{"id":3836,"date":"2020-02-09T18:47:09","date_gmt":"2020-02-09T18:47:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/mathforlibscoreq\/?post_type=chapter&#038;p=3836"},"modified":"2021-02-05T23:55:29","modified_gmt":"2021-02-05T23:55:29","slug":"solving-equations-using-the-subtraction-and-addition-properties-of-equality","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/chapter\/solving-equations-using-the-subtraction-and-addition-properties-of-equality\/","title":{"raw":"Solving Equations Using the Subtraction and Addition Properties of Equality","rendered":"Solving Equations Using the Subtraction and Addition Properties of Equality"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify the subtraction property of equality<\/li>\r\n \t<li>Solve a linear equation using the subtraction property of equality<\/li>\r\n \t<li>Solve equations using the addition property of equality<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Model the Subtraction Property of Equality<\/h2>\r\nWe will use a model to help you 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&nbsp;\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 Solution[\/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 Solution[\/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\n<h2>Solve Equations Using the Subtraction Property of Equality<\/h2>\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>Subtraction Property of Equality<\/h3>\r\nFor any numbers [latex]a,b[\/latex], and [latex]c[\/latex],\r\n\r\nif [latex]a=b[\/latex]\r\n\r\nthen [latex]a-c=b-c[\/latex]\r\n\r\n<\/div>\r\nThink about twin brothers Andy and Bobby. They are [latex]17[\/latex] years old. How old was Andy [latex]3[\/latex] years ago? He was [latex]3[\/latex] years less than [latex]17[\/latex], so his age was [latex]17 - 3[\/latex], or [latex]14[\/latex]. What about Bobby\u2019s age [latex]3[\/latex] years ago? Of course, he was [latex]14[\/latex] also. Their ages are equal now, and subtracting the same quantity from both of them resulted in equal ages [latex]3[\/latex] years ago.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}a=b\\\\ a - 3=b - 3\\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3>Solve an equation using the Subtraction Property of Equality<\/h3>\r\n<ol id=\"eip-id1168469785635\" class=\"stepwise\">\r\n \t<li>Use the Subtraction Property of Equality to isolate the variable.<\/li>\r\n \t<li>Simplify the expressions on both sides of the equation.<\/li>\r\n \t<li>Check the solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]x+8=17[\/latex].\r\n[reveal-answer q=\"494917\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"494917\"]\r\n\r\nSolution\r\nWe will use the Subtraction Property of Equality to isolate [latex]x[\/latex].\r\n<table id=\"eip-id1168467165114\" class=\"unnumbered unstyled\" summary=\"The image shows the equation, x plus 8 equal to 17. Take 8 away from both sides of the equation to get x plus 8 minus 8 equal to 17 minus 8. On the left, plus 8 and minus 8 cancel out to leave just x. On the right 17 minus 8 is 9. The equation becomes x equal to 9.Back to the original equation, x plus 8 equal to 17. Substitute 9 in for x to check. The equation becomes 9 plus 8 equal to 17. Is this true? The left side simplifies by adding 9 and 8 to get 17. Both sides of the equal symbol are 17.\">\r\n<tbody>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px\"><\/td>\r\n<td style=\"height: 15px\">[latex]x+8=17[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px\">\r\n<td style=\"height: 15px\">Subtract [latex]8[\/latex] from both sides.<\/td>\r\n<td style=\"height: 15px\">[latex]x+8\\color{red}{--8}=17\\color{red}{--8}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 20px\">\r\n<td style=\"height: 20px\">Simplify.<\/td>\r\n<td style=\"height: 20px\">[latex]x=9[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 21.71875px\">\r\n<td style=\"height: 21.71875px\"><\/td>\r\n<td style=\"height: 21.71875px\">[latex]x+8=17[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px\">\r\n<td style=\"height: 22px\"><\/td>\r\n<td style=\"height: 22px\">[latex]\\color{red}{9}+8=17[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 22px\">\r\n<td style=\"height: 22px\"><\/td>\r\n<td style=\"height: 22px\">[latex]17=17\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince [latex]x=9[\/latex] makes [latex]x+8=17[\/latex] a true statement, we know [latex]9[\/latex] is the solution to the equation.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]141712[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]100=y+74[\/latex].\r\n[reveal-answer q=\"410828\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"410828\"]\r\n\r\nSolution\r\nTo solve an equation, we must always isolate the variable\u2014it doesn\u2019t matter which side it is on. To isolate [latex]y[\/latex], we will subtract [latex]74[\/latex] from both sides.\r\n<table id=\"eip-id1168467375696\" class=\"unnumbered unstyled\" summary=\"The image shows the given equation, 100 equal to y plus 74. Take 74 away from both sides of the equation to get 100 minus 74 equal to y plus 74 minus 74. On the left 100 minus 74 is 26. On the right plus 74 and minus 74 cancel out to leave just y. The equation becomes 26 equal to y.\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]100=y+74[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract [latex]74[\/latex] from both sides.<\/td>\r\n<td>[latex]100\\color{red}{--74}=y+74\\color{red}{--74}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]26=y[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]26[\/latex] for [latex]y[\/latex] to check.\r\n\r\n[latex]100=y+74[\/latex]\r\n\r\n[latex]100=\\color{red}{26}+74[\/latex]\r\n\r\n[latex]100=100\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSince [latex]y=26[\/latex] makes [latex]100=y+74[\/latex] a true statement, we have found the solution to this equation.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]146457[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solve Equations Using the Addition Property of Equality<\/h2>\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>Addition Property of Equality<\/h3>\r\nFor any numbers [latex]a,b[\/latex] , and [latex]c[\/latex] ,\r\n\r\nif [latex]a=b[\/latex]\r\n\r\nthen [latex]a+c=b+c[\/latex]\r\n\r\n<\/div>\r\nRemember the [latex]17\\text{-year-old}[\/latex] twins, Andy and Bobby? In ten years, Andy\u2019s age will still equal Bobby\u2019s age. They will both be [latex]27[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}a=b\\\\ a+10=b+10\\end{array}[\/latex]\r\nWe can add the same number to both sides and still keep the equality.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3>Solve an equation using the Addition Property of Equality<\/h3>\r\n<ol id=\"eip-id1168469683666\" class=\"stepwise\">\r\n \t<li>Use the Addition Property of Equality to isolate the variable.<\/li>\r\n \t<li>Simplify the expressions on both sides of the equation.<\/li>\r\n \t<li>Check the solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]x - 5=8[\/latex].\r\n\r\nSolution\r\nWe will use the Addition Property of Equality to isolate the variable.\r\n<table id=\"eip-id1168467172615\" class=\"unnumbered unstyled\" summary=\"The image shows the given equation, x minus 5 equal to 8. Add 5 to both sides of the equation to get x minus 5 plus 5 equal to 8 plus 5. On the left, minus 5 plus 5 cancel out to leave just x. On the right 8 plus 5 is 13. The equation becomes x equal to 13. Back to the original equation, x minus 5 equal to 8. Substitute 13 in for x to check. The equation becomes 13 minus 5 equal to 8. Is this true? The left side simplifies by subtracting 5 from 13 to get 8. Both sides of the equal symbol are 8.\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]x--5=8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add [latex]5[\/latex] to both sides.<\/td>\r\n<td>[latex]x--5\\color{red}{+5}=8\\color{red}{+5}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]x=13[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Now we can check. Let [latex]x=\\color{red}{13}[\/latex].<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x--5=8[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\color{red}{13}--5=8[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]8=8\\quad\\checkmark[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]146458[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSolve: [latex]27=a - 16[\/latex].\r\n[reveal-answer q=\"360315\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"360315\"]\r\n\r\nSolution\r\nWe will add [latex]16[\/latex] to each side to isolate the variable.\r\n<table id=\"eip-id1168468296088\" class=\"unnumbered unstyled\" summary=\"The image shows the given equation, 27 equal to a minus 16. Add 16 to both sides of the equation to get 27 plus 16 equal to a minus 16 plus 16. On the left 27 plus 16 is 43. On the right, minus 16 plus 16 cancel out to leave just a. The equation becomes 43 equal to a. Back to the original equation,27 equal to a minus 16. Substitute 43 in for a to check. The equation becomes 27 equal to 43 minus 16. Is this true? The right side simplifies by subtracting 16 from 43 to get 27. Both sides of the equal symbol are 27.\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]27=a--16[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add [latex]16[\/latex] to each side.<\/td>\r\n<td>[latex]27\\color{red}{+16}=a--16\\color{red}{+16}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]43=a[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Now we can check. Let [latex]a=\\color{red}{43}[\/latex].<\/td>\r\n<td>[latex]27=a--16[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]27=\\color{red}{43}--16[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]27=27\\quad\\checkmark[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe solution to [latex]27=a - 16[\/latex] is [latex]a=43[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]146459[\/ohm_question]\r\n\r\n<\/div>\r\nIn the following video we show more examples of how to use the addition and subtraction properties of equality to solve one step linear equations involving whole numbers.\r\n\r\nhttps:\/\/youtu.be\/yqdlj0lv7Cc","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify the subtraction property of equality<\/li>\n<li>Solve a linear equation using the subtraction property of equality<\/li>\n<li>Solve equations using the addition property of equality<\/li>\n<\/ul>\n<\/div>\n<h2>Model the Subtraction Property of Equality<\/h2>\n<p>We will use a model to help you 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<p>&nbsp;<\/p>\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 Solution<\/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 Solution<\/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<h2>Solve Equations Using the Subtraction Property of Equality<\/h2>\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>Subtraction Property of Equality<\/h3>\n<p>For any numbers [latex]a,b[\/latex], and [latex]c[\/latex],<\/p>\n<p>if [latex]a=b[\/latex]<\/p>\n<p>then [latex]a-c=b-c[\/latex]<\/p>\n<\/div>\n<p>Think about twin brothers Andy and Bobby. They are [latex]17[\/latex] years old. How old was Andy [latex]3[\/latex] years ago? He was [latex]3[\/latex] years less than [latex]17[\/latex], so his age was [latex]17 - 3[\/latex], or [latex]14[\/latex]. What about Bobby\u2019s age [latex]3[\/latex] years ago? Of course, he was [latex]14[\/latex] also. Their ages are equal now, and subtracting the same quantity from both of them resulted in equal ages [latex]3[\/latex] years ago.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}a=b\\\\ a - 3=b - 3\\end{array}[\/latex]<\/p>\n<div class=\"textbox shaded\">\n<h3>Solve an equation using the Subtraction Property of Equality<\/h3>\n<ol id=\"eip-id1168469785635\" class=\"stepwise\">\n<li>Use the Subtraction Property of Equality to isolate the variable.<\/li>\n<li>Simplify the expressions on both sides of the equation.<\/li>\n<li>Check the solution.<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]x+8=17[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q494917\">Show Solution<\/span><\/p>\n<div id=\"q494917\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution<br \/>\nWe will use the Subtraction Property of Equality to isolate [latex]x[\/latex].<\/p>\n<table id=\"eip-id1168467165114\" class=\"unnumbered unstyled\" summary=\"The image shows the equation, x plus 8 equal to 17. Take 8 away from both sides of the equation to get x plus 8 minus 8 equal to 17 minus 8. On the left, plus 8 and minus 8 cancel out to leave just x. On the right 17 minus 8 is 9. The equation becomes x equal to 9.Back to the original equation, x plus 8 equal to 17. Substitute 9 in for x to check. The equation becomes 9 plus 8 equal to 17. Is this true? The left side simplifies by adding 9 and 8 to get 17. Both sides of the equal symbol are 17.\">\n<tbody>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px\"><\/td>\n<td style=\"height: 15px\">[latex]x+8=17[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 15px\">\n<td style=\"height: 15px\">Subtract [latex]8[\/latex] from both sides.<\/td>\n<td style=\"height: 15px\">[latex]x+8\\color{red}{--8}=17\\color{red}{--8}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 20px\">\n<td style=\"height: 20px\">Simplify.<\/td>\n<td style=\"height: 20px\">[latex]x=9[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 21.71875px\">\n<td style=\"height: 21.71875px\"><\/td>\n<td style=\"height: 21.71875px\">[latex]x+8=17[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 22px\">\n<td style=\"height: 22px\"><\/td>\n<td style=\"height: 22px\">[latex]\\color{red}{9}+8=17[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 22px\">\n<td style=\"height: 22px\"><\/td>\n<td style=\"height: 22px\">[latex]17=17\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since [latex]x=9[\/latex] makes [latex]x+8=17[\/latex] a true statement, we know [latex]9[\/latex] is the solution to the equation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm141712\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141712&theme=oea&iframe_resize_id=ohm141712&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]100=y+74[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q410828\">Show Solution<\/span><\/p>\n<div id=\"q410828\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution<br \/>\nTo solve an equation, we must always isolate the variable\u2014it doesn\u2019t matter which side it is on. To isolate [latex]y[\/latex], we will subtract [latex]74[\/latex] from both sides.<\/p>\n<table id=\"eip-id1168467375696\" class=\"unnumbered unstyled\" summary=\"The image shows the given equation, 100 equal to y plus 74. Take 74 away from both sides of the equation to get 100 minus 74 equal to y plus 74 minus 74. On the left 100 minus 74 is 26. On the right plus 74 and minus 74 cancel out to leave just y. The equation becomes 26 equal to y.\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]100=y+74[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract [latex]74[\/latex] from both sides.<\/td>\n<td>[latex]100\\color{red}{--74}=y+74\\color{red}{--74}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]26=y[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]26[\/latex] for [latex]y[\/latex] to check.<\/p>\n<p>[latex]100=y+74[\/latex]<\/p>\n<p>[latex]100=\\color{red}{26}+74[\/latex]<\/p>\n<p>[latex]100=100\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Since [latex]y=26[\/latex] makes [latex]100=y+74[\/latex] a true statement, we have found the solution to this equation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146457\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146457&theme=oea&iframe_resize_id=ohm146457&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solve Equations Using the Addition Property of Equality<\/h2>\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>Addition Property of Equality<\/h3>\n<p>For any numbers [latex]a,b[\/latex] , and [latex]c[\/latex] ,<\/p>\n<p>if [latex]a=b[\/latex]<\/p>\n<p>then [latex]a+c=b+c[\/latex]<\/p>\n<\/div>\n<p>Remember the [latex]17\\text{-year-old}[\/latex] twins, Andy and Bobby? In ten years, Andy\u2019s age will still equal Bobby\u2019s age. They will both be [latex]27[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}a=b\\\\ a+10=b+10\\end{array}[\/latex]<br \/>\nWe can add the same number to both sides and still keep the equality.<\/p>\n<div class=\"textbox shaded\">\n<h3>Solve an equation using the Addition Property of Equality<\/h3>\n<ol id=\"eip-id1168469683666\" class=\"stepwise\">\n<li>Use the Addition Property of Equality to isolate the variable.<\/li>\n<li>Simplify the expressions on both sides of the equation.<\/li>\n<li>Check the solution.<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]x - 5=8[\/latex].<\/p>\n<p>Solution<br \/>\nWe will use the Addition Property of Equality to isolate the variable.<\/p>\n<table id=\"eip-id1168467172615\" class=\"unnumbered unstyled\" summary=\"The image shows the given equation, x minus 5 equal to 8. Add 5 to both sides of the equation to get x minus 5 plus 5 equal to 8 plus 5. On the left, minus 5 plus 5 cancel out to leave just x. On the right 8 plus 5 is 13. The equation becomes x equal to 13. Back to the original equation, x minus 5 equal to 8. Substitute 13 in for x to check. The equation becomes 13 minus 5 equal to 8. Is this true? The left side simplifies by subtracting 5 from 13 to get 8. Both sides of the equal symbol are 8.\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]x--5=8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add [latex]5[\/latex] to both sides.<\/td>\n<td>[latex]x--5\\color{red}{+5}=8\\color{red}{+5}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]x=13[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Now we can check. Let [latex]x=\\color{red}{13}[\/latex].<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]x--5=8[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]\\color{red}{13}--5=8[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]8=8\\quad\\checkmark[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146458\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146458&theme=oea&iframe_resize_id=ohm146458&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Solve: [latex]27=a - 16[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q360315\">Show Solution<\/span><\/p>\n<div id=\"q360315\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution<br \/>\nWe will add [latex]16[\/latex] to each side to isolate the variable.<\/p>\n<table id=\"eip-id1168468296088\" class=\"unnumbered unstyled\" summary=\"The image shows the given equation, 27 equal to a minus 16. Add 16 to both sides of the equation to get 27 plus 16 equal to a minus 16 plus 16. On the left 27 plus 16 is 43. On the right, minus 16 plus 16 cancel out to leave just a. The equation becomes 43 equal to a. Back to the original equation,27 equal to a minus 16. Substitute 43 in for a to check. The equation becomes 27 equal to 43 minus 16. Is this true? The right side simplifies by subtracting 16 from 43 to get 27. Both sides of the equal symbol are 27.\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]27=a--16[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add [latex]16[\/latex] to each side.<\/td>\n<td>[latex]27\\color{red}{+16}=a--16\\color{red}{+16}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]43=a[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Now we can check. Let [latex]a=\\color{red}{43}[\/latex].<\/td>\n<td>[latex]27=a--16[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]27=\\color{red}{43}--16[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]27=27\\quad\\checkmark[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The solution to [latex]27=a - 16[\/latex] is [latex]a=43[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146459\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146459&theme=oea&iframe_resize_id=ohm146459&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following video we show more examples of how to use the addition and subtraction properties of equality to solve one step linear equations involving whole numbers.<\/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\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-3836\">\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>Question ID 146459, 146458, 146457. <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>. <strong>License Terms<\/strong>: IMathAS Community License<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex: Solve One Step Equations By Add and Subtract Whole Numbers (Variable on Left). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/yqdlj0lv7Cc\">https:\/\/youtu.be\/yqdlj0lv7Cc<\/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, Specific attribution<\/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 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