{"id":845,"date":"2021-09-20T17:36:18","date_gmt":"2021-09-20T17:36:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/?post_type=chapter&#038;p=845"},"modified":"2023-10-10T19:30:57","modified_gmt":"2023-10-10T19:30:57","slug":"4-1-3-solving-linear-equations-that-require-simplification","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/4-1-3-solving-linear-equations-that-require-simplification\/","title":{"raw":"4.1.2 Solving Multi-Step Linear Equations","rendered":"4.1.2 Solving Multi-Step Linear Equations"},"content":{"raw":"<nav style=\"background-color: #007fab;\" role=\"navigation\">\r\n<div class=\"book-title-wrapper\">\r\n<div class=\"bombadil-logo\"><a href=\"https:\/\/courses.lumenlearning.com\/\"><img src=\"https:\/\/courses.lumenlearning.com\/boundless-algebra\/wp-content\/themes\/bombadil\/assets\/images\/LumenOnDark-150x69.png\" alt=\"Lumen\" \/><\/a><\/div>\r\n<\/div>\r\n<\/nav>\r\n<h1 style=\"text-align: center;\">Solving Linear Equations that Require Simplification<\/h1>\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve equations that need to be simplified<\/li>\r\n \t<li>Solve a linear equation that requires multiple steps and a combination of the properties of equality<\/li>\r\n \t<li>Use the distributive property to solve equations containing parentheses<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Using the Properties of Equality to Solve Linear Equations<\/h2>\r\nSome linear equations can be solved in one or two steps, while others are more complicated. \u00a0Although multi-step equations take more time and more operations to solve, they can still be simplified and solved by applying basic algebraic rules.\u00a0 In this section, we will look at equations that require some additional steps before they can be solved.\r\n<h3>Combining Like Terms<\/h3>\r\nMany equations start out more complicated than the ones we\u2019ve just solved in the previous section. Let's work through some examples that will employ simplifying by combining like terms.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve: [latex]8x+9x - 5x=-3+15[\/latex]\r\n\r\nSolution:\r\n\r\nFirst, we need to simplify both sides of the equation as much as possible.\r\n\r\nStart by combining like terms to simplify each side.\r\n<table id=\"eip-id1168466098204\" class=\"unnumbered unstyled\" summary=\"The first line says \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]8x+9x-5x=-3+15[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Combine like terms.<\/td>\r\n<td>[latex]12x=12[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide both sides by 12 to isolate x.<\/td>\r\n<td>[latex]\\Large\\frac{12x}{\\color{red}{12}}\\normalsize =\\Large\\frac{12}{\\color{red}{12}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]x=1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check your answer. Let [latex]x=1[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]8x+9x-5x=-3+15[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]8\\cdot\\color{red}{1}+9\\cdot\\color{red}{1}-5\\cdot\\color{red}{1}\\stackrel{\\text{?}}{=}-3+15[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]8+9-5\\stackrel{\\text{?}}{=}-3+15[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]12=12\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve [latex]3x+5x+4-x+7=88[\/latex]\r\n\r\n[reveal-answer q=\"455516\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455516\"]\r\n\r\nThere are three \u201clike\u201d terms [latex]3x[\/latex], [latex]5x[\/latex],\u00a0and\u00a0[latex]\u2013x[\/latex]\u00a0involving a variable.\u00a0Combine these \u201clike\u201d terms. [latex]4[\/latex] and [latex]7[\/latex] are also \u201clike\u201d terms and can be added.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,3x+5x+4-x+7=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x+4+7=\\,\\,\\,88\\end{array}[\/latex]<\/p>\r\nThe equation is now in the form\u00a0[latex]ax+b=c[\/latex], so we can solve as before.\r\n<p style=\"text-align: center;\">[latex]7x+11\\,\\,\\,=\\,\\,\\,88[\/latex]<\/p>\r\nSubtract \u00a0[latex]11[\/latex] from both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}7x+11\\,\\,\\,=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-11\\,\\,\\,\\,\\,\\,\\,-11}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x\\,\\,\\,=\\,\\,\\,77\\end{array}[\/latex]<\/p>\r\nDivide both sides by [latex]7[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{7x}\\,\\,\\,=\\,\\,\\,\\underline{77}\\\\7\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,=\\,\\,\\,11\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=11[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try\u00a0it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141884&amp;theme=oea&amp;iframe_resize_id=mom2[\/embed]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nThe video shows an example of solving a linear equation that requires combining like terms.\r\n\r\nhttps:\/\/youtu.be\/ez_sP2OTGjU\r\n<h3><\/h3>\r\n<h3>Variables on the right side of the equation<\/h3>\r\nIt is not necessary to always have the variables on the left side of the equation. The Property of Reflection tells us that, if [latex]a=b[\/latex], then [latex]b=a[\/latex]. Let's consider an example with variables on the right side.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]11 - 20=17y - 8y - 6y[\/latex]\r\n<h4>Solution:<\/h4>\r\nSimplify each side by combining like terms.\r\n<table id=\"eip-id1168466111452\" class=\"unnumbered unstyled\" summary=\"The first line shows 11 minus 20 equals 17y minus 8y minus 6y. The next line says \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]11-20=17y-8y-6y[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify each side.<\/td>\r\n<td>[latex]-9=3y[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide both sides by 3 to isolate y.<\/td>\r\n<td>[latex]\\Large\\frac{-9}{\\color{red}{3}}\\normalsize =\\Large\\frac{3y}{\\color{red}{3}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]-3=y[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check your answer. Let [latex]y=-3[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]11-20=17y-8y-6y[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]11-20\\stackrel{\\text{?}}{=}17(\r\n\\color{red}{-3})-8(\\color{red}{-3})-6(\\color{red}{-3})[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]11-20\\stackrel{\\text{?}}{=}-51+24+18[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-9=-9\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nNotice that the variable ended up on the right side of the equals sign when we solved the equation. It is usually preferable to take one more step to write the solution with the variable on the left side of the equals sign.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try\u00a0it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141901&amp;theme=oea&amp;iframe_resize_id=mom23[\/embed]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/qe89pkRKzRw\r\n<h3><\/h3>\r\n<h3>Variables on both sides of the equation<\/h3>\r\nSo far we have solved equations with variables on only one\u00a0side of the equation. This does not happen all the time\u2014so now we\u2019ll see how to solve equations where there are variable terms on both sides of the equation, as in this equation: [latex]4x-6=2x+10[\/latex]. We will start by choosing a variable side and a constant side, and then use the Subtraction and Addition Properties of Equality to collect all variables on one side and all constants on the other side. Remember, what we do to the left side of the equation, we must do to the right side as well.\r\n\r\nTo solve [latex]4x-6=2x+10[\/latex], we need to \u201cmove\u201d one of the variable terms.\u00a0This can make it difficult to decide which side to work with. It doesn\u2019t matter which term gets moved, [latex]4x[\/latex] or [latex]2x[\/latex], however, to avoid negative coefficients, we can move the smaller term.\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nSolve:\u00a0[latex]4x-6=2x+10[\/latex]\r\n<h4>Solution<\/h4>\r\nChoose the variable term to \"move\"\u2014to avoid negative terms subtract [latex]2x[\/latex] from both sides:\r\n<p style=\"text-align: center;\">[latex]\\,\\,\\,4x-6=2x+10\\\\\\underline{\\color{blue}{-2x}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\color{blue}{-2x}}\\\\\\,\\,\\,2x-6=10[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Add\u00a06 to both\u00a0sides to isolate the variable term:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x-6=10\\\\\\underline{\\,\\,\\,\\,\\color{blue}{+6}\\,\\,\\,\\color{blue}{+6}}\\\\2x=16\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Divide each side by [latex]2[\/latex] to isolate the variable [latex]x[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{2x}{\\color{blue}{2}}=\\frac{16}{\\color{blue}{2}}\\\\\\\\x=8\\end{array}[\/latex]<\/p>\r\nCheck the solution:\u00a0[latex]4x-6=2x+10 \\\\ 4\\color{blue}{(8)}-6 = 2\\color{blue}{(8)}+10 \\\\ 32-6=16+10 \\\\ 26=26[\/latex] True\r\n<h4>Answer<\/h4>\r\n[latex]x=8[\/latex]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nThis video shows an example of solving equations that have variables on both sides of the equals sign.\r\n\r\nhttps:\/\/youtu.be\/f3ujWNPL0Bw\r\n\r\n&nbsp;\r\n\r\nIn the next example,\u00a0the variable, [latex]x[\/latex], is on both sides, but the constants appear only on the right side, so we'll make the right side the \"constant\" side. Then the left side will be the \"variable\" side.\r\n<div class=\"textbox exercises\">\r\n<h3>ExampleS<\/h3>\r\nSolve:\r\n\r\n1. [latex]5x=4x+7[\/latex]\r\n\r\n2.\u00a0[latex]7x=-x+24[\/latex]\r\n<h4>Solution 1:<\/h4>\r\n<table id=\"eip-id1168465095460\" class=\"unnumbered unstyled\" summary=\"The first line says 5x equals 4x plus 7. The left side is labeled \">\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\">[latex]5x[\/latex] is the side containing only a <span style=\"color: #000000;\"><span style=\"color: #ff0000;\">variable<\/span>.<\/span><span style=\"color: #000000;\">[latex]4x+7[\/latex] is the side containing a <span style=\"color: #ff0000;\">constant<\/span>.<\/span><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">We don't want any variables on the right, so subtract the [latex]4x[\/latex] .<\/td>\r\n<td>[latex]5x\\color{red}{-4x}=4x\\color{red}{-4x}+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Simplify.<\/td>\r\n<td>[latex]x=7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">We have all the variables on one side and the constants on the other. We have solved the equation.<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td><\/td>\r\n<td>[latex]5x=4x+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]7[\/latex] for [latex]x[\/latex] .<\/td>\r\n<td><\/td>\r\n<td>[latex]5(\\color{red}{7})\\stackrel{\\text{?}}{=}4(\\color{red}{7})+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td><\/td>\r\n<td>[latex]35\\stackrel{\\text{?}}{=}28+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td><\/td>\r\n<td>[latex]35=35\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n<h4>Solution 2:<\/h4>\r\nThe only constant, [latex]24[\/latex], is on the right, so let the left side be the variable side.\r\n<table id=\"eip-id1168467377760\" class=\"unnumbered unstyled\" summary=\"The first line says 7x equals negative x plus 24. The left side is labeled \">\r\n<tbody>\r\n<tr>\r\n<td>[latex]7x[\/latex] is the side containing only a <span style=\"color: #000000;\"><span style=\"color: #ff0000;\">variable<\/span>.<\/span><span style=\"color: #000000;\">[latex]-x+24[\/latex] is the side containing a <span style=\"color: #ff0000;\">constant<\/span>.<\/span><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Remove the [latex]-x[\/latex] from the right side by adding [latex]x[\/latex] to both sides.<\/td>\r\n<td>[latex]7x\\color{red}{+x}=-x\\color{red}{+x}+24[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]8x=24[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>All the variables are on the left and the constants are on the right. Divide both sides by [latex]8[\/latex].<\/td>\r\n<td>[latex]\\Large\\frac{8x}{\\color{red}{8}}\\normalsize =\\Large\\frac{24}{\\color{red}{8}}[\/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>Check:<\/td>\r\n<td>[latex]7x=-x+24[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]x=3[\/latex].<\/td>\r\n<td>[latex]7(\\color{red}{3})\\stackrel{\\text{?}}{=}-(\\color{red}{3})+24[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]21=21\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try\u00a0it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142129&amp;theme=oea&amp;iframe_resize_id=mom3[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142132&amp;theme=oea&amp;iframe_resize_id=mom4[\/embed]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nSometimes it is beneficial to have the variable term on the right side of the equation. There is no \"correct\" side to have the variable term, but the choice can help to avoid working with negative signs.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]5y - 8=7y[\/latex]\r\n<h4>Solution:<\/h4>\r\nThe only constant, [latex]-8[\/latex], is on the left side of the equation, and the variable, [latex]y[\/latex], is on both sides. Let\u2019s leave the constant on the left and collect the variables to the right.\r\n<table id=\"eip-id1168468768462\" class=\"unnumbered unstyled\" summary=\"The first line says 5y minus 8 equals 7y. The left side is labeled \">\r\n<tbody>\r\n<tr>\r\n<td>[latex]5y-8[\/latex] is the side containing a <span style=\"color: #000000;\"><span style=\"color: #ff0000;\">constant<\/span>.<\/span><span style=\"color: #000000;\">[latex]7y[\/latex] is the side containing only a <\/span><span style=\"color: #ff0000;\">variable<\/span><span style=\"color: #000000;\">.<\/span><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract [latex]5y[\/latex] from both sides.<\/td>\r\n<td>[latex]5y\\color{red}{-5y}-8=7y\\color{red}{-5y}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]-8=2y[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>We have the variables on the right and the constants on the left. Divide both sides by [latex]2[\/latex].<\/td>\r\n<td>[latex]\\Large\\frac{-8}{\\color{red}{2}}\\normalsize =\\Large\\frac{2y}{\\color{red}{2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]-4=y[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite with the variable on the left.<\/td>\r\n<td>[latex]y=-4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td>[latex]5y-8=7y[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Let [latex]y=-4[\/latex].<\/td>\r\n<td>[latex]5(\\color{red}{-4})-8\\stackrel{\\text{?}}{=}7(\\color{red}{-4})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]-20-8\\stackrel{\\text{?}}{=}-28[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]-28=-28\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try\u00a0it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142125&amp;theme=oea&amp;iframe_resize_id=mom2[\/embed]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<h3>Variables and Constants on Both Sides<\/h3>\r\nThe next example will be the first to have variables <em>and<\/em> constants on both sides of the equation. As we did before, we\u2019ll collect the variable terms to one side and the constants to the other side. As the number of variable and constant terms increase, so do the number of steps it takes to solve the equation.\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nSolve:\r\n\r\n1. [latex]7x+5=6x+2[\/latex]\r\n\r\n2.\u00a0[latex]6n - 2=-3n+7[\/latex]\r\n<h4>1. Solution:<\/h4>\r\nStart 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<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><\/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>[latex]7x\\color{red}{-6x}+5=6x\\color{red}{-6x}+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}=2\\color{red}{-5}[\/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><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td>[latex]7x+5=6x+2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Let [latex]x=-3[\/latex].<\/td>\r\n<td>[latex]7(\\color{red}{-3})+5\\stackrel{\\text{?}}{=}6(\\color{red}{-3})+2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]-21+5\\stackrel{\\text{?}}{=}-18+2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]16=16\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAnswer: [latex]x=3[\/latex]\r\n<h4>2. Solution<\/h4>\r\nSolve: [latex]6n - 2=-3n+7[\/latex]\r\n\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<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><\/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>[latex]6n\\color{red}{+3n}-2=-3n\\color{red}{+3n}+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}=7\\color{red}{+2}[\/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}}\\normalsize =\\Large\\frac{9}{\\color{red}{9}}[\/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>[latex]6n-2=-3n+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]1[\/latex] for [latex]n[\/latex].<\/td>\r\n<td>[latex]6(\\color{red}{1})-2\\stackrel{\\text{?}}{=}-3(\\color{red}{1})+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]4=4\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAnswer: [latex]n=1[\/latex]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nThe following video shows an example of how to solve a multi-step equation. It doesn't matter which side we choose to be the variable side; we can get the correct answer either way.\r\n\r\nhttps:\/\/youtu.be\/_hBoWoctfAo\r\n\r\n&nbsp;\r\n\r\nIn the next example, we move the variable terms to the right side, to keep a positive coefficient on the variable.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nSolve: [latex]2a - 7=5a+8[\/latex]\r\n\r\n&nbsp;\r\n<h4>Solution:<\/h4>\r\nThis equation has [latex]2a[\/latex] on the left and [latex]5a[\/latex] on the right. Since [latex]5&gt;2[\/latex], make the right side the variable side and the left side the constant side.\r\n<table id=\"eip-id1168466004451\" 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><\/td>\r\n<td>[latex]2a-7=5a+8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract [latex]2a[\/latex] from both sides to remove the variable term from the left.<\/td>\r\n<td>[latex]2a\\color{red}{-2a}-7=5a\\color{red}{-2a}+8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Combine like terms.<\/td>\r\n<td>[latex]-7=3a+8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract [latex]8[\/latex] from both sides to remove the constant from the right.<\/td>\r\n<td>[latex]-7\\color{red}{-8}=3a+8\\color{red}{-8}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]-15=3a[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide both sides by [latex]3[\/latex] to make [latex]1[\/latex] the coefficient of [latex]a[\/latex] .<\/td>\r\n<td>[latex]\\Large\\frac{-15}{\\color{red}{3}}\\normalsize =\\Large\\frac{3a}{\\color{red}{3}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]-5=a[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td>[latex]2a-7=5a+8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Let [latex]a=-5[\/latex]<\/td>\r\n<td>[latex]2(\\color{red}{-5})-7\\stackrel{\\text{?}}{=}5(\\color{red}{-5})+8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]-10-7\\stackrel{\\text{?}}{=}-25+8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]-17=-17\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n&nbsp;\r\n\r\nThe following video shows another example of solving a multi-step\u00a0equation.\r\n\r\nhttps:\/\/youtu.be\/kiYPW6hrTS4\r\n\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try\u00a0it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142134&amp;theme=oea&amp;iframe_resize_id=mom20[\/embed]\r\n\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142136&amp;theme=oea&amp;iframe_resize_id=mom200[\/embed]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nWe just showed a lot of examples of different kinds of linear equations that may be encountered. There are some good habits to develop that will help us solve all kinds of linear equations. We\u2019ll summarize the steps we took for easy reference.\r\n<div class=\"textbox shaded\">\r\n<h3>Solve an equation with variables and constants on both sides<\/h3>\r\n<ol id=\"eip-id1168468371331\" class=\"stepwise\">\r\n \t<li>Choose one side to be the variable side, and the other to be the constant side.<\/li>\r\n \t<li>Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.<\/li>\r\n \t<li>Collect the constants to the other side, using the Addition or Subtraction Property of Equality.<\/li>\r\n \t<li>Make the coefficient of the variable [latex]1[\/latex], using the Multiplication or Division Property of Equality.<\/li>\r\n \t<li>Check the solution by substituting it into the original equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2>No Solution and Infinite Solutions<\/h2>\r\nEquations that have a solution are called\u00a0<em><strong>conditional equations<\/strong><\/em><strong>.\u00a0<\/strong>The truth of the equation is found only when the solution is entered in the equation. For example, [latex]2x-5=3[\/latex] is true only when the solution [latex]x=4[\/latex] is put into the equation. Any other value of [latex]x[\/latex] makes the equation false. The truth of the equation is conditional upon the value of the variable.\r\n\r\nIt is possible for a linear equation in one variable to have an infinite number of solutions because it is always true. Equations that are always true are called\u00a0<strong><em>identities<\/em><\/strong><i>.\u00a0<\/i>For example the equation [latex]3x+5x=8x[\/latex] is always true, no matter the value of [latex]x[\/latex]. Consequently, the equation has an infinite number of solutions. All real numbers are solutions of this equation.\r\n\r\nIt is also possible for a linear equation to have no solution. Such equations are called\u00a0<strong><i>contradictions<\/i>. <\/strong>For example, [latex]3x+5=3x-2[\/latex] simplifies to [latex]5=-2[\/latex], which is a contradiction. Consequently, this equation has no solution.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nSolve the equation:\r\n\r\n1. [latex]4x-3=2x+5+2x[\/latex]\r\n\r\n2. [latex]5x-4-3x=2x-4[\/latex]\r\n\r\nSolution:\r\n<table id=\"eip-id1168466426761\" class=\"unnumbered unstyled\" summary=\"The top line says 4y plus 3 equals 8y. Beside this is \">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 244px;\">1.<\/td>\r\n<td style=\"width: 400px;\">[latex]4x-3=2x+5+2x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 244px;\">Simplify the right side by adding like terms<\/td>\r\n<td style=\"width: 400px;\">\u00a0[latex]4x-3=4x+5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 244px;\">Subtract [latex]4x[\/latex] from both sides<\/td>\r\n<td style=\"width: 400px;\">\u00a0[latex]-3=5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 244px;\">Contradiction<\/td>\r\n<td style=\"width: 400px;\">No solution<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table id=\"eip-id1168466426761\" class=\"unnumbered unstyled\" summary=\"The top line says 4y plus 3 equals 8y. Beside this is \">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 244px;\">2.<\/td>\r\n<td style=\"width: 400px;\">[latex]5x-4-3x=2x-4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 244px;\">Simplify the right side by adding like terms<\/td>\r\n<td style=\"width: 400px;\">\u00a0[latex]2x-4=2x-4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 244px;\">Subtract [latex]4x[\/latex] from both sides<\/td>\r\n<td style=\"width: 400px;\">\u00a0[latex]-4=-4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 244px;\">Identity<\/td>\r\n<td style=\"width: 400px;\">Solution is all real numbers<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\nSolve the equation.\r\n\r\n1. [latex]5x+2=5x[\/latex]\r\n\r\n2. [latex]3x-4+9=5x+3-2x+2[\/latex]\r\n\r\n[reveal-answer q=\"H00067\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"H00067\"]\r\n\r\n1. Contradiction. There is no solution.\r\n\r\n2.\u00a0Identity.\u00a0Solution is all real numbers.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWhen solving a linear equation in one variable, if we end up with a statement that is always true, we have an identity. A identity is true no matter the value of the variable, so the solution is the set of all real numbers. On the other hand, if we end up with an statement that is always false, we have a contradiction. A contradiction is never true, so the equation has no solution.\r\n<h2 id=\"title2\">The Distributive Property<\/h2>\r\nAs we solve linear equations, we often need to do some work to write\u00a0the linear equations in a form we are familiar with solving.\u00a0This section will focus on manipulating an equation we are asked to solve in such a way that we can use the skills we learned for solving multi-step equations to ultimately arrive at the solution.\r\n\r\nThe distributive property can come in handy to simplify an equation. Using this property, we multiply the number in front of the parentheses by each term inside of the parentheses.\r\n<div class=\"textbox shaded\">\r\n<h3>The Distributive Property of Multiplication over addition<\/h3>\r\nFor all real numbers [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex],\u00a0[latex]a(b+c)=ab+ac[\/latex].\r\n\r\nWhen a number multiplies an expression inside parentheses, we can distribute the multiplication to each term of the expression individually.\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for [latex]a[\/latex].\r\n\r\n[latex]4\\left(2a+3\\right)=28[\/latex]\r\n<h4>Solution:<\/h4>\r\nApply the distributive property to expand [latex]4\\left(2a+3\\right)[\/latex] to [latex]8a+12[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4\\left(2a+3\\right)=28\\\\ 8a+12=28\\end{array}[\/latex]<\/p>\r\nSubtract [latex]12[\/latex]\u00a0from both sides to isolate\u00a0the variable term.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8a+12\\,\\,\\,=\\,\\,\\,28\\\\ \\underline{-12\\,\\,\\,\\,\\,\\,-12}\\\\ 8a\\,\\,\\,=\\,\\,\\,16\\end{array}[\/latex]<\/p>\r\nDivide both terms by [latex]8[\/latex] to get a coefficient of [latex]1[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\underline{8a}=\\underline{16}\\\\8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,8\\\\a\\,=\\,\\,2\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]a=2[\/latex]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn our next example, we will use the distributive property of multiplication over addition first, simplify, then use the division property to finally solve.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]-3\\left(n - 2\\right)-6=21[\/latex]\r\n\r\n&nbsp;\r\n\r\nSolution:\r\n<table id=\"eip-id1168468278405\" class=\"unnumbered unstyled\" summary=\"The first line shows negative 3 times parentheses n minus 2 minus 6 equals 21. The next line says \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]-3(n-2)-6=21[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute.<\/td>\r\n<td>[latex]-3n+6-6=21[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]-3n=21[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide both sides by -3 to isolate n.<\/td>\r\n<td>[latex]\\Large\\frac{-3n}{\\color{red}{-3}}\\normalsize =\\Large\\frac{21}{\\color{red}{-3}}[\/latex][latex]n=-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check your answer. Let [latex]n=-7[\/latex] .<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3(n-2)-6=21[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3(\\color{red}{-7}-2)-6\\stackrel{\\text{?}}{=}21[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3(-9)-6\\stackrel{\\text{?}}{=}21[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]27-6\\stackrel{\\text{?}}{=}21[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]21=21\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141911&amp;theme=oea&amp;iframe_resize_id=mom27[\/embed]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nThe video that follows shows another example of how to use the distributive property to solve a multi-step linear equation.\r\n\r\nhttps:\/\/youtu.be\/aQOkD8L57V0\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]3\\left(n - 4\\right)-2n=-3[\/latex]\r\n\r\nSolution:\r\nThe left side of the equation has an expression that we should simplify.\r\n<table id=\"eip-id1168468254328\" class=\"unnumbered unstyled\" summary=\"The top line says \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]3(n-4)-2n=-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute on the left.<\/td>\r\n<td>[latex]3n-12-2n=-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the Commutative Property to rearrange terms.<\/td>\r\n<td>[latex]3n-2n-12=-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Combine like terms.<\/td>\r\n<td>[latex]n-12=-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Isolate <em>n<\/em> using the Addition Property of Equality.<\/td>\r\n<td>[latex]n-12\\color{red}{+12}=-3\\color{red}{+12}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]n=9[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check.Substitute [latex]n=9[\/latex] into the original equation.\r\n[latex]3(n-4)-2n=-3[\/latex]\r\n[latex]3(\\color{red}{9}-4)-2\\cdot\\color{red}{9}=-3[\/latex]\r\n[latex]3(5)-18=-3[\/latex]\r\n[latex]15-18=-3[\/latex]\r\n[latex]-3=-3\\quad\\checkmark[\/latex]\r\nThe solution checks.<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY\u00a0IT<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141737&amp;theme=oea&amp;iframe_resize_id=mom22[\/embed]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\nThe next example has expressions on both sides that need to be simplified.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]2\\left(3k - 1\\right)-5k=-2 - 7[\/latex]\r\n\r\nSolution:\r\nBoth sides of the equation have expressions that we should simplify before we isolate the variable.\r\n<table id=\"eip-id1168469785088\" class=\"unnumbered unstyled\" summary=\"The top line says 2 times parentheses 3k minus 1 minus 5k equals negative 2 minus 7. The next line says \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]2(3k-1)-5k=-2-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute on the left, subtract on the right.<\/td>\r\n<td>[latex]6k-2-5k=-9[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the Commutative Property of Addition.<\/td>\r\n<td>[latex]6k-5k-2=-9[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Combine like terms.<\/td>\r\n<td>[latex]k-2=-9[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Undo subtraction by using the Addition Property of Equality.<\/td>\r\n<td>[latex]k-2\\color{red}{+2}=-9\\color{red}{+2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]k=-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check.Let [latex]k=-7[\/latex].\r\n[latex]2(3k-1)-5k=-2-7[\/latex]\r\n[latex]2(3(\\color{red}{-7}-1)-5(\\color{red}{-7})=-2-7[\/latex]\r\n[latex]2(-21-1)-5(-7)=-9[\/latex]\r\n[latex]2(-22)+35=-9[\/latex]\r\n[latex]-44+35=-9[\/latex]\r\n[latex]-9=-9\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe solution checks.\r\n\r\nAnswer: [latex]k=-7[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY\u00a0IT<\/h3>\r\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141739&amp;theme=oea&amp;iframe_resize_id=mom220[\/embed]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nThe following video presents another example of how to solve an equation that requires simplifying before using the addition and subtraction properties.\r\n\r\nhttps:\/\/youtu.be\/shGKzDBA5kQ\r\n<h3><\/h3>\r\nIn the next example, there are parentheses on both sides of the equals sign, so we need to use the distributive property twice.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for [latex]t[\/latex].\r\n\r\n[latex]2\\left(4t-5\\right)=-3\\left(2t+1\\right)[\/latex]\r\n\r\nSolution:\r\n\r\nApply the distributive property to expand [latex]2\\left(4t-5\\right)[\/latex] to [latex]8t-10[\/latex] and [latex]-3\\left(2t+1\\right)[\/latex] to [latex]-6t-3[\/latex]. Be careful in this step\u2014you are distributing a negative number, so keep track of the sign of each number after you multiply.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(4t-5\\right)=-3\\left(2t+1\\right)\\,\\,\\,\\,\\,\\, \\\\ 8t-10=-6t-3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nAdd [latex]-6t[\/latex] to both sides to begin combining like terms.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8t-10=-6t-3\\\\ \\underline{+6t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+6t}\\,\\,\\,\\,\\,\\,\\,\\\\ 14t-10=\\,\\,\\,\\,-3\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nAdd [latex]10[\/latex] to both sides of the equation to isolate [latex]t[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}14t-10=-3\\\\ \\underline{+10\\,\\,\\,+10}\\\\ 14t=\\,\\,\\,7\\,\\end{array}[\/latex]<\/p>\r\nThe last step is to divide both sides by [latex]14[\/latex] to completely isolate [latex]t[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}14t=7\\,\\,\\,\\,\\\\\\frac{14t}{14}=\\frac{7}{14}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]t=\\frac{1}{2}[\/latex]\r\n\r\nWe simplified the fraction [latex]\\frac{7}{14}[\/latex] into [latex]\\frac{1}{2}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]39431[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nThe following video solves another multi-step equation with two sets of parentheses.\r\n\r\nhttps:\/\/youtu.be\/StomYTb7Xb8\r\n<h2><\/h2>","rendered":"<nav style=\"background-color: #007fab;\" role=\"navigation\">\n<div class=\"book-title-wrapper\">\n<div class=\"bombadil-logo\"><a href=\"https:\/\/courses.lumenlearning.com\/\"><img decoding=\"async\" src=\"https:\/\/courses.lumenlearning.com\/boundless-algebra\/wp-content\/themes\/bombadil\/assets\/images\/LumenOnDark-150x69.png\" alt=\"Lumen\" \/><\/a><\/div>\n<\/div>\n<\/nav>\n<h1 style=\"text-align: center;\">Solving Linear Equations that Require Simplification<\/h1>\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve equations that need to be simplified<\/li>\n<li>Solve a linear equation that requires multiple steps and a combination of the properties of equality<\/li>\n<li>Use the distributive property to solve equations containing parentheses<\/li>\n<\/ul>\n<\/div>\n<h2>Using the Properties of Equality to Solve Linear Equations<\/h2>\n<p>Some linear equations can be solved in one or two steps, while others are more complicated. \u00a0Although multi-step equations take more time and more operations to solve, they can still be simplified and solved by applying basic algebraic rules.\u00a0 In this section, we will look at equations that require some additional steps before they can be solved.<\/p>\n<h3>Combining Like Terms<\/h3>\n<p>Many equations start out more complicated than the ones we\u2019ve just solved in the previous section. Let&#8217;s work through some examples that will employ simplifying by combining like terms.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve: [latex]8x+9x - 5x=-3+15[\/latex]<\/p>\n<p>Solution:<\/p>\n<p>First, we need to simplify both sides of the equation as much as possible.<\/p>\n<p>Start by combining like terms to simplify each side.<\/p>\n<table id=\"eip-id1168466098204\" class=\"unnumbered unstyled\" summary=\"The first line says\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]8x+9x-5x=-3+15[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Combine like terms.<\/td>\n<td>[latex]12x=12[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide both sides by 12 to isolate x.<\/td>\n<td>[latex]\\Large\\frac{12x}{\\color{red}{12}}\\normalsize =\\Large\\frac{12}{\\color{red}{12}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]x=1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check your answer. Let [latex]x=1[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]8x+9x-5x=-3+15[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]8\\cdot\\color{red}{1}+9\\cdot\\color{red}{1}-5\\cdot\\color{red}{1}\\stackrel{\\text{?}}{=}-3+15[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]8+9-5\\stackrel{\\text{?}}{=}-3+15[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]12=12\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve [latex]3x+5x+4-x+7=88[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q455516\">Show Solution<\/span><\/p>\n<div id=\"q455516\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are three \u201clike\u201d terms [latex]3x[\/latex], [latex]5x[\/latex],\u00a0and\u00a0[latex]\u2013x[\/latex]\u00a0involving a variable.\u00a0Combine these \u201clike\u201d terms. [latex]4[\/latex] and [latex]7[\/latex] are also \u201clike\u201d terms and can be added.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,3x+5x+4-x+7=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x+4+7=\\,\\,\\,88\\end{array}[\/latex]<\/p>\n<p>The equation is now in the form\u00a0[latex]ax+b=c[\/latex], so we can solve as before.<\/p>\n<p style=\"text-align: center;\">[latex]7x+11\\,\\,\\,=\\,\\,\\,88[\/latex]<\/p>\n<p>Subtract \u00a0[latex]11[\/latex] from both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}7x+11\\,\\,\\,=\\,\\,\\,88\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-11\\,\\,\\,\\,\\,\\,\\,-11}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7x\\,\\,\\,=\\,\\,\\,77\\end{array}[\/latex]<\/p>\n<p>Divide both sides by [latex]7[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{7x}\\,\\,\\,=\\,\\,\\,\\underline{77}\\\\7\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,7\\,\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x\\,\\,\\,=\\,\\,\\,11\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=11[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try\u00a0it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm141884\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141884&#38;theme=oea&#38;iframe_resize_id=ohm141884&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The video shows an example of solving a linear equation that requires combining like terms.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Solving an Equation that Requires Combining Like Terms\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ez_sP2OTGjU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3><\/h3>\n<h3>Variables on the right side of the equation<\/h3>\n<p>It is not necessary to always have the variables on the left side of the equation. The Property of Reflection tells us that, if [latex]a=b[\/latex], then [latex]b=a[\/latex]. Let&#8217;s consider an example with variables on the right side.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]11 - 20=17y - 8y - 6y[\/latex]<\/p>\n<h4>Solution:<\/h4>\n<p>Simplify each side by combining like terms.<\/p>\n<table id=\"eip-id1168466111452\" class=\"unnumbered unstyled\" summary=\"The first line shows 11 minus 20 equals 17y minus 8y minus 6y. The next line says\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]11-20=17y-8y-6y[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify each side.<\/td>\n<td>[latex]-9=3y[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide both sides by 3 to isolate y.<\/td>\n<td>[latex]\\Large\\frac{-9}{\\color{red}{3}}\\normalsize =\\Large\\frac{3y}{\\color{red}{3}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]-3=y[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check your answer. Let [latex]y=-3[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]11-20=17y-8y-6y[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]11-20\\stackrel{\\text{?}}{=}17(  \\color{red}{-3})-8(\\color{red}{-3})-6(\\color{red}{-3})[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]11-20\\stackrel{\\text{?}}{=}-51+24+18[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]-9=-9\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>Notice that the variable ended up on the right side of the equals sign when we solved the equation. It is usually preferable to take one more step to write the solution with the variable on the left side of the equals sign.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try\u00a0it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm141901\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141901&#38;theme=oea&#38;iframe_resize_id=ohm141901&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Solve Linear Equations in One Variable with Simplifying (One-Step Mult\/Div)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/qe89pkRKzRw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3><\/h3>\n<h3>Variables on both sides of the equation<\/h3>\n<p>So far we have solved equations with variables on only one\u00a0side of the equation. This does not happen all the time\u2014so now we\u2019ll see how to solve equations where there are variable terms on both sides of the equation, as in this equation: [latex]4x-6=2x+10[\/latex]. We will start by choosing a variable side and a constant side, and then use the Subtraction and Addition Properties of Equality to collect all variables on one side and all constants on the other side. Remember, what we do to the left side of the equation, we must do to the right side as well.<\/p>\n<p>To solve [latex]4x-6=2x+10[\/latex], we need to \u201cmove\u201d one of the variable terms.\u00a0This can make it difficult to decide which side to work with. It doesn\u2019t matter which term gets moved, [latex]4x[\/latex] or [latex]2x[\/latex], however, to avoid negative coefficients, we can move the smaller term.<\/p>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>Solve:\u00a0[latex]4x-6=2x+10[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>Choose the variable term to &#8220;move&#8221;\u2014to avoid negative terms subtract [latex]2x[\/latex] from both sides:<\/p>\n<p style=\"text-align: center;\">[latex]\\,\\,\\,4x-6=2x+10\\\\\\underline{\\color{blue}{-2x}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\color{blue}{-2x}}\\\\\\,\\,\\,2x-6=10[\/latex]<\/p>\n<p style=\"text-align: left;\">Add\u00a06 to both\u00a0sides to isolate the variable term:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x-6=10\\\\\\underline{\\,\\,\\,\\,\\color{blue}{+6}\\,\\,\\,\\color{blue}{+6}}\\\\2x=16\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Divide each side by [latex]2[\/latex] to isolate the variable [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{2x}{\\color{blue}{2}}=\\frac{16}{\\color{blue}{2}}\\\\\\\\x=8\\end{array}[\/latex]<\/p>\n<p>Check the solution:\u00a0[latex]4x-6=2x+10 \\\\ 4\\color{blue}{(8)}-6 = 2\\color{blue}{(8)}+10 \\\\ 32-6=16+10 \\\\ 26=26[\/latex] True<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=8[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>This video shows an example of solving equations that have variables on both sides of the equals sign.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Solve an Equation with Variable on Both Sides\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/f3ujWNPL0Bw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p>In the next example,\u00a0the variable, [latex]x[\/latex], is on both sides, but the constants appear only on the right side, so we&#8217;ll make the right side the &#8220;constant&#8221; side. Then the left side will be the &#8220;variable&#8221; side.<\/p>\n<div class=\"textbox exercises\">\n<h3>ExampleS<\/h3>\n<p>Solve:<\/p>\n<p>1. [latex]5x=4x+7[\/latex]<\/p>\n<p>2.\u00a0[latex]7x=-x+24[\/latex]<\/p>\n<h4>Solution 1:<\/h4>\n<table id=\"eip-id1168465095460\" class=\"unnumbered unstyled\" summary=\"The first line says 5x equals 4x plus 7. The left side is labeled\">\n<tbody>\n<tr>\n<td colspan=\"2\">[latex]5x[\/latex] is the side containing only a <span style=\"color: #000000;\"><span style=\"color: #ff0000;\">variable<\/span>.<\/span><span style=\"color: #000000;\">[latex]4x+7[\/latex] is the side containing a <span style=\"color: #ff0000;\">constant<\/span>.<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">We don&#8217;t want any variables on the right, so subtract the [latex]4x[\/latex] .<\/td>\n<td>[latex]5x\\color{red}{-4x}=4x\\color{red}{-4x}+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Simplify.<\/td>\n<td>[latex]x=7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">We have all the variables on one side and the constants on the other. We have solved the equation.<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td><\/td>\n<td>[latex]5x=4x+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]7[\/latex] for [latex]x[\/latex] .<\/td>\n<td><\/td>\n<td>[latex]5(\\color{red}{7})\\stackrel{\\text{?}}{=}4(\\color{red}{7})+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><\/td>\n<td>[latex]35\\stackrel{\\text{?}}{=}28+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><\/td>\n<td>[latex]35=35\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<h4>Solution 2:<\/h4>\n<p>The only constant, [latex]24[\/latex], is on the right, so let the left side be the variable side.<\/p>\n<table id=\"eip-id1168467377760\" class=\"unnumbered unstyled\" summary=\"The first line says 7x equals negative x plus 24. The left side is labeled\">\n<tbody>\n<tr>\n<td>[latex]7x[\/latex] is the side containing only a <span style=\"color: #000000;\"><span style=\"color: #ff0000;\">variable<\/span>.<\/span><span style=\"color: #000000;\">[latex]-x+24[\/latex] is the side containing a <span style=\"color: #ff0000;\">constant<\/span>.<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Remove the [latex]-x[\/latex] from the right side by adding [latex]x[\/latex] to both sides.<\/td>\n<td>[latex]7x\\color{red}{+x}=-x\\color{red}{+x}+24[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]8x=24[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>All the variables are on the left and the constants are on the right. Divide both sides by [latex]8[\/latex].<\/td>\n<td>[latex]\\Large\\frac{8x}{\\color{red}{8}}\\normalsize =\\Large\\frac{24}{\\color{red}{8}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]x=3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>[latex]7x=-x+24[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]x=3[\/latex].<\/td>\n<td>[latex]7(\\color{red}{3})\\stackrel{\\text{?}}{=}-(\\color{red}{3})+24[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]21=21\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try\u00a0it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm142129\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142129&#38;theme=oea&#38;iframe_resize_id=ohm142129&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm142132\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142132&#38;theme=oea&#38;iframe_resize_id=ohm142132&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Sometimes it is beneficial to have the variable term on the right side of the equation. There is no &#8220;correct&#8221; side to have the variable term, but the choice can help to avoid working with negative signs.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]5y - 8=7y[\/latex]<\/p>\n<h4>Solution:<\/h4>\n<p>The only constant, [latex]-8[\/latex], is on the left side of the equation, and the variable, [latex]y[\/latex], is on both sides. Let\u2019s leave the constant on the left and collect the variables to the right.<\/p>\n<table id=\"eip-id1168468768462\" class=\"unnumbered unstyled\" summary=\"The first line says 5y minus 8 equals 7y. The left side is labeled\">\n<tbody>\n<tr>\n<td>[latex]5y-8[\/latex] is the side containing a <span style=\"color: #000000;\"><span style=\"color: #ff0000;\">constant<\/span>.<\/span><span style=\"color: #000000;\">[latex]7y[\/latex] is the side containing only a <\/span><span style=\"color: #ff0000;\">variable<\/span><span style=\"color: #000000;\">.<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Subtract [latex]5y[\/latex] from both sides.<\/td>\n<td>[latex]5y\\color{red}{-5y}-8=7y\\color{red}{-5y}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]-8=2y[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>We have the variables on the right and the constants on the left. Divide both sides by [latex]2[\/latex].<\/td>\n<td>[latex]\\Large\\frac{-8}{\\color{red}{2}}\\normalsize =\\Large\\frac{2y}{\\color{red}{2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]-4=y[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Rewrite with the variable on the left.<\/td>\n<td>[latex]y=-4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>[latex]5y-8=7y[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Let [latex]y=-4[\/latex].<\/td>\n<td>[latex]5(\\color{red}{-4})-8\\stackrel{\\text{?}}{=}7(\\color{red}{-4})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]-20-8\\stackrel{\\text{?}}{=}-28[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]-28=-28\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>[\/hidden-answer]<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try\u00a0it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm142125\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142125&#38;theme=oea&#38;iframe_resize_id=ohm142125&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h3><\/h3>\n<h3>Variables and Constants on Both Sides<\/h3>\n<p>The next example will be the first to have variables <em>and<\/em> constants on both sides of the equation. As we did before, we\u2019ll collect the variable terms to one side and the constants to the other side. As the number of variable and constant terms increase, so do the number of steps it takes to solve the equation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>Solve:<\/p>\n<p>1. [latex]7x+5=6x+2[\/latex]<\/p>\n<p>2.\u00a0[latex]6n - 2=-3n+7[\/latex]<\/p>\n<h4>1. Solution:<\/h4>\n<p>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><\/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>[latex]7x\\color{red}{-6x}+5=6x\\color{red}{-6x}+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}=2\\color{red}{-5}[\/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><\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>[latex]7x+5=6x+2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Let [latex]x=-3[\/latex].<\/td>\n<td>[latex]7(\\color{red}{-3})+5\\stackrel{\\text{?}}{=}6(\\color{red}{-3})+2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]-21+5\\stackrel{\\text{?}}{=}-18+2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]16=16\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Answer: [latex]x=3[\/latex]<\/p>\n<h4>2. Solution<\/h4>\n<p>Solve: [latex]6n - 2=-3n+7[\/latex]<\/p>\n<p>We 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><\/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>[latex]6n\\color{red}{+3n}-2=-3n\\color{red}{+3n}+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}=7\\color{red}{+2}[\/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}}\\normalsize =\\Large\\frac{9}{\\color{red}{9}}[\/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>[latex]6n-2=-3n+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]1[\/latex] for [latex]n[\/latex].<\/td>\n<td>[latex]6(\\color{red}{1})-2\\stackrel{\\text{?}}{=}-3(\\color{red}{1})+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]4=4\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Answer: [latex]n=1[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The following video shows an example of how to solve a multi-step equation. It doesn&#8217;t matter which side we choose to be the variable side; we can get the correct answer either way.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Solve a Linear Equation in One Variable with Variables on Both Sides: 2x+8=-2x-24\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/_hBoWoctfAo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p>In the next example, we move the variable terms to the right side, to keep a positive coefficient on the variable.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Solve: [latex]2a - 7=5a+8[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<h4>Solution:<\/h4>\n<p>This equation has [latex]2a[\/latex] on the left and [latex]5a[\/latex] on the right. Since [latex]5>2[\/latex], make the right side the variable side and the left side the constant side.<\/p>\n<table id=\"eip-id1168466004451\" 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><\/td>\n<td>[latex]2a-7=5a+8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract [latex]2a[\/latex] from both sides to remove the variable term from the left.<\/td>\n<td>[latex]2a\\color{red}{-2a}-7=5a\\color{red}{-2a}+8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Combine like terms.<\/td>\n<td>[latex]-7=3a+8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract [latex]8[\/latex] from both sides to remove the constant from the right.<\/td>\n<td>[latex]-7\\color{red}{-8}=3a+8\\color{red}{-8}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]-15=3a[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide both sides by [latex]3[\/latex] to make [latex]1[\/latex] the coefficient of [latex]a[\/latex] .<\/td>\n<td>[latex]\\Large\\frac{-15}{\\color{red}{3}}\\normalsize =\\Large\\frac{3a}{\\color{red}{3}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]-5=a[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>[latex]2a-7=5a+8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Let [latex]a=-5[\/latex]<\/td>\n<td>[latex]2(\\color{red}{-5})-7\\stackrel{\\text{?}}{=}5(\\color{red}{-5})+8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]-10-7\\stackrel{\\text{?}}{=}-25+8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]-17=-17\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The following video shows another example of solving a multi-step\u00a0equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Solve a Linear Equation in One Variable with Variables on Both Sides: 2m-9=6m-17\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/kiYPW6hrTS4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try\u00a0it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm142134\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142134&#38;theme=oea&#38;iframe_resize_id=ohm142134&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm142136\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=142136&#38;theme=oea&#38;iframe_resize_id=ohm142136&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>We just showed a lot of examples of different kinds of linear equations that may be encountered. There are some good habits to develop that will help us solve all kinds of linear equations. We\u2019ll summarize the steps we took for easy reference.<\/p>\n<div class=\"textbox shaded\">\n<h3>Solve an equation with variables and constants on both sides<\/h3>\n<ol id=\"eip-id1168468371331\" class=\"stepwise\">\n<li>Choose one side to be the variable side, and the other to be the constant side.<\/li>\n<li>Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.<\/li>\n<li>Collect the constants to the other side, using the Addition or Subtraction Property of Equality.<\/li>\n<li>Make the coefficient of the variable [latex]1[\/latex], using the Multiplication or Division Property of Equality.<\/li>\n<li>Check the solution by substituting it into the original equation.<\/li>\n<\/ol>\n<\/div>\n<h2>No Solution and Infinite Solutions<\/h2>\n<p>Equations that have a solution are called\u00a0<em><strong>conditional equations<\/strong><\/em><strong>.\u00a0<\/strong>The truth of the equation is found only when the solution is entered in the equation. For example, [latex]2x-5=3[\/latex] is true only when the solution [latex]x=4[\/latex] is put into the equation. Any other value of [latex]x[\/latex] makes the equation false. The truth of the equation is conditional upon the value of the variable.<\/p>\n<p>It is possible for a linear equation in one variable to have an infinite number of solutions because it is always true. Equations that are always true are called\u00a0<strong><em>identities<\/em><\/strong><i>.\u00a0<\/i>For example the equation [latex]3x+5x=8x[\/latex] is always true, no matter the value of [latex]x[\/latex]. Consequently, the equation has an infinite number of solutions. All real numbers are solutions of this equation.<\/p>\n<p>It is also possible for a linear equation to have no solution. Such equations are called\u00a0<strong><i>contradictions<\/i>. <\/strong>For example, [latex]3x+5=3x-2[\/latex] simplifies to [latex]5=-2[\/latex], which is a contradiction. Consequently, this equation has no solution.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Solve the equation:<\/p>\n<p>1. [latex]4x-3=2x+5+2x[\/latex]<\/p>\n<p>2. [latex]5x-4-3x=2x-4[\/latex]<\/p>\n<p>Solution:<\/p>\n<table id=\"eip-id1168466426761\" class=\"unnumbered unstyled\" summary=\"The top line says 4y plus 3 equals 8y. Beside this is\">\n<tbody>\n<tr>\n<td style=\"width: 244px;\">1.<\/td>\n<td style=\"width: 400px;\">[latex]4x-3=2x+5+2x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 244px;\">Simplify the right side by adding like terms<\/td>\n<td style=\"width: 400px;\">\u00a0[latex]4x-3=4x+5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 244px;\">Subtract [latex]4x[\/latex] from both sides<\/td>\n<td style=\"width: 400px;\">\u00a0[latex]-3=5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 244px;\">Contradiction<\/td>\n<td style=\"width: 400px;\">No solution<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table id=\"eip-id1168466426761\" class=\"unnumbered unstyled\" summary=\"The top line says 4y plus 3 equals 8y. Beside this is\">\n<tbody>\n<tr>\n<td style=\"width: 244px;\">2.<\/td>\n<td style=\"width: 400px;\">[latex]5x-4-3x=2x-4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 244px;\">Simplify the right side by adding like terms<\/td>\n<td style=\"width: 400px;\">\u00a0[latex]2x-4=2x-4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 244px;\">Subtract [latex]4x[\/latex] from both sides<\/td>\n<td style=\"width: 400px;\">\u00a0[latex]-4=-4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 244px;\">Identity<\/td>\n<td style=\"width: 400px;\">Solution is all real numbers<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p>Solve the equation.<\/p>\n<p>1. [latex]5x+2=5x[\/latex]<\/p>\n<p>2. [latex]3x-4+9=5x+3-2x+2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qH00067\">Show Answer<\/span><\/p>\n<div id=\"qH00067\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. Contradiction. There is no solution.<\/p>\n<p>2.\u00a0Identity.\u00a0Solution is all real numbers.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>When solving a linear equation in one variable, if we end up with a statement that is always true, we have an identity. A identity is true no matter the value of the variable, so the solution is the set of all real numbers. On the other hand, if we end up with an statement that is always false, we have a contradiction. A contradiction is never true, so the equation has no solution.<\/p>\n<h2 id=\"title2\">The Distributive Property<\/h2>\n<p>As we solve linear equations, we often need to do some work to write\u00a0the linear equations in a form we are familiar with solving.\u00a0This section will focus on manipulating an equation we are asked to solve in such a way that we can use the skills we learned for solving multi-step equations to ultimately arrive at the solution.<\/p>\n<p>The distributive property can come in handy to simplify an equation. Using this property, we multiply the number in front of the parentheses by each term inside of the parentheses.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Distributive Property of Multiplication over addition<\/h3>\n<p>For all real numbers [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex],\u00a0[latex]a(b+c)=ab+ac[\/latex].<\/p>\n<p>When a number multiplies an expression inside parentheses, we can distribute the multiplication to each term of the expression individually.<\/p>\n<\/div>\n<h3><\/h3>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for [latex]a[\/latex].<\/p>\n<p>[latex]4\\left(2a+3\\right)=28[\/latex]<\/p>\n<h4>Solution:<\/h4>\n<p>Apply the distributive property to expand [latex]4\\left(2a+3\\right)[\/latex] to [latex]8a+12[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4\\left(2a+3\\right)=28\\\\ 8a+12=28\\end{array}[\/latex]<\/p>\n<p>Subtract [latex]12[\/latex]\u00a0from both sides to isolate\u00a0the variable term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8a+12\\,\\,\\,=\\,\\,\\,28\\\\ \\underline{-12\\,\\,\\,\\,\\,\\,-12}\\\\ 8a\\,\\,\\,=\\,\\,\\,16\\end{array}[\/latex]<\/p>\n<p>Divide both terms by [latex]8[\/latex] to get a coefficient of [latex]1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\underline{8a}=\\underline{16}\\\\8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,8\\\\a\\,=\\,\\,2\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]a=2[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In our next example, we will use the distributive property of multiplication over addition first, simplify, then use the division property to finally solve.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]-3\\left(n - 2\\right)-6=21[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Solution:<\/p>\n<table id=\"eip-id1168468278405\" class=\"unnumbered unstyled\" summary=\"The first line shows negative 3 times parentheses n minus 2 minus 6 equals 21. The next line says\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]-3(n-2)-6=21[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute.<\/td>\n<td>[latex]-3n+6-6=21[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]-3n=21[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide both sides by -3 to isolate n.<\/td>\n<td>[latex]\\Large\\frac{-3n}{\\color{red}{-3}}\\normalsize =\\Large\\frac{21}{\\color{red}{-3}}[\/latex][latex]n=-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check your answer. Let [latex]n=-7[\/latex] .<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]-3(n-2)-6=21[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]-3(\\color{red}{-7}-2)-6\\stackrel{\\text{?}}{=}21[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]-3(-9)-6\\stackrel{\\text{?}}{=}21[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]27-6\\stackrel{\\text{?}}{=}21[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]21=21\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm141911\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141911&#38;theme=oea&#38;iframe_resize_id=ohm141911&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The video that follows shows another example of how to use the distributive property to solve a multi-step linear equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Solving an Equation with One Set of Parentheses\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/aQOkD8L57V0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]3\\left(n - 4\\right)-2n=-3[\/latex]<\/p>\n<p>Solution:<br \/>\nThe left side of the equation has an expression that we should simplify.<\/p>\n<table id=\"eip-id1168468254328\" class=\"unnumbered unstyled\" summary=\"The top line says\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]3(n-4)-2n=-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute on the left.<\/td>\n<td>[latex]3n-12-2n=-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the Commutative Property to rearrange terms.<\/td>\n<td>[latex]3n-2n-12=-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Combine like terms.<\/td>\n<td>[latex]n-12=-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Isolate <em>n<\/em> using the Addition Property of Equality.<\/td>\n<td>[latex]n-12\\color{red}{+12}=-3\\color{red}{+12}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]n=9[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check.Substitute [latex]n=9[\/latex] into the original equation.<br \/>\n[latex]3(n-4)-2n=-3[\/latex]<br \/>\n[latex]3(\\color{red}{9}-4)-2\\cdot\\color{red}{9}=-3[\/latex]<br \/>\n[latex]3(5)-18=-3[\/latex]<br \/>\n[latex]15-18=-3[\/latex]<br \/>\n[latex]-3=-3\\quad\\checkmark[\/latex]<br \/>\nThe solution checks.<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY\u00a0IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm141737\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141737&#38;theme=oea&#38;iframe_resize_id=ohm141737&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h3><\/h3>\n<p>The next example has expressions on both sides that need to be simplified.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]2\\left(3k - 1\\right)-5k=-2 - 7[\/latex]<\/p>\n<p>Solution:<br \/>\nBoth sides of the equation have expressions that we should simplify before we isolate the variable.<\/p>\n<table id=\"eip-id1168469785088\" class=\"unnumbered unstyled\" summary=\"The top line says 2 times parentheses 3k minus 1 minus 5k equals negative 2 minus 7. The next line says\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]2(3k-1)-5k=-2-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute on the left, subtract on the right.<\/td>\n<td>[latex]6k-2-5k=-9[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the Commutative Property of Addition.<\/td>\n<td>[latex]6k-5k-2=-9[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Combine like terms.<\/td>\n<td>[latex]k-2=-9[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Undo subtraction by using the Addition Property of Equality.<\/td>\n<td>[latex]k-2\\color{red}{+2}=-9\\color{red}{+2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]k=-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check.Let [latex]k=-7[\/latex].<br \/>\n[latex]2(3k-1)-5k=-2-7[\/latex]<br \/>\n[latex]2(3(\\color{red}{-7}-1)-5(\\color{red}{-7})=-2-7[\/latex]<br \/>\n[latex]2(-21-1)-5(-7)=-9[\/latex]<br \/>\n[latex]2(-22)+35=-9[\/latex]<br \/>\n[latex]-44+35=-9[\/latex]<br \/>\n[latex]-9=-9\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The solution checks.<\/p>\n<p>Answer: [latex]k=-7[\/latex]<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY\u00a0IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm141739\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141739&#38;theme=oea&#38;iframe_resize_id=ohm141739&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The following video presents another example of how to solve an equation that requires simplifying before using the addition and subtraction properties.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Solve Linear Equations in One Variable with Simplifying (One-Step Add\/Subtract)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/shGKzDBA5kQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3><\/h3>\n<p>In the next example, there are parentheses on both sides of the equals sign, so we need to use the distributive property twice.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for [latex]t[\/latex].<\/p>\n<p>[latex]2\\left(4t-5\\right)=-3\\left(2t+1\\right)[\/latex]<\/p>\n<p>Solution:<\/p>\n<p>Apply the distributive property to expand [latex]2\\left(4t-5\\right)[\/latex] to [latex]8t-10[\/latex] and [latex]-3\\left(2t+1\\right)[\/latex] to [latex]-6t-3[\/latex]. Be careful in this step\u2014you are distributing a negative number, so keep track of the sign of each number after you multiply.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(4t-5\\right)=-3\\left(2t+1\\right)\\,\\,\\,\\,\\,\\, \\\\ 8t-10=-6t-3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Add [latex]-6t[\/latex] to both sides to begin combining like terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8t-10=-6t-3\\\\ \\underline{+6t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+6t}\\,\\,\\,\\,\\,\\,\\,\\\\ 14t-10=\\,\\,\\,\\,-3\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Add [latex]10[\/latex] to both sides of the equation to isolate [latex]t[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}14t-10=-3\\\\ \\underline{+10\\,\\,\\,+10}\\\\ 14t=\\,\\,\\,7\\,\\end{array}[\/latex]<\/p>\n<p>The last step is to divide both sides by [latex]14[\/latex] to completely isolate [latex]t[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}14t=7\\,\\,\\,\\,\\\\\\frac{14t}{14}=\\frac{7}{14}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]t=\\frac{1}{2}[\/latex]<\/p>\n<p>We simplified the fraction [latex]\\frac{7}{14}[\/latex] into [latex]\\frac{1}{2}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm39431\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=39431&theme=oea&iframe_resize_id=ohm39431&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The following video solves another multi-step equation with two sets of parentheses.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Solving an Equation with Parentheses on Both Sides\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/StomYTb7Xb8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2><\/h2>\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-845\">\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. Provided by: Lumen Learning. . <strong>Authored 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><li>Solving Two Step Equations (Basic). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/fCyxSVQKeRw\">https:\/\/youtu.be\/fCyxSVQKeRw<\/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>Solving an Equation that Requires Combining Like Terms. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ez_sP2OTGjU\">https:\/\/youtu.be\/ez_sP2OTGjU<\/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>Solve an Equation with Variable on Both Sides. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/f3ujWNPL0Bw\">https:\/\/youtu.be\/f3ujWNPL0Bw<\/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>No Solutions and Infinite Solutions. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Solving an Equation with One Set of Parentheses. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/aQOkD8L57V0\">https:\/\/youtu.be\/aQOkD8L57V0<\/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>Solving an Equation with Parentheses on Both Sides. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/StomYTb7Xb8\">https:\/\/youtu.be\/StomYTb7Xb8<\/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 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 and Education. <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><li>Ex 4: Solving Absolute Value Equations (Requires Isolating Abs. Value). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/-HrOMkIiSfU\">https:\/\/youtu.be\/-HrOMkIiSfU<\/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":370291,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation. Provided by: Lumen Learning. \",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Solving Two Step Equations (Basic)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/fCyxSVQKeRw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Solving an Equation that Requires Combining Like Terms\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ez_sP2OTGjU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Solve an Equation with Variable on Both Sides\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/f3ujWNPL0Bw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 4: Solving Absolute Value Equations (Requires Isolating Abs. Value)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/-HrOMkIiSfU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"No Solutions and Infinite Solutions\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Solving an Equation with One Set of Parentheses\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/aQOkD8L57V0\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Solving an Equation with Parentheses on Both Sides\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/StomYTb7Xb8\",\"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-845","chapter","type-chapter","status-publish","hentry"],"part":642,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/845","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/users\/370291"}],"version-history":[{"count":15,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/845\/revisions"}],"predecessor-version":[{"id":2745,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/845\/revisions\/2745"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/parts\/642"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/845\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/media?parent=845"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=845"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/contributor?post=845"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/license?post=845"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}