text{or}\\,r - 8 = 0 &amp; &amp; \\,\\text{Solve}\\,\\,\\text{each}\\,<br \/>\n\\,\\text{equation}\\\\<br \/>\n\\underline{- 1 - 1} \\underline{+ 8 + 8} &amp; &amp; \\\\<br \/>\n2 r = - 1 \\,\\text{or}\\,r = 8 &amp; &amp; \\\\<br \/>\n\\overline{2} \\overline{2} &amp; &amp; \\\\<br \/>\nr = - \\frac{1}{2} \\,\\text{or}\\, r = 8 &amp; &amp; \\,\\text{Can}' t \\,\\text{have}\\,a<br \/>\n\\,\\text{negative}\\,\\,\\text{rate}\\,\\\\<br \/>\n8 \\,\\text{mph}\\,&amp; &amp; \\,\\text{Our}\\,\\,\\text{Solution}<br \/>\n\\end{eqnarray*}<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>8 mph<\/p>\n<p style=\"text-align: left\"><\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex 3:  Rational Equation Application - Plane and Car Travelling the Same Time\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Zjx2hCjKtiQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Ex 4:  Rational Equation Application - Two Bikers Riding Different Distances\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/BpFVUXrChLM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Concentration of a mixture problems<\/h3>\n<p>Mixtures are made of\u00a0ratios of different substances that may include chemicals, foods, water, or gases. There are many different situations where mixtures may occur both in nature and as a means to produce a desired product or outcome. \u00a0For example, chemical spills, manufacturing and even biochemical reactions involve mixtures. \u00a0The thing that can make mixtures interesting mathematically is when components of the mixture are\u00a0added at different rates and concentrations. In our last example we will define an equation that models the\u00a0concentration \u00a0- or ratio of sugar to water - in a large mixing tank over time. You are asked whether the final concentration of sugar is greater than the concentration at the beginning.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q332373\">Show Answer<\/span><\/p>\n<div id=\"q332373\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]t[\/latex]\u00a0be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is a linear equation, as is the amount of sugar in the tank. We can write an equation independently for each:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{cases}\\text{water: }W\\left(t\\right)=100+10t\\text{ in gallons}\\\\ \\text{sugar: }S\\left(t\\right)=5+1t\\text{ in pounds}\\end{cases}[\/latex]<\/p>\n<p>The concentration, [latex]C[\/latex], will be the ratio of pounds of sugar to gallons of water<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle C\\left(t\\right)=\\frac{5+t}{100+10t}[\/latex]<\/p>\n<p>The concentration after 12 minutes is given by evaluating [latex]C\\left(t\\right)[\/latex] at [latex]t=\\text{ }12[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle C\\left(12\\right)=\\frac{5+12}{100+10\\left(12\\right)} =\\frac{17}{220}[\/latex]<\/p>\n<p>This means the concentration is 17 pounds of sugar to 220 gallons of water.<\/p>\n<p>At the beginning, the concentration is<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle C\\left(0\\right)=\\frac{5+0}{100+10\\left(0\\right)}=\\frac{1}{20}[\/latex]<\/p>\n<p>Since [latex]\\displaystyle \\frac{17}{220}\\approx 0.08>\\frac{1}{20}=0.05[\/latex], the concentration is greater after 12 minutes than at the beginning.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show another example of how to use rational functions\u00a0to model mixing.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Rational Function Application - Concentration of a Mixture\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/GD6H7BE_0EI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2><\/h2>\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-2934\">\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>Solve Basic Rational Equations. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/R9y2D9VFw0I\">https:\/\/youtu.be\/R9y2D9VFw0I<\/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>Solve Rational Equations with Like Denominators. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/gGA-dF_aQQQ\">https:\/\/youtu.be\/gGA-dF_aQQQ<\/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>Solve Basic Rational Equations. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/R9y2D9VFw0I\">https:\/\/youtu.be\/R9y2D9VFw0I<\/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>Screenshot: map with scale factor. <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>Screenshot: so many cars, so many tires. <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>Screenshot: Water temperature in the ocean varies inversely with depth. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/Lumen%20Learning\">http:\/\/Lumen%20Learning<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Direct Variation Application - Aluminum Can Usage. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/DLPKiMD_ZZw\">https:\/\/youtu.be\/DLPKiMD_ZZw<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Inverse Variation Application - Number of Workers and Job Time. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/y9wqI6Uo6_M\">https:\/\/youtu.be\/y9wqI6Uo6_M<\/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>Joint Variation: Determine the Variation Constant (Volume of a Cone). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/JREPATMScbM\">https:\/\/youtu.be\/JREPATMScbM<\/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>Screenshot: A Good Day&#039;s Work. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><li>Ex 1: Rational Equation Application - Painting Together. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/SzSasnDF7Ms\">https:\/\/youtu.be\/SzSasnDF7Ms<\/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>Rational Function Application - Concentration of a Mixture. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/GD6H7BE_0EI\">https:\/\/youtu.be\/GD6H7BE_0EI<\/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>Unit 15: Rational Expressions, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Solve a Literal Equation for a Variable. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=ecEUUbRLDQs&#038;feature=youtu.be\">https:\/\/www.youtube.com\/watch?v=ecEUUbRLDQs&#038;feature=youtu.be<\/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>Screenshot: Matroyshka, or nesting dolls.. <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>Ex: Proportion Applications - Mixtures  . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/yGid1a_x38g\">https:\/\/youtu.be\/yGid1a_x38g<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Rational Equation App - Find Individual Working Time Given Time Working Together. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=kbRSYb8UYqU&#038;feature=youtu.be\">https:\/\/www.youtube.com\/watch?v=kbRSYb8UYqU&#038;feature=youtu.be<\/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: Mixture Problem. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at   http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/li><li>Quadratics - Revenue and Distance - Motion Examples. <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>Project<\/strong>: Beginning and Intermediate Algebra. <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":21,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 15: Rational Expressions, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Solve Basic Rational Equations\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/R9y2D9VFw0I\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Solve a Literal Equation for a Variable\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/watch?v=ecEUUbRLDQs&feature=youtu.be\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Solve Rational Equations with Like Denominators\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/gGA-dF_aQQQ\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Solve Basic Rational Equations\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/R9y2D9VFw0I\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Screenshot: Matroyshka, or nesting dolls.\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Screenshot: map with scale factor\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Proportion Applications - Mixtures  \",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/yGid1a_x38g\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Screenshot: so many cars, so many tires\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Screenshot: Water temperature in the ocean varies inversely with depth\",\"author\":\"\",\"organization\":\"\",\"url\":\"Lumen Learning\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex: Direct Variation Application - Aluminum Can Usage\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/DLPKiMD_ZZw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex: Inverse Variation Application - 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