{"id":95,"date":"2023-06-21T13:22:32","date_gmt":"2023-06-21T13:22:32","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/translate-and-solve-word-problems\/"},"modified":"2024-01-08T18:47:42","modified_gmt":"2024-01-08T18:47:42","slug":"translate-and-solve-word-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/translate-and-solve-word-problems\/","title":{"raw":"R1.4   Using and Manipulating Formulas","rendered":"R1.4   Using and Manipulating Formulas"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Recognize formulas for perimeter, area, and volume<\/li>\r\n \t<li>Recognize the formula that relates distance, rate, and time<\/li>\r\n \t<li>Rearrange formulas to isolate specific variables<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div style=\"text-align: left;\"><\/div>\r\nMany real-world applications (problems that can be modeled by a mathematical equation and solved for an unknown quantity) involve formulas that describe relationships between quantities. For example, wall-to-wall carpeting is sold by the square foot. To find out how much carpeting would cover the floor of a 12' x 15' foot room, we can multiply the length and width of the room to discover how many square feet of carpeting is required.\u00a0 [latex]12 \\times 15 = 180[\/latex]. So, we would need to purchase 180 square feet of carpeting. Examples of formulas that commonly occur in applications include:\r\n<ul>\r\n \t<li>the <strong>perimeter<\/strong> of a rectangle of length L and width W\r\n<ul>\r\n \t<li>[latex]P=2L+2W[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>the <strong>area<\/strong> of a rectangular region of length L and width W\r\n<ul>\r\n \t<li>[latex]A=LW[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>the <strong>volume<\/strong> of a rectangular solid with length L, width W, and height H\r\n<ul>\r\n \t<li>[latex]V=LWH[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>the\u00a0<strong>distance<\/strong> [latex]d[\/latex] covered when traveling at a constant\u00a0<strong>rate<\/strong> [latex]r[\/latex] for some\u00a0<strong>time\u00a0<\/strong>[latex]t[\/latex]\r\n<ul>\r\n \t<li>[latex]d=rt[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\nFormulas such as these may be used to solve problems by substituting known values and solving for an unknown value. You should know these formulas and be able to recognize when to apply them to a problem.\r\n<h2>Isolate Variables in Formulas<\/h2>\r\nSometimes, we need to isolate one of the variables in a formula in order to solve for the unknown. In the carpeting example above, we found that it would take 180 square feet of carpet to cover a 12' x 15' room. We knew the dimensions of the room and solved for how many square feet of carpet we needed, But what if, instead, we had tiles that we could use to cover the floor and we knew the length of the room, but needed to know what width room could be fully tiled?\r\n\r\nLet's say you have a banquet hall with one wall that can slide to create a room of variable width. The length of the room (the length of the movable wall) is 85 feet. If you have 4,590 square feet of tile, what must the width be to ensure the entire room is fully tiled?\r\n\r\nUse the formula [latex]A=LW[\/latex], but solve it for W, width.\u00a0[latex]\\dfrac{A}{L}=W[\/latex]. Then, substitute what you know:\r\n\r\n[latex]\\dfrac{4590}{85} = W = 54[\/latex]. It looks like we need to slide the wall out to create a width of 54 feet in order to use all the tile.\u00a0.\r\n\r\nThis technique, isolating a variable of choice in any formula, is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly, for different values of the unknown variable.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: isolate a variable in a formula<\/h3>\r\nIsolate the variable for width [latex]w[\/latex]<i>\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\\\\ \\,\\,\\,\\Large\\frac{P-2l}{2}\\,\\,=\\,\\,w\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\Large\\frac{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=\\Large\\frac{P-2l}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0isolate a variable in a formula<\/h3>\r\n&nbsp;\r\n\r\nUse the multiplication and division properties of equality to isolate the variable <em>b<\/em>\u00a0given [latex]A=\\Large\\frac{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=\\Large\\frac{1}{2}\\normalsize bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\Large\\frac{1}{2}\\normalsize bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\Large\\frac{2A}{h}=\\frac{bh}{h}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\Large\\frac{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=\\Large\\frac{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=\\Large\\frac{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=\\Large\\frac{1}{2}\\normalsize bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\Large\\frac{1}{2}\\normalsize bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\Large\\frac{2A}{b}=\\frac{bh}{b}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\Large\\frac{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=\\Large\\frac{2A}{b}[\/latex]\r\n[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nIsolate the variable for height, [latex]h[\/latex], from the formula for the surface area of a cylinder, [latex]s=2\\pi rh+2\\pi r^{2}[\/latex].\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}\\Large\\frac{S-2\\pi r^{2}}{2\\pi r}=\\frac{2\\pi rh}{2\\pi r} \\\\\\\\\\Large\\frac{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=\\Large\\frac{S-2\\pi r^{2}}{2\\pi r}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nHopefully, you can see the value in being able to isolate a variable of interest! Here is another formula for which the skill is particularly useful.\r\n\r\nThe formula for converting from the Fahrenheit temperature scale to the Celsius scale is given by\u00a0[latex]C=\\left(F--32\\right)\\cdot\\Large\\frac{5}{9}[\/latex]. This formula takes a temperature in Fahrenheit, [latex]F[\/latex], and coverts it to an equivalent temperature in Celsius, [latex]C[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: use a formula to convert fahrenheit to celsius<\/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\\Large\\frac{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\\Large\\frac{5}{9}\\\\\\\\\\left(\\Large\\frac{9}{5}\\normalsize\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\left(\\Large\\frac{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\nBut what if we have the temperature in Celsius already, and wish to convert it to Fahrenheit? We can solve the formula for [latex]F[\/latex]. This will give us a formula that takes a temperature on the Celsius scale and converts it to an equivalent Fahrenheit temperature.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: isolate a variable to convert celsius to fahrenheit<\/h3>\r\nSolve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F. Then use the new formula to convert\u00a0a temperature of [latex]21^{\\circ}{C}[\/latex] to Fahrenheit.\r\n\r\n[latex]C=\\left(F--32\\right)\\cdot\\Large\\frac{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(\\Large\\frac{9}{5}\\normalsize\\right)C=\\left(F-32\\right)\\left(\\Large\\frac{5}{9}\\normalsize\\right)\\left(\\Large\\frac{9}{5}\\normalsize\\right)\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\Large\\frac{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}\\Large\\frac{9}{5}\\normalsize\\,C+32=F-32+32\\\\\\\\\\Large\\frac{9}{5}\\normalsize\\,C+32=F\\\\F=\\Large\\frac{9}{5}\\normalsize C+32\\end{array}[\/latex]<\/p>\r\nAnd we can convert\u00a0[latex]21^{\\circ}{C}[\/latex]\u00a0 by substituting it into the new formula:\r\n\r\n[latex]F=\\dfrac{9}{5}(21)+32[\/latex]\r\n\r\n[latex]21^{\\circ}{C}=69.8^{\\circ}{F}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Recognize formulas for perimeter, area, and volume<\/li>\n<li>Recognize the formula that relates distance, rate, and time<\/li>\n<li>Rearrange formulas to isolate specific variables<\/li>\n<\/ul>\n<\/div>\n<div style=\"text-align: left;\"><\/div>\n<p>Many real-world applications (problems that can be modeled by a mathematical equation and solved for an unknown quantity) involve formulas that describe relationships between quantities. For example, wall-to-wall carpeting is sold by the square foot. To find out how much carpeting would cover the floor of a 12&#8242; x 15&#8242; foot room, we can multiply the length and width of the room to discover how many square feet of carpeting is required.\u00a0 [latex]12 \\times 15 = 180[\/latex]. So, we would need to purchase 180 square feet of carpeting. Examples of formulas that commonly occur in applications include:<\/p>\n<ul>\n<li>the <strong>perimeter<\/strong> of a rectangle of length L and width W\n<ul>\n<li>[latex]P=2L+2W[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>the <strong>area<\/strong> of a rectangular region of length L and width W\n<ul>\n<li>[latex]A=LW[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>the <strong>volume<\/strong> of a rectangular solid with length L, width W, and height H\n<ul>\n<li>[latex]V=LWH[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>the\u00a0<strong>distance<\/strong> [latex]d[\/latex] covered when traveling at a constant\u00a0<strong>rate<\/strong> [latex]r[\/latex] for some\u00a0<strong>time\u00a0<\/strong>[latex]t[\/latex]\n<ul>\n<li>[latex]d=rt[\/latex].<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>Formulas such as these may be used to solve problems by substituting known values and solving for an unknown value. You should know these formulas and be able to recognize when to apply them to a problem.<\/p>\n<h2>Isolate Variables in Formulas<\/h2>\n<p>Sometimes, we need to isolate one of the variables in a formula in order to solve for the unknown. In the carpeting example above, we found that it would take 180 square feet of carpet to cover a 12&#8242; x 15&#8242; room. We knew the dimensions of the room and solved for how many square feet of carpet we needed, But what if, instead, we had tiles that we could use to cover the floor and we knew the length of the room, but needed to know what width room could be fully tiled?<\/p>\n<p>Let&#8217;s say you have a banquet hall with one wall that can slide to create a room of variable width. The length of the room (the length of the movable wall) is 85 feet. If you have 4,590 square feet of tile, what must the width be to ensure the entire room is fully tiled?<\/p>\n<p>Use the formula [latex]A=LW[\/latex], but solve it for W, width.\u00a0[latex]\\dfrac{A}{L}=W[\/latex]. Then, substitute what you know:<\/p>\n<p>[latex]\\dfrac{4590}{85} = W = 54[\/latex]. It looks like we need to slide the wall out to create a width of 54 feet in order to use all the tile.\u00a0.<\/p>\n<p>This technique, isolating a variable of choice in any formula, is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly, for different values of the unknown variable.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: isolate a variable in a formula<\/h3>\n<p>Isolate the variable for width [latex]w[\/latex]<i>\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\\\\ \\,\\,\\,\\Large\\frac{P-2l}{2}\\,\\,=\\,\\,w\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\Large\\frac{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=\\Large\\frac{P-2l}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0isolate a variable in a formula<\/h3>\n<p>&nbsp;<\/p>\n<p>Use the multiplication and division properties of equality to isolate the variable <em>b<\/em>\u00a0given [latex]A=\\Large\\frac{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=\\Large\\frac{1}{2}\\normalsize bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\Large\\frac{1}{2}\\normalsize bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\Large\\frac{2A}{h}=\\frac{bh}{h}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\Large\\frac{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=\\Large\\frac{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=\\Large\\frac{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=\\Large\\frac{1}{2}\\normalsize bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\Large\\frac{1}{2}\\normalsize bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\Large\\frac{2A}{b}=\\frac{bh}{b}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\Large\\frac{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=\\Large\\frac{2A}{b}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Isolate the variable for height, [latex]h[\/latex], from the formula for the surface area of a cylinder, [latex]s=2\\pi rh+2\\pi r^{2}[\/latex].<\/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}\\Large\\frac{S-2\\pi r^{2}}{2\\pi r}=\\frac{2\\pi rh}{2\\pi r} \\\\\\\\\\Large\\frac{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=\\Large\\frac{S-2\\pi r^{2}}{2\\pi r}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>Hopefully, you can see the value in being able to isolate a variable of interest! Here is another formula for which the skill is particularly useful.<\/p>\n<p>The formula for converting from the Fahrenheit temperature scale to the Celsius scale is given by\u00a0[latex]C=\\left(F--32\\right)\\cdot\\Large\\frac{5}{9}[\/latex]. This formula takes a temperature in Fahrenheit, [latex]F[\/latex], and coverts it to an equivalent temperature in Celsius, [latex]C[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: use a formula to convert fahrenheit to celsius<\/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\\Large\\frac{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\\Large\\frac{5}{9}\\\\\\\\\\left(\\Large\\frac{9}{5}\\normalsize\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\left(\\Large\\frac{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>But what if we have the temperature in Celsius already, and wish to convert it to Fahrenheit? We can solve the formula for [latex]F[\/latex]. This will give us a formula that takes a temperature on the Celsius scale and converts it to an equivalent Fahrenheit temperature.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: isolate a variable to convert celsius to fahrenheit<\/h3>\n<p>Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F. Then use the new formula to convert\u00a0a temperature of [latex]21^{\\circ}{C}[\/latex] to Fahrenheit.<\/p>\n<p>[latex]C=\\left(F--32\\right)\\cdot\\Large\\frac{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(\\Large\\frac{9}{5}\\normalsize\\right)C=\\left(F-32\\right)\\left(\\Large\\frac{5}{9}\\normalsize\\right)\\left(\\Large\\frac{9}{5}\\normalsize\\right)\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\Large\\frac{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}\\Large\\frac{9}{5}\\normalsize\\,C+32=F-32+32\\\\\\\\\\Large\\frac{9}{5}\\normalsize\\,C+32=F\\\\F=\\Large\\frac{9}{5}\\normalsize C+32\\end{array}[\/latex]<\/p>\n<p>And we can convert\u00a0[latex]21^{\\circ}{C}[\/latex]\u00a0 by substituting it into the new formula:<\/p>\n<p>[latex]F=\\dfrac{9}{5}(21)+32[\/latex]<\/p>\n<p>[latex]21^{\\circ}{C}=69.8^{\\circ}{F}[\/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-95\">\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><\/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":9,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-95","chapter","type-chapter","status-publish","hentry"],"part":91,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/95","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/95\/revisions"}],"predecessor-version":[{"id":1484,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/95\/revisions\/1484"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/91"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/95\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=95"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=95"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=95"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=95"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}