{"id":16114,"date":"2019-10-01T17:41:10","date_gmt":"2019-10-01T17:41:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-solving-word-problems-with-geometry\/"},"modified":"2020-10-22T09:11:59","modified_gmt":"2020-10-22T09:11:59","slug":"read-solving-word-problems-with-geometry","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/chapter\/read-solving-word-problems-with-geometry\/","title":{"raw":"7.4.b - Problems Involving Formulas II","rendered":"7.4.b &#8211; Problems Involving Formulas II"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve a formula for a specific variables<\/li>\r\n \t<li>Evaluate a formula using substitution<\/li>\r\n \t<li>Solve area and perimeter words problems<\/li>\r\n \t<li>Solve temperature conversion problems<\/li>\r\n<\/ul>\r\n<\/div>\r\nFormulas come up in many different areas of life. We have seen the formula that relates distance, rate, and time, and the formula for simple interest on an investment. In this section, we will look further at formulas, and see examples of formulas for dimensions of geometric shapes, as well as the formula for converting temperature between Fahrenheit and Celsius.\r\n<h2>Geometry<\/h2>\r\nThere are many geometric shapes that have been well studied over the years. We know quite a bit about circles, rectangles, and triangles. Mathematicians have proven many formulas that\u00a0describe the dimensions of geometric shapes including area, perimeter, surface area, and volume.\r\n<h3>Perimeter<\/h3>\r\nPerimeter is the distance around an object. For example, consider a rectangle with a length of [latex]8[\/latex] and a width of [latex]3[\/latex]. There are two lengths and two widths in a rectangle (opposite sides), so we add [latex]8+8+3+3=22[\/latex]. Since\u00a0there are two lengths and two widths in a rectangle, you may find the perimeter of a rectangle using\u00a0the formula [latex]{P}=2\\left({L}\\right)+2\\left({W}\\right)[\/latex] where\r\n\r\n[latex]L[\/latex] = Length\r\n\r\n[latex]W[\/latex] = Width\r\n\r\nIn the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nYou want to make another garden box the same size as the one you already have. You write down the dimensions of the box and go to the lumber store to buy some boards. When you get there, you realize you didn't write down the width dimension\u2014only the perimeter and length. You want the exact dimensions so you can have the store cut the lumber for you.\r\n\r\nHere is what you have written down:\r\n\r\nPerimeter = [latex]16.4[\/latex] feet\r\nLength = [latex]4.7[\/latex] feet\r\n\r\nCan you find the dimensions you need to have your boards cut at the lumber store? If so, how many boards do you need and what lengths should they be?\r\n\r\n[reveal-answer q=\"719712\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"719712\"]\r\n\r\n<strong>Read and Understand:<\/strong>\u00a0We know perimeter = [latex]16.4[\/latex] feet and length = [latex]4.7[\/latex] feet, and we want to find width.\r\n\r\n<strong>Define and Translate:<\/strong>\r\n\r\nDefine the known and unknown dimensions:\r\n\r\n[latex]W[\/latex] = width\r\n\r\n[latex]P[\/latex] = [latex]16.4[\/latex]\r\n\r\n[latex]L[\/latex] = [latex]4.7[\/latex]\r\n\r\n<strong>Write and Solve:<\/strong>\r\n\r\nFirst, we will substitute the dimensions we know into the formula for perimeter:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{L}\\,\\,\\,\\,\\,P=2{W}+2{L}\\\\\\\\16.4=2\\left(W\\right)+2\\left(4.7\\right)\\end{array}[\/latex]<\/p>\r\nThen, we will isolate [latex]W[\/latex] to find the unknown width.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{L}16.4=2\\left(W\\right)+2\\left(4.7\\right)\\\\16.4=2{W}+9.4\\\\\\underline{-9.4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-9.4}\\\\\\,\\,\\,\\,\\,\\,\\,7=2\\left(W\\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{7}{2}=\\frac{2\\left(W\\right)}{2}\\\\\\,\\,\\,\\,3.5=W\\end{array}[\/latex]<\/p>\r\nWrite the width as a decimal to make cutting the boards easier and replace the units on the measurement, or you won't get the right size of board!\r\n\r\n<strong>Check and Interpret:<\/strong>\r\n\r\nIf we replace the width we found, [latex]W=3.5\\text{ feet }[\/latex] into the formula for perimeter with the dimensions we wrote down, we can check our work:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{L}\\,\\,\\,\\,\\,{P}=2\\left({L}\\right)+2\\left({W}\\right)\\\\\\\\{16.4}=2\\left({4.7}\\right)+2\\left({3.5}\\right)\\\\\\\\{16.4}=9.4+7\\\\\\\\{16.4}=16.4\\end{array}[\/latex]<\/p>\r\nOur calculation for width checks out. We need to ask for [latex]2[\/latex] boards cut to [latex]3.5[\/latex] feet and [latex]2[\/latex] boards cut to [latex]4.7[\/latex] feet so we can make the new garden box.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThis video shows a similar garden box problem.\r\n\r\nhttps:\/\/youtu.be\/jlxPgKQfhQs\r\n\r\nWe could have isolated the [latex]W[\/latex] in the formula for perimeter before we solved the equation, and if we were going to use the formula many times, it could save a lot of time. The next example shows how to isolate a variable in a formula before substituting known dimensions or values into the formula.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIsolate the term containing the variable, [latex]W[\/latex], from the formula for the perimeter of a rectangle<em>: <\/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[latex]W[\/latex] by subtracting [latex]2L[\/latex]\u00a0from 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 [latex]W[\/latex] 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\\\\ \\,\\,\\,\\frac{P-2L}{2}\\,\\,=\\,\\,W\\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=\\frac{P-2L}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Area<\/h2>\r\nThe area of a triangle is given by\u00a0[latex] A=\\frac{1}{2}bh[\/latex] where\r\n\r\n[latex]A[\/latex] = area\r\n[latex]b[\/latex] = the length of the base\r\n[latex]h[\/latex] = the height of the triangle\r\n\r\nRemember, that when two variables, or a number and a variable, are sitting next to each other without a mathematical operator between them, you can assume they are being multiplied. This can seem frustrating, but you can think of it like mathematical slang. Over the years, people who use math frequently have just made that shortcut enough times that it has been adopted as convention.\r\n\r\nIn the next example, we will use the formula for area of a triangle to find a missing dimension, as well as use substitution to solve for the base of a triangle given the area and height.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFind the base ([latex]b[\/latex]) of a triangle with an area ([latex]A[\/latex]) of [latex]20[\/latex] square feet and a height ([latex]h[\/latex]) of [latex]8[\/latex] feet.\r\n[reveal-answer q=\"698967\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"698967\"]\r\n\r\nUse the formula for the area of a triangle, [latex] {A}=\\frac{{1}}{{2}}{bh}[\/latex].\r\n\r\nSubstitute the given lengths into the formula and solve for\u00a0[latex]b[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{l}\\,\\,A=\\frac{1}{2}bh\\\\\\\\20=\\frac{1}{2}b\\cdot 8\\\\\\\\20=\\frac{8}{2}b\\\\\\\\20=4b\\\\\\\\\\frac{20}{4}=\\frac{4b}{4}\\\\\\\\ \\,\\,\\,5=b\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe base of the triangle measures [latex]5[\/latex] feet.[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can rewrite the formula in terms of [latex]b[\/latex] or [latex]h[\/latex] as we did with the perimeter formula previously.\u00a0This probably seems abstract, but it can help you develop your equation-solving skills, as well as help you get more comfortable with working with all kinds of variables, not just [latex]x[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\n1. Use the multiplication and division properties of equality to isolate the variable [latex]b[\/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=\\frac{1}{2}bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\frac{1}{2}bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{h}=\\frac{bh}{h}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\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=\\frac{2A}{h}[\/latex]\r\n[\/hidden-answer]<\/p>\r\n2. Use the multiplication and division properties of equality to isolate the variable [latex]h[\/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=\\frac{1}{2}bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\frac{1}{2}bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{b}=\\frac{bh}{b}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\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=\\frac{2A}{b}[\/latex]\r\n[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nThe following video shows another example of finding the base of a triangle, given area and height.\r\n\r\nhttps:\/\/youtu.be\/VQZQvJ3rXYg\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, [latex]h[\/latex].\r\n\r\nIn this example, the variable [latex]h[\/latex] is buried pretty deep in the formula for the 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 [latex]h[\/latex].\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, [latex]h[\/latex], 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 [latex]h[\/latex] by dividing both sides of the equation by [latex]2\\pi r[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\frac{S-2\\pi r^{2}}{2\\pi r}=\\frac{2\\pi rh}{2\\pi r} \\\\\\\\ \\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=\\frac{S-2\\pi r^{2}}{2\\pi r}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<h2><span style=\"line-height: 1.5\">Temperature<\/span><\/h2>\r\nLet\u2019s 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 \\frac{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 \\frac{5}{9}[\/latex]<\/p>\r\nIsolate the variable [latex]F[\/latex] to obtain the equivalent temperature.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}12=\\left(F-32\\right)\\cdot \\frac{5}{9}\\\\\\\\\\left(\\frac{9}{5}\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\left(\\frac{108}{5}\\right)=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 [latex]F[\/latex] 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 [latex]F[\/latex].\r\n\r\n[latex]C=\\left(F--32\\right)\\cdot \\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 [latex]F[\/latex], it would be best to clear the fraction involving [latex]F[\/latex] 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(\\frac{9}{5}\\right)C=\\left(F-32\\right)\\left(\\frac{5}{9}\\right)\\left(\\frac{9}{5}\\right)\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{9}{5}C=F-32\\end{array}[\/latex]<\/p>\r\nAdd [latex]32[\/latex] to both sides.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\frac{9}{5}\\,C+32=F-32+32\\\\\\\\\\frac{9}{5}\\,C+32=F\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]F=\\frac{9}{5}C+32[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nIn the last video, we show how to convert from Celsius to Fahrenheit.\r\nhttps:\/\/youtu.be\/DRydX8V-JwY\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]2379[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Practice Isolating a Variable in a Formula<\/h2>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[ohm_question]196857[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve a formula for a specific variables<\/li>\n<li>Evaluate a formula using substitution<\/li>\n<li>Solve area and perimeter words problems<\/li>\n<li>Solve temperature conversion problems<\/li>\n<\/ul>\n<\/div>\n<p>Formulas come up in many different areas of life. We have seen the formula that relates distance, rate, and time, and the formula for simple interest on an investment. In this section, we will look further at formulas, and see examples of formulas for dimensions of geometric shapes, as well as the formula for converting temperature between Fahrenheit and Celsius.<\/p>\n<h2>Geometry<\/h2>\n<p>There are many geometric shapes that have been well studied over the years. We know quite a bit about circles, rectangles, and triangles. Mathematicians have proven many formulas that\u00a0describe the dimensions of geometric shapes including area, perimeter, surface area, and volume.<\/p>\n<h3>Perimeter<\/h3>\n<p>Perimeter is the distance around an object. For example, consider a rectangle with a length of [latex]8[\/latex] and a width of [latex]3[\/latex]. There are two lengths and two widths in a rectangle (opposite sides), so we add [latex]8+8+3+3=22[\/latex]. Since\u00a0there are two lengths and two widths in a rectangle, you may find the perimeter of a rectangle using\u00a0the formula [latex]{P}=2\\left({L}\\right)+2\\left({W}\\right)[\/latex] where<\/p>\n<p>[latex]L[\/latex] = Length<\/p>\n<p>[latex]W[\/latex] = Width<\/p>\n<p>In the following example, we will use the problem-solving method we developed to find an unknown width using the formula for the perimeter of a rectangle. By substituting the dimensions we know into the formula, we will be able to isolate the unknown width and find our solution.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>You want to make another garden box the same size as the one you already have. You write down the dimensions of the box and go to the lumber store to buy some boards. When you get there, you realize you didn&#8217;t write down the width dimension\u2014only the perimeter and length. You want the exact dimensions so you can have the store cut the lumber for you.<\/p>\n<p>Here is what you have written down:<\/p>\n<p>Perimeter = [latex]16.4[\/latex] feet<br \/>\nLength = [latex]4.7[\/latex] feet<\/p>\n<p>Can you find the dimensions you need to have your boards cut at the lumber store? If so, how many boards do you need and what lengths should they be?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q719712\">Show Solution<\/span><\/p>\n<div id=\"q719712\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:<\/strong>\u00a0We know perimeter = [latex]16.4[\/latex] feet and length = [latex]4.7[\/latex] feet, and we want to find width.<\/p>\n<p><strong>Define and Translate:<\/strong><\/p>\n<p>Define the known and unknown dimensions:<\/p>\n<p>[latex]W[\/latex] = width<\/p>\n<p>[latex]P[\/latex] = [latex]16.4[\/latex]<\/p>\n<p>[latex]L[\/latex] = [latex]4.7[\/latex]<\/p>\n<p><strong>Write and Solve:<\/strong><\/p>\n<p>First, we will substitute the dimensions we know into the formula for perimeter:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{L}\\,\\,\\,\\,\\,P=2{W}+2{L}\\\\\\\\16.4=2\\left(W\\right)+2\\left(4.7\\right)\\end{array}[\/latex]<\/p>\n<p>Then, we will isolate [latex]W[\/latex] to find the unknown width.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{L}16.4=2\\left(W\\right)+2\\left(4.7\\right)\\\\16.4=2{W}+9.4\\\\\\underline{-9.4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-9.4}\\\\\\,\\,\\,\\,\\,\\,\\,7=2\\left(W\\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{7}{2}=\\frac{2\\left(W\\right)}{2}\\\\\\,\\,\\,\\,3.5=W\\end{array}[\/latex]<\/p>\n<p>Write the width as a decimal to make cutting the boards easier and replace the units on the measurement, or you won&#8217;t get the right size of board!<\/p>\n<p><strong>Check and Interpret:<\/strong><\/p>\n<p>If we replace the width we found, [latex]W=3.5\\text{ feet }[\/latex] into the formula for perimeter with the dimensions we wrote down, we can check our work:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{L}\\,\\,\\,\\,\\,{P}=2\\left({L}\\right)+2\\left({W}\\right)\\\\\\\\{16.4}=2\\left({4.7}\\right)+2\\left({3.5}\\right)\\\\\\\\{16.4}=9.4+7\\\\\\\\{16.4}=16.4\\end{array}[\/latex]<\/p>\n<p>Our calculation for width checks out. We need to ask for [latex]2[\/latex] boards cut to [latex]3.5[\/latex] feet and [latex]2[\/latex] boards cut to [latex]4.7[\/latex] feet so we can make the new garden box.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>This video shows a similar garden box problem.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Find the Width of a Rectangle Given the Perimeter \/ Literal Equation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/jlxPgKQfhQs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We could have isolated the [latex]W[\/latex] in the formula for perimeter before we solved the equation, and if we were going to use the formula many times, it could save a lot of time. The next example shows how to isolate a variable in a formula before substituting known dimensions or values into the formula.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Isolate the term containing the variable, [latex]W[\/latex], from the formula for the perimeter of a rectangle<em>: <\/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[latex]W[\/latex] by subtracting [latex]2L[\/latex]\u00a0from 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 [latex]W[\/latex] 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\\\\ \\,\\,\\,\\frac{P-2L}{2}\\,\\,=\\,\\,W\\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=\\frac{P-2L}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Area<\/h2>\n<p>The area of a triangle is given by\u00a0[latex]A=\\frac{1}{2}bh[\/latex] where<\/p>\n<p>[latex]A[\/latex] = area<br \/>\n[latex]b[\/latex] = the length of the base<br \/>\n[latex]h[\/latex] = the height of the triangle<\/p>\n<p>Remember, that when two variables, or a number and a variable, are sitting next to each other without a mathematical operator between them, you can assume they are being multiplied. This can seem frustrating, but you can think of it like mathematical slang. Over the years, people who use math frequently have just made that shortcut enough times that it has been adopted as convention.<\/p>\n<p>In the next example, we will use the formula for area of a triangle to find a missing dimension, as well as use substitution to solve for the base of a triangle given the area and height.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Find the base ([latex]b[\/latex]) of a triangle with an area ([latex]A[\/latex]) of [latex]20[\/latex] square feet and a height ([latex]h[\/latex]) of [latex]8[\/latex] feet.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q698967\">Show Solution<\/span><\/p>\n<div id=\"q698967\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the formula for the area of a triangle, [latex]{A}=\\frac{{1}}{{2}}{bh}[\/latex].<\/p>\n<p>Substitute the given lengths into the formula and solve for\u00a0[latex]b[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{l}\\,\\,A=\\frac{1}{2}bh\\\\\\\\20=\\frac{1}{2}b\\cdot 8\\\\\\\\20=\\frac{8}{2}b\\\\\\\\20=4b\\\\\\\\\\frac{20}{4}=\\frac{4b}{4}\\\\\\\\ \\,\\,\\,5=b\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The base of the triangle measures [latex]5[\/latex] feet.<\/p><\/div>\n<\/div>\n<\/div>\n<p>We can rewrite the formula in terms of [latex]b[\/latex] or [latex]h[\/latex] as we did with the perimeter formula previously.\u00a0This probably seems abstract, but it can help you develop your equation-solving skills, as well as help you get more comfortable with working with all kinds of variables, not just [latex]x[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>1. Use the multiplication and division properties of equality to isolate the variable [latex]b[\/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=\\frac{1}{2}bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\frac{1}{2}bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{h}=\\frac{bh}{h}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\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=\\frac{2A}{h}[\/latex]\n<\/div>\n<\/div>\n<p>2. Use the multiplication and division properties of equality to isolate the variable [latex]h[\/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=\\frac{1}{2}bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\frac{1}{2}bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{b}=\\frac{bh}{b}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\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=\\frac{2A}{b}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows another example of finding the base of a triangle, given area and height.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Find the Base of a Triangle Given Area \/ Literal Equation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/VQZQvJ3rXYg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\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, [latex]h[\/latex].<\/p>\n<p>In this example, the variable [latex]h[\/latex] is buried pretty deep in the formula for the 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 [latex]h[\/latex].<\/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, [latex]h[\/latex], 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 [latex]h[\/latex] by dividing both sides of the equation by [latex]2\\pi r[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}\\frac{S-2\\pi r^{2}}{2\\pi r}=\\frac{2\\pi rh}{2\\pi r} \\\\\\\\ \\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=\\frac{S-2\\pi r^{2}}{2\\pi r}[\/latex]<\/div>\n<\/div>\n<\/div>\n<h2><span style=\"line-height: 1.5\">Temperature<\/span><\/h2>\n<p>Let\u2019s 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 \\frac{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 \\frac{5}{9}[\/latex]<\/p>\n<p>Isolate the variable [latex]F[\/latex] to obtain the equivalent temperature.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}12=\\left(F-32\\right)\\cdot \\frac{5}{9}\\\\\\\\\\left(\\frac{9}{5}\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\left(\\frac{108}{5}\\right)=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 [latex]F[\/latex] 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 [latex]F[\/latex].<\/p>\n<p>[latex]C=\\left(F--32\\right)\\cdot \\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 [latex]F[\/latex], it would be best to clear the fraction involving [latex]F[\/latex] 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(\\frac{9}{5}\\right)C=\\left(F-32\\right)\\left(\\frac{5}{9}\\right)\\left(\\frac{9}{5}\\right)\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{9}{5}C=F-32\\end{array}[\/latex]<\/p>\n<p>Add [latex]32[\/latex] to both sides.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\frac{9}{5}\\,C+32=F-32+32\\\\\\\\\\frac{9}{5}\\,C+32=F\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]F=\\frac{9}{5}C+32[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>In the last video, we show how to convert from Celsius to Fahrenheit.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-3\" title=\"Convert Celsius to Fahrenheit \/ Literal Equation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/DRydX8V-JwY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm2379\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2379&theme=oea&iframe_resize_id=ohm2379&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Practice Isolating a Variable in a Formula<\/h2>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm196857\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=196857&theme=oea&iframe_resize_id=ohm196857&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-16114\">\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>Find the Width of a Rectangle Given the Perimeter \/ Literal Equation. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/jlxPgKQfhQs\">https:\/\/youtu.be\/jlxPgKQfhQs<\/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 Base of a Triangle Given Area \/ Literal Equation. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/VQZQvJ3rXYg\">https:\/\/youtu.be\/VQZQvJ3rXYg<\/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>Convert Celsius to Fahrenheit \/ Literal Equation. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/DRydX8V-JwY\">https:\/\/youtu.be\/DRydX8V-JwY<\/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>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>Unit 10: Solving Equations and Inequalities, 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\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/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":169554,"menu_order":23,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Find the Width of a Rectangle Given the Perimeter \/ Literal Equation\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/jlxPgKQfhQs\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Find the Base of a Triangle Given Area \/ Literal Equation\",\"author\":\"James Sousa 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