{"id":373,"date":"2016-06-01T20:49:48","date_gmt":"2016-06-01T20:49:48","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=373"},"modified":"2019-07-24T21:01:16","modified_gmt":"2019-07-24T21:01:16","slug":"read-rates-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-rates-2\/","title":{"raw":"Solve an Application Using a Formula","rendered":"Solve an Application Using a Formula"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning OUTCOMES<\/h3>\r\n<ul>\r\n \t<li>Solve\u00a0distance, rate, and time problems<\/li>\r\n \t<li>Solve area, volume, and perimeter problems<\/li>\r\n \t<li>Solve temperature conversion problems<\/li>\r\n \t<li>Rearrange formulas to isolate specific variables<\/li>\r\n \t<li>Identify an unknown given a formula<\/li>\r\n<\/ul>\r\n<\/div>\r\nMany applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem\u2019s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the <strong>area<\/strong> of a rectangular region, [latex]A=LW[\/latex]; the <strong>perimeter<\/strong> of a rectangle, [latex]P=2L+2W[\/latex]; and the <strong>volume<\/strong> of a rectangular solid, [latex]V=LWH[\/latex]. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.\r\n<h2>Distance, Rate, and Time<\/h2>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIt takes Andrew [latex]30[\/latex] min to drive to work in the morning. He drives home using the same route, but it takes [latex]10[\/latex] min longer and he averages [latex]10 mi\/h[\/latex] less than in the morning. How far does Andrew drive to work?\r\n[reveal-answer q=\"61824\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"61824\"]\r\n\r\nThis is a distance problem, so we can use the formula [latex]d=rt[\/latex], where distance equals rate multiplied by time. Note that when rate is given in mi\/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.\r\n\r\nFirst, we identify the known and unknown quantities. Andrew\u2019s morning drive to work takes [latex]30[\/latex] min, or [latex]\\dfrac{1}{2}[\/latex] h at rate [latex]r[\/latex]. His drive home takes [latex]40[\/latex] min, or [latex]\\dfrac{2}{3}[\/latex] h, and his speed averages [latex]10 mi\/h[\/latex] less than the morning drive. Both trips cover distance [latex]d[\/latex]. A table, such as the one below, is often helpful for keeping track of information in these types of problems.\r\n<table summary=\"A table with 3 rows and 4 columns. The first entry in the first row is blank, the rest are: d, r, and t. The entries in the second row are: To Work, d, r, and . The entries in the third row are: To Home, d, r 10, and 2\/3.\">\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>[latex]d[\/latex]<\/th>\r\n<th>[latex]r[\/latex]<\/th>\r\n<th>[latex]t[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td><strong>To Work<\/strong><\/td>\r\n<td>[latex]d[\/latex]<\/td>\r\n<td>[latex]r[\/latex]<\/td>\r\n<td>[latex]\\dfrac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>To Home<\/strong><\/td>\r\n<td>[latex]d[\/latex]<\/td>\r\n<td>[latex]r - 10[\/latex]<\/td>\r\n<td>[latex]\\dfrac{2}{3}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWrite two equations, one for each trip.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}d=r\\left(\\dfrac{1}{2}\\normalsize\\right)\\hfill &amp; \\text{To work}\\hfill \\\\ d=\\left(r - 10\\right)\\left(\\dfrac{2}{3}\\normalsize\\right)\\hfill &amp; \\text{To home}\\hfill \\end{array}[\/latex]<\/div>\r\nAs both equations equal the same distance, we set them equal to each other and solve for <em>r<\/em>.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}r\\left(\\dfrac{1}{2}\\right)&amp;=\\left(r - 10\\right)\\left(\\dfrac{2}{3}\\right)\\hfill \\\\ \\frac{1}{2} r&amp;=\\frac{2}{3} r-\\frac{20}{3}\\hfill \\\\\\frac{1}{2} r-\\frac{2}{3} r&amp;=-\\frac{20}{3}\\hfill \\\\ -\\frac{1}{6} r&amp;=-\\frac{20}{3}\\hfill \\\\ r&amp;=-\\frac{20}{3}\\left(-6\\right)\\hfill \\\\ r&amp;=40\\hfill \\end{array}[\/latex]<\/div>\r\nWe have solved for the rate of speed to work, [latex]40 mph[\/latex]. Substituting [latex]40[\/latex] into the rate on the return trip yields [latex]30 mi\/h[\/latex]. Now we can answer the question. Substitute the rate back into either equation and solve for <em>d.<\/em>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d\\hfill&amp;=40\\left(\\dfrac{1}{2}\\normalsize\\right)\\hfill \\\\ \\hfill&amp;=20\\hfill \\end{array}[\/latex]<\/div>\r\nThe distance between home and work is [latex]20 mi[\/latex].Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for [latex]r[\/latex].\r\n<div style=\"text-align: center;\">\r\n\r\n[latex]\\begin{array}{rl}r\\left(\\dfrac{1}{2}\\normalsize\\right)&amp;=\\left(r - 10\\right)\\left(\\dfrac{2}{3}\\normalsize\\right)\\hfill \\\\ 6\\times r\\left(\\dfrac{1}{2}\\normalsize\\right)&amp; =6\\times \\left(r - 10\\right)\\left(\\dfrac{2}{3}\\normalsize\\right)\\hfill \\\\ 3r&amp; =4\\left(r - 10\\right)\\hfill \\\\ 3r&amp; =4r - 40\\hfill \\\\ -r&amp; =-40\\hfill \\\\ r&amp; =40\\hfill \\end{array}[\/latex]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will find the length and width of a rectangular field given its perimeter and a relationship between its sides.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe perimeter of a rectangular outdoor patio is [latex]54[\/latex] ft. The length is [latex]3[\/latex] ft greater than the width. What are the dimensions of the patio?\r\n[reveal-answer q=\"918170\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"918170\"]\r\n\r\nThe perimeter formula is standard: [latex]P=2L+2W[\/latex]. We have two unknown quantities, length and width. However, we can write the length in terms of the width as [latex]L=W+3[\/latex]. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200341\/CNX_CAT_Figure_02_03_003.jpg\" alt=\"A rectangle with the length labeled as: L = W + 3 and the width labeled as: W.\" width=\"487\" height=\"166\" \/>\r\n\r\nNow we can solve for the width and then calculate the length.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 54=2\\left(W+3\\right)+2W\\hfill \\\\ 54=2W+6+2W\\hfill \\\\ 54=4W+6\\hfill \\\\ 48=4W\\hfill \\\\ 12=W\\hfill \\\\ \\left(12+3\\right)=L\\hfill \\\\ 15=L\\hfill \\end{array}[\/latex]<\/div>\r\nThe dimensions are [latex]L=15[\/latex] ft and [latex]W=12[\/latex] ft.[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows an example of finding the dimensions of a rectangular field given its perimeter.\r\n\r\n[embed]https:\/\/youtu.be\/VyK-HQr02iQ[\/embed]\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe perimeter of a tablet of graph paper is [latex]48[\/latex] in. The length is [latex]6[\/latex] in. more than the width. Find the area of the graph paper.\r\n\r\n[reveal-answer q=\"859423\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"859423\"]\r\n\r\nThe standard formula for area is [latex]A=LW[\/latex]; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.\r\n\r\nWe know that the length is [latex]6[\/latex] in. more than the width, so we can write length as [latex]L=W+6[\/latex]. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}P&amp;=2L+2W\\hfill \\\\ 48&amp;=2\\left(W+6\\right)+2W\\hfill \\\\ 48&amp;=2W+12+2W\\hfill \\\\ 48&amp;=4W+12\\hfill \\\\ 36&amp;=4W\\hfill \\\\ 9&amp;=W\\hfill \\\\ \\left(9+6\\right)&amp;=L\\hfill \\\\ 15&amp;=L\\hfill \\end{array}[\/latex]<\/div>\r\nNow, we find the area given the dimensions of [latex]L=15[\/latex] in. and [latex]W=9[\/latex] in.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}A\\hfill&amp;=LW\\hfill \\\\ A\\hfill&amp;=15\\left(9\\right)\\hfill \\\\ \\hfill&amp;=135\\text{ in}^{2}\\hfill \\end{array}[\/latex]\r\nThe area is [latex]135[\/latex] in<sup>2<\/sup>.[\/hidden-answer]<\/div>\r\n<\/div>\r\nThe following video shows another example of finding the area\u00a0of a rectangle given its perimeter and the relationship between its side lengths.\r\n\r\nhttps:\/\/youtu.be\/zUlU64Umnq4\r\n<h2>Volume<\/h2>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the dimensions of a shipping box given that the length is twice the width, the height is [latex]8[\/latex] inches, and the volume is [latex]1,600[\/latex] in.<sup>3<\/sup>.\r\n\r\n[reveal-answer q=\"686872\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"686872\"]\r\n\r\nThe formula for the volume of a box is given as [latex]V=LWH[\/latex], the product of length, width, and height. We are given that [latex]L=2W[\/latex], and [latex]H=8[\/latex]. The volume is [latex]1,600[\/latex] cubic inches.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}V=LWH\\hfill \\\\ 1,600=\\left(2W\\right)W\\left(8\\right)\\hfill \\\\ 1,600=16{W}^{2}\\hfill \\\\ 100={W}^{2}\\hfill \\\\ 10=W\\hfill \\end{array}[\/latex]<\/div>\r\nThe dimensions are [latex]L=20[\/latex] in., [latex]W=10[\/latex] in., and [latex]H=8[\/latex] in. Note that the square root of [latex]{W}^{2}[\/latex] would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nExpress the formula for the surface area of a cylinder, [latex]s=2\\pi rh+2\\pi r^{2}[\/latex], in terms of the height, <em>h<\/em>.\r\n\r\nIn this example, the variable <em>h<\/em> is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to\u00a0write down what you think is the best first step to take to isolate <em>h<\/em>.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"194805\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"194805\"]\r\n\r\nIsolate the term containing the variable\u00a0<em>h\u00a0<\/em>by subtracting [latex]2\\pi r^{2}[\/latex]from both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}S\\,\\,=2\\pi rh+2\\pi r^{2} \\\\ \\underline{-2\\pi r^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2\\pi r^{2}}\\\\S-2\\pi r^{2}\\,\\,\\,\\,=\\,\\,\\,\\,2\\pi rh\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nNext, isolate the variable h by dividing both sides of the equation by [latex]2\\pi r[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\dfrac{S-2\\pi r^{2}}{2\\pi r}=\\frac{2\\pi rh}{2\\pi r} \\\\\\\\\\dfrac{S-2\\pi r^{2}}{2\\pi r}=h\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nYou can rewrite the equation so the isolated variable is on the left side.\r\n<p style=\"text-align: center;\">[latex]h=\\dfrac{S-2\\pi r^{2}}{2\\pi r}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nIn the following video we show how to f<span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Find the Volume of a Right Circular Cylinder Formed from a Given Rectangle\">ind the volume of a right circular cylinder formed from a given rectangle.<\/span>\r\n\r\nhttps:\/\/youtu.be\/KS1pGO_g3vM\r\n<h2>Isolate Variables in Formulas<\/h2>\r\nSometimes, it is easier to isolate the variable you are solving for when you are using a formula. This is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly. In the next examples, we will use algebraic properties to isolate a variable in a formula.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIsolate the term containing the variable\u00a0<i>w\u00a0<\/i>from the formula for the perimeter of a rectangle<em>: \u00a0<\/em>\r\n<p style=\"text-align: center;\">[latex]{P}=2\\left({L}\\right)+2\\left({W}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"967601\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"967601\"]\r\n\r\nFirst, isolate the term with\u00a0<i>w<\/i> by subtracting 2<em>l<\/em> from both sides of the equation.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,P\\,=\\,\\,\\,\\,2l+2w\\\\\\underline{\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,}\\\\P-2l=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2w\\end{array}[\/latex]<i> <\/i><\/p>\r\nNext, clear the coefficient of <i>w <\/i>by dividing both sides of the equation by [latex]2[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}\\underline{P-2l}=\\underline{2w}\\\\\\,\\,\\,\\,\\,\\,2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\\\ \\,\\,\\,\\dfrac{P-2l}{2}\\,\\,=\\,\\,w\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\dfrac{P-2l}{2}\\end{array}[\/latex]<\/p>\r\nYou can rewrite the equation so the isolated variable is on the left side.\r\n<p style=\"text-align: center;\">[latex]w=\\dfrac{P-2l}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse the multiplication and division properties of equality to isolate the variable <em>b<\/em>\u00a0given [latex]A=\\dfrac{1}{2}\\normalsize bh[\/latex]\r\n\r\n[reveal-answer q=\"291790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"291790\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,A=\\dfrac{1}{2}\\normalsize bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\dfrac{1}{2}\\normalsize bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\dfrac{2A}{h}=\\frac{bh}{h}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\dfrac{2A}{h}=\\frac{b\\cancel{h}}{\\cancel{h}}\\end{array}[\/latex]<\/p>\r\nWrite the equation with the desired variable on the left-hand side as a matter of convention:\r\n<p style=\"text-align: center;\">[latex]b=\\dfrac{2A}{h}[\/latex]\r\n[\/hidden-answer]<\/p>\r\nUse the multiplication and division properties of equality to isolate the variable <em>h\u00a0<\/em>given [latex]A=\\dfrac{1}{2}\\normalsize bh[\/latex]\r\n[reveal-answer q=\"595790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"595790\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,A=\\dfrac{1}{2}\\normalsize bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\dfrac{1}{2}\\normalsize bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\dfrac{2A}{b}=\\frac{bh}{b}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\dfrac{2A}{b}=\\frac{h\\cancel{b}}{\\cancel{b}}\\end{array}[\/latex]<\/p>\r\nWrite the equation with the desired variable on the left-hand side as a matter of convention:\r\n<p style=\"text-align: center;\">[latex]h=\\dfrac{2A}{b}[\/latex]\r\n[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<h2>Temperature<\/h2>\r\nLet us look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.\r\n<p style=\"text-align: center;\">[latex]C=\\left(F--32\\right)\\cdot\\dfrac{5}{9}[\/latex]<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGiven a temperature of [latex]12^{\\circ}{C}[\/latex], find the equivalent in [latex]{}^{\\circ}{F}[\/latex].\r\n[reveal-answer q=\"594254\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"594254\"]\r\n\r\nSubstitute the given temperature in[latex]{}^{\\circ}{C}[\/latex]\u00a0into the conversion formula:\r\n<p style=\"text-align: center;\">[latex]12=\\left(F-32\\right)\\cdot\\dfrac{5}{9}[\/latex]<\/p>\r\nIsolate the variable F to obtain the equivalent temperature.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}12=\\left(F-32\\right)\\cdot\\dfrac{5}{9}\\\\\\\\\\left(\\dfrac{9}{5}\\normalsize\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\left(\\dfrac{108}{5}\\normalsize\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\21.6=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{+32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+32}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\53.6={}^{\\circ}{F}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAs with the other formulas we have worked with, we could have isolated the variable F first then substituted in the given temperature in Celsius.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.\r\n\r\n[latex]C=\\left(F--32\\right)\\cdot\\dfrac{5}{9}[\/latex]\r\n\r\n[reveal-answer q=\"591790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"591790\"]\r\n\r\nTo isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex] \\displaystyle \\frac{9}{5}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\\\,\\,\\,\\,\\left(\\dfrac{9}{5}\\normalsize\\right)C=\\left(F-32\\right)\\left(\\dfrac{5}{9}\\normalsize\\right)\\left(\\dfrac{9}{5}\\normalsize\\right)\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\dfrac{9}{5}\\normalsize C=F-32\\end{array}[\/latex]<\/p>\r\nAdd 32 to both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\dfrac{9}{5}\\normalsize\\,C+32=F-32+32\\\\\\\\\\dfrac{9}{5}\\normalsize\\,C+32=F\\\\F=\\dfrac{9}{5}\\normalsize C+32\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning OUTCOMES<\/h3>\n<ul>\n<li>Solve\u00a0distance, rate, and time problems<\/li>\n<li>Solve area, volume, and perimeter problems<\/li>\n<li>Solve temperature conversion problems<\/li>\n<li>Rearrange formulas to isolate specific variables<\/li>\n<li>Identify an unknown given a formula<\/li>\n<\/ul>\n<\/div>\n<p>Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem\u2019s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the <strong>area<\/strong> of a rectangular region, [latex]A=LW[\/latex]; the <strong>perimeter<\/strong> of a rectangle, [latex]P=2L+2W[\/latex]; and the <strong>volume<\/strong> of a rectangular solid, [latex]V=LWH[\/latex]. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.<\/p>\n<h2>Distance, Rate, and Time<\/h2>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>It takes Andrew [latex]30[\/latex] min to drive to work in the morning. He drives home using the same route, but it takes [latex]10[\/latex] min longer and he averages [latex]10 mi\/h[\/latex] less than in the morning. How far does Andrew drive to work?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q61824\">Show Solution<\/span><\/p>\n<div id=\"q61824\" class=\"hidden-answer\" style=\"display: none\">\n<p>This is a distance problem, so we can use the formula [latex]d=rt[\/latex], where distance equals rate multiplied by time. Note that when rate is given in mi\/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.<\/p>\n<p>First, we identify the known and unknown quantities. Andrew\u2019s morning drive to work takes [latex]30[\/latex] min, or [latex]\\dfrac{1}{2}[\/latex] h at rate [latex]r[\/latex]. His drive home takes [latex]40[\/latex] min, or [latex]\\dfrac{2}{3}[\/latex] h, and his speed averages [latex]10 mi\/h[\/latex] less than the morning drive. Both trips cover distance [latex]d[\/latex]. A table, such as the one below, is often helpful for keeping track of information in these types of problems.<\/p>\n<table summary=\"A table with 3 rows and 4 columns. The first entry in the first row is blank, the rest are: d, r, and t. The entries in the second row are: To Work, d, r, and . The entries in the third row are: To Home, d, r 10, and 2\/3.\">\n<thead>\n<tr>\n<th><\/th>\n<th>[latex]d[\/latex]<\/th>\n<th>[latex]r[\/latex]<\/th>\n<th>[latex]t[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><strong>To Work<\/strong><\/td>\n<td>[latex]d[\/latex]<\/td>\n<td>[latex]r[\/latex]<\/td>\n<td>[latex]\\dfrac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>To Home<\/strong><\/td>\n<td>[latex]d[\/latex]<\/td>\n<td>[latex]r - 10[\/latex]<\/td>\n<td>[latex]\\dfrac{2}{3}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Write two equations, one for each trip.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}d=r\\left(\\dfrac{1}{2}\\normalsize\\right)\\hfill & \\text{To work}\\hfill \\\\ d=\\left(r - 10\\right)\\left(\\dfrac{2}{3}\\normalsize\\right)\\hfill & \\text{To home}\\hfill \\end{array}[\/latex]<\/div>\n<p>As both equations equal the same distance, we set them equal to each other and solve for <em>r<\/em>.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}r\\left(\\dfrac{1}{2}\\right)&=\\left(r - 10\\right)\\left(\\dfrac{2}{3}\\right)\\hfill \\\\ \\frac{1}{2} r&=\\frac{2}{3} r-\\frac{20}{3}\\hfill \\\\\\frac{1}{2} r-\\frac{2}{3} r&=-\\frac{20}{3}\\hfill \\\\ -\\frac{1}{6} r&=-\\frac{20}{3}\\hfill \\\\ r&=-\\frac{20}{3}\\left(-6\\right)\\hfill \\\\ r&=40\\hfill \\end{array}[\/latex]<\/div>\n<p>We have solved for the rate of speed to work, [latex]40 mph[\/latex]. Substituting [latex]40[\/latex] into the rate on the return trip yields [latex]30 mi\/h[\/latex]. Now we can answer the question. Substitute the rate back into either equation and solve for <em>d.<\/em><\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d\\hfill&=40\\left(\\dfrac{1}{2}\\normalsize\\right)\\hfill \\\\ \\hfill&=20\\hfill \\end{array}[\/latex]<\/div>\n<p>The distance between home and work is [latex]20 mi[\/latex].Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for [latex]r[\/latex].<\/p>\n<div style=\"text-align: center;\">\n<p>[latex]\\begin{array}{rl}r\\left(\\dfrac{1}{2}\\normalsize\\right)&=\\left(r - 10\\right)\\left(\\dfrac{2}{3}\\normalsize\\right)\\hfill \\\\ 6\\times r\\left(\\dfrac{1}{2}\\normalsize\\right)& =6\\times \\left(r - 10\\right)\\left(\\dfrac{2}{3}\\normalsize\\right)\\hfill \\\\ 3r& =4\\left(r - 10\\right)\\hfill \\\\ 3r& =4r - 40\\hfill \\\\ -r& =-40\\hfill \\\\ r& =40\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will find the length and width of a rectangular field given its perimeter and a relationship between its sides.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The perimeter of a rectangular outdoor patio is [latex]54[\/latex] ft. The length is [latex]3[\/latex] ft greater than the width. What are the dimensions of the patio?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q918170\">Show Solution<\/span><\/p>\n<div id=\"q918170\" class=\"hidden-answer\" style=\"display: none\">\n<p>The perimeter formula is standard: [latex]P=2L+2W[\/latex]. We have two unknown quantities, length and width. However, we can write the length in terms of the width as [latex]L=W+3[\/latex]. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200341\/CNX_CAT_Figure_02_03_003.jpg\" alt=\"A rectangle with the length labeled as: L = W + 3 and the width labeled as: W.\" width=\"487\" height=\"166\" \/><\/p>\n<p>Now we can solve for the width and then calculate the length.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}P=2L+2W\\hfill \\\\ 54=2\\left(W+3\\right)+2W\\hfill \\\\ 54=2W+6+2W\\hfill \\\\ 54=4W+6\\hfill \\\\ 48=4W\\hfill \\\\ 12=W\\hfill \\\\ \\left(12+3\\right)=L\\hfill \\\\ 15=L\\hfill \\end{array}[\/latex]<\/div>\n<p>The dimensions are [latex]L=15[\/latex] ft and [latex]W=12[\/latex] ft.<\/p><\/div>\n<\/div>\n<\/div>\n<p>The following video shows an example of finding the dimensions of a rectangular field given its perimeter.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Find the Dimensions and Area of a Field Given the Perimeter\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/VyK-HQr02iQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The perimeter of a tablet of graph paper is [latex]48[\/latex] in. The length is [latex]6[\/latex] in. more than the width. Find the area of the graph paper.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q859423\">Show Solution<\/span><\/p>\n<div id=\"q859423\" class=\"hidden-answer\" style=\"display: none\">\n<p>The standard formula for area is [latex]A=LW[\/latex]; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.<\/p>\n<p>We know that the length is [latex]6[\/latex] in. more than the width, so we can write length as [latex]L=W+6[\/latex]. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}P&=2L+2W\\hfill \\\\ 48&=2\\left(W+6\\right)+2W\\hfill \\\\ 48&=2W+12+2W\\hfill \\\\ 48&=4W+12\\hfill \\\\ 36&=4W\\hfill \\\\ 9&=W\\hfill \\\\ \\left(9+6\\right)&=L\\hfill \\\\ 15&=L\\hfill \\end{array}[\/latex]<\/div>\n<p>Now, we find the area given the dimensions of [latex]L=15[\/latex] in. and [latex]W=9[\/latex] in.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}A\\hfill&=LW\\hfill \\\\ A\\hfill&=15\\left(9\\right)\\hfill \\\\ \\hfill&=135\\text{ in}^{2}\\hfill \\end{array}[\/latex]<br \/>\nThe area is [latex]135[\/latex] in<sup>2<\/sup>.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows another example of finding the area\u00a0of a rectangle given its perimeter and the relationship between its side lengths.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex: Find the Area of a Rectangle Given the Perimeter\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/zUlU64Umnq4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Volume<\/h2>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the dimensions of a shipping box given that the length is twice the width, the height is [latex]8[\/latex] inches, and the volume is [latex]1,600[\/latex] in.<sup>3<\/sup>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q686872\">Show Solution<\/span><\/p>\n<div id=\"q686872\" class=\"hidden-answer\" style=\"display: none\">\n<p>The formula for the volume of a box is given as [latex]V=LWH[\/latex], the product of length, width, and height. We are given that [latex]L=2W[\/latex], and [latex]H=8[\/latex]. The volume is [latex]1,600[\/latex] cubic inches.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}V=LWH\\hfill \\\\ 1,600=\\left(2W\\right)W\\left(8\\right)\\hfill \\\\ 1,600=16{W}^{2}\\hfill \\\\ 100={W}^{2}\\hfill \\\\ 10=W\\hfill \\end{array}[\/latex]<\/div>\n<p>The dimensions are [latex]L=20[\/latex] in., [latex]W=10[\/latex] in., and [latex]H=8[\/latex] in. Note that the square root of [latex]{W}^{2}[\/latex] would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Express the formula for the surface area of a cylinder, [latex]s=2\\pi rh+2\\pi r^{2}[\/latex], in terms of the height, <em>h<\/em>.<\/p>\n<p>In this example, the variable <em>h<\/em> is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to\u00a0write down what you think is the best first step to take to isolate <em>h<\/em>.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q194805\">Show Solution<\/span><\/p>\n<div id=\"q194805\" class=\"hidden-answer\" style=\"display: none\">\n<p>Isolate the term containing the variable\u00a0<em>h\u00a0<\/em>by subtracting [latex]2\\pi r^{2}[\/latex]from both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}S\\,\\,=2\\pi rh+2\\pi r^{2} \\\\ \\underline{-2\\pi r^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2\\pi r^{2}}\\\\S-2\\pi r^{2}\\,\\,\\,\\,=\\,\\,\\,\\,2\\pi rh\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Next, isolate the variable h by dividing both sides of the equation by [latex]2\\pi r[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\dfrac{S-2\\pi r^{2}}{2\\pi r}=\\frac{2\\pi rh}{2\\pi r} \\\\\\\\\\dfrac{S-2\\pi r^{2}}{2\\pi r}=h\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>You can rewrite the equation so the isolated variable is on the left side.<\/p>\n<p style=\"text-align: center;\">[latex]h=\\dfrac{S-2\\pi r^{2}}{2\\pi r}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show how to f<span id=\"eow-title\" class=\"watch-title\" dir=\"ltr\" title=\"Find the Volume of a Right Circular Cylinder Formed from a Given Rectangle\">ind the volume of a right circular cylinder formed from a given rectangle.<\/span><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Find the Volume of a Right Circular Cylinder Formed from a Given Rectangle\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/KS1pGO_g3vM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Isolate Variables in Formulas<\/h2>\n<p>Sometimes, it is easier to isolate the variable you are solving for when you are using a formula. This is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly. In the next examples, we will use algebraic properties to isolate a variable in a formula.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Isolate the term containing the variable\u00a0<i>w\u00a0<\/i>from the formula for the perimeter of a rectangle<em>: \u00a0<\/em><\/p>\n<p style=\"text-align: center;\">[latex]{P}=2\\left({L}\\right)+2\\left({W}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q967601\">Show Solution<\/span><\/p>\n<div id=\"q967601\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, isolate the term with\u00a0<i>w<\/i> by subtracting 2<em>l<\/em> from both sides of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,P\\,=\\,\\,\\,\\,2l+2w\\\\\\underline{\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,}\\\\P-2l=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2w\\end{array}[\/latex]<i> <\/i><\/p>\n<p>Next, clear the coefficient of <i>w <\/i>by dividing both sides of the equation by [latex]2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}\\underline{P-2l}=\\underline{2w}\\\\\\,\\,\\,\\,\\,\\,2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\\\ \\,\\,\\,\\dfrac{P-2l}{2}\\,\\,=\\,\\,w\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\dfrac{P-2l}{2}\\end{array}[\/latex]<\/p>\n<p>You can rewrite the equation so the isolated variable is on the left side.<\/p>\n<p style=\"text-align: center;\">[latex]w=\\dfrac{P-2l}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use the multiplication and division properties of equality to isolate the variable <em>b<\/em>\u00a0given [latex]A=\\dfrac{1}{2}\\normalsize bh[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q291790\">Show Solution<\/span><\/p>\n<div id=\"q291790\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,A=\\dfrac{1}{2}\\normalsize bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\dfrac{1}{2}\\normalsize bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\dfrac{2A}{h}=\\frac{bh}{h}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\dfrac{2A}{h}=\\frac{b\\cancel{h}}{\\cancel{h}}\\end{array}[\/latex]<\/p>\n<p>Write the equation with the desired variable on the left-hand side as a matter of convention:<\/p>\n<p style=\"text-align: center;\">[latex]b=\\dfrac{2A}{h}[\/latex]\n<\/div>\n<\/div>\n<p>Use the multiplication and division properties of equality to isolate the variable <em>h\u00a0<\/em>given [latex]A=\\dfrac{1}{2}\\normalsize bh[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q595790\">Show Solution<\/span><\/p>\n<div id=\"q595790\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,A=\\dfrac{1}{2}\\normalsize bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\dfrac{1}{2}\\normalsize bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\dfrac{2A}{b}=\\frac{bh}{b}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\dfrac{2A}{b}=\\frac{h\\cancel{b}}{\\cancel{b}}\\end{array}[\/latex]<\/p>\n<p>Write the equation with the desired variable on the left-hand side as a matter of convention:<\/p>\n<p style=\"text-align: center;\">[latex]h=\\dfrac{2A}{b}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<h2>Temperature<\/h2>\n<p>Let us look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.<\/p>\n<p style=\"text-align: center;\">[latex]C=\\left(F--32\\right)\\cdot\\dfrac{5}{9}[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Given a temperature of [latex]12^{\\circ}{C}[\/latex], find the equivalent in [latex]{}^{\\circ}{F}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q594254\">Show Solution<\/span><\/p>\n<div id=\"q594254\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the given temperature in[latex]{}^{\\circ}{C}[\/latex]\u00a0into the conversion formula:<\/p>\n<p style=\"text-align: center;\">[latex]12=\\left(F-32\\right)\\cdot\\dfrac{5}{9}[\/latex]<\/p>\n<p>Isolate the variable F to obtain the equivalent temperature.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}12=\\left(F-32\\right)\\cdot\\dfrac{5}{9}\\\\\\\\\\left(\\dfrac{9}{5}\\normalsize\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\left(\\dfrac{108}{5}\\normalsize\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\21.6=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{+32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+32}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\53.6={}^{\\circ}{F}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>As with the other formulas we have worked with, we could have isolated the variable F first then substituted in the given temperature in Celsius.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F.<\/p>\n<p>[latex]C=\\left(F--32\\right)\\cdot\\dfrac{5}{9}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q591790\">Show Solution<\/span><\/p>\n<div id=\"q591790\" class=\"hidden-answer\" style=\"display: none\">\n<p>To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex]\\displaystyle \\frac{9}{5}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\\\,\\,\\,\\,\\left(\\dfrac{9}{5}\\normalsize\\right)C=\\left(F-32\\right)\\left(\\dfrac{5}{9}\\normalsize\\right)\\left(\\dfrac{9}{5}\\normalsize\\right)\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\dfrac{9}{5}\\normalsize C=F-32\\end{array}[\/latex]<\/p>\n<p>Add 32 to both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\dfrac{9}{5}\\normalsize\\,C+32=F-32+32\\\\\\\\\\dfrac{9}{5}\\normalsize\\,C+32=F\\\\F=\\dfrac{9}{5}\\normalsize C+32\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\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-373\">\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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@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>Ex: Find the Dimensions and Area of a Field Given the Perimeter Mathispower4u . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/VyK-HQr02iQ\">https:\/\/youtu.be\/VyK-HQr02iQ<\/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: Find the Area of a Rectangle Given the Perimeter. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/zUlU64Umnq4\">https:\/\/youtu.be\/zUlU64Umnq4<\/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>Find the Volume of a Right Circular Cylinder Formed from a Given Rectangle. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/KS1pGO_g3vM\">https:\/\/youtu.be\/KS1pGO_g3vM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\"},{\"type\":\"original\",\"description\":\"Revision 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