{"id":1222,"date":"2024-05-07T01:42:40","date_gmt":"2024-05-07T01:42:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1222"},"modified":"2024-12-03T18:03:52","modified_gmt":"2024-12-03T18:03:52","slug":"module-4-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/module-4-background-youll-need-2\/","title":{"raw":"Non-Linear Equations: Background You'll Need 2","rendered":"Non-Linear Equations: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n \t<li><span data-sheets-root=\"1\">Rearrange formulas to solve for one variable<\/span><\/li>\r\n<\/ul>\r\n<\/section>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.\r\n\r\nExamples 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\r\nWe often need to rearrange formulas to isolate a particular variable. This process is crucial for solving equations and expressing relationships between variables in a more useful form.\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to Isolate a Variable:\r\n<\/strong>\r\n<ol>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Identify the target variable<\/strong>: Determine which variable you want to isolate.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Use inverse operations<\/strong>: Apply the opposite of each operation affecting your target variable, working from the outside in.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Perform the same operations on both sides<\/strong>: Remember, what you do to one side of the equation, you must do to the other to maintain equality.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Simplify<\/strong>: Combine like terms and simplify expressions as you go.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">Isolate 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=\"679301\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"679301\"]\r\n\r\n<center>[latex]\\begin{align*} \\text{Perimeter formula} &amp; : &amp; P &amp;= 2l + 2w &amp; \\\\ \\text{Subtract \\( 2l \\) from both sides} &amp; : &amp; P - 2l &amp;= 2w &amp; \\\\ \\text{Divide both sides by 2} &amp; : &amp; w &amp;= \\frac{P - 2l}{2} &amp; \\end{align*}[\/latex]<\/center>This results in [latex]w[\/latex], the width, being expressed solely in terms of the perimeter [latex]P[\/latex] and the length [latex]l[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">Isolate the variable for height, [latex]h[\/latex], from the formula for the surface area of a cylinder,<center>[latex]s=2\\pi rh+2\\pi r^{2}[\/latex]<\/center>[reveal-answer q=\"589392\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"589392\"]\r\n\r\n<center>[latex]\\begin{align*} \\text{Surface area formula} &amp; : &amp; S &amp;= 2\\pi rh + 2\\pi r^2 &amp; \\\\ \\text{Subtract the area of the bases} &amp; : &amp; S - 2\\pi r^2 &amp;= 2\\pi rh &amp; \\\\ \\text{Divide both sides by } 2\\pi r &amp; : &amp; h &amp;= \\frac{S - 2\\pi r^2}{2\\pi r} &amp; \\end{align*}[\/latex]<\/center>This equation now expresses [latex]h[\/latex], the height of the cylinder, in terms of the surface area [latex]S[\/latex] and the radius [latex]r[\/latex].[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18997[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18998[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li><span data-sheets-root=\"1\">Rearrange formulas to solve for one variable<\/span><\/li>\n<\/ul>\n<\/section>\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.<\/p>\n<p>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<p>We often need to rearrange formulas to isolate a particular variable. This process is crucial for solving equations and expressing relationships between variables in a more useful form.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to Isolate a Variable:<br \/>\n<\/strong><\/p>\n<ol>\n<li class=\"whitespace-normal break-words\"><strong>Identify the target variable<\/strong>: Determine which variable you want to isolate.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Use inverse operations<\/strong>: Apply the opposite of each operation affecting your target variable, working from the outside in.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Perform the same operations on both sides<\/strong>: Remember, what you do to one side of the equation, you must do to the other to maintain equality.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Simplify<\/strong>: Combine like terms and simplify expressions as you go.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q679301\">Show Answer<\/button><\/p>\n<div id=\"q679301\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\begin{align*} \\text{Perimeter formula} & : & P &= 2l + 2w & \\\\ \\text{Subtract \\( 2l \\) from both sides} & : & P - 2l &= 2w & \\\\ \\text{Divide both sides by 2} & : & w &= \\frac{P - 2l}{2} & \\end{align*}[\/latex]<\/div>\n<p>This results in [latex]w[\/latex], the width, being expressed solely in terms of the perimeter [latex]P[\/latex] and the length [latex]l[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Isolate the variable for height, [latex]h[\/latex], from the formula for the surface area of a cylinder,<\/p>\n<div style=\"text-align: center;\">[latex]s=2\\pi rh+2\\pi r^{2}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q589392\">Show Answer<\/button><\/p>\n<div id=\"q589392\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\begin{align*} \\text{Surface area formula} & : & S &= 2\\pi rh + 2\\pi r^2 & \\\\ \\text{Subtract the area of the bases} & : & S - 2\\pi r^2 &= 2\\pi rh & \\\\ \\text{Divide both sides by } 2\\pi r & : & h &= \\frac{S - 2\\pi r^2}{2\\pi r} & \\end{align*}[\/latex]<\/div>\n<p>This equation now expresses [latex]h[\/latex], the height of the cylinder, in terms of the surface area [latex]S[\/latex] and the radius [latex]r[\/latex].<\/p><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18997\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18997&theme=lumen&iframe_resize_id=ohm18997&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18998\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18998&theme=lumen&iframe_resize_id=ohm18998&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":92,"module-header":"background_you_need","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1222"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1222\/revisions"}],"predecessor-version":[{"id":6583,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1222\/revisions\/6583"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/92"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1222\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1222"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1222"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1222"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1222"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}