{"id":357,"date":"2018-04-17T02:20:14","date_gmt":"2018-04-17T02:20:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/?post_type=chapter&#038;p=357"},"modified":"2024-04-26T22:04:56","modified_gmt":"2024-04-26T22:04:56","slug":"solving-single-step-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/chapter\/solving-single-step-equations\/","title":{"raw":"Solving Single-Step Equations","rendered":"Solving Single-Step Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcome<\/h3>\r\n<ul>\r\n \t<li>Use the addition, subtraction, multiplication, and division properties to solve single-step equations<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe will begin solving equations below, but first we need to understand what a solution to an equation looks like. The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same \u2014 it is the answer to the puzzle.\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Solution of an Equation<\/h3>\r\nA solution of an equation is a value of a variable that makes a true statement when substituted into the equation.\r\n\r\nDetermine whether a number is a solution to an equation.\r\n<ol id=\"eip-95\" class=\"stepwise\">\r\n \t<li>Substitute the number for the variable in the equation.<\/li>\r\n \t<li>Simplify the expressions on both sides of the equation.<\/li>\r\n \t<li>Determine whether the resulting equation is true.\r\n<ul id=\"fs-id1703984\">\r\n \t<li>If it is true, the number is a solution.<\/li>\r\n \t<li>If it is not true, the number is not a solution.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn the following example, we will show how to determine whether a number is a solution to an equation that contains addition and subtraction. You can use this idea to check your work later when you are solving equations.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nDetermine whether [latex]y=\\Large\\frac{3}{4}[\/latex] is a solution for [latex]4y+3=8y[\/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><\/td>\r\n<td>[latex]4y+3=8y[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]\\color{red}{\\Large\\frac{3}{4}}[\/latex] for [latex]y[\/latex]<\/td>\r\n<td>[latex]4(\\color{red}{\\Large\\frac{3}{4}}\\normalsize)+3\\stackrel{\\text{?}}{=}8(\\color{red}{\\Large\\frac{3}{4}})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]3+3\\stackrel{\\text{?}}{=}6[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add.<\/td>\r\n<td>[latex]6=6\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince [latex]y=\\Large\\frac{3}{4}[\/latex] results in a true equation, [latex]\\Large\\frac{3}{4}[\/latex] is a solution to the equation [latex]4y+3=8y[\/latex].\r\n\r\n<\/div>\r\nNow it is your turn to determine whether a fraction is the solution to an equation.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141719&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"380\"><\/iframe>\r\n\r\n<\/div>\r\nLet's use a model to help understand how the process of solving an equation is like solving a puzzle. An envelope represents the variable \u2013 since its contents are unknown \u2013 and each counter represents one.\r\n\r\nSuppose a desk has an imaginary line dividing it in half. We place three counters and an envelope on the left side of desk, and eight counters on the right side of the desk.\u00a0Both sides of the desk have the same number of counters, but some counters are hidden in the envelope. Can you tell how many counters are in the envelope?\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215900\/CNX_BMath_Figure_02_03_001.png\" alt=\"The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 8 counters.\" \/>\r\nWhat steps are you taking in your mind to figure out how many counters are in the envelope? Perhaps you are thinking \"I need to remove the [latex]3[\/latex] counters from the left side to get the envelope by itself. Those [latex]3[\/latex] counters on the left match with [latex]3[\/latex] on the right, so I can take them away from both sides. That leaves five counters on the right, so there must be [latex]5[\/latex] counters in the envelope.\"\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215129\/CNX_BMath_Figure_02_03_002.png\" alt=\"The image is in two parts. On the left is a rectangle divided in half vertically. On the left side of the rectangle is an envelope with three counters below it. The 3 counters are circled in red with an arrow pointing out of the rectangle. On the right side is 8 counters. The bottom 3 counters are circled in red with an arrow pointing out of the rectangle. The 3 circled counters are removed from both sides of the rectangle, creating the new rectangle on the right of the image which is also divided in half vertically. On the left side of the rectangle is just an envelope. On the right side is 5 counters.\" \/>\r\nWhat algebraic equation is modeled by this situation? Each side of the desk represents an expression and the center line takes the place of the equal sign. We will call the contents of the envelope [latex]x[\/latex], so the number of counters on the left side of the desk is [latex]x+3[\/latex]. On the right side of the desk are [latex]8[\/latex] counters. We are told that [latex]x+3[\/latex] is equal to [latex]8[\/latex] so our equation is [latex]x+3=8[\/latex].\r\n<p style=\"text-align: center;\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215900\/CNX_BMath_Figure_02_03_001.png\" alt=\"The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 8 counters.\" \/>\r\n[latex]x+3=8[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Let\u2019s write algebraically the steps we took to discover how many counters were in the envelope.<\/p>\r\n\r\n<table id=\"eip-id1168467162379\" class=\"unnumbered unstyled\" summary=\"In this image, the first line shows x plus three equals eight. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]x+3=8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>First, we took away three from each side.<\/td>\r\n<td>[latex]x+3\\color{red}{--3}=8\\color {red}{--3}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Then we were left with five.<\/td>\r\n<td>[latex]x=5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow let\u2019s check our solution. We substitute [latex]5[\/latex] for [latex]x[\/latex] in the original equation and see if we get a true statement.\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]x+3=8[\/latex]<\/p>\r\n<p style=\"text-align: left; padding-left: 60px;\">[latex]\\color{red}{5}+3=8[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">[latex]8=8\\quad\\checkmark[\/latex]\u00a0 Our solution is correct. Five counters in the envelope plus three more equals eight.<\/p>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nWrite an equation modeled by the envelopes and counters, and then solve the equation:\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215904\/CNX_BMath_Figure_02_03_003_img.png\" alt=\"The image is divided in half vertically. On the left side is an envelope with 4 counters below it. On the right side is 5 counters.\" \/>\r\n\r\nSolution\r\n<table id=\"eip-id1168468230261\" class=\"unnumbered unstyled\" summary=\".\">\r\n<tbody>\r\n<tr>\r\n<td>On the left, write [latex]x[\/latex] for the contents of the envelope, add the [latex]4[\/latex] counters, so we have [latex]x+4[\/latex] .<\/td>\r\n<td>[latex]x+4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>On the right, there are [latex]5[\/latex] counters.<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>The two sides are equal.<\/td>\r\n<td>[latex]x+4=5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Solve the equation by subtracting [latex]4[\/latex] counters from each side.<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215905\/CNX_BMath_Figure_02_03_005_img.png\" alt=\"The image is in two parts. On the left is a rectangle divided in half vertically. On the left side of the rectangle is an envelope with 4 counters below it. The 4 counters are circled in red with an arrow pointing out of the rectangle. On the right side is 5 counters. The bottom 4 counters are circled in red with an arrow pointing out of the rectangle. The 4 circled counters are removed from both sides of the rectangle, creating the new rectangle on the right of the image which is also divided in half vertically. On the left side of the rectangle is just an envelope. On the right side is 1 counter.\" \/>\r\nWe can see that there is one counter in the envelope. This can be shown algebraically as:\r\n<p style=\"padding-left: 60px;\">[latex]x+4=5[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">[latex]x+4\\color{red}{--4}=5\\color{red}{--4}[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">[latex]x=1[\/latex]<\/p>\r\nSubstitute [latex]1[\/latex] for [latex]x[\/latex] in the equation to check.\r\n<p style=\"padding-left: 60px;\">[latex]x+4=5[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">[latex]\\color{red}{1}+4=5[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">[latex]5=5\\quad\\checkmark[\/latex]<\/p>\r\nSince [latex]x=1[\/latex] makes the statement true, we know that [latex]1[\/latex] is indeed a solution.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nWrite the equation modeled by the envelopes and counters, and then solve the equation:\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215908\/CNX_BMath_Figure_02_03_004_img.png\" alt=\"The image is divided in half vertically. On the left side is an envelope with one counter below it. On the right side is 7 counters.\" \/>\r\n\r\n[reveal-answer q=\"7845\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"7845\"]\r\n\r\n[latex]x+1=7[\/latex]\r\n\r\n[latex]x=6[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nWrite the equation modeled by the envelopes and counters, and then solve the equation:\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215909\/CNX_BMath_Figure_02_03_006_img.png\" alt=\"The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 4 counters.\" \/>\r\n\r\n[reveal-answer q=\"9034\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"9034\"]\r\n\r\n[latex]x+3=4[\/latex]\r\n\r\n[latex]x=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOur puzzle has given us an idea of what we need to do to solve an equation. The goal is to isolate the variable by itself on one side of the equation. In the previous examples, we used the Subtraction Property of Equality, which states that when we subtract the same quantity from both sides of an equation, we still have equality.\r\n<div class=\"textbox shaded\">\r\n<h3 id=\"fs-id1956928\"><strong>Subtraction Property of Equality<\/strong><\/h3>\r\nFor all real numbers [latex]a,b[\/latex], and [latex]c[\/latex], if [latex]a=b[\/latex], then [latex]a-c=b-c[\/latex].\r\n\r\n<\/div>\r\nIn all the equations we have solved so far, a number was added to the variable on one side of the equation. We used subtraction to \"undo\" the addition in order to isolate the variable.\r\n\r\nBut suppose we have an equation with a number subtracted from the variable, such as [latex]x - 5=8[\/latex]. We want to isolate the variable, so to \"undo\" the subtraction we will add the number to both sides.\r\n\r\nWe use the Addition Property of Equality, which says we can add the same number to both sides of the equation without changing the equality. Notice how it mirrors the Subtraction Property of Equality.\r\n<div class=\"textbox shaded\">\r\n<h3 id=\"fs-id1166490863495\"><strong>Addition Property of Equality<\/strong><\/h3>\r\nFor all real numbers [latex]a,b[\/latex], and [latex]c[\/latex], if [latex]a=b[\/latex], then [latex]a+c=b+c[\/latex].\r\n\r\n<\/div>\r\nWhen you add or subtract the same quantity from both sides of an equation, you still have equality.\r\n\r\nWe introduced the Subtraction Property of Equality earlier by modeling equations with envelopes and counters. The image below\u00a0models the equation [latex]x+3=8[\/latex].\r\n<figure id=\"CNX_BMath_Figure_08_01_026\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24222533\/CNX_BMath_Figure_08_01_026.png\" alt=\"An envelope and three yellow counters are shown on the left side. On the right side are eight yellow counters.\" width=\"218\" height=\"133\" \/><\/figure>\r\nThe goal is to isolate the variable on one side of the equation. So we \"took away\" [latex]3[\/latex] from both sides of the equation and found the solution [latex]x=5[\/latex].\r\n\r\nSome people picture a balance scale, as in the image below, when they solve equations.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24222535\/CNX_BMath_Figure_08_01_001.png\" alt=\"Three balance scales are shown. The top scale has one mass on each side and is balanced. The middle scale has 2 masses on each side and is balanced. The bottom scale has 1 mass on one side and 2 masses on the other and is unbalanced.\" width=\"386\" height=\"406\" \/>\r\nThe quantities on both sides of the equal sign in an equation are equal, or balanced. Just as with the balance scale, whatever you do to one side of the equation you must also do to the other to keep it balanced.\r\n\r\nIn the following examples we\u00a0review how to use Subtraction and Addition Properties of Equality to solve equations. We need to isolate the variable on one side of the equation. You can check your solutions by substituting the value into the original equation to make sure you have a true statement.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nSolve: [latex]x+11=-3[\/latex]\r\n[reveal-answer q=\"190834\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"190834\"]\r\n\r\nSolution:\r\nTo isolate [latex]x[\/latex], we undo the addition of [latex]11[\/latex] by using the Subtraction Property of Equality.\r\n<table id=\"eip-id1168468291100\" class=\"unnumbered unstyled\" summary=\"The top line says x plus 11 equals negative 3. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\"><\/td>\r\n<td>[latex]x+11=-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Subtract 11 from each side to \"undo\" the addition.<\/td>\r\n<td>[latex]x+11\\color{red}{-11}=-3\\color{red}{-11}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Simplify.<\/td>\r\n<td>[latex]x=-14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td>[latex]x+11=-3[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]x=-14[\/latex] .<\/td>\r\n<td>[latex]\\color{red}{-14}+11\\stackrel{\\text{?}}{=}-3[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]-3=-3\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince [latex]x=-14[\/latex] makes [latex]x+11=-3[\/latex] a true statement, we know that it is a solution to the equation.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow you can try solving an equation that requires using the addition property.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY IT<\/h3>\r\n<iframe id=\"mom200\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141721&amp;theme=oea&amp;iframe_resize_id=mom200\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\nIn the original equation in the previous example, [latex]11[\/latex] was added to the [latex]x[\/latex] , so we subtracted [latex]11[\/latex] to \"undo\" the addition. In the next example, we will need to \"undo\" subtraction by using the Addition Property of Equality.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nSolve: [latex]m - 4=-5[\/latex]\r\n[reveal-answer q=\"604060\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"604060\"]\r\n\r\nSolution:\r\n<table id=\"eip-id1168467157298\" class=\"unnumbered unstyled\" summary=\"The first line says m minus 4 equals negative 5. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td colspan=\"2\"><\/td>\r\n<td>[latex]m-4=-5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Add 4 to each side to \"undo\" the subtraction.<\/td>\r\n<td>[latex]m-4\\color{red}{+4}=-5\\color{red}{+4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"2\">Simplify.<\/td>\r\n<td>[latex]m=-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check:<\/td>\r\n<td>[latex]m-4=-5[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]m=-1[\/latex] .<\/td>\r\n<td>[latex]\\color{red}{-1}+4\\stackrel{\\text{?}}{=}-5[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]-5=-5\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td><\/td>\r\n<td>The solution to [latex]m - 4=-5[\/latex] is [latex]m=-1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow you can try using the addition property to solve an equation.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY\u00a0IT<\/h3>\r\n<iframe id=\"mom27\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141723&amp;theme=oea&amp;iframe_resize_id=mom27\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\nIn the following video, we present more examples of solving equations using the addition and subtraction properties.\r\n\r\nhttps:\/\/youtu.be\/yqdlj0lv7Cc\r\n\r\nYou may encounter equations that contain fractions or decimals \u2014 especially in financial applications \u2014 so let's practice solving those in the following problems.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY\u00a0IT<\/h3>\r\n<iframe id=\"mom25\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141732&amp;theme=oea&amp;iframe_resize_id=mom25\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY\u00a0IT<\/h3>\r\n[ohm_question]156990[\/ohm_question]\r\n\r\n<\/div>\r\nAll of the equations we have solved so far have been of the form [latex]x+a=b[\/latex] or [latex]x-a=b[\/latex]. We were able to isolate the variable by adding or subtracting the constant term. Now we\u2019ll see how to solve equations that involve division.\r\n\r\nWe will model an equation with envelopes and counters.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220604\/CNX_BMath_Figure_03_05_001.png\" alt=\"This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed.\" \/>\r\nHere, there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are \"hidden\" in the envelopes. So how many counters are in each envelope?\r\n\r\nTo determine the number, separate the counters on the right side into [latex]2[\/latex] groups of the same size. So [latex]6[\/latex] counters divided into [latex]2[\/latex] groups means there must be [latex]3[\/latex] counters in each group (since [latex]6\\div2=3[\/latex]).\r\n\r\nWhat equation models the situation shown in the figure below? There are two envelopes, and each contains [latex]x[\/latex] counters. Together, the two envelopes must contain a total of [latex]6[\/latex] counters. So the equation that models the situation is [latex]2x=6[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220605\/CNX_BMath_Figure_03_05_002.png\" alt=\"This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed. Under the figure is two times x equals 6.\" \/>\r\nWe can divide both sides of the equation by [latex]2[\/latex] as we did with the envelopes and counters.\r\n<p style=\"text-align: center;\">[latex]\\Large{\\frac{2x}{\\color{red}{2}}=\\frac{6}{\\color{red}{2}}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=3[\/latex]<\/p>\r\nWe found that each envelope contains [latex]\\text{3 counters.}[\/latex] Does this check? We know [latex]2\\cdot 3=6[\/latex], so it works. Three counters in each of two envelopes does equal six.\r\n\r\nAnother example is shown below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220607\/CNX_BMath_Figure_03_05_003.png\" alt=\"This image has two columns. In the first column are three envelopes. In the second column there are four rows of three blue circles. Underneath the image is the equation 3x equals 12.\" \/>\r\nNow we have [latex]3[\/latex] identical envelopes and [latex]\\text{12 counters.}[\/latex] How many counters are in each envelope? We have to separate the [latex]\\text{12 counters}[\/latex] into [latex]\\text{3 groups.}[\/latex] Since [latex]12\\div 3=4[\/latex], there must be [latex]\\text{4 counters}[\/latex] in each envelope.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220608\/CNX_BMath_Figure_03_05_004.png\" alt=\"This image has two columns. In the first column are four envelopes. In the second column there are twelve blue circles.\" \/>\r\nThe equation that models the situation is [latex]3x=12[\/latex]. We can divide both sides of the equation by [latex]3[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\Large{\\frac{3x}{\\color{red}{3}}=\\frac{12}{\\color{red}{3}}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=4[\/latex]<\/p>\r\nDoes this check? It does because [latex]3\\cdot 4=12[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Division Property of Equality<\/h3>\r\nFor all real numbers [latex]a,b,c[\/latex], and [latex]c\\ne 0[\/latex], if [latex]a=b[\/latex], then [latex]\\Large\\frac{a}{c}\\normalsize =\\Large\\frac{b}{c}[\/latex].\r\n\r\nStated simply, when you divide both sides of an equation by the same quantity, you still have equality.\r\n\r\n<\/div>\r\nRemember, the goal is to \"undo\" the operation on the variable. In the example below the variable is multiplied by [latex]4[\/latex], so we will divide both sides by [latex]4[\/latex] to \"undo\" the multiplication.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]4x=-28[\/latex]\r\n\r\nSolution:\r\n\r\nTo solve this equation, we use the Division Property of Equality to divide both sides by [latex]4[\/latex].\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]4x=-28[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide both sides by 4 to undo the multiplication.<\/td>\r\n<td>[latex]\\Large\\frac{4x}{\\color{red}4}\\normalsize =\\Large\\frac{-28}{\\color{red}4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]x =-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check your answer.<\/td>\r\n<td>[latex]4x=-28[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Let [latex]x=-7[\/latex]. Substitute [latex]-7[\/latex] for x.<\/td>\r\n<td>[latex]4(\\color{red}{-7})\\stackrel{\\text{?}}{=}-28[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>\u00a0[latex]-28=-28[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince this is a true statement, [latex]x=-7[\/latex] is a solution to [latex]4x=-28[\/latex].\r\n\r\n<\/div>\r\nNow you can try to solve an equation that requires division and\u00a0includes negative numbers.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try\u00a0it<\/h3>\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141857&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\nNow, consider the equation [latex]\\Large\\frac{x}{4}\\normalsize=3[\/latex]. We want to know what number divided by [latex]4[\/latex] gives [latex]3[\/latex]. So to \"undo\" the division, we will need to multiply by [latex]4[\/latex]. The <em data-effect=\"italics\">Multiplication Property of Equality<\/em> will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Multiplication Property of Equality<\/h3>\r\nFor all real numbers [latex]a,b,c[\/latex], if [latex]a=b[\/latex], then [latex]ac=bc[\/latex].\r\n\r\nStated simply, when you multiply both sides of an equation by the same quantity, you still have equality.\r\n\r\n<\/div>\r\nPreviously we learned how to \"undo\" multiplication by dividing. How do you think we \"undo\" division?\r\n\r\nNext, we will show an example that requires us to use multiplication to undo division.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]\\Large\\frac{a}{-7}\\normalsize =-42[\/latex]\r\n\r\n[reveal-answer q=\"399032\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"399032\"]\r\n\r\nSolution:\r\nHere [latex]a[\/latex] is divided by [latex]-7[\/latex]. We can multiply both sides by [latex]-7[\/latex] to isolate [latex]a[\/latex].\r\n<table id=\"eip-id1168468288515\" class=\"unnumbered unstyled\" summary=\"The top shows a over negative 7 equals negative 42. The next line says \">\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\Large\\frac{a}{-7}\\normalsize =-42[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply both sides by [latex]-7[\/latex] .<\/td>\r\n<td>[latex]\\color{red}{-7}(\r\n\r\n\\Large\\frac{a}{-7}\\normalsize)=\\color{red}{-7}(-42)[\/latex]\r\n\r\n[latex]\r\n\r\n\\Large\\frac{-7a}{-7}\\normalsize=294[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]a=294[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check your answer.<\/td>\r\n<td>[latex]\\Large\\frac{a}{-7}\\normalsize=-42[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Let [latex]a=294[\/latex] .<\/td>\r\n<td>[latex]\\Large\\frac{\\color{red}{294}}{-7}\\normalsize\\stackrel{\\text{?}}{=}-42[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]-42=-42\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow see if you can solve a\u00a0problem that requires multiplication to undo division. Recall the rules for multiplying two negative numbers \u2014 two negatives give a positive when they are multiplied.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try\u00a0it<\/h3>\r\n<iframe id=\"mom21\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141868&amp;theme=oea&amp;iframe_resize_id=mom21\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\nAs you begin to solve equations that require several steps you may find that you end up with an equation that looks like the one in the next example, with a negative variable. \u00a0As a standard practice, it is good to ensure that variables are positive when you are solving equations. The next example will show you how.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]-r=2[\/latex]\r\n\r\n[reveal-answer q=\"388033\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"388033\"]\r\n\r\nSolution:\r\nRemember [latex]-r[\/latex] is equivalent to [latex]-1r[\/latex].\r\n<table id=\"eip-id1168469604717\" class=\"unnumbered unstyled\" summary=\"The first line says negative r equals 2. The next line says \">\r\n<tbody>\r\n<tr>\r\n<td>[latex]-r=2[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Rewrite [latex]-r[\/latex] as [latex]-1r[\/latex] .<\/td>\r\n<td>[latex]-1r=2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Divide both sides by [latex]-1[\/latex] .<\/td>\r\n<td>[latex]\\Large\\frac{-1r}{\\color{red}{-1}}\\normalsize =\\Large\\frac{2}{\\color{red}{-1}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]r=-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Check.<\/td>\r\n<td>[latex]-r=2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]r=-2[\/latex]<\/td>\r\n<td>[latex]-(\\color{red}{-2})\\stackrel{\\text{?}}{=}2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]2=2\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow you can try to solve an equation with a negative variable.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try\u00a0it<\/h3>\r\n<iframe id=\"mom22\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141865&amp;theme=oea&amp;iframe_resize_id=mom22\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\nThe next video includes examples of using the division and multiplication properties to solve basic equations.\r\n\r\nhttps:\/\/youtu.be\/BN7iVWWl2y0","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcome<\/h3>\n<ul>\n<li>Use the addition, subtraction, multiplication, and division properties to solve single-step equations<\/li>\n<\/ul>\n<\/div>\n<p>We will begin solving equations below, but first we need to understand what a solution to an equation looks like. The purpose in solving an equation is to find the value or values of the variable that make each side of the equation the same \u2014 it is the answer to the puzzle.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Solution of an Equation<\/h3>\n<p>A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.<\/p>\n<p>Determine whether a number is a solution to an equation.<\/p>\n<ol id=\"eip-95\" class=\"stepwise\">\n<li>Substitute the number for the variable in the equation.<\/li>\n<li>Simplify the expressions on both sides of the equation.<\/li>\n<li>Determine whether the resulting equation is true.\n<ul id=\"fs-id1703984\">\n<li>If it is true, the number is a solution.<\/li>\n<li>If it is not true, the number is not a solution.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<p>In the following example, we will show how to determine whether a number is a solution to an equation that contains addition and subtraction. You can use this idea to check your work later when you are solving equations.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Determine whether [latex]y=\\Large\\frac{3}{4}[\/latex] is a solution for [latex]4y+3=8y[\/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><\/td>\n<td>[latex]4y+3=8y[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]\\color{red}{\\Large\\frac{3}{4}}[\/latex] for [latex]y[\/latex]<\/td>\n<td>[latex]4(\\color{red}{\\Large\\frac{3}{4}}\\normalsize)+3\\stackrel{\\text{?}}{=}8(\\color{red}{\\Large\\frac{3}{4}})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]3+3\\stackrel{\\text{?}}{=}6[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add.<\/td>\n<td>[latex]6=6\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since [latex]y=\\Large\\frac{3}{4}[\/latex] results in a true equation, [latex]\\Large\\frac{3}{4}[\/latex] is a solution to the equation [latex]4y+3=8y[\/latex].<\/p>\n<\/div>\n<p>Now it is your turn to determine whether a fraction is the solution to an equation.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141719&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"380\"><\/iframe><\/p>\n<\/div>\n<p>Let&#8217;s use a model to help understand how the process of solving an equation is like solving a puzzle. An envelope represents the variable \u2013 since its contents are unknown \u2013 and each counter represents one.<\/p>\n<p>Suppose a desk has an imaginary line dividing it in half. We place three counters and an envelope on the left side of desk, and eight counters on the right side of the desk.\u00a0Both sides of the desk have the same number of counters, but some counters are hidden in the envelope. Can you tell how many counters are in the envelope?<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215900\/CNX_BMath_Figure_02_03_001.png\" alt=\"The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 8 counters.\" \/><br \/>\nWhat steps are you taking in your mind to figure out how many counters are in the envelope? Perhaps you are thinking &#8220;I need to remove the [latex]3[\/latex] counters from the left side to get the envelope by itself. Those [latex]3[\/latex] counters on the left match with [latex]3[\/latex] on the right, so I can take them away from both sides. That leaves five counters on the right, so there must be [latex]5[\/latex] counters in the envelope.&#8221;<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215129\/CNX_BMath_Figure_02_03_002.png\" alt=\"The image is in two parts. On the left is a rectangle divided in half vertically. On the left side of the rectangle is an envelope with three counters below it. The 3 counters are circled in red with an arrow pointing out of the rectangle. On the right side is 8 counters. The bottom 3 counters are circled in red with an arrow pointing out of the rectangle. The 3 circled counters are removed from both sides of the rectangle, creating the new rectangle on the right of the image which is also divided in half vertically. On the left side of the rectangle is just an envelope. On the right side is 5 counters.\" \/><br \/>\nWhat algebraic equation is modeled by this situation? Each side of the desk represents an expression and the center line takes the place of the equal sign. We will call the contents of the envelope [latex]x[\/latex], so the number of counters on the left side of the desk is [latex]x+3[\/latex]. On the right side of the desk are [latex]8[\/latex] counters. We are told that [latex]x+3[\/latex] is equal to [latex]8[\/latex] so our equation is [latex]x+3=8[\/latex].<\/p>\n<p style=\"text-align: center;\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215900\/CNX_BMath_Figure_02_03_001.png\" alt=\"The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 8 counters.\" \/><br \/>\n[latex]x+3=8[\/latex]<\/p>\n<p style=\"text-align: left;\">Let\u2019s write algebraically the steps we took to discover how many counters were in the envelope.<\/p>\n<table id=\"eip-id1168467162379\" class=\"unnumbered unstyled\" summary=\"In this image, the first line shows x plus three equals eight. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]x+3=8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>First, we took away three from each side.<\/td>\n<td>[latex]x+3\\color{red}{--3}=8\\color {red}{--3}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Then we were left with five.<\/td>\n<td>[latex]x=5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now let\u2019s check our solution. We substitute [latex]5[\/latex] for [latex]x[\/latex] in the original equation and see if we get a true statement.<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]x+3=8[\/latex]<\/p>\n<p style=\"text-align: left; padding-left: 60px;\">[latex]\\color{red}{5}+3=8[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">[latex]8=8\\quad\\checkmark[\/latex]\u00a0 Our solution is correct. Five counters in the envelope plus three more equals eight.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Write an equation modeled by the envelopes and counters, and then solve the equation:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215904\/CNX_BMath_Figure_02_03_003_img.png\" alt=\"The image is divided in half vertically. On the left side is an envelope with 4 counters below it. On the right side is 5 counters.\" \/><\/p>\n<p>Solution<\/p>\n<table id=\"eip-id1168468230261\" class=\"unnumbered unstyled\" summary=\".\">\n<tbody>\n<tr>\n<td>On the left, write [latex]x[\/latex] for the contents of the envelope, add the [latex]4[\/latex] counters, so we have [latex]x+4[\/latex] .<\/td>\n<td>[latex]x+4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>On the right, there are [latex]5[\/latex] counters.<\/td>\n<td>[latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>The two sides are equal.<\/td>\n<td>[latex]x+4=5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Solve the equation by subtracting [latex]4[\/latex] counters from each side.<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215905\/CNX_BMath_Figure_02_03_005_img.png\" alt=\"The image is in two parts. On the left is a rectangle divided in half vertically. On the left side of the rectangle is an envelope with 4 counters below it. The 4 counters are circled in red with an arrow pointing out of the rectangle. On the right side is 5 counters. The bottom 4 counters are circled in red with an arrow pointing out of the rectangle. The 4 circled counters are removed from both sides of the rectangle, creating the new rectangle on the right of the image which is also divided in half vertically. On the left side of the rectangle is just an envelope. On the right side is 1 counter.\" \/><br \/>\nWe can see that there is one counter in the envelope. This can be shown algebraically as:<\/p>\n<p style=\"padding-left: 60px;\">[latex]x+4=5[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">[latex]x+4\\color{red}{--4}=5\\color{red}{--4}[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">[latex]x=1[\/latex]<\/p>\n<p>Substitute [latex]1[\/latex] for [latex]x[\/latex] in the equation to check.<\/p>\n<p style=\"padding-left: 60px;\">[latex]x+4=5[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">[latex]\\color{red}{1}+4=5[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">[latex]5=5\\quad\\checkmark[\/latex]<\/p>\n<p>Since [latex]x=1[\/latex] makes the statement true, we know that [latex]1[\/latex] is indeed a solution.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Write the equation modeled by the envelopes and counters, and then solve the equation:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215908\/CNX_BMath_Figure_02_03_004_img.png\" alt=\"The image is divided in half vertically. On the left side is an envelope with one counter below it. On the right side is 7 counters.\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q7845\">Show Answer<\/span><\/p>\n<div id=\"q7845\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x+1=7[\/latex]<\/p>\n<p>[latex]x=6[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>Write the equation modeled by the envelopes and counters, and then solve the equation:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24215909\/CNX_BMath_Figure_02_03_006_img.png\" alt=\"The image is divided in half vertically. On the left side is an envelope with three counters below it. On the right side is 4 counters.\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q9034\">Show Answer<\/span><\/p>\n<div id=\"q9034\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x+3=4[\/latex]<\/p>\n<p>[latex]x=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Our puzzle has given us an idea of what we need to do to solve an equation. The goal is to isolate the variable by itself on one side of the equation. In the previous examples, we used the Subtraction Property of Equality, which states that when we subtract the same quantity from both sides of an equation, we still have equality.<\/p>\n<div class=\"textbox shaded\">\n<h3 id=\"fs-id1956928\"><strong>Subtraction Property of Equality<\/strong><\/h3>\n<p>For all real numbers [latex]a,b[\/latex], and [latex]c[\/latex], if [latex]a=b[\/latex], then [latex]a-c=b-c[\/latex].<\/p>\n<\/div>\n<p>In all the equations we have solved so far, a number was added to the variable on one side of the equation. We used subtraction to &#8220;undo&#8221; the addition in order to isolate the variable.<\/p>\n<p>But suppose we have an equation with a number subtracted from the variable, such as [latex]x - 5=8[\/latex]. We want to isolate the variable, so to &#8220;undo&#8221; the subtraction we will add the number to both sides.<\/p>\n<p>We use the Addition Property of Equality, which says we can add the same number to both sides of the equation without changing the equality. Notice how it mirrors the Subtraction Property of Equality.<\/p>\n<div class=\"textbox shaded\">\n<h3 id=\"fs-id1166490863495\"><strong>Addition Property of Equality<\/strong><\/h3>\n<p>For all real numbers [latex]a,b[\/latex], and [latex]c[\/latex], if [latex]a=b[\/latex], then [latex]a+c=b+c[\/latex].<\/p>\n<\/div>\n<p>When you add or subtract the same quantity from both sides of an equation, you still have equality.<\/p>\n<p>We introduced the Subtraction Property of Equality earlier by modeling equations with envelopes and counters. The image below\u00a0models the equation [latex]x+3=8[\/latex].<\/p>\n<figure id=\"CNX_BMath_Figure_08_01_026\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24222533\/CNX_BMath_Figure_08_01_026.png\" alt=\"An envelope and three yellow counters are shown on the left side. On the right side are eight yellow counters.\" width=\"218\" height=\"133\" \/><\/figure>\n<p>The goal is to isolate the variable on one side of the equation. So we &#8220;took away&#8221; [latex]3[\/latex] from both sides of the equation and found the solution [latex]x=5[\/latex].<\/p>\n<p>Some people picture a balance scale, as in the image below, when they solve equations.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24222535\/CNX_BMath_Figure_08_01_001.png\" alt=\"Three balance scales are shown. The top scale has one mass on each side and is balanced. The middle scale has 2 masses on each side and is balanced. The bottom scale has 1 mass on one side and 2 masses on the other and is unbalanced.\" width=\"386\" height=\"406\" \/><br \/>\nThe quantities on both sides of the equal sign in an equation are equal, or balanced. Just as with the balance scale, whatever you do to one side of the equation you must also do to the other to keep it balanced.<\/p>\n<p>In the following examples we\u00a0review how to use Subtraction and Addition Properties of Equality to solve equations. We need to isolate the variable on one side of the equation. You can check your solutions by substituting the value into the original equation to make sure you have a true statement.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Solve: [latex]x+11=-3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q190834\">Show Answer<\/span><\/p>\n<div id=\"q190834\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<br \/>\nTo isolate [latex]x[\/latex], we undo the addition of [latex]11[\/latex] by using the Subtraction Property of Equality.<\/p>\n<table id=\"eip-id1168468291100\" class=\"unnumbered unstyled\" summary=\"The top line says x plus 11 equals negative 3. The next line says,\">\n<tbody>\n<tr>\n<td colspan=\"2\"><\/td>\n<td>[latex]x+11=-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Subtract 11 from each side to &#8220;undo&#8221; the addition.<\/td>\n<td>[latex]x+11\\color{red}{-11}=-3\\color{red}{-11}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Simplify.<\/td>\n<td>[latex]x=-14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>[latex]x+11=-3[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]x=-14[\/latex] .<\/td>\n<td>[latex]\\color{red}{-14}+11\\stackrel{\\text{?}}{=}-3[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]-3=-3\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since [latex]x=-14[\/latex] makes [latex]x+11=-3[\/latex] a true statement, we know that it is a solution to the equation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Now you can try solving an equation that requires using the addition property.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>TRY IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom200\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141721&amp;theme=oea&amp;iframe_resize_id=mom200\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>In the original equation in the previous example, [latex]11[\/latex] was added to the [latex]x[\/latex] , so we subtracted [latex]11[\/latex] to &#8220;undo&#8221; the addition. In the next example, we will need to &#8220;undo&#8221; subtraction by using the Addition Property of Equality.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Solve: [latex]m - 4=-5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q604060\">Show Answer<\/span><\/p>\n<div id=\"q604060\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<\/p>\n<table id=\"eip-id1168467157298\" class=\"unnumbered unstyled\" summary=\"The first line says m minus 4 equals negative 5. The next line says,\">\n<tbody>\n<tr>\n<td colspan=\"2\"><\/td>\n<td>[latex]m-4=-5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Add 4 to each side to &#8220;undo&#8221; the subtraction.<\/td>\n<td>[latex]m-4\\color{red}{+4}=-5\\color{red}{+4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td colspan=\"2\">Simplify.<\/td>\n<td>[latex]m=-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check:<\/td>\n<td>[latex]m-4=-5[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]m=-1[\/latex] .<\/td>\n<td>[latex]\\color{red}{-1}+4\\stackrel{\\text{?}}{=}-5[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]-5=-5\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><\/td>\n<td>The solution to [latex]m - 4=-5[\/latex] is [latex]m=-1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<p>Now you can try using the addition property to solve an equation.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>TRY\u00a0IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom27\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141723&amp;theme=oea&amp;iframe_resize_id=mom27\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>In the following video, we present more examples of solving equations using the addition and subtraction properties.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Solve One Step Equations By Add and Subtract Whole Numbers (Variable on Left)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/yqdlj0lv7Cc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You may encounter equations that contain fractions or decimals \u2014 especially in financial applications \u2014 so let&#8217;s practice solving those in the following problems.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>TRY\u00a0IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom25\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141732&amp;theme=oea&amp;iframe_resize_id=mom25\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY\u00a0IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm156990\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=156990&theme=oea&iframe_resize_id=ohm156990&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>All of the equations we have solved so far have been of the form [latex]x+a=b[\/latex] or [latex]x-a=b[\/latex]. We were able to isolate the variable by adding or subtracting the constant term. Now we\u2019ll see how to solve equations that involve division.<\/p>\n<p>We will model an equation with envelopes and counters.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220604\/CNX_BMath_Figure_03_05_001.png\" alt=\"This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed.\" \/><br \/>\nHere, there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are &#8220;hidden&#8221; in the envelopes. So how many counters are in each envelope?<\/p>\n<p>To determine the number, separate the counters on the right side into [latex]2[\/latex] groups of the same size. So [latex]6[\/latex] counters divided into [latex]2[\/latex] groups means there must be [latex]3[\/latex] counters in each group (since [latex]6\\div2=3[\/latex]).<\/p>\n<p>What equation models the situation shown in the figure below? There are two envelopes, and each contains [latex]x[\/latex] counters. Together, the two envelopes must contain a total of [latex]6[\/latex] counters. So the equation that models the situation is [latex]2x=6[\/latex].<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220605\/CNX_BMath_Figure_03_05_002.png\" alt=\"This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed. Under the figure is two times x equals 6.\" \/><br \/>\nWe can divide both sides of the equation by [latex]2[\/latex] as we did with the envelopes and counters.<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{\\frac{2x}{\\color{red}{2}}=\\frac{6}{\\color{red}{2}}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=3[\/latex]<\/p>\n<p>We found that each envelope contains [latex]\\text{3 counters.}[\/latex] Does this check? We know [latex]2\\cdot 3=6[\/latex], so it works. Three counters in each of two envelopes does equal six.<\/p>\n<p>Another example is shown below.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220607\/CNX_BMath_Figure_03_05_003.png\" alt=\"This image has two columns. In the first column are three envelopes. In the second column there are four rows of three blue circles. Underneath the image is the equation 3x equals 12.\" \/><br \/>\nNow we have [latex]3[\/latex] identical envelopes and [latex]\\text{12 counters.}[\/latex] How many counters are in each envelope? We have to separate the [latex]\\text{12 counters}[\/latex] into [latex]\\text{3 groups.}[\/latex] Since [latex]12\\div 3=4[\/latex], there must be [latex]\\text{4 counters}[\/latex] in each envelope.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24220608\/CNX_BMath_Figure_03_05_004.png\" alt=\"This image has two columns. In the first column are four envelopes. In the second column there are twelve blue circles.\" \/><br \/>\nThe equation that models the situation is [latex]3x=12[\/latex]. We can divide both sides of the equation by [latex]3[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{\\frac{3x}{\\color{red}{3}}=\\frac{12}{\\color{red}{3}}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=4[\/latex]<\/p>\n<p>Does this check? It does because [latex]3\\cdot 4=12[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Division Property of Equality<\/h3>\n<p>For all real numbers [latex]a,b,c[\/latex], and [latex]c\\ne 0[\/latex], if [latex]a=b[\/latex], then [latex]\\Large\\frac{a}{c}\\normalsize =\\Large\\frac{b}{c}[\/latex].<\/p>\n<p>Stated simply, when you divide both sides of an equation by the same quantity, you still have equality.<\/p>\n<\/div>\n<p>Remember, the goal is to &#8220;undo&#8221; the operation on the variable. In the example below the variable is multiplied by [latex]4[\/latex], so we will divide both sides by [latex]4[\/latex] to &#8220;undo&#8221; the multiplication.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]4x=-28[\/latex]<\/p>\n<p>Solution:<\/p>\n<p>To solve this equation, we use the Division Property of Equality to divide both sides by [latex]4[\/latex].<\/p>\n<table>\n<tbody>\n<tr>\n<td>[latex]4x=-28[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide both sides by 4 to undo the multiplication.<\/td>\n<td>[latex]\\Large\\frac{4x}{\\color{red}4}\\normalsize =\\Large\\frac{-28}{\\color{red}4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]x =-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check your answer.<\/td>\n<td>[latex]4x=-28[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Let [latex]x=-7[\/latex]. Substitute [latex]-7[\/latex] for x.<\/td>\n<td>[latex]4(\\color{red}{-7})\\stackrel{\\text{?}}{=}-28[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>\u00a0[latex]-28=-28[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since this is a true statement, [latex]x=-7[\/latex] is a solution to [latex]4x=-28[\/latex].<\/p>\n<\/div>\n<p>Now you can try to solve an equation that requires division and\u00a0includes negative numbers.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try\u00a0it<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141857&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<p>Now, consider the equation [latex]\\Large\\frac{x}{4}\\normalsize=3[\/latex]. We want to know what number divided by [latex]4[\/latex] gives [latex]3[\/latex]. So to &#8220;undo&#8221; the division, we will need to multiply by [latex]4[\/latex]. The <em data-effect=\"italics\">Multiplication Property of Equality<\/em> will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Multiplication Property of Equality<\/h3>\n<p>For all real numbers [latex]a,b,c[\/latex], if [latex]a=b[\/latex], then [latex]ac=bc[\/latex].<\/p>\n<p>Stated simply, when you multiply both sides of an equation by the same quantity, you still have equality.<\/p>\n<\/div>\n<p>Previously we learned how to &#8220;undo&#8221; multiplication by dividing. How do you think we &#8220;undo&#8221; division?<\/p>\n<p>Next, we will show an example that requires us to use multiplication to undo division.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]\\Large\\frac{a}{-7}\\normalsize =-42[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q399032\">Show Answer<\/span><\/p>\n<div id=\"q399032\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<br \/>\nHere [latex]a[\/latex] is divided by [latex]-7[\/latex]. We can multiply both sides by [latex]-7[\/latex] to isolate [latex]a[\/latex].<\/p>\n<table id=\"eip-id1168468288515\" class=\"unnumbered unstyled\" summary=\"The top shows a over negative 7 equals negative 42. The next line says\">\n<tbody>\n<tr>\n<td>[latex]\\Large\\frac{a}{-7}\\normalsize =-42[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply both sides by [latex]-7[\/latex] .<\/td>\n<td>[latex]\\color{red}{-7}(    \\Large\\frac{a}{-7}\\normalsize)=\\color{red}{-7}(-42)[\/latex]<\/p>\n<p>[latex]\\Large\\frac{-7a}{-7}\\normalsize=294[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]a=294[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check your answer.<\/td>\n<td>[latex]\\Large\\frac{a}{-7}\\normalsize=-42[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Let [latex]a=294[\/latex] .<\/td>\n<td>[latex]\\Large\\frac{\\color{red}{294}}{-7}\\normalsize\\stackrel{\\text{?}}{=}-42[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]-42=-42\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<p>Now see if you can solve a\u00a0problem that requires multiplication to undo division. Recall the rules for multiplying two negative numbers \u2014 two negatives give a positive when they are multiplied.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try\u00a0it<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom21\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141868&amp;theme=oea&amp;iframe_resize_id=mom21\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p>As you begin to solve equations that require several steps you may find that you end up with an equation that looks like the one in the next example, with a negative variable. \u00a0As a standard practice, it is good to ensure that variables are positive when you are solving equations. The next example will show you how.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]-r=2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q388033\">Show Answer<\/span><\/p>\n<div id=\"q388033\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution:<br \/>\nRemember [latex]-r[\/latex] is equivalent to [latex]-1r[\/latex].<\/p>\n<table id=\"eip-id1168469604717\" class=\"unnumbered unstyled\" summary=\"The first line says negative r equals 2. The next line says\">\n<tbody>\n<tr>\n<td>[latex]-r=2[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>Rewrite [latex]-r[\/latex] as [latex]-1r[\/latex] .<\/td>\n<td>[latex]-1r=2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Divide both sides by [latex]-1[\/latex] .<\/td>\n<td>[latex]\\Large\\frac{-1r}{\\color{red}{-1}}\\normalsize =\\Large\\frac{2}{\\color{red}{-1}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]r=-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Check.<\/td>\n<td>[latex]-r=2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]r=-2[\/latex]<\/td>\n<td>[latex]-(\\color{red}{-2})\\stackrel{\\text{?}}{=}2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]2=2\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<p>Now you can try to solve an equation with a negative variable.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try\u00a0it<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom22\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141865&amp;theme=oea&amp;iframe_resize_id=mom22\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<p>The next video includes examples of using the division and multiplication properties to solve basic equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Solving One Step Equations Using Multiplication and Division (Basic)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/BN7iVWWl2y0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-357\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Prealgebra. <strong>Provided by<\/strong>: OpenStax. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757<\/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":17533,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Prealgebra\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757\"}]","CANDELA_OUTCOMES_GUID":"8326a2db-5c35-4add-bcd8-f2f6fc41c836, 83cc8391-9ba3-4f84-b70e-6c932db4cebf","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-357","chapter","type-chapter","status-publish","hentry"],"part":26,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapters\/357","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":14,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapters\/357\/revisions"}],"predecessor-version":[{"id":3995,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapters\/357\/revisions\/3995"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/parts\/26"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapters\/357\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/wp\/v2\/media?parent=357"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/pressbooks\/v2\/chapter-type?post=357"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/wp\/v2\/contributor?post=357"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-accountingformanagers\/wp-json\/wp\/v2\/license?post=357"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}