{"id":182,"date":"2023-11-08T16:10:17","date_gmt":"2023-11-08T16:10:17","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/15-3-1-rational-formulas-and-variation\/"},"modified":"2026-02-13T19:16:47","modified_gmt":"2026-02-13T19:16:47","slug":"4-7-solving-applications-involving-rational-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/4-7-solving-applications-involving-rational-equations\/","title":{"raw":"4.7 Solving Applications Involving Rational Equations","rendered":"4.7 Solving Applications Involving Rational Equations"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve motion applications using rational equations.<\/li>\r\n \t<li>Solve work applications using rational equations.<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe reasoning behind both work and (uniform) motion applications is very similar. You should already be familiar with the formula [latex]d=r\\cdot t[\/latex] relating the quantities of distance, rate, and time.\u00a0The rate, [latex]r[\/latex], in the formula, is the <em>rate of change<\/em> of distance, [latex]d[\/latex], with respect to time, [latex]t[\/latex]. For example, if an object moves at a (constant) speed of [latex]30[\/latex] miles per hour, then it travels 30 miles in one hour, [latex]30\\cdot 2 = 60[\/latex] miles in [latex]2[\/latex] hours, [latex]30\\cdot 3.5=105[\/latex] miles in [latex]3.5[\/latex] hours, etc. Note that to compute the distance traveled, we multiply the rate (expressed in a unit of distance by a unit of time) by the number of (compatible) units of time. A similar\r\n<p style=\"text-align: center;\">\u201cquantity\u201d [latex]=[\/latex] \u201cthe rate of change of the quantity with respect to a counting unit\u201d [latex]\\cdot[\/latex] \u201cthe number of counting units\u201d<\/p>\r\nprinciple is used in the following relationships:\r\n<ul>\r\n \t<li>[latex]R=x\\cdot p[\/latex] between the revenue, [latex]R[\/latex] (the amount of money that comes into the business), the quantity, [latex]x[\/latex], of a product sold, and the unit price, [latex]p[\/latex] (i.e., dollars per unit sold). Note that the price per unit is the <em>rate of change<\/em> of revenue with respect to the quantity of a product sold. (Refer to <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-1-applications-of-systems-of-2-linear-equations\/#CRP\" target=\"_blank\" rel=\"noopener\">this description<\/a>.)<\/li>\r\n \t<li>[latex]m=C\\cdot V[\/latex]\u00a0between the mass, [latex]m[\/latex], of a solvent, the volume, [latex]V[\/latex], of a solution, and the concentration, [latex]C[\/latex], the amount of solvent per unit volume of solution.\u00a0Note that the concentration is the <em>rate of change<\/em> of mass with respect to volume. (Refer to <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-1-applications-of-systems-of-2-linear-equations\/#mixture\" target=\"_blank\" rel=\"noopener\">this description<\/a>.)<\/li>\r\n<\/ul>\r\n<h2>Motion Applications<\/h2>\r\nMotion applications have already been introduced in <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-1-applications-of-systems-of-2-linear-equations\/#motion\" target=\"_blank\" rel=\"noopener\">Sec. 2.1<\/a>. We continue working with uniform motion\u00a0with a constant speed in this section. Equivalent forms of the [latex]d=r\\cdot t[\/latex] are [latex]r=\\dfrac{d}{t}[\/latex] and [latex]t=\\dfrac{d}{r}[\/latex] (the latter form has been discussed in <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/solving-a-formula-for-a-specific-variable\/#drtforms\" target=\"_blank\" rel=\"noopener\">Sec. 4.6<\/a>).\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAvery can paddle her kayak [latex]4[\/latex] miles per hour in still water. It takes her as long to paddle [latex]10[\/latex] miles upstream as it takes her to travel [latex]22[\/latex] miles downstream. Determine the speed of the river's current.\r\n\r\n[reveal-answer q=\"462935\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"462935\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>\r\n<p id=\"fs-id1167834098208\">Recall that when the kayak is going downstream, in the same direction as the river's current, the current helps push the kayak, so the kayak\u2019s actual speed is faster than its speed in still water. The actual speed at which the kayak is moving is the sum of the speeds in still water and the river's current. When<span style=\"font-size: 1rem; text-align: initial;\">\u00a0the kayak is going upstream, opposite to the river current though, the current is going against the kayak, so the kayak\u2019s actual speed is slower than its speed in still water. The actual speed of the kayak is the difference between the speed in still water and the speed of the river's current.<\/span><\/p>\r\nThe speed of the kayak in still water is known while the speed of the river's current is to be found. We do know the distances Avery paddles her kayak upstream and downstream. We do not know the times but we do know the two times are the same.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>Let [latex]c[\/latex] be the unknown speed of the river's current.\u00a0Going upstream, the actual rate of the kayak is [latex]4-c[\/latex]. Going downstream, the actual rate is [latex]4+c[\/latex].\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know. The table is not necessary, but can help you organize.\r\n<table style=\"font-size: 110%;\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td><strong>Distance<\/strong><\/td>\r\n<td><strong>Rate (or speed)<\/strong><\/td>\r\n<td><strong>Time<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Upstream<\/strong><\/td>\r\n<td>[latex]10[\/latex]<\/td>\r\n<td>[latex]4-c[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Downstream<\/strong><\/td>\r\n<td>[latex]22[\/latex]<\/td>\r\n<td>[latex]4+c[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\\dfrac{d}{r}[\/latex]:\r\n<table style=\"font-size: 110%;\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td><strong>Distance<\/strong><\/td>\r\n<td><strong>Rate (or speed)<\/strong><\/td>\r\n<td><strong>Time<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Upstream<\/strong><\/td>\r\n<td>[latex]10[\/latex]<\/td>\r\n<td>[latex]4-c[\/latex]<\/td>\r\n<td>[latex]\\dfrac{10}{4-c}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Downstream<\/strong><\/td>\r\n<td>[latex]22[\/latex]<\/td>\r\n<td>[latex]4+c[\/latex]<\/td>\r\n<td>[latex]\\dfrac{22}{4+c}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince the two times are <em>the same<\/em> (mathematically, <em>equal<\/em>), w<span style=\"font-size: 1rem; text-align: initial;\">e obtain the following rational equation:<\/span>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{10}{4-c}=\\dfrac{22}{4+c}[\/latex]<\/p>\r\nLet's acknowledge that the restricted values of [latex]c[\/latex] are [latex]4[\/latex] and [latex]-4[\/latex]. (Can you explain why?) Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\require{color}\r\n\\begin{align}\r\n\\frac{10}{4-c}{\\color{red}\\: \\cdot\\: (4-c)\\cdot (4+c)}&amp;=\\frac{22}{4+c}{\\color{red}\\: \\cdot\\: (4-c)\\cdot (4+c)} &amp;&amp; \\color{blue}{\\textsf{multiply by the LCD}}\\\\[5pt]\r\n\\frac{10}{{\\color{red}\\cancel{\\color{black}{(4-c)}}}}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{(4-c)}}}\\cdot (4+c)}{1} &amp;= \\frac{22}{{\\color{red}\\cancel{\\color{black}{(4+c)}}}}\\cdot \\frac{(4-c)\\cdot {\\color{red}\\cancel{\\color{black}{(4+c)}}}}{1} &amp;&amp; \\color{blue}{\\textsf{clear the fractions}}\\\\[5pt]10\\cdot (4+c) &amp;= 22\\cdot (4-c)\\\\[5pt]\r\n\\underset{\\large{\\color{red}{-40}}}{40}+\\underset{\\large{\\color{red}{+22c}}}{10c} &amp;= \\underset{\\large{\\color{red}{-40}}}{88}-\\underset{\\large{\\color{red}{+22c}}}{22c} &amp;&amp; \\color{blue}{\\textsf{isolate $c$-terms}}\\\\[5pt]\r\n{\\color{red}{\\frac{\\cancel{\\color{black}{32}}{\\color{black}{\\: \\cdot\\: c}}}{\\cancel{32} \\cdot 1}}} &amp;= {\\color{red}{\\frac{{\\color{black}{48}}}{32}}} &amp;&amp; \\color{blue}{\\textsf{divide to isolate $c$}}\\\\[5pt]\r\nc &amp;= \\frac{{\\color{red}\\cancel{\\color{black}{16}}}\\cdot 3}{{\\color{red}\\cancel{\\color{black}{16}}}\\cdot 2} &amp;&amp; \\color{blue}{\\textsf{simplify}}\\\\[5pt]\r\nc &amp;= \\frac{3}{2}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n<strong>Answer:\u00a0<\/strong>The river's current's speed is [latex]\\dfrac{3}{2}[\/latex]\u00a0miles per hour (or [latex]1.5[\/latex] miles per hour).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nHassan can bicycle [latex]72[\/latex] kilometers in the same time as it takes him to walk [latex]30[\/latex] kilometers. He can ride [latex]7[\/latex] km\/h faster than he can walk. How fast can he walk?\r\n\r\n[reveal-answer q=\"318933\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"318933\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>\r\n\r\nNeither the speed of bicycling nor walking is known but we know Hassan can ride [latex]7[\/latex] km\/h faster than he can walk. The walking speed is to be found. We do know the distances Hassan bicycles and walks. We do not know the times but we do know the two times are the same.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>Let [latex]w[\/latex] be the unknown speed of Hassan walking.\u00a0Since he can ride [latex]7[\/latex] km\/h <strong>faster<\/strong> than he can walk, his bicycling speed is [latex]w+7[\/latex].\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know. The table is not necessary, but can help you organize.\r\n<table style=\"font-size: 110%;\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td><strong>Distance<\/strong><\/td>\r\n<td><strong>Rate (or speed)<\/strong><\/td>\r\n<td><strong>Time<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Bicycling<\/strong><\/td>\r\n<td>[latex]72[\/latex]<\/td>\r\n<td>[latex]w+7[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Walking<\/strong><\/td>\r\n<td>[latex]30[\/latex]<\/td>\r\n<td>[latex]w[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\\dfrac{d}{r}[\/latex]:\r\n<table style=\"font-size: 110%; width: 555px;\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 130px;\"><\/td>\r\n<td style=\"width: 95px;\"><strong>Distance<\/strong><\/td>\r\n<td style=\"width: 157px;\"><strong>Rate (or speed)<\/strong><\/td>\r\n<td style=\"width: 173px;\"><strong>Time<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 130px;\"><strong>Bicycling<\/strong><\/td>\r\n<td style=\"width: 95px;\">[latex]72[\/latex]<\/td>\r\n<td style=\"width: 157px;\">[latex]w+7[\/latex]<\/td>\r\n<td style=\"width: 173px;\">[latex]\\dfrac{72}{w+7}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 130px;\"><strong>Walking<\/strong><\/td>\r\n<td style=\"width: 95px;\">[latex]30[\/latex]<\/td>\r\n<td style=\"width: 157px;\">[latex]w[\/latex]<\/td>\r\n<td style=\"width: 173px;\">[latex]\\dfrac{30}{w}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince the two times are <em>the same<\/em> (mathematically, <em>equal<\/em>), w<span style=\"font-size: 1rem; text-align: initial;\">e obtain the following rational equation:<\/span>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{72}{w+7}=\\dfrac{30}{w}[\/latex]<\/p>\r\nLet's acknowledge that the restricted values of [latex]w[\/latex] are [latex]-7[\/latex] and [latex]0[\/latex]. Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\require{color}\r\n\\begin{align}\r\n\\frac{72}{w+7}&amp;=\\frac{30}{w}\\\\[5pt]\r\n72\\cdot w &amp;= (w+7)\\cdot 30 &amp;&amp; \\color{blue}{\\textsf{clear the fractions}}\\\\[5pt]\r\n\\underset{\\large{\\color{red}{-30w}}}{72w} &amp;= \\underset{\\large{\\color{red}{-30w}}}{30w}+210\\\\[5pt]\r\n{\\color{red}{\\frac{\\cancel{\\color{black}{42}}{\\color{black}{\\: \\cdot\\: w}}}{\\cancel{42} \\cdot 1}}} &amp;= {\\color{red}{\\frac{{\\color{black}{210}}}{42}}}\\\\[5pt]\r\nw &amp;= 5\r\n\\end{align}\r\n[\/latex]<\/p>\r\n<strong>Answer:\u00a0<\/strong>Hassan\u00a0can walk [latex]5[\/latex]\u00a0kilometers per hour.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe Vistula river current's speed is [latex]4[\/latex] miles per hour. A river tram in Krak\u00f3w travels to a point [latex]6[\/latex] miles upstream and back again in [latex]2[\/latex] hours. What is the speed of the river tram\u00a0in still water (without a current)?\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/plikimpi.krakow.pl\/\/zalacznik\/36001\/4.jpg\" alt=\"decorative image\" width=\"300\" height=\"300\" \/>\r\n\r\n[reveal-answer q=\"361230\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"361230\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>\r\n\r\nThe speed of the river's current is known while the speed\u00a0of the river tram in still water is to be found. We do know the distances the river tram goes upstream and downstream. We do not know the individual times but we do know the roundtrip total time.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>Let [latex]b[\/latex] be the unknown speed of the river tram.\u00a0Going upstream, the actual rate of the river tram is [latex]b-4[\/latex]. Going downstream, the actual rate is [latex]b+4[\/latex].\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know. The table is not necessary, but can help you organize. Note that sometimes the variable is first and other times it is last, like in our previous example. The boat or vehicle number\/variable is always first.\r\n<table style=\"font-size: 110%;\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td><strong>Distance<\/strong><\/td>\r\n<td><strong>Rate (or speed)<\/strong><\/td>\r\n<td><strong>Time<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Upstream<\/strong><\/td>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]b-4[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Downstream<\/strong><\/td>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]b+4[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\\dfrac{d}{r}[\/latex]:\r\n<table style=\"font-size: 110%;\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td><strong>Distance<\/strong><\/td>\r\n<td><strong>Rate (or speed)<\/strong><\/td>\r\n<td><strong>Time<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Upstream<\/strong><\/td>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]b-4[\/latex]<\/td>\r\n<td>[latex]\\dfrac{6}{b-4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Downstream<\/strong><\/td>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]b+4[\/latex]<\/td>\r\n<td>[latex]\\dfrac{6}{b+4}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince the two times <em>total<\/em> (mathematically, <em>their sum is<\/em>) 2 hours, w<span style=\"font-size: 1rem; text-align: initial;\">e obtain the following rational equation:<\/span>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{6}{b-4}+\\dfrac{6}{b+4}=2[\/latex]<\/p>\r\nLet's acknowledge that the restricted values of [latex]b[\/latex] are [latex]4[\/latex] and [latex]-4[\/latex]. Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\require{color}\r\n\\begin{align}\r\n\\frac{6}{b-4}+\\frac{6}{b+4}&amp;=2\\\\[5pt]\r\n{\\color{red}{(b-4)\\cdot(b+4)\\cdot\\: }}\\left(\\frac{6}{b-4}+\\frac{6}{b+4}\\right)&amp;=2{\\color{red}{\\: \\cdot\\: (b-4)\\cdot(b+4)}} &amp;&amp; \\color{blue}{\\textsf{multiply by the LCD}}\\\\[5pt]\r\n\\frac{{\\color{red}\\cancel{\\color{black}{(b-4)}}}\\cdot(b+4)}{1}\\cdot \\frac{6}{{\\color{red}\\cancel{\\color{black}{(b-4)}}}\\cdot 1} + \\frac{(b-4)\\cdot{\\color{red}\\cancel{\\color{black}{(b+4)}}}}{1}\\cdot \\frac{6}{{\\color{red}\\cancel{\\color{black}{(b+4)}}}\\cdot 1}&amp;= 2\\cdot (b^2-16) &amp;&amp; \\color{blue}{\\textsf{distribute and multiply}}\\\\[5pt]\r\n6b+24+\\underset{\\large{\\color{red}{-12b}}}{6b}-24 &amp;= \\underset{\\large{\\color{red}{-12b}}}{2b^2}-32 &amp;&amp; \\color{blue}{\\textsf{simplify and isolate $0$}}\\\\[5pt]\r\n{\\color{red}{\\frac{{\\color{black}{0}}}{2}}} &amp;= {\\color{red}{\\frac{{\\color{black}{2b^2-12b-32}}}{2}}} &amp;&amp; \\color{blue}{\\textsf{divide out the GCF}}\\\\[5pt]\r\n0 &amp;= b^2-6b-16\\\\[5pt]\r\n0 &amp;= (b-8)(b+2) &amp;&amp; \\color{blue}{\\textsf{factor}}\\\\[5pt]\r\nb-\\underset{\\large{\\color{red}{+8}}}{8} &amp;= \\underset{\\large{\\color{red}{+8}}}{0}\\ \\ \\textsf{or}\\ \\ b+\\underset{\\large{\\color{red}{-2}}}{2} = \\underset{\\large{\\color{red}{-2}}}{0} &amp;&amp; \\color{blue}{\\textsf{Zero-Product Property}}\\\\[5pt]\r\nb &amp;= 8 \\ \\ \\textsf{or}\\ \\ b = -2\r\n\\end{align}\r\n[\/latex]<\/p>\r\nOf course, the speed of the river tram in still water cannot be negative, so we discard [latex]b = -2[\/latex].\r\n\r\n<strong>Answer:\u00a0<\/strong>The river tram's speed in still water is [latex]8[\/latex]\u00a0miles per hour.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLogan jogged to his grandma\u2019s house [latex]5[\/latex] miles away and then purchased a ride back home. It took him [latex]50[\/latex] minutes longer to jog there than ride back. If his jogging rate was [latex]25[\/latex] mph slower than the rate when he was riding, what was his jogging rate?\r\n\r\n[reveal-answer q=\"24043\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"24043\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>\r\n\r\nNeither the speed of jogging nor riding is known but we know Logan's jogging rate was [latex]25[\/latex] mph slower than the rate of the ride. Logan's jogging rate is to be found. We do know the distances both ways. We do not know the times but we do know it took him [latex]50[\/latex] minutes (five sixths of an hour) longer to jog there than ride back.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>Let [latex]j[\/latex] be the unknown speed of Logan jogging.\u00a0Since his jogging rate was [latex]25[\/latex] mph slower than the rate of the ride, meaning that the ride speed was [latex]25[\/latex] mph faster, the ride speed was [latex]j+25[\/latex].\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know. The table is not necessary, but can help you organize.\r\n<table style=\"font-size: 110%;\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td><strong>Distance<\/strong><\/td>\r\n<td><strong>Rate (or speed)<\/strong><\/td>\r\n<td><strong>Time<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>There<\/strong><\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<td>[latex]j[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Back<\/strong><\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<td>[latex]j+25[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\\dfrac{d}{r}[\/latex]:\r\n<table style=\"font-size: 110%; width: 555px;\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 130px;\"><\/td>\r\n<td style=\"width: 95px;\"><strong>Distance<\/strong><\/td>\r\n<td style=\"width: 157px;\"><strong>Rate (or speed)<\/strong><\/td>\r\n<td style=\"width: 173px;\"><strong>Time<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 130px;\"><strong>There<\/strong><\/td>\r\n<td style=\"width: 95px;\">[latex]5[\/latex]<\/td>\r\n<td style=\"width: 157px;\">[latex]j[\/latex]<\/td>\r\n<td style=\"width: 173px;\">[latex]\\dfrac{5}{j}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 130px;\"><strong>Back<\/strong><\/td>\r\n<td style=\"width: 95px;\">[latex]5[\/latex]<\/td>\r\n<td style=\"width: 157px;\">[latex]j+25[\/latex]<\/td>\r\n<td style=\"width: 173px;\">[latex]\\dfrac{5}{j+25}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince it took him five sixths of an hour longer to jog there than ride back, meaning that the time There (the longer time) was five sixths of an hour more than the time Back (the shorter time), w<span style=\"font-size: 1rem; text-align: initial;\">e obtain the following rational equation:<\/span>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{5}{j}=\\dfrac{5}{j+25}+\\dfrac{5}{6}[\/latex]<\/p>\r\nLet's acknowledge that the restricted values of [latex]j[\/latex] are [latex]0[\/latex] and [latex]-25[\/latex]. Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\require{color}\r\n\\begin{align}\r\n{\\color{red}{6\\cdot j\\cdot (j+25)\\: \\cdot\\: }}\\frac{5}{j}&amp;=\\left(\\frac{5}{j+25}+\\frac{5}{6}\\right){\\color{red}{\\: \\cdot\\: 6\\cdot j\\cdot (j+25)}}\\\\[5pt]\r\n\\frac{6\\cdot {\\color{red}\\cancel{\\color{black}{j}}}\\cdot (j+25)}{1}\\cdot\\frac{5}{{\\color{red}\\cancel{\\color{black}{j}}}\\cdot 1}&amp;=\\frac{5}{{\\color{red}\\cancel{\\color{black}{(j+25)}}}\\cdot 1}\\cdot\\frac{6\\cdot j\\cdot {\\color{red}\\cancel{\\color{black}{(j+25)}}}}{1}+\\frac{5}{{\\color{red}\\cancel{\\color{black}{6}}}\\cdot 1}\\cdot\\frac{{\\color{red}\\cancel{\\color{black}{6}}}\\cdot j\\cdot (j+25)}{1}\\\\[5pt]\r\n\\underset{\\large{\\color{red}{-30j}}}{30j}+\\underset{\\large{\\color{red}{-750}}}{750} &amp;= \\underset{\\large{\\color{red}{-30j}}}{30j}+5j^2+\\underset{\\large{\\color{red}{-750}}}{125j}\\\\[5pt]\r\n{\\color{red}{\\frac{{\\color{black}{0}}}{5}}}&amp;={\\color{red}{\\frac{{\\color{black}{5j^2+125j-750}}}{5}}}\\\\[5pt]\r\n0&amp;=j^2+25j-150\\\\[5pt]\r\n0&amp;=(j+30)(j-5)\\\\[5pt]\r\nj+\\underset{\\large{\\color{red}{-30}}}{30} &amp;= \\underset{\\large{\\color{red}{-30}}}{0}\\ \\ \\textsf{or}\\ \\ j-\\underset{\\large{\\color{red}{+5}}}{5} = \\underset{\\large{\\color{red}{+5}}}{0}\\\\[5pt]\r\nj &amp;= -30 \\ \\ \\textsf{or}\\ \\ j = 5\r\n\\end{align}\r\n[\/latex]<\/p>\r\nOf course, jogging rate cannot be negative, so we discard [latex]j = -30[\/latex].\r\n\r\n<strong>Answer:\u00a0<\/strong>Logan's jogging rate was [latex]5[\/latex]\u00a0miles per hour.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote that in the example above, we could have presented the equation\u00a0[latex]\\dfrac{5}{j}=\\dfrac{5}{j+25}+\\dfrac{5}{6}[\/latex] in equivalent form\u00a0[latex]\\dfrac{5}{j}-\\dfrac{5}{j+25}=\\dfrac{5}{6}[\/latex]. Such setup can be perceived as one\u00a0of the three common situations:\r\n<ul>\r\n \t<li>\u201cset equal\u201d for \u201cthe same time\u201d problems,<\/li>\r\n \t<li>\u201cadd together\u201d for \u201cthe total time\u201d problems, and<\/li>\r\n \t<li>\u201csubtract\u201d for \u201cthe leaves earlier, or takes longer\u201d problems.<\/li>\r\n<\/ul>\r\n<h2>Work Applications<\/h2>\r\nWork problems often ask you to calculate how long it will take different people or machines working at different speeds to finish a task. Let [latex]W[\/latex] represent a completed portion of a job (as a fraction or percentage) and [latex]t[\/latex] the time an entity works on the job. For example, consider a math professor grading an exam. The job is to grade all exams. If it takes the professor [latex]5[\/latex] hours to complete that job, we can say that [latex]100\\%[\/latex] of the job (or [latex]1[\/latex] fully completed job) is completed in [latex]5[\/latex] hours. We define the <strong>rate of work<\/strong>, [latex]r[\/latex], expressing what portion of the job is completed in a unit of time, or mathematically, the <em>rate of change<\/em> of the completed portion of a job with respect to time, in a similar way as rates of change in the examples above. We divide the completed portion of the job by the time it took the entity to complete that portion. Assuming a <strong>constant<\/strong> rate of work, we can say that the rate of grading the exam by our math professor can be computed as [latex]1[\/latex] (fully graded exam) divided by [latex]5[\/latex] hours, or mathematically, [latex]\\dfrac{1\\ \\text{job}}{5\\ \\text{hours}}=\\dfrac{1}{5}\\ \\text{job per hour}[\/latex]. That leads to a relationship based on the principle\r\n<p style=\"text-align: center;\">\u201cquantity\u201d [latex]=[\/latex] \u201cthe rate of change of the quantity with respect to a counting unit\u201d [latex]\\cdot[\/latex] \u201cthe number of counting units\u201d<\/p>\r\nof the form\r\n<p style=\"text-align: center;\">\u201ccompleted portion of a job\u201d [latex]=[\/latex] \u201cthe rate of change of the completed portion of a job with respect to time\u201d [latex]\\cdot[\/latex] \u201cthe time an entity works on the job\u201d,<\/p>\r\nor in simpler terms,\r\n<p style=\"text-align: center;\">\u201ccompleted portion of a job\u201d [latex]=[\/latex] \u201cthe rate of work\u201d [latex]\\cdot[\/latex] \u201cthe time spent on the job\u201d,<\/p>\r\nand mathematically,\r\n<p style=\"text-align: center;\">[latex]W=r\\cdot t[\/latex].<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">Since we usually refer to completing [latex]100\\%[\/latex] of a task, [latex]W=1[\/latex] in the equation above, resulting in a <em>reciprocal relationship<\/em> between the work rate and time to complete the task since we have [latex]1 = \\textsf{rate}\\cdot \\textsf{time}[\/latex] and hence\u00a0[latex]\\textsf{rate} = \\dfrac{1}{\\textsf{time}}[\/latex] and\u00a0[latex]\\textsf{time} = \\dfrac{1}{\\textsf{rate}}[\/latex].<\/span>\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">Some work problems include multiple machines or people working on a task together for the same amount of time but at different rates. In that case, we can add their individual work rates together to get a total work rate. Let us look at an example.<\/span>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nMyra takes\u00a0[latex]2[\/latex] hours to plant\u00a0[latex]50[\/latex] flower bulbs. Francis takes\u00a0[latex]3[\/latex] hours to plant\u00a0[latex]45[\/latex] flower bulbs. Working together, how long should it take them to plant [latex]150[\/latex] bulbs?\r\n\r\n<img class=\"aligncenter wp-image-1631 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/07\/male-and-female-planting-flower-bulbs-together-300x300.jpg\" alt=\"decorative image\" width=\"300\" height=\"300\" \/>\r\n\r\n[reveal-answer q=\"550322\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"550322\"]\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">If two entities with individual rates of work do a task together, we can add their individual work rates together to get a combined work rate.\u00a0<\/span>\r\n\r\nIn our example,\u00a0Myra takes\u00a0[latex]2[\/latex] hours to plant\u00a0[latex]50[\/latex] flower bulbs. Hence, she can plant [latex]25[\/latex] bulbs in [latex]1[\/latex] hour. The full task is to plant [latex]150[\/latex] bulbs. Thus, she can do [latex]\\dfrac{25}{150}=\\dfrac{1}{6}[\/latex] of the task in [latex]1[\/latex] hour, so her individual rate of work is [latex]\\dfrac{1}{6}[\/latex].\r\n\r\nSimilarly,\u00a0Francis takes\u00a0[latex]3[\/latex] hours to plant\u00a0[latex]45[\/latex] bulbs and hence he can plant\u00a0[latex]15[\/latex] bulbs in [latex]1[\/latex] hour, so\u00a0his individual rate of work is\u00a0[latex]\\dfrac{15}{150}=\\dfrac{1}{10}[\/latex].\r\n\r\nTheir combined rate of work (when they work together) is then [latex]\\dfrac{1}{6}+\\dfrac{1}{10}[\/latex]. Therefore, due to the\u00a0<span style=\"font-size: 1rem; text-align: initial;\"><em>reciprocal relationship<\/em> between the work rate and time to complete the task, it will take them<\/span>\r\n<p style=\"text-align: center;\">[latex]\\dfrac{1}{\\; \\dfrac{1}{6}+\\dfrac{1}{10}\\; }[\/latex]<\/p>\r\nhours\u00a0<span style=\"font-size: 1rem; text-align: initial;\">to complete the task working together. We recognize this expression as a complex fraction. Refer to <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/4-4-simplifying-complex-rational-expressions\/\" target=\"_blank\" rel=\"noopener\">Sec. 4.4<\/a> how to simplify complex fractions. Using the second method and noting that [latex]LCM(6,10)=LCM(2^1\\cdot 3^1,2^1\\cdot 5^1)=2^1\\cdot 3^1\\cdot 5^1=30[\/latex], we obtain<\/span>\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{1}{\\; \\dfrac{1}{6}+\\dfrac{1}{10}\\; }&amp;=\\frac{1{\\color{red}{\\: \\cdot\\: 30}}}{\\; \\left(\\dfrac{1}{6}+\\dfrac{1}{10}\\right){\\color{red}{\\: \\cdot\\: 30}}\\; }\\\\[5pt]\r\n&amp;=\\frac{30}{\\; \\dfrac{1}{{\\color{red}\\cancel{\\color{black}{6}}}\\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{6}}}\\cdot 5}{1}+\\dfrac{1}{{\\color{red}\\cancel{\\color{black}{10}}}\\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{10}}}\\cdot 3}{1}\\; }\\\\[5pt]\r\n&amp;=\\frac{30}{5+3}\\\\[5pt]\r\n&amp;=\\frac{30}{8}\\\\[5pt]\r\n&amp;=\\frac{{\\color{red}\\cancel{\\color{black}{2}}}\\cdot 15}{{\\color{red}\\cancel{\\color{black}{2}}}\\cdot 4}\\\\[5pt]\r\n&amp;=\\frac{15}{4}\r\n\\end{align}\r\n[\/latex]<\/p>\r\n<strong>Answer:\u00a0<\/strong>Working together, it should take them [latex]\\dfrac{15}{4}[\/latex] hours (or [latex]3.75[\/latex] hours, or [latex]3[\/latex] hours and [latex]45[\/latex] minutes).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote that in the example above, if we let [latex]t[\/latex] denote the time it should take Myra and Francis to plant [latex]150[\/latex] bulbs together, there are two equivalent ways of expressing their combined rate of work:\r\n<ul>\r\n \t<li>as the sum of their individual rates,\u00a0[latex]\\dfrac{1}{6}+\\dfrac{1}{10}[\/latex], and<\/li>\r\n \t<li>as the reciprocal of the time [latex]t[\/latex], [latex]\\dfrac{1}{t}[\/latex].<\/li>\r\n<\/ul>\r\nThat leads to the rational equation\u00a0[latex]\\dfrac{1}{6}+\\dfrac{1}{10}=\\dfrac{1}{t}[\/latex] that can also be used to solve the problem.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=SzSasnDF7Ms&amp;feature=youtu.be\r\n\r\nThere are other types of work problems. 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 and their relative times.\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\u00a0[latex]3[\/latex] times as long as it would take Joe to paint the entire house. Working together, they can complete the job in\u00a0[latex]24[\/latex] 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\"]\r\n\r\nLet [latex]t[\/latex] be the unknown time it takes Joe to complete the job alone. Then, since\u00a0it would take John [latex]3[\/latex] times as long as it would take Joe\u00a0to complete the job alone, John's time working alone to\u00a0complete the job, is [latex]3t[\/latex].\u00a0Therefore, due to the\u00a0<em>reciprocal relationship<\/em>\u00a0between the\u00a0time and work rate to complete the job, Joe's rate of work is [latex]\\dfrac{1}{t}[\/latex] and\u00a0John's rate of work is [latex]\\dfrac{1}{3t}[\/latex]. Their combined rate of work is then\u00a0[latex]\\dfrac{1}{t}+\\dfrac{1}{3t}[\/latex]. Another way of obtaining that combined rate is to use the fact that\u00a0they can complete the job in\u00a0[latex]24[\/latex] hours\u00a0working together.\u00a0Due to the\u00a0<em>reciprocal relationship<\/em>\u00a0between the\u00a0time and work rate to complete the job, their combined rate of work is then\u00a0[latex]\\dfrac{1}{24}[\/latex]. Of course, both expressions describing the same quantity must be equal, so we obtain the following rational equation:\r\n<p style=\"text-align: center;\">[latex]\\dfrac{1}{t}+\\dfrac{1}{3t}=\\dfrac{1}{24}[\/latex]<\/p>\r\nLet's acknowledge that the only restricted value of [latex]t[\/latex] is [latex]0[\/latex]. Now,\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{align}\r\n\\frac{1{\\color{red}{\\: \\cdot\\: 3}}}{t{\\color{red}{\\: \\cdot\\: 3}}}+\\frac{1}{3t}&amp;=\\frac{1}{24}\\\\[5pt]\r\n\\frac{4}{3t}&amp;=\\frac{1}{24}\\\\[5pt]\r\n4\\cdot 24&amp;=3t\\cdot 1 &amp;&amp; \\color{blue}{\\textsf{clear the fractions}}\\\\[5pt]\r\n{\\color{red}{\\frac{{\\color{black}{96}}}{3}}}&amp;={\\color{red}{\\frac{\\cancel{\\color{black}{3}}{\\color{black}{\\: \\cdot\\: t}}}{\\cancel{3} \\cdot 1}}}\\\\[5pt]\r\n32 &amp;= t\r\n\\end{align}\r\n[\/latex]<\/p>\r\nSince\u00a0John's time working alone to\u00a0complete the job is [latex]3t[\/latex], it's [latex]3\\cdot 32 = 96[\/latex] hours.\r\n\r\n<strong>Answer:\u00a0<\/strong>It would take Joe [latex]32[\/latex] hours working alone while it would take John [latex]96[\/latex] hours working alone.\r\n\r\n&nbsp;\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\r\n<p style=\"text-align: center;\">[latex] \\dfrac{1}{a}+\\dfrac{1}{b}=\\dfrac{1}{t}[\/latex],<\/p>\r\nwhere [latex]t[\/latex] is the time to do the job together, [latex]a[\/latex] is the time it takes entity A to do the job alone, and [latex]b[\/latex] is the time it takes entity B to do the job alone. The key idea here is to figure out each entity\u2019s individual rate of work. Then, once those rates are identified, add them together, set it equal to the reciprocal of the time [latex]t[\/latex], and solve the rational equation.\r\n\r\nWe conclude with another example of two people painting at different rates in the following video.\r\n\r\nhttps:\/\/youtu.be\/SzSasnDF7Ms","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve motion applications using rational equations.<\/li>\n<li>Solve work applications using rational equations.<\/li>\n<\/ul>\n<\/div>\n<p>The reasoning behind both work and (uniform) motion applications is very similar. You should already be familiar with the formula [latex]d=r\\cdot t[\/latex] relating the quantities of distance, rate, and time.\u00a0The rate, [latex]r[\/latex], in the formula, is the <em>rate of change<\/em> of distance, [latex]d[\/latex], with respect to time, [latex]t[\/latex]. For example, if an object moves at a (constant) speed of [latex]30[\/latex] miles per hour, then it travels 30 miles in one hour, [latex]30\\cdot 2 = 60[\/latex] miles in [latex]2[\/latex] hours, [latex]30\\cdot 3.5=105[\/latex] miles in [latex]3.5[\/latex] hours, etc. Note that to compute the distance traveled, we multiply the rate (expressed in a unit of distance by a unit of time) by the number of (compatible) units of time. A similar<\/p>\n<p style=\"text-align: center;\">\u201cquantity\u201d [latex]=[\/latex] \u201cthe rate of change of the quantity with respect to a counting unit\u201d [latex]\\cdot[\/latex] \u201cthe number of counting units\u201d<\/p>\n<p>principle is used in the following relationships:<\/p>\n<ul>\n<li>[latex]R=x\\cdot p[\/latex] between the revenue, [latex]R[\/latex] (the amount of money that comes into the business), the quantity, [latex]x[\/latex], of a product sold, and the unit price, [latex]p[\/latex] (i.e., dollars per unit sold). Note that the price per unit is the <em>rate of change<\/em> of revenue with respect to the quantity of a product sold. (Refer to <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-1-applications-of-systems-of-2-linear-equations\/#CRP\" target=\"_blank\" rel=\"noopener\">this description<\/a>.)<\/li>\n<li>[latex]m=C\\cdot V[\/latex]\u00a0between the mass, [latex]m[\/latex], of a solvent, the volume, [latex]V[\/latex], of a solution, and the concentration, [latex]C[\/latex], the amount of solvent per unit volume of solution.\u00a0Note that the concentration is the <em>rate of change<\/em> of mass with respect to volume. (Refer to <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-1-applications-of-systems-of-2-linear-equations\/#mixture\" target=\"_blank\" rel=\"noopener\">this description<\/a>.)<\/li>\n<\/ul>\n<h2>Motion Applications<\/h2>\n<p>Motion applications have already been introduced in <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-1-applications-of-systems-of-2-linear-equations\/#motion\" target=\"_blank\" rel=\"noopener\">Sec. 2.1<\/a>. We continue working with uniform motion\u00a0with a constant speed in this section. Equivalent forms of the [latex]d=r\\cdot t[\/latex] are [latex]r=\\dfrac{d}{t}[\/latex] and [latex]t=\\dfrac{d}{r}[\/latex] (the latter form has been discussed in <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/solving-a-formula-for-a-specific-variable\/#drtforms\" target=\"_blank\" rel=\"noopener\">Sec. 4.6<\/a>).<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Avery can paddle her kayak [latex]4[\/latex] miles per hour in still water. It takes her as long to paddle [latex]10[\/latex] miles upstream as it takes her to travel [latex]22[\/latex] miles downstream. Determine the speed of the river&#8217;s current.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q462935\">Show Solution<\/span><\/p>\n<div id=\"q462935\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong><\/p>\n<p id=\"fs-id1167834098208\">Recall that when the kayak is going downstream, in the same direction as the river&#8217;s current, the current helps push the kayak, so the kayak\u2019s actual speed is faster than its speed in still water. The actual speed at which the kayak is moving is the sum of the speeds in still water and the river&#8217;s current. When<span style=\"font-size: 1rem; text-align: initial;\">\u00a0the kayak is going upstream, opposite to the river current though, the current is going against the kayak, so the kayak\u2019s actual speed is slower than its speed in still water. The actual speed of the kayak is the difference between the speed in still water and the speed of the river&#8217;s current.<\/span><\/p>\n<p>The speed of the kayak in still water is known while the speed of the river&#8217;s current is to be found. We do know the distances Avery paddles her kayak upstream and downstream. We do not know the times but we do know the two times are the same.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>Let [latex]c[\/latex] be the unknown speed of the river&#8217;s current.\u00a0Going upstream, the actual rate of the kayak is [latex]4-c[\/latex]. Going downstream, the actual rate is [latex]4+c[\/latex].<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know. The table is not necessary, but can help you organize.<\/p>\n<table style=\"font-size: 110%;\">\n<tbody>\n<tr>\n<td><\/td>\n<td><strong>Distance<\/strong><\/td>\n<td><strong>Rate (or speed)<\/strong><\/td>\n<td><strong>Time<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Upstream<\/strong><\/td>\n<td>[latex]10[\/latex]<\/td>\n<td>[latex]4-c[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><strong>Downstream<\/strong><\/td>\n<td>[latex]22[\/latex]<\/td>\n<td>[latex]4+c[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\\dfrac{d}{r}[\/latex]:<\/p>\n<table style=\"font-size: 110%;\">\n<tbody>\n<tr>\n<td><\/td>\n<td><strong>Distance<\/strong><\/td>\n<td><strong>Rate (or speed)<\/strong><\/td>\n<td><strong>Time<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Upstream<\/strong><\/td>\n<td>[latex]10[\/latex]<\/td>\n<td>[latex]4-c[\/latex]<\/td>\n<td>[latex]\\dfrac{10}{4-c}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Downstream<\/strong><\/td>\n<td>[latex]22[\/latex]<\/td>\n<td>[latex]4+c[\/latex]<\/td>\n<td>[latex]\\dfrac{22}{4+c}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since the two times are <em>the same<\/em> (mathematically, <em>equal<\/em>), w<span style=\"font-size: 1rem; text-align: initial;\">e obtain the following rational equation:<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{10}{4-c}=\\dfrac{22}{4+c}[\/latex]<\/p>\n<p>Let&#8217;s acknowledge that the restricted values of [latex]c[\/latex] are [latex]4[\/latex] and [latex]-4[\/latex]. (Can you explain why?) Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\require{color}  \\begin{align}  \\frac{10}{4-c}{\\color{red}\\: \\cdot\\: (4-c)\\cdot (4+c)}&=\\frac{22}{4+c}{\\color{red}\\: \\cdot\\: (4-c)\\cdot (4+c)} && \\color{blue}{\\textsf{multiply by the LCD}}\\\\[5pt]  \\frac{10}{{\\color{red}\\cancel{\\color{black}{(4-c)}}}}\\cdot \\frac{{\\color{red}\\cancel{\\color{black}{(4-c)}}}\\cdot (4+c)}{1} &= \\frac{22}{{\\color{red}\\cancel{\\color{black}{(4+c)}}}}\\cdot \\frac{(4-c)\\cdot {\\color{red}\\cancel{\\color{black}{(4+c)}}}}{1} && \\color{blue}{\\textsf{clear the fractions}}\\\\[5pt]10\\cdot (4+c) &= 22\\cdot (4-c)\\\\[5pt]  \\underset{\\large{\\color{red}{-40}}}{40}+\\underset{\\large{\\color{red}{+22c}}}{10c} &= \\underset{\\large{\\color{red}{-40}}}{88}-\\underset{\\large{\\color{red}{+22c}}}{22c} && \\color{blue}{\\textsf{isolate $c$-terms}}\\\\[5pt]  {\\color{red}{\\frac{\\cancel{\\color{black}{32}}{\\color{black}{\\: \\cdot\\: c}}}{\\cancel{32} \\cdot 1}}} &= {\\color{red}{\\frac{{\\color{black}{48}}}{32}}} && \\color{blue}{\\textsf{divide to isolate $c$}}\\\\[5pt]  c &= \\frac{{\\color{red}\\cancel{\\color{black}{16}}}\\cdot 3}{{\\color{red}\\cancel{\\color{black}{16}}}\\cdot 2} && \\color{blue}{\\textsf{simplify}}\\\\[5pt]  c &= \\frac{3}{2}  \\end{align}[\/latex]<\/p>\n<p><strong>Answer:\u00a0<\/strong>The river&#8217;s current&#8217;s speed is [latex]\\dfrac{3}{2}[\/latex]\u00a0miles per hour (or [latex]1.5[\/latex] miles per hour).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Hassan can bicycle [latex]72[\/latex] kilometers in the same time as it takes him to walk [latex]30[\/latex] kilometers. He can ride [latex]7[\/latex] km\/h faster than he can walk. How fast can he walk?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q318933\">Show Solution<\/span><\/p>\n<div id=\"q318933\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong><\/p>\n<p>Neither the speed of bicycling nor walking is known but we know Hassan can ride [latex]7[\/latex] km\/h faster than he can walk. The walking speed is to be found. We do know the distances Hassan bicycles and walks. We do not know the times but we do know the two times are the same.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>Let [latex]w[\/latex] be the unknown speed of Hassan walking.\u00a0Since he can ride [latex]7[\/latex] km\/h <strong>faster<\/strong> than he can walk, his bicycling speed is [latex]w+7[\/latex].<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know. The table is not necessary, but can help you organize.<\/p>\n<table style=\"font-size: 110%;\">\n<tbody>\n<tr>\n<td><\/td>\n<td><strong>Distance<\/strong><\/td>\n<td><strong>Rate (or speed)<\/strong><\/td>\n<td><strong>Time<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Bicycling<\/strong><\/td>\n<td>[latex]72[\/latex]<\/td>\n<td>[latex]w+7[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><strong>Walking<\/strong><\/td>\n<td>[latex]30[\/latex]<\/td>\n<td>[latex]w[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\\dfrac{d}{r}[\/latex]:<\/p>\n<table style=\"font-size: 110%; width: 555px;\">\n<tbody>\n<tr>\n<td style=\"width: 130px;\"><\/td>\n<td style=\"width: 95px;\"><strong>Distance<\/strong><\/td>\n<td style=\"width: 157px;\"><strong>Rate (or speed)<\/strong><\/td>\n<td style=\"width: 173px;\"><strong>Time<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 130px;\"><strong>Bicycling<\/strong><\/td>\n<td style=\"width: 95px;\">[latex]72[\/latex]<\/td>\n<td style=\"width: 157px;\">[latex]w+7[\/latex]<\/td>\n<td style=\"width: 173px;\">[latex]\\dfrac{72}{w+7}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 130px;\"><strong>Walking<\/strong><\/td>\n<td style=\"width: 95px;\">[latex]30[\/latex]<\/td>\n<td style=\"width: 157px;\">[latex]w[\/latex]<\/td>\n<td style=\"width: 173px;\">[latex]\\dfrac{30}{w}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since the two times are <em>the same<\/em> (mathematically, <em>equal<\/em>), w<span style=\"font-size: 1rem; text-align: initial;\">e obtain the following rational equation:<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{72}{w+7}=\\dfrac{30}{w}[\/latex]<\/p>\n<p>Let&#8217;s acknowledge that the restricted values of [latex]w[\/latex] are [latex]-7[\/latex] and [latex]0[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\require{color}  \\begin{align}  \\frac{72}{w+7}&=\\frac{30}{w}\\\\[5pt]  72\\cdot w &= (w+7)\\cdot 30 && \\color{blue}{\\textsf{clear the fractions}}\\\\[5pt]  \\underset{\\large{\\color{red}{-30w}}}{72w} &= \\underset{\\large{\\color{red}{-30w}}}{30w}+210\\\\[5pt]  {\\color{red}{\\frac{\\cancel{\\color{black}{42}}{\\color{black}{\\: \\cdot\\: w}}}{\\cancel{42} \\cdot 1}}} &= {\\color{red}{\\frac{{\\color{black}{210}}}{42}}}\\\\[5pt]  w &= 5  \\end{align}[\/latex]<\/p>\n<p><strong>Answer:\u00a0<\/strong>Hassan\u00a0can walk [latex]5[\/latex]\u00a0kilometers per hour.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The Vistula river current&#8217;s speed is [latex]4[\/latex] miles per hour. A river tram in Krak\u00f3w travels to a point [latex]6[\/latex] miles upstream and back again in [latex]2[\/latex] hours. What is the speed of the river tram\u00a0in still water (without a current)?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/plikimpi.krakow.pl\/\/zalacznik\/36001\/4.jpg\" alt=\"decorative image\" width=\"300\" height=\"300\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q361230\">Show Solution<\/span><\/p>\n<div id=\"q361230\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong><\/p>\n<p>The speed of the river&#8217;s current is known while the speed\u00a0of the river tram in still water is to be found. We do know the distances the river tram goes upstream and downstream. We do not know the individual times but we do know the roundtrip total time.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>Let [latex]b[\/latex] be the unknown speed of the river tram.\u00a0Going upstream, the actual rate of the river tram is [latex]b-4[\/latex]. Going downstream, the actual rate is [latex]b+4[\/latex].<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know. The table is not necessary, but can help you organize. Note that sometimes the variable is first and other times it is last, like in our previous example. The boat or vehicle number\/variable is always first.<\/p>\n<table style=\"font-size: 110%;\">\n<tbody>\n<tr>\n<td><\/td>\n<td><strong>Distance<\/strong><\/td>\n<td><strong>Rate (or speed)<\/strong><\/td>\n<td><strong>Time<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Upstream<\/strong><\/td>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]b-4[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><strong>Downstream<\/strong><\/td>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]b+4[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\\dfrac{d}{r}[\/latex]:<\/p>\n<table style=\"font-size: 110%;\">\n<tbody>\n<tr>\n<td><\/td>\n<td><strong>Distance<\/strong><\/td>\n<td><strong>Rate (or speed)<\/strong><\/td>\n<td><strong>Time<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>Upstream<\/strong><\/td>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]b-4[\/latex]<\/td>\n<td>[latex]\\dfrac{6}{b-4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Downstream<\/strong><\/td>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]b+4[\/latex]<\/td>\n<td>[latex]\\dfrac{6}{b+4}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since the two times <em>total<\/em> (mathematically, <em>their sum is<\/em>) 2 hours, w<span style=\"font-size: 1rem; text-align: initial;\">e obtain the following rational equation:<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{6}{b-4}+\\dfrac{6}{b+4}=2[\/latex]<\/p>\n<p>Let&#8217;s acknowledge that the restricted values of [latex]b[\/latex] are [latex]4[\/latex] and [latex]-4[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\require{color}  \\begin{align}  \\frac{6}{b-4}+\\frac{6}{b+4}&=2\\\\[5pt]  {\\color{red}{(b-4)\\cdot(b+4)\\cdot\\: }}\\left(\\frac{6}{b-4}+\\frac{6}{b+4}\\right)&=2{\\color{red}{\\: \\cdot\\: (b-4)\\cdot(b+4)}} && \\color{blue}{\\textsf{multiply by the LCD}}\\\\[5pt]  \\frac{{\\color{red}\\cancel{\\color{black}{(b-4)}}}\\cdot(b+4)}{1}\\cdot \\frac{6}{{\\color{red}\\cancel{\\color{black}{(b-4)}}}\\cdot 1} + \\frac{(b-4)\\cdot{\\color{red}\\cancel{\\color{black}{(b+4)}}}}{1}\\cdot \\frac{6}{{\\color{red}\\cancel{\\color{black}{(b+4)}}}\\cdot 1}&= 2\\cdot (b^2-16) && \\color{blue}{\\textsf{distribute and multiply}}\\\\[5pt]  6b+24+\\underset{\\large{\\color{red}{-12b}}}{6b}-24 &= \\underset{\\large{\\color{red}{-12b}}}{2b^2}-32 && \\color{blue}{\\textsf{simplify and isolate $0$}}\\\\[5pt]  {\\color{red}{\\frac{{\\color{black}{0}}}{2}}} &= {\\color{red}{\\frac{{\\color{black}{2b^2-12b-32}}}{2}}} && \\color{blue}{\\textsf{divide out the GCF}}\\\\[5pt]  0 &= b^2-6b-16\\\\[5pt]  0 &= (b-8)(b+2) && \\color{blue}{\\textsf{factor}}\\\\[5pt]  b-\\underset{\\large{\\color{red}{+8}}}{8} &= \\underset{\\large{\\color{red}{+8}}}{0}\\ \\ \\textsf{or}\\ \\ b+\\underset{\\large{\\color{red}{-2}}}{2} = \\underset{\\large{\\color{red}{-2}}}{0} && \\color{blue}{\\textsf{Zero-Product Property}}\\\\[5pt]  b &= 8 \\ \\ \\textsf{or}\\ \\ b = -2  \\end{align}[\/latex]<\/p>\n<p>Of course, the speed of the river tram in still water cannot be negative, so we discard [latex]b = -2[\/latex].<\/p>\n<p><strong>Answer:\u00a0<\/strong>The river tram&#8217;s speed in still water is [latex]8[\/latex]\u00a0miles per hour.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Logan jogged to his grandma\u2019s house [latex]5[\/latex] miles away and then purchased a ride back home. It took him [latex]50[\/latex] minutes longer to jog there than ride back. If his jogging rate was [latex]25[\/latex] mph slower than the rate when he was riding, what was his jogging rate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q24043\">Show Solution<\/span><\/p>\n<div id=\"q24043\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong><\/p>\n<p>Neither the speed of jogging nor riding is known but we know Logan&#8217;s jogging rate was [latex]25[\/latex] mph slower than the rate of the ride. Logan&#8217;s jogging rate is to be found. We do know the distances both ways. We do not know the times but we do know it took him [latex]50[\/latex] minutes (five sixths of an hour) longer to jog there than ride back.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>Let [latex]j[\/latex] be the unknown speed of Logan jogging.\u00a0Since his jogging rate was [latex]25[\/latex] mph slower than the rate of the ride, meaning that the ride speed was [latex]25[\/latex] mph faster, the ride speed was [latex]j+25[\/latex].<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know. The table is not necessary, but can help you organize.<\/p>\n<table style=\"font-size: 110%;\">\n<tbody>\n<tr>\n<td><\/td>\n<td><strong>Distance<\/strong><\/td>\n<td><strong>Rate (or speed)<\/strong><\/td>\n<td><strong>Time<\/strong><\/td>\n<\/tr>\n<tr>\n<td><strong>There<\/strong><\/td>\n<td>[latex]5[\/latex]<\/td>\n<td>[latex]j[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><strong>Back<\/strong><\/td>\n<td>[latex]5[\/latex]<\/td>\n<td>[latex]j+25[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\\dfrac{d}{r}[\/latex]:<\/p>\n<table style=\"font-size: 110%; width: 555px;\">\n<tbody>\n<tr>\n<td style=\"width: 130px;\"><\/td>\n<td style=\"width: 95px;\"><strong>Distance<\/strong><\/td>\n<td style=\"width: 157px;\"><strong>Rate (or speed)<\/strong><\/td>\n<td style=\"width: 173px;\"><strong>Time<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 130px;\"><strong>There<\/strong><\/td>\n<td style=\"width: 95px;\">[latex]5[\/latex]<\/td>\n<td style=\"width: 157px;\">[latex]j[\/latex]<\/td>\n<td style=\"width: 173px;\">[latex]\\dfrac{5}{j}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 130px;\"><strong>Back<\/strong><\/td>\n<td style=\"width: 95px;\">[latex]5[\/latex]<\/td>\n<td style=\"width: 157px;\">[latex]j+25[\/latex]<\/td>\n<td style=\"width: 173px;\">[latex]\\dfrac{5}{j+25}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since it took him five sixths of an hour longer to jog there than ride back, meaning that the time There (the longer time) was five sixths of an hour more than the time Back (the shorter time), w<span style=\"font-size: 1rem; text-align: initial;\">e obtain the following rational equation:<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{5}{j}=\\dfrac{5}{j+25}+\\dfrac{5}{6}[\/latex]<\/p>\n<p>Let&#8217;s acknowledge that the restricted values of [latex]j[\/latex] are [latex]0[\/latex] and [latex]-25[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\require{color}  \\begin{align}  {\\color{red}{6\\cdot j\\cdot (j+25)\\: \\cdot\\: }}\\frac{5}{j}&=\\left(\\frac{5}{j+25}+\\frac{5}{6}\\right){\\color{red}{\\: \\cdot\\: 6\\cdot j\\cdot (j+25)}}\\\\[5pt]  \\frac{6\\cdot {\\color{red}\\cancel{\\color{black}{j}}}\\cdot (j+25)}{1}\\cdot\\frac{5}{{\\color{red}\\cancel{\\color{black}{j}}}\\cdot 1}&=\\frac{5}{{\\color{red}\\cancel{\\color{black}{(j+25)}}}\\cdot 1}\\cdot\\frac{6\\cdot j\\cdot {\\color{red}\\cancel{\\color{black}{(j+25)}}}}{1}+\\frac{5}{{\\color{red}\\cancel{\\color{black}{6}}}\\cdot 1}\\cdot\\frac{{\\color{red}\\cancel{\\color{black}{6}}}\\cdot j\\cdot (j+25)}{1}\\\\[5pt]  \\underset{\\large{\\color{red}{-30j}}}{30j}+\\underset{\\large{\\color{red}{-750}}}{750} &= \\underset{\\large{\\color{red}{-30j}}}{30j}+5j^2+\\underset{\\large{\\color{red}{-750}}}{125j}\\\\[5pt]  {\\color{red}{\\frac{{\\color{black}{0}}}{5}}}&={\\color{red}{\\frac{{\\color{black}{5j^2+125j-750}}}{5}}}\\\\[5pt]  0&=j^2+25j-150\\\\[5pt]  0&=(j+30)(j-5)\\\\[5pt]  j+\\underset{\\large{\\color{red}{-30}}}{30} &= \\underset{\\large{\\color{red}{-30}}}{0}\\ \\ \\textsf{or}\\ \\ j-\\underset{\\large{\\color{red}{+5}}}{5} = \\underset{\\large{\\color{red}{+5}}}{0}\\\\[5pt]  j &= -30 \\ \\ \\textsf{or}\\ \\ j = 5  \\end{align}[\/latex]<\/p>\n<p>Of course, jogging rate cannot be negative, so we discard [latex]j = -30[\/latex].<\/p>\n<p><strong>Answer:\u00a0<\/strong>Logan&#8217;s jogging rate was [latex]5[\/latex]\u00a0miles per hour.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Note that in the example above, we could have presented the equation\u00a0[latex]\\dfrac{5}{j}=\\dfrac{5}{j+25}+\\dfrac{5}{6}[\/latex] in equivalent form\u00a0[latex]\\dfrac{5}{j}-\\dfrac{5}{j+25}=\\dfrac{5}{6}[\/latex]. Such setup can be perceived as one\u00a0of the three common situations:<\/p>\n<ul>\n<li>\u201cset equal\u201d for \u201cthe same time\u201d problems,<\/li>\n<li>\u201cadd together\u201d for \u201cthe total time\u201d problems, and<\/li>\n<li>\u201csubtract\u201d for \u201cthe leaves earlier, or takes longer\u201d problems.<\/li>\n<\/ul>\n<h2>Work Applications<\/h2>\n<p>Work problems often ask you to calculate how long it will take different people or machines working at different speeds to finish a task. Let [latex]W[\/latex] represent a completed portion of a job (as a fraction or percentage) and [latex]t[\/latex] the time an entity works on the job. For example, consider a math professor grading an exam. The job is to grade all exams. If it takes the professor [latex]5[\/latex] hours to complete that job, we can say that [latex]100\\%[\/latex] of the job (or [latex]1[\/latex] fully completed job) is completed in [latex]5[\/latex] hours. We define the <strong>rate of work<\/strong>, [latex]r[\/latex], expressing what portion of the job is completed in a unit of time, or mathematically, the <em>rate of change<\/em> of the completed portion of a job with respect to time, in a similar way as rates of change in the examples above. We divide the completed portion of the job by the time it took the entity to complete that portion. Assuming a <strong>constant<\/strong> rate of work, we can say that the rate of grading the exam by our math professor can be computed as [latex]1[\/latex] (fully graded exam) divided by [latex]5[\/latex] hours, or mathematically, [latex]\\dfrac{1\\ \\text{job}}{5\\ \\text{hours}}=\\dfrac{1}{5}\\ \\text{job per hour}[\/latex]. That leads to a relationship based on the principle<\/p>\n<p style=\"text-align: center;\">\u201cquantity\u201d [latex]=[\/latex] \u201cthe rate of change of the quantity with respect to a counting unit\u201d [latex]\\cdot[\/latex] \u201cthe number of counting units\u201d<\/p>\n<p>of the form<\/p>\n<p style=\"text-align: center;\">\u201ccompleted portion of a job\u201d [latex]=[\/latex] \u201cthe rate of change of the completed portion of a job with respect to time\u201d [latex]\\cdot[\/latex] \u201cthe time an entity works on the job\u201d,<\/p>\n<p>or in simpler terms,<\/p>\n<p style=\"text-align: center;\">\u201ccompleted portion of a job\u201d [latex]=[\/latex] \u201cthe rate of work\u201d [latex]\\cdot[\/latex] \u201cthe time spent on the job\u201d,<\/p>\n<p>and mathematically,<\/p>\n<p style=\"text-align: center;\">[latex]W=r\\cdot t[\/latex].<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">Since we usually refer to completing [latex]100\\%[\/latex] of a task, [latex]W=1[\/latex] in the equation above, resulting in a <em>reciprocal relationship<\/em> between the work rate and time to complete the task since we have [latex]1 = \\textsf{rate}\\cdot \\textsf{time}[\/latex] and hence\u00a0[latex]\\textsf{rate} = \\dfrac{1}{\\textsf{time}}[\/latex] and\u00a0[latex]\\textsf{time} = \\dfrac{1}{\\textsf{rate}}[\/latex].<\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">Some work problems include multiple machines or people working on a task together for the same amount of time but at different rates. In that case, we can add their individual work rates together to get a total work rate. Let us look at an example.<\/span><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Myra takes\u00a0[latex]2[\/latex] hours to plant\u00a0[latex]50[\/latex] flower bulbs. Francis takes\u00a0[latex]3[\/latex] hours to plant\u00a0[latex]45[\/latex] flower bulbs. Working together, how long should it take them to plant [latex]150[\/latex] bulbs?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1631 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/07\/male-and-female-planting-flower-bulbs-together-300x300.jpg\" alt=\"decorative image\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/07\/male-and-female-planting-flower-bulbs-together-300x300.jpg 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/07\/male-and-female-planting-flower-bulbs-together-150x150.jpg 150w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/07\/male-and-female-planting-flower-bulbs-together-768x768.jpg 768w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/07\/male-and-female-planting-flower-bulbs-together-65x65.jpg 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/07\/male-and-female-planting-flower-bulbs-together-225x225.jpg 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/07\/male-and-female-planting-flower-bulbs-together-350x350.jpg 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2024\/07\/male-and-female-planting-flower-bulbs-together.jpg 1024w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/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\">\n<p><span style=\"font-size: 1rem; text-align: initial;\">If two entities with individual rates of work do a task together, we can add their individual work rates together to get a combined work rate.\u00a0<\/span><\/p>\n<p>In our example,\u00a0Myra takes\u00a0[latex]2[\/latex] hours to plant\u00a0[latex]50[\/latex] flower bulbs. Hence, she can plant [latex]25[\/latex] bulbs in [latex]1[\/latex] hour. The full task is to plant [latex]150[\/latex] bulbs. Thus, she can do [latex]\\dfrac{25}{150}=\\dfrac{1}{6}[\/latex] of the task in [latex]1[\/latex] hour, so her individual rate of work is [latex]\\dfrac{1}{6}[\/latex].<\/p>\n<p>Similarly,\u00a0Francis takes\u00a0[latex]3[\/latex] hours to plant\u00a0[latex]45[\/latex] bulbs and hence he can plant\u00a0[latex]15[\/latex] bulbs in [latex]1[\/latex] hour, so\u00a0his individual rate of work is\u00a0[latex]\\dfrac{15}{150}=\\dfrac{1}{10}[\/latex].<\/p>\n<p>Their combined rate of work (when they work together) is then [latex]\\dfrac{1}{6}+\\dfrac{1}{10}[\/latex]. Therefore, due to the\u00a0<span style=\"font-size: 1rem; text-align: initial;\"><em>reciprocal relationship<\/em> between the work rate and time to complete the task, it will take them<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{1}{\\; \\dfrac{1}{6}+\\dfrac{1}{10}\\; }[\/latex]<\/p>\n<p>hours\u00a0<span style=\"font-size: 1rem; text-align: initial;\">to complete the task working together. We recognize this expression as a complex fraction. Refer to <a href=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/4-4-simplifying-complex-rational-expressions\/\" target=\"_blank\" rel=\"noopener\">Sec. 4.4<\/a> how to simplify complex fractions. Using the second method and noting that [latex]LCM(6,10)=LCM(2^1\\cdot 3^1,2^1\\cdot 5^1)=2^1\\cdot 3^1\\cdot 5^1=30[\/latex], we obtain<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{1}{\\; \\dfrac{1}{6}+\\dfrac{1}{10}\\; }&=\\frac{1{\\color{red}{\\: \\cdot\\: 30}}}{\\; \\left(\\dfrac{1}{6}+\\dfrac{1}{10}\\right){\\color{red}{\\: \\cdot\\: 30}}\\; }\\\\[5pt]  &=\\frac{30}{\\; \\dfrac{1}{{\\color{red}\\cancel{\\color{black}{6}}}\\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{6}}}\\cdot 5}{1}+\\dfrac{1}{{\\color{red}\\cancel{\\color{black}{10}}}\\cdot 1}\\cdot\\dfrac{{\\color{red}\\cancel{\\color{black}{10}}}\\cdot 3}{1}\\; }\\\\[5pt]  &=\\frac{30}{5+3}\\\\[5pt]  &=\\frac{30}{8}\\\\[5pt]  &=\\frac{{\\color{red}\\cancel{\\color{black}{2}}}\\cdot 15}{{\\color{red}\\cancel{\\color{black}{2}}}\\cdot 4}\\\\[5pt]  &=\\frac{15}{4}  \\end{align}[\/latex]<\/p>\n<p><strong>Answer:\u00a0<\/strong>Working together, it should take them [latex]\\dfrac{15}{4}[\/latex] hours (or [latex]3.75[\/latex] hours, or [latex]3[\/latex] hours and [latex]45[\/latex] minutes).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Note that in the example above, if we let [latex]t[\/latex] denote the time it should take Myra and Francis to plant [latex]150[\/latex] bulbs together, there are two equivalent ways of expressing their combined rate of work:<\/p>\n<ul>\n<li>as the sum of their individual rates,\u00a0[latex]\\dfrac{1}{6}+\\dfrac{1}{10}[\/latex], and<\/li>\n<li>as the reciprocal of the time [latex]t[\/latex], [latex]\\dfrac{1}{t}[\/latex].<\/li>\n<\/ul>\n<p>That leads to the rational equation\u00a0[latex]\\dfrac{1}{6}+\\dfrac{1}{10}=\\dfrac{1}{t}[\/latex] that can also be used to solve the problem.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" 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>There are other types of work problems. 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 and their relative times.<\/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\u00a0[latex]3[\/latex] times as long as it would take Joe to paint the entire house. Working together, they can complete the job in\u00a0[latex]24[\/latex] 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\">\n<p>Let [latex]t[\/latex] be the unknown time it takes Joe to complete the job alone. Then, since\u00a0it would take John [latex]3[\/latex] times as long as it would take Joe\u00a0to complete the job alone, John&#8217;s time working alone to\u00a0complete the job, is [latex]3t[\/latex].\u00a0Therefore, due to the\u00a0<em>reciprocal relationship<\/em>\u00a0between the\u00a0time and work rate to complete the job, Joe&#8217;s rate of work is [latex]\\dfrac{1}{t}[\/latex] and\u00a0John&#8217;s rate of work is [latex]\\dfrac{1}{3t}[\/latex]. Their combined rate of work is then\u00a0[latex]\\dfrac{1}{t}+\\dfrac{1}{3t}[\/latex]. Another way of obtaining that combined rate is to use the fact that\u00a0they can complete the job in\u00a0[latex]24[\/latex] hours\u00a0working together.\u00a0Due to the\u00a0<em>reciprocal relationship<\/em>\u00a0between the\u00a0time and work rate to complete the job, their combined rate of work is then\u00a0[latex]\\dfrac{1}{24}[\/latex]. Of course, both expressions describing the same quantity must be equal, so we obtain the following rational equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{1}{t}+\\dfrac{1}{3t}=\\dfrac{1}{24}[\/latex]<\/p>\n<p>Let&#8217;s acknowledge that the only restricted value of [latex]t[\/latex] is [latex]0[\/latex]. Now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}  \\frac{1{\\color{red}{\\: \\cdot\\: 3}}}{t{\\color{red}{\\: \\cdot\\: 3}}}+\\frac{1}{3t}&=\\frac{1}{24}\\\\[5pt]  \\frac{4}{3t}&=\\frac{1}{24}\\\\[5pt]  4\\cdot 24&=3t\\cdot 1 && \\color{blue}{\\textsf{clear the fractions}}\\\\[5pt]  {\\color{red}{\\frac{{\\color{black}{96}}}{3}}}&={\\color{red}{\\frac{\\cancel{\\color{black}{3}}{\\color{black}{\\: \\cdot\\: t}}}{\\cancel{3} \\cdot 1}}}\\\\[5pt]  32 &= t  \\end{align}[\/latex]<\/p>\n<p>Since\u00a0John&#8217;s time working alone to\u00a0complete the job is [latex]3t[\/latex], it&#8217;s [latex]3\\cdot 32 = 96[\/latex] hours.<\/p>\n<p><strong>Answer:\u00a0<\/strong>It would take Joe [latex]32[\/latex] hours working alone while it would take John [latex]96[\/latex] hours working alone.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video that follows, we show another example of finding one person&#8217;s work rate given a combined work rate.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" 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<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{1}{a}+\\dfrac{1}{b}=\\dfrac{1}{t}[\/latex],<\/p>\n<p>where [latex]t[\/latex] is the time to do the job together, [latex]a[\/latex] is the time it takes entity A to do the job alone, and [latex]b[\/latex] is the time it takes entity B to do the job alone. The key idea here is to figure out each entity\u2019s individual rate of work. Then, once those rates are identified, add them together, set it equal to the reciprocal of the time [latex]t[\/latex], and solve the rational equation.<\/p>\n<p>We conclude with another example of two people painting at different rates in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" 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\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-182\">\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>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\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><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>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. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/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 1: Rational Equation Application - Painting Together. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <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>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><\/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":395986,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Screenshot: A Good Day\\'s Work\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 15: Rational Expressions, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Rational Equation Application - 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