{"id":307,"date":"2023-06-21T13:22:54","date_gmt":"2023-06-21T13:22:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/the-substitution-and-addition-methods\/"},"modified":"2023-07-31T17:12:07","modified_gmt":"2023-07-31T17:12:07","slug":"the-substitution-and-addition-methods","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/the-substitution-and-addition-methods\/","title":{"raw":"\u25aa   The Substitution and Addition Methods","rendered":"\u25aa   The Substitution and Addition Methods"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the substitution method to find solution(s) to a system of two linear equations.<\/li>\r\n \t<li>Use the addition method to find solution(s) to a system of linear equations.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Solving Systems of Equations by Substitution<\/h2>\r\nSolving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a <strong>system of linear equations<\/strong> that are more precise than graphing. One such method is solving a system of equations by the <strong>substitution method<\/strong>, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of two equations in two variables, solve using the substitution method.<\/h3>\r\n<ol>\r\n \t<li>Solve one of the two equations for one of the variables in terms of the other.<\/li>\r\n \t<li>Substitute the expression for this variable into the second equation, then solve for the remaining variable.<\/li>\r\n \t<li>Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.<\/li>\r\n \t<li>Check the solution in both equations.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a System of Equations in Two Variables by Substitution<\/h3>\r\nSolve the following system of equations by substitution.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=-5 \\\\ 2x-5y&amp;=1 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"786744\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"786744\"]\r\n\r\nFirst, we will solve the first equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=-5 \\\\ y&amp;=x - 5 \\end{align}[\/latex]<\/p>\r\nNow we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 5y&amp;=1 \\\\ 2x - 5\\left(x - 5\\right)&amp;=1 \\\\ 2x - 5x+25&amp;=1 \\\\ -3x&amp;=-24 \\\\ x&amp;=8 \\end{align}[\/latex]<\/p>\r\nNow, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-\\left(8\\right)+y&amp;=-5 \\\\ y&amp;=3 \\end{align}[\/latex]<\/p>\r\nOur solution is [latex]\\left(8,3\\right)[\/latex].\r\n\r\nCheck the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=-5 \\\\ -\\left(8\\right)+\\left(3\\right)&amp;=-5 &amp;&amp; \\text{True} \\\\[3mm] 2x - 5y&amp;=1 \\\\ 2\\left(8\\right)-5\\left(3\\right)&amp;=1 &amp;&amp; \\text{True} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will be given an example of solving a system of two equations using the substitution method.\r\n\r\nhttps:\/\/youtu.be\/MIXL35YRzRw\r\n\r\nIf you had chosen the other equation to start with in the previous example, you would still be able to find the same solution. It is really a matter of preference, because sometimes solving for a variable will result in having to work with fractions. As you become more experienced with algebra, you will be able to anticipate what choices will lead to more desirable outcomes.\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nSince the concepts in this module rely heavily upon those from earlier modules, feel free to go back to review information you may not have fully obtained the first time. Math is best digested over time with repetition that leads to deeper appreciation of the ideas involved.\r\n\r\nAdditional practice will also help you build skills using learned concepts in creative ways to make good choices of technique in different situations.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nYou can use an online graphing calculator to help you solve a\u00a0system of equations by substitution. We will use the following system to show you how:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x&amp;=y+3 \\\\ 4&amp;=3x - 2y \\end{align}[\/latex]<\/p>\r\nFirst, solve both equations for y:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}y&amp;=x-3 \\\\ y&amp;=\\frac{3}{2}x - 2 \\end{align}[\/latex]<\/p>\r\nNow enter [latex]x-3=\\frac{3}{2}x - 2[\/latex] into the calculator. You will see that calculator has provided you with [latex]x = -2[\/latex].\r\n\r\nYou now can substitute\u00a0[latex]x = -2[\/latex] into both equations. \u00a0If you get the same result for both, you have found an ordered pair solution. Give it a try.\r\n\r\n[reveal-answer q=\"783260\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"783260\"]\r\n\r\n[latex]\\left(-2,-5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[ohm_question height=\"320\" ]115164[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h4>Can the substitution method be used to solve any linear system in two variables?<\/h4>\r\n<em>Yes, but the method works best if one of the equations contains a coefficient of 1 or \u20131 so that we do not have to deal with fractions.<\/em>\r\n\r\n<\/div>\r\nThe following video is ~10 minutes long and provides a mini-lesson on using the substitution method to solve a system of linear equations. \u00a0We present three different examples, and also use a graphing tool to help summarize the solution for each example.\r\n\r\nhttps:\/\/youtu.be\/HxhacvH49o8\r\n<h2>Solving Systems of Equations in Two Variables by the Addition Method<\/h2>\r\nA third method of <strong>solving systems of linear equations<\/strong> is the <strong>addition method,\u00a0<\/strong>this method is also called the\u00a0<strong>elimination method<\/strong>. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of equations, solve using the addition method.<\/h3>\r\n<ol>\r\n \t<li>Write both equations with <em>x<\/em>- and <em>y<\/em>-variables on the left side of the equal sign and constants on the right.<\/li>\r\n \t<li>Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.<\/li>\r\n \t<li>Solve the resulting equation for the remaining variable.<\/li>\r\n \t<li>Substitute that value into one of the original equations and solve for the second variable.<\/li>\r\n \t<li>Check the solution by substituting the values into the other equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a System by the Addition Method<\/h3>\r\nSolve the given system of equations by addition.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&amp;=-1 \\\\ -x+y&amp;=3 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"924657\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924657\"]\r\n\r\nBoth equations are already set equal to a constant. Notice that the coefficient of [latex]x[\/latex] in the second equation, \u20131, is the opposite of the coefficient of [latex]x[\/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} x+2y&amp;=-1 \\\\ -x+y&amp;=3 \\\\ \\hline 3y&amp;=2\\end{align}[\/latex]<\/p>\r\nNow that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3y&amp;=2 \\\\ y&amp;=\\dfrac{2}{3} \\end{align}[\/latex]<\/p>\r\nThen, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=3 \\\\ -x+\\frac{2}{3}&amp;=3 \\\\ -x&amp;=3-\\frac{2}{3} \\\\ -x&amp;=\\frac{7}{3} \\\\ x&amp;=-\\frac{7}{3} \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The solution to this system is [latex]\\left(-\\frac{7}{3},\\frac{2}{3}\\right)[\/latex].<\/p>\r\nCheck the solution in the first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&amp;=-1 \\\\ \\left(-\\frac{7}{3}\\right)+2\\left(\\frac{2}{3}\\right)&amp;= \\\\ -\\frac{7}{3}+\\frac{4}{3}&amp;= \\\\ -\\frac{3}{3}&amp;= \\\\ -1&amp;=-1&amp;&amp; \\text{True} \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWe gain an important perspective on systems of equations by looking at the graphical representation. See the graph below\u00a0to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183607\/CNX_Precalc_Figure_09_01_0042.jpg\" alt=\"A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.\" width=\"487\" height=\"291\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will see another example of how to use the method of elimination to solve a system of linear equations.\r\nhttps:\/\/youtu.be\/M4IEmwcqR3c\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[ohm_question height=\"280\"]115130[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Addition Method When Multiplication of One Equation Is Required<\/h3>\r\nSolve the given system of equations by the <strong>addition method<\/strong>.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&amp;=-11 \\\\ x - 2y&amp;=11 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"883001\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"883001\"]\r\n\r\nAdding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[\/latex] in it and the second equation has [latex]x[\/latex]. So if we multiply the second equation by [latex]-3,\\text{}[\/latex] the <em>x<\/em>-terms will add to zero.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&amp;=11 \\\\ -3\\left(x - 2y\\right)&amp;=-3\\left(11\\right) &amp;&amp; \\text{Multiply both sides by }-3 \\\\ -3x+6y&amp;=-33 &amp;&amp; \\text{Use the distributive property}. \\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now, let\u2019s add them.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&amp;=\u221211 \\\\ \u22123x+6y&amp;=\u221233 \\\\ \\hline 11y&amp;=\u221244 \\\\ y&amp;=\u22124 \\end{align}[\/latex]<\/p>\r\nFor the last step, we substitute [latex]y=-4[\/latex] into one of the original equations and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&amp;=-11\\\\ 3x+5\\left(-4\\right)&amp;=-11\\\\ 3x - 20&amp;=-11\\\\ 3x&amp;=9\\\\ x&amp;=3\\end{align}[\/latex]<\/p>\r\nOur solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&amp;=11 \\\\ \\left(3\\right)-2\\left(-4\\right)&amp;=3+8 \\\\ &amp;=11 &amp;&amp; \\text{True} \\end{align}[\/latex]<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183609\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nBelow is another video example of using the elimination method to solve a system of linear equations when multiplication of one equation is required.\r\nhttps:\/\/youtu.be\/_liDhKops2w\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nRemember to work through the given examples and practice problems on paper and more than once to deepen your understanding.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the system of equations by addition.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 7y&amp;=2\\\\ 3x+y&amp;=-20\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"609174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"609174\"]\r\n\r\n[latex]\\left(-6,-2\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[ohm_question height=\"295\"]115120[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Addition Method When Multiplication of Both Equations Is Required<\/h3>\r\nSolve the given system of equations in two variables by addition.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+3y&amp;=-16 \\\\ 5x - 10y&amp;=30\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"114755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"114755\"]\r\n\r\nOne equation has [latex]2x[\/latex] and the other has [latex]5x[\/latex]. The least common multiple is [latex]10x[\/latex] so we will have to multiply both equations by a constant in order to eliminate one variable. Let\u2019s eliminate [latex]x[\/latex] by multiplying the first equation by [latex]-5[\/latex] and the second equation by [latex]2[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} -5\\left(2x+3y\\right)&amp;=-5\\left(-16\\right) \\\\ -10x - 15y&amp;=80 \\\\[3mm] 2\\left(5x - 10y\\right)&amp;=2\\left(30\\right) \\\\ 10x - 20y&amp;=60 \\end{align}[\/latex]<\/p>\r\nThen, we add the two equations together.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \u221210x\u221215y&amp;=80 \\\\ 10x\u221220y&amp;=60 \\\\ \\hline \u221235y&amp;=140 \\\\ y&amp;=\u22124 \\end{align}[\/latex]<\/p>\r\nSubstitute [latex]y=-4[\/latex] into the original first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+3\\left(-4\\right)&amp;=-16\\\\ 2x - 12&amp;=-16\\\\ 2x&amp;=-4\\\\ x&amp;=-2\\end{align}[\/latex]<\/p>\r\nThe solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the other equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 5x - 10y&amp;=30\\\\ 5\\left(-2\\right)-10\\left(-4\\right)&amp;=30\\\\ -10+40&amp;=30\\\\ 30&amp;=30\\end{align}[\/latex]<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183611\/CNX_Precalc_Figure_09_01_0062.jpg\" alt=\"A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.\" width=\"487\" height=\"366\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Addition Method in Systems of Equations Containing Fractions<\/h3>\r\nSolve the given system of equations in two variables by addition.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{x}{3}+\\frac{y}{6}&amp;=3 \\\\[1mm] \\frac{x}{2}-\\frac{y}{4}&amp;=1 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"359287\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"359287\"]\r\n\r\nFirst clear each equation of fractions by multiplying both sides of the equation by the least common denominator.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}6\\left(\\frac{x}{3}+\\frac{y}{6}\\right)&amp;=6\\left(3\\right) \\\\[1mm] 2x+y&amp;=18 \\\\[3mm] 4\\left(\\frac{x}{2}-\\frac{y}{4}\\right)&amp;=4\\left(1\\right) \\\\[1mm] 2x-y&amp;=4 \\end{align}[\/latex]<\/p>\r\nNow multiply the second equation by [latex]-1[\/latex] so that we can eliminate\u00a0<em>x<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}-1\\left(2x-y\\right)&amp;=-1\\left(4\\right) \\\\[1mm] -2x+y&amp;=-4 \\end{align}[\/latex]<\/p>\r\nAdd the two equations to eliminate\u00a0<em>x<\/em>\u00a0and solve the resulting equation for <em>y<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x+y&amp;=18 \\\\ \u22122x+y&amp;=\u22124 \\\\ \\hline 2y&amp;=14 \\\\ y&amp;=7 \\end{align}[\/latex]<\/p>\r\nSubstitute [latex]y=7[\/latex] into the first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+\\left(7\\right)&amp;=18 \\\\ 2x&amp;=11 \\\\ x&amp;=\\frac{11}{2} \\\\ &amp;=7.5 \\end{align}[\/latex]<\/p>\r\nThe solution is [latex]\\left(\\frac{11}{2},7\\right)[\/latex]. Check it in the other equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{x}{2}-\\frac{y}{4}&amp;=1\\\\[1mm] \\frac{\\frac{11}{2}}{2}-\\frac{7}{4}&amp;=1\\\\[1mm] \\frac{11}{4}-\\frac{7}{4}&amp;=1\\\\[1mm] \\frac{4}{4}&amp;=1\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will find one more example of using the elimination method to solve a system; this one has coefficients that are fractions.\r\nhttps:\/\/youtu.be\/s3S64b1DrtQ\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the system of equations by addition.\r\n\r\n[latex]\\begin{align}2x+3y&amp;=8\\\\ 3x+5y&amp;=10\\end{align}[\/latex]\r\n\r\n[reveal-answer q=\"326265\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"326265\"]\r\n\r\n[latex]\\left(10,-4\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[ohm_question height=\"275\"]115110[\/ohm_question]\r\n\r\n<\/div>\r\nin the following video we present more examples of how to use the addition (elimination) method to solve a system of two linear equations.\r\n\r\nhttps:\/\/youtu.be\/ova8GSmPV4o","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the substitution method to find solution(s) to a system of two linear equations.<\/li>\n<li>Use the addition method to find solution(s) to a system of linear equations.<\/li>\n<\/ul>\n<\/div>\n<h2>Solving Systems of Equations by Substitution<\/h2>\n<p>Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a <strong>system of linear equations<\/strong> that are more precise than graphing. One such method is solving a system of equations by the <strong>substitution method<\/strong>, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a system of two equations in two variables, solve using the substitution method.<\/h3>\n<ol>\n<li>Solve one of the two equations for one of the variables in terms of the other.<\/li>\n<li>Substitute the expression for this variable into the second equation, then solve for the remaining variable.<\/li>\n<li>Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.<\/li>\n<li>Check the solution in both equations.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a System of Equations in Two Variables by Substitution<\/h3>\n<p>Solve the following system of equations by substitution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=-5 \\\\ 2x-5y&=1 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q786744\">Show Solution<\/span><\/p>\n<div id=\"q786744\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we will solve the first equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=-5 \\\\ y&=x - 5 \\end{align}[\/latex]<\/p>\n<p>Now we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 5y&=1 \\\\ 2x - 5\\left(x - 5\\right)&=1 \\\\ 2x - 5x+25&=1 \\\\ -3x&=-24 \\\\ x&=8 \\end{align}[\/latex]<\/p>\n<p>Now, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-\\left(8\\right)+y&=-5 \\\\ y&=3 \\end{align}[\/latex]<\/p>\n<p>Our solution is [latex]\\left(8,3\\right)[\/latex].<\/p>\n<p>Check the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=-5 \\\\ -\\left(8\\right)+\\left(3\\right)&=-5 && \\text{True} \\\\[3mm] 2x - 5y&=1 \\\\ 2\\left(8\\right)-5\\left(3\\right)&=1 && \\text{True} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will be given an example of solving a system of two equations using the substitution method.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 2:  Solve a System of Equations Using Substitution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/MIXL35YRzRw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>If you had chosen the other equation to start with in the previous example, you would still be able to find the same solution. It is really a matter of preference, because sometimes solving for a variable will result in having to work with fractions. As you become more experienced with algebra, you will be able to anticipate what choices will lead to more desirable outcomes.<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Since the concepts in this module rely heavily upon those from earlier modules, feel free to go back to review information you may not have fully obtained the first time. Math is best digested over time with repetition that leads to deeper appreciation of the ideas involved.<\/p>\n<p>Additional practice will also help you build skills using learned concepts in creative ways to make good choices of technique in different situations.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>You can use an online graphing calculator to help you solve a\u00a0system of equations by substitution. We will use the following system to show you how:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x&=y+3 \\\\ 4&=3x - 2y \\end{align}[\/latex]<\/p>\n<p>First, solve both equations for y:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}y&=x-3 \\\\ y&=\\frac{3}{2}x - 2 \\end{align}[\/latex]<\/p>\n<p>Now enter [latex]x-3=\\frac{3}{2}x - 2[\/latex] into the calculator. You will see that calculator has provided you with [latex]x = -2[\/latex].<\/p>\n<p>You now can substitute\u00a0[latex]x = -2[\/latex] into both equations. \u00a0If you get the same result for both, you have found an ordered pair solution. Give it a try.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q783260\">Show Solution<\/span><\/p>\n<div id=\"q783260\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(-2,-5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm115164\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115164&theme=oea&iframe_resize_id=ohm115164&show_question_numbers\" width=\"100%\" height=\"320\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>Can the substitution method be used to solve any linear system in two variables?<\/h4>\n<p><em>Yes, but the method works best if one of the equations contains a coefficient of 1 or \u20131 so that we do not have to deal with fractions.<\/em><\/p>\n<\/div>\n<p>The following video is ~10 minutes long and provides a mini-lesson on using the substitution method to solve a system of linear equations. \u00a0We present three different examples, and also use a graphing tool to help summarize the solution for each example.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Solving Systems of Equations using Substitution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/HxhacvH49o8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Solving Systems of Equations in Two Variables by the Addition Method<\/h2>\n<p>A third method of <strong>solving systems of linear equations<\/strong> is the <strong>addition method,\u00a0<\/strong>this method is also called the\u00a0<strong>elimination method<\/strong>. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, solve using the addition method.<\/h3>\n<ol>\n<li>Write both equations with <em>x<\/em>&#8211; and <em>y<\/em>-variables on the left side of the equal sign and constants on the right.<\/li>\n<li>Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.<\/li>\n<li>Solve the resulting equation for the remaining variable.<\/li>\n<li>Substitute that value into one of the original equations and solve for the second variable.<\/li>\n<li>Check the solution by substituting the values into the other equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a System by the Addition Method<\/h3>\n<p>Solve the given system of equations by addition.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&=-1 \\\\ -x+y&=3 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924657\">Show Solution<\/span><\/p>\n<div id=\"q924657\" class=\"hidden-answer\" style=\"display: none\">\n<p>Both equations are already set equal to a constant. Notice that the coefficient of [latex]x[\/latex] in the second equation, \u20131, is the opposite of the coefficient of [latex]x[\/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} x+2y&=-1 \\\\ -x+y&=3 \\\\ \\hline 3y&=2\\end{align}[\/latex]<\/p>\n<p>Now that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3y&=2 \\\\ y&=\\dfrac{2}{3} \\end{align}[\/latex]<\/p>\n<p>Then, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-x+y&=3 \\\\ -x+\\frac{2}{3}&=3 \\\\ -x&=3-\\frac{2}{3} \\\\ -x&=\\frac{7}{3} \\\\ x&=-\\frac{7}{3} \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">The solution to this system is [latex]\\left(-\\frac{7}{3},\\frac{2}{3}\\right)[\/latex].<\/p>\n<p>Check the solution in the first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&=-1 \\\\ \\left(-\\frac{7}{3}\\right)+2\\left(\\frac{2}{3}\\right)&= \\\\ -\\frac{7}{3}+\\frac{4}{3}&= \\\\ -\\frac{3}{3}&= \\\\ -1&=-1&& \\text{True} \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We gain an important perspective on systems of equations by looking at the graphical representation. See the graph below\u00a0to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183607\/CNX_Precalc_Figure_09_01_0042.jpg\" alt=\"A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.\" width=\"487\" height=\"291\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will see another example of how to use the method of elimination to solve a system of linear equations.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 1:  Solve a System of Equations Using the Elimination Method\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/M4IEmwcqR3c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm115130\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115130&theme=oea&iframe_resize_id=ohm115130&show_question_numbers\" width=\"100%\" height=\"280\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Addition Method When Multiplication of One Equation Is Required<\/h3>\n<p>Solve the given system of equations by the <strong>addition method<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&=-11 \\\\ x - 2y&=11 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q883001\">Show Solution<\/span><\/p>\n<div id=\"q883001\" class=\"hidden-answer\" style=\"display: none\">\n<p>Adding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[\/latex] in it and the second equation has [latex]x[\/latex]. So if we multiply the second equation by [latex]-3,\\text{}[\/latex] the <em>x<\/em>-terms will add to zero.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&=11 \\\\ -3\\left(x - 2y\\right)&=-3\\left(11\\right) && \\text{Multiply both sides by }-3 \\\\ -3x+6y&=-33 && \\text{Use the distributive property}. \\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now, let\u2019s add them.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&=\u221211 \\\\ \u22123x+6y&=\u221233 \\\\ \\hline 11y&=\u221244 \\\\ y&=\u22124 \\end{align}[\/latex]<\/p>\n<p>For the last step, we substitute [latex]y=-4[\/latex] into one of the original equations and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&=-11\\\\ 3x+5\\left(-4\\right)&=-11\\\\ 3x - 20&=-11\\\\ 3x&=9\\\\ x&=3\\end{align}[\/latex]<\/p>\n<p>Our solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x - 2y&=11 \\\\ \\left(3\\right)-2\\left(-4\\right)&=3+8 \\\\ &=11 && \\text{True} \\end{align}[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183609\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Below is another video example of using the elimination method to solve a system of linear equations when multiplication of one equation is required.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex 2:  Solve a System of Equations Using the Elimination Method\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/_liDhKops2w?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Remember to work through the given examples and practice problems on paper and more than once to deepen your understanding.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the system of equations by addition.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x - 7y&=2\\\\ 3x+y&=-20\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q609174\">Show Solution<\/span><\/p>\n<div id=\"q609174\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(-6,-2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm115120\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115120&theme=oea&iframe_resize_id=ohm115120&show_question_numbers\" width=\"100%\" height=\"295\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Addition Method When Multiplication of Both Equations Is Required<\/h3>\n<p>Solve the given system of equations in two variables by addition.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+3y&=-16 \\\\ 5x - 10y&=30\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q114755\">Show Solution<\/span><\/p>\n<div id=\"q114755\" class=\"hidden-answer\" style=\"display: none\">\n<p>One equation has [latex]2x[\/latex] and the other has [latex]5x[\/latex]. The least common multiple is [latex]10x[\/latex] so we will have to multiply both equations by a constant in order to eliminate one variable. Let\u2019s eliminate [latex]x[\/latex] by multiplying the first equation by [latex]-5[\/latex] and the second equation by [latex]2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} -5\\left(2x+3y\\right)&=-5\\left(-16\\right) \\\\ -10x - 15y&=80 \\\\[3mm] 2\\left(5x - 10y\\right)&=2\\left(30\\right) \\\\ 10x - 20y&=60 \\end{align}[\/latex]<\/p>\n<p>Then, we add the two equations together.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \u221210x\u221215y&=80 \\\\ 10x\u221220y&=60 \\\\ \\hline \u221235y&=140 \\\\ y&=\u22124 \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]y=-4[\/latex] into the original first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+3\\left(-4\\right)&=-16\\\\ 2x - 12&=-16\\\\ 2x&=-4\\\\ x&=-2\\end{align}[\/latex]<\/p>\n<p>The solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the other equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 5x - 10y&=30\\\\ 5\\left(-2\\right)-10\\left(-4\\right)&=30\\\\ -10+40&=30\\\\ 30&=30\\end{align}[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183611\/CNX_Precalc_Figure_09_01_0062.jpg\" alt=\"A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.\" width=\"487\" height=\"366\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Addition Method in Systems of Equations Containing Fractions<\/h3>\n<p>Solve the given system of equations in two variables by addition.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{x}{3}+\\frac{y}{6}&=3 \\\\[1mm] \\frac{x}{2}-\\frac{y}{4}&=1 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q359287\">Show Solution<\/span><\/p>\n<div id=\"q359287\" class=\"hidden-answer\" style=\"display: none\">\n<p>First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}6\\left(\\frac{x}{3}+\\frac{y}{6}\\right)&=6\\left(3\\right) \\\\[1mm] 2x+y&=18 \\\\[3mm] 4\\left(\\frac{x}{2}-\\frac{y}{4}\\right)&=4\\left(1\\right) \\\\[1mm] 2x-y&=4 \\end{align}[\/latex]<\/p>\n<p>Now multiply the second equation by [latex]-1[\/latex] so that we can eliminate\u00a0<em>x<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}-1\\left(2x-y\\right)&=-1\\left(4\\right) \\\\[1mm] -2x+y&=-4 \\end{align}[\/latex]<\/p>\n<p>Add the two equations to eliminate\u00a0<em>x<\/em>\u00a0and solve the resulting equation for <em>y<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x+y&=18 \\\\ \u22122x+y&=\u22124 \\\\ \\hline 2y&=14 \\\\ y&=7 \\end{align}[\/latex]<\/p>\n<p>Substitute [latex]y=7[\/latex] into the first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+\\left(7\\right)&=18 \\\\ 2x&=11 \\\\ x&=\\frac{11}{2} \\\\ &=7.5 \\end{align}[\/latex]<\/p>\n<p>The solution is [latex]\\left(\\frac{11}{2},7\\right)[\/latex]. Check it in the other equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\frac{x}{2}-\\frac{y}{4}&=1\\\\[1mm] \\frac{\\frac{11}{2}}{2}-\\frac{7}{4}&=1\\\\[1mm] \\frac{11}{4}-\\frac{7}{4}&=1\\\\[1mm] \\frac{4}{4}&=1\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will find one more example of using the elimination method to solve a system; this one has coefficients that are fractions.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex: Solve a System of Equations Using Eliminations (Fractions)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/s3S64b1DrtQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the system of equations by addition.<\/p>\n<p>[latex]\\begin{align}2x+3y&=8\\\\ 3x+5y&=10\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q326265\">Show Solution<\/span><\/p>\n<div id=\"q326265\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(10,-4\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"ohm115110\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115110&theme=oea&iframe_resize_id=ohm115110&show_question_numbers\" width=\"100%\" height=\"275\"><\/iframe><\/p>\n<\/div>\n<p>in the following video we present more examples of how to use the addition (elimination) method to solve a system of two linear equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Solving Systems of Equations using Elimination\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ova8GSmPV4o?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-307\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Solving Systems of Equations using Elimination. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ova8GSmPV4o\">https:\/\/youtu.be\/ova8GSmPV4o<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Solving Systems of Equations using Substitution . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/HxhacvH49o8\">https:\/\/youtu.be\/HxhacvH49o8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 115164, 115120, 115110. <strong>Authored by<\/strong>: Shabazian, Roy. <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 CC-BY + GPL<\/li><li>Ex 1: Solve a System of Equations Using the Elimination Method.. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/M4IEmwcqR3c\">https:\/\/youtu.be\/M4IEmwcqR3c<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Solve a System of Equations Using the Elimination Method. . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/_liDhKops2w\">https:\/\/youtu.be\/_liDhKops2w<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Solve a System of Equations Using Eliminations (Fractions).. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/s3S64b1DrtQ\">https:\/\/youtu.be\/s3S64b1DrtQ<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Solving Systems of Equations using Elimination\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ova8GSmPV4o\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Solving Systems of Equations using Substitution \",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/HxhacvH49o8\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 115164, 115120, 115110\",\"author\":\"Shabazian, 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