{"id":2036,"date":"2023-04-21T17:44:59","date_gmt":"2023-04-21T17:44:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=2036"},"modified":"2024-10-18T20:55:42","modified_gmt":"2024-10-18T20:55:42","slug":"exponential-quadratic-and-logarithmic-functions-background-youll-need-page-2","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/exponential-quadratic-and-logarithmic-functions-background-youll-need-page-2\/","title":{"raw":"Personal Finance - Common Scenarios: Background You'll Need - Page 2","rendered":"Personal Finance &#8211; Common Scenarios: Background You&#8217;ll Need &#8211; Page 2"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Solve complex equations&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:7041,&quot;3&quot;:{&quot;1&quot;:0},&quot;10&quot;:2,&quot;11&quot;:3,&quot;12&quot;:0,&quot;14&quot;:{&quot;1&quot;:2,&quot;2&quot;:0},&quot;15&quot;:&quot;Calibri&quot;}\">Solve complex equations<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Solving Complex Equations<\/h2>\r\n<p>You are likely to encounter equations that are more challenging than the basic linear ones. These complex equations can involve multiple variables, higher powers, or intricate combinations of terms. The key to mastering these equations lies in recognizing patterns and understanding how to manipulate the terms to reveal the solutions.<\/p>\r\n<section class=\"textbox questionHelp\">\r\n<p>When solving complex equations:<\/p>\r\n<ul>\r\n\t<li>Identify the structure of the equation and plan your approach.<\/li>\r\n\t<li>Look for ways to simplify the equation, such as combining like terms or factoring.<\/li>\r\n\t<li>Be methodical in your application of algebraic rules to avoid common pitfalls.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox proTip\">Our strategy will involve choosing one side of the equation to be the variable side, and the other side of the equation to be the constant side. This will help us with organization. Then, we will use the [pb_glossary id=\"13103\"]Subtraction and Addition Properties of Equality[\/pb_glossary], step by step, to isolate\u00a0the variable terms on one side of the equation.<\/section>\r\n<section class=\"textbox example\">Solve the following for [latex]x[\/latex]: <center>[latex]7x+5=6x+2[\/latex]<\/center>[reveal-answer q=\"683835\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"683835\"]Start by choosing which side will be the variable side and which side will be the constant side. The variable terms are [latex]7x[\/latex] and [latex]6x[\/latex]. Since [latex]7[\/latex] is greater than [latex]6[\/latex], make the left side the variable side and so the right side will be the constant side.\r\n\r\n\r\n<table id=\"eip-id1168468709344\" class=\"unnumbered unstyled\" summary=\"The first line says 7x plus 5 equals 6x plus 2. The next line says \">\r\n<tbody>\r\n<tr>\r\n<td>&nbsp;<\/td>\r\n<td>[latex]7x+5=6x+2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Collect the variable terms to the left side by subtracting [latex]6x[\/latex] from both sides.<\/td>\r\n<td style=\"width: 50%;\">[latex]7x(\\color{red}{-6x}\\color{black})+5=6x(\\color{red}{-6x}\\color{black})+2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]x+5=2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Now, collect the constants to the right side by subtracting [latex]5[\/latex] from both sides.<\/td>\r\n<td>[latex]x+5(\\color{red}{-5}\\color{black})=2(\\color{red}{-5}\\color{black})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]x=-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The solution is [latex]x=-3[\/latex] .<\/td>\r\n<td>&nbsp;<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td>\r\n<p>[latex]7x+5=6x+2[\/latex]<\/p>\r\n<p>Let [latex]x=-3[\/latex]<\/p>\r\n<p>[latex]7(\\color{red}{-3}\\color{black})+5\\stackrel{\\text{?}}{=}6(\\color{red}{-3}\\color{black})+2[\/latex]<\/p>\r\n<p>[latex]-21+5\\stackrel{\\text{?}}{=}-18+2[\/latex]<\/p>\r\n<p>[latex]16=16\\quad\\checkmark[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">Solve the following for [latex]n[\/latex]: <center>[latex]6n - 2=-3n+7[\/latex]<\/center>[reveal-answer q=\"73462\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"73462\"]<br \/>\r\nWe have [latex]6n[\/latex] on the left and [latex]-3n[\/latex] on the right. Since [latex]6&gt;-3[\/latex], make the left side the \"variable\" side.\r\n\r\n\r\n<table id=\"eip-id1168467335489\" class=\"unnumbered unstyled\" summary=\"The top line says 6n minus 2 equals negative 3n plus 7. The next line says \">\r\n<tbody>\r\n<tr>\r\n<td>&nbsp;<\/td>\r\n<td>[latex]6n-2=-3n+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>We don't want variables on the right side\u2014add [latex]3n[\/latex] to both sides to leave only constants on the right.<\/td>\r\n<td style=\"width: 50%;\">[latex]6n(\\color{red}{+3n}\\color{black})-2=-3n(\\color{red}{+3n}\\color{black})+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Combine like terms.<\/td>\r\n<td>[latex]9n-2=7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>We don't want any constants on the left side, so add [latex]2[\/latex] to both sides.<\/td>\r\n<td>[latex]9n-2(\\color{red}{+2}\\color{black})=7(\\color{red}{+2}\\color{black})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]9n=9[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The variable term is on the left and the constant term is on the right. To get the coefficient of [latex]n[\/latex] to be one, divide both sides by [latex]9[\/latex].<\/td>\r\n<td>[latex]\\Large\\frac{9n}{\\color{red}{9}\\color{black}}\\normalsize =\\Large\\frac{9}{\\color{red}{9}\\color{black}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]n=1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td>\r\n<p>[latex]6n-2=-3n+7[\/latex]<\/p>\r\n<p>Substitute [latex]1[\/latex] for [latex]n[\/latex]<\/p>\r\n<p>[latex]6(\\color{red}{1}\\color{black})-2\\stackrel{\\text{?}}{=}-3(\\color{red}{1}\\color{black})+7[\/latex]<\/p>\r\n<p>[latex]4=4\\quad\\checkmark[\/latex]<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]4483[\/ohm2_question]<\/section>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]4485[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Solve complex equations&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:7041,&quot;3&quot;:{&quot;1&quot;:0},&quot;10&quot;:2,&quot;11&quot;:3,&quot;12&quot;:0,&quot;14&quot;:{&quot;1&quot;:2,&quot;2&quot;:0},&quot;15&quot;:&quot;Calibri&quot;}\">Solve complex equations<\/span><\/li>\n<\/ul>\n<\/section>\n<h2>Solving Complex Equations<\/h2>\n<p>You are likely to encounter equations that are more challenging than the basic linear ones. These complex equations can involve multiple variables, higher powers, or intricate combinations of terms. The key to mastering these equations lies in recognizing patterns and understanding how to manipulate the terms to reveal the solutions.<\/p>\n<section class=\"textbox questionHelp\">\n<p>When solving complex equations:<\/p>\n<ul>\n<li>Identify the structure of the equation and plan your approach.<\/li>\n<li>Look for ways to simplify the equation, such as combining like terms or factoring.<\/li>\n<li>Be methodical in your application of algebraic rules to avoid common pitfalls.<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox proTip\">Our strategy will involve choosing one side of the equation to be the variable side, and the other side of the equation to be the constant side. This will help us with organization. Then, we will use the <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_2036_13103\">Subtraction and Addition Properties of Equality<\/a>, step by step, to isolate\u00a0the variable terms on one side of the equation.<\/section>\n<section class=\"textbox example\">Solve the following for [latex]x[\/latex]: <\/p>\n<div style=\"text-align: center;\">[latex]7x+5=6x+2[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q683835\">Show Answer<\/button><\/p>\n<div id=\"q683835\" class=\"hidden-answer\" style=\"display: none\">Start by choosing which side will be the variable side and which side will be the constant side. The variable terms are [latex]7x[\/latex] and [latex]6x[\/latex]. Since [latex]7[\/latex] is greater than [latex]6[\/latex], make the left side the variable side and so the right side will be the constant side.<\/p>\n<table id=\"eip-id1168468709344\" class=\"unnumbered unstyled\" summary=\"The first line says 7x plus 5 equals 6x plus 2. The next line says\">\n<tbody>\n<tr>\n<td>&nbsp;<\/td>\n<td>[latex]7x+5=6x+2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Collect the variable terms to the left side by subtracting [latex]6x[\/latex] from both sides.<\/td>\n<td style=\"width: 50%;\">[latex]7x(\\color{red}{-6x}\\color{black})+5=6x(\\color{red}{-6x}\\color{black})+2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]x+5=2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Now, collect the constants to the right side by subtracting [latex]5[\/latex] from both sides.<\/td>\n<td>[latex]x+5(\\color{red}{-5}\\color{black})=2(\\color{red}{-5}\\color{black})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]x=-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The solution is [latex]x=-3[\/latex] .<\/td>\n<td>&nbsp;<\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>\n[latex]7x+5=6x+2[\/latex]<\/p>\n<p>Let [latex]x=-3[\/latex]<\/p>\n<p>[latex]7(\\color{red}{-3}\\color{black})+5\\stackrel{\\text{?}}{=}6(\\color{red}{-3}\\color{black})+2[\/latex]<br \/>\n[latex]-21+5\\stackrel{\\text{?}}{=}-18+2[\/latex]<br \/>\n[latex]16=16\\quad\\checkmark[\/latex]\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Solve the following for [latex]n[\/latex]: <\/p>\n<div style=\"text-align: center;\">[latex]6n - 2=-3n+7[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q73462\">Show Answer<\/button><\/p>\n<div id=\"q73462\" class=\"hidden-answer\" style=\"display: none\">\nWe have [latex]6n[\/latex] on the left and [latex]-3n[\/latex] on the right. Since [latex]6>-3[\/latex], make the left side the &#8220;variable&#8221; side.<\/p>\n<table id=\"eip-id1168467335489\" class=\"unnumbered unstyled\" summary=\"The top line says 6n minus 2 equals negative 3n plus 7. The next line says\">\n<tbody>\n<tr>\n<td>&nbsp;<\/td>\n<td>[latex]6n-2=-3n+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>We don&#8217;t want variables on the right side\u2014add [latex]3n[\/latex] to both sides to leave only constants on the right.<\/td>\n<td style=\"width: 50%;\">[latex]6n(\\color{red}{+3n}\\color{black})-2=-3n(\\color{red}{+3n}\\color{black})+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Combine like terms.<\/td>\n<td>[latex]9n-2=7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>We don&#8217;t want any constants on the left side, so add [latex]2[\/latex] to both sides.<\/td>\n<td>[latex]9n-2(\\color{red}{+2}\\color{black})=7(\\color{red}{+2}\\color{black})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]9n=9[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The variable term is on the left and the constant term is on the right. To get the coefficient of [latex]n[\/latex] to be one, divide both sides by [latex]9[\/latex].<\/td>\n<td>[latex]\\Large\\frac{9n}{\\color{red}{9}\\color{black}}\\normalsize =\\Large\\frac{9}{\\color{red}{9}\\color{black}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]n=1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>\n[latex]6n-2=-3n+7[\/latex]<\/p>\n<p>Substitute [latex]1[\/latex] for [latex]n[\/latex]<\/p>\n<p>[latex]6(\\color{red}{1}\\color{black})-2\\stackrel{\\text{?}}{=}-3(\\color{red}{1}\\color{black})+7[\/latex]<br \/>\n[latex]4=4\\quad\\checkmark[\/latex]\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm4483\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=4483&theme=lumen&iframe_resize_id=ohm4483&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm4485\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=4485&theme=lumen&iframe_resize_id=ohm4485&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<div class=\"glossary\"><span class=\"screen-reader-text\" id=\"definition\">definition<\/span><template id=\"term_2036_13103\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_2036_13103\"><div tabindex=\"-1\"><p>Addition Property of Equality: This principle states that if you add the same number to both sides of an equation, the equation remains true.<\/p>\n<p>Subtraction Property of Equality: Similarly, this property tells us that if you subtract the same number from both sides of an equation, the equation continues to be true. <\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><\/div>","protected":false},"author":16,"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":4885,"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\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2036"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/users\/16"}],"version-history":[{"count":18,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2036\/revisions"}],"predecessor-version":[{"id":14000,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2036\/revisions\/14000"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/parts\/4885"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/2036\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=2036"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=2036"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=2036"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=2036"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}