{"id":3388,"date":"2017-02-22T20:42:38","date_gmt":"2017-02-22T20:42:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=3388"},"modified":"2017-04-21T21:35:53","modified_gmt":"2017-04-21T21:35:53","slug":"the-substitution-and-addition-methods","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/the-substitution-and-addition-methods\/","title":{"raw":"The Substitution and Addition Methods","rendered":"The Substitution and Addition Methods"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/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\r\n[latex]\\begin{array}{l}\\text{ }-x+y=-5\\hfill \\\\ \\text{ }2x - 5y=1\\hfill \\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"786744\"]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\r\n[latex]\\begin{array}{l}-x+y=-5\\hfill \\\\ \\text{ }y=x - 5\\hfill \\end{array}[\/latex]\r\n\r\nNow we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.\r\n\r\n[latex]\\begin{array}{l}\\text{ }2x - 5y=1\\hfill \\\\ 2x - 5\\left(x - 5\\right)=1\\hfill \\\\ \\text{ }2x - 5x+25=1\\hfill \\\\ \\text{ }-3x=-24\\hfill \\\\ \\text{ }x=8\\hfill \\end{array}[\/latex]\r\n\r\nNow, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].\r\n\r\n[latex]\\begin{array}{l}-\\left(8\\right)+y=-5\\hfill \\\\ \\text{ }y=3\\hfill \\end{array}[\/latex]\r\n\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\r\n[latex]\\begin{array}{llll}-x+y=-5\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ -\\left(8\\right)+\\left(3\\right)=-5\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{True}\\hfill \\\\ 2x - 5y=1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ 2\\left(8\\right)-5\\left(3\\right)=1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nYou can use Desmos to help you solve a\u00a0system of equations by substitution. We will use the following system to show you how:\r\n\r\n[latex]\\begin{array}{l}x=y+3\\hfill \\\\ 4=3x - 2y\\hfill \\end{array}[\/latex]\r\n\r\nFirst, solve both equations for y:\r\n\r\n[latex]\\begin{array}{l}y=x-3\\hfill \\\\ y=\\frac{3}{2}x - 2\\hfill \\end{array}[\/latex]\r\n\r\nNow enter [latex]x-3=\\frac{3}{2}x - 2[\/latex] into Desmos. You will see that Desmos 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<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n\r\n[reveal-answer q=\"783260\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"783260\"][latex]\\left(-2,-5\\right)[\/latex][\/hidden-answer]\r\n\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\n&nbsp;\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\r\n[latex]\\begin{array}{l}x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"924657\"]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\r\n[latex]\\frac{\\begin{array}{l}\\hfill \\\\ x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}}{\\text{}\\text{}\\text{}\\text{}\\text{}3y=2}[\/latex]\r\n\r\nNow that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].\r\n\r\n[latex]\\begin{array}{l}3y=2\\hfill \\\\ \\text{ }y=\\frac{2}{3}\\hfill \\end{array}[\/latex]\r\n\r\nThen, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex].\r\n\r\n[latex]\\begin{array}{l}\\text{ }-x+y=3\\hfill \\\\ \\text{ }-x+\\frac{2}{3}=3\\hfill \\\\ \\text{ }-x=3-\\frac{2}{3}\\hfill \\\\ \\text{ }-x=\\frac{7}{3}\\hfill \\\\ \\text{ }x=-\\frac{7}{3}\\hfill \\end{array}[\/latex]\r\nThe solution to this system is [latex]\\left(-\\frac{7}{3},\\frac{2}{3}\\right)[\/latex].\r\n\r\nCheck the solution in the first equation.\r\n\r\n[latex]\\begin{array}{llll}\\text{ }x+2y=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }\\left(-\\frac{7}{3}\\right)+2\\left(\\frac{2}{3}\\right)=\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-\\frac{7}{3}+\\frac{4}{3}=\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-\\frac{3}{3}=\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-1=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]\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\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\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\r\n[latex]\\begin{array}{l}3x+5y=-11\\hfill \\\\ \\hfill \\\\ x - 2y=11\\hfill \\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"883001\"]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\r\n[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ -3\\left(x - 2y\\right)=-3\\left(11\\right)\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Multiply both sides by }-3.\\hfill \\\\ \\text{ }-3x+6y=-33\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Use the distributive property}.\\hfill \\end{array}[\/latex]\r\nNow, let\u2019s add them.\r\n\r\n[latex]\\begin{array}\\ \\hfill 3x+5y=\u221211 \\\\ \\hfill \u22123x+6y=\u221233 \\\\ \\text{_____________} \\\\ \\hfill 11y=\u221244 \\\\ \\hfill y=\u22124 \\end{array}[\/latex]\r\n\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\r\n[latex]\\begin{array}{c}3x+5y=-11\\\\ 3x+5\\left(-4\\right)=-11\\\\ 3x - 20=-11\\\\ 3x=9\\\\ x=3\\end{array}[\/latex]\r\n\r\nOur solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.\r\n\r\n[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\left(3\\right)-2\\left(-4\\right)=3+8\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]\r\n\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\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\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\r\n[latex]\\begin{array}{c}2x - 7y=2\\\\ 3x+y=-20\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"609174\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"609174\"][latex]\\left(-6,-2\\right)[\/latex][\/hidden-answer]\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\r\n[latex]\\begin{array}{c}2x+3y=-16\\\\ 5x - 10y=30\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"114755\"]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\r\n[latex]\\begin{array}{l} -5\\left(2x+3y\\right)=-5\\left(-16\\right)\\hfill \\\\ \\text{ }-10x - 15y=80\\hfill \\\\ \\text{ }2\\left(5x - 10y\\right)=2\\left(30\\right)\\hfill \\\\ \\text{ }10x - 20y=60\\hfill \\end{array}[\/latex]\r\n\r\nThen, we add the two equations together.\r\n\r\n[latex]\\begin{array}\\ \u221210x\u221215y=80 \\\\ 10x\u221220y=60 \\\\ \\text{______________} \\\\ \\text{ }\u221235y=140 \\\\ y=\u22124 \\end{array}[\/latex]\r\n\r\nSubstitute [latex]y=-4[\/latex] into the original first equation.\r\n\r\n[latex]\\begin{array}{c}2x+3\\left(-4\\right)=-16\\\\ 2x - 12=-16\\\\ 2x=-4\\\\ x=-2\\end{array}[\/latex]\r\n\r\nThe solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the other equation.\r\n\r\n[latex]\\begin{array}{r}\\hfill \\text{ }5x - 10y=30\\\\ \\hfill 5\\left(-2\\right)-10\\left(-4\\right)=30\\\\ \\hfill \\text{ }-10+40=30\\\\ \\hfill \\text{ }30=30\\end{array}[\/latex]\r\n\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\" data-media-type=\"image\/jpg\" \/>\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\r\n[latex]\\begin{array}{l}\\frac{x}{3}+\\frac{y}{6}=3\\hfill \\\\ \\frac{x}{2}-\\frac{y}{4}=\\text{ }1\\hfill \\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"359287\"]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\r\n[latex]\\begin{array}{l}6\\left(\\frac{x}{3}+\\frac{y}{6}\\right)=6\\left(3\\right)\\hfill \\\\ \\text{ }2x+y=18\\hfill \\\\ 4\\left(\\frac{x}{2}-\\frac{y}{4}\\right)=4\\left(1\\right)\\hfill \\\\ \\text{ }2x-y=4\\hfill \\end{array}[\/latex]\r\n\r\nNow multiply the second equation by [latex]-1[\/latex] so that we can eliminate the <em>x<\/em>-variable.\r\n\r\n[latex]\\begin{array}{l}-1\\left(2x-y\\right)=-1\\left(4\\right)\\hfill \\\\ \\text{ }-2x+y=-4\\hfill \\end{array}[\/latex]\r\n\r\nAdd the two equations to eliminate the <em>x<\/em>-variable and solve the resulting equation.\r\n\r\n[latex]\\begin{array}\\ \\hfill 2x+y=18 \\\\ \\hfill\u22122x+y=\u22124 \\\\ \\text{_____________} \\\\ \\hfill 2y=14 \\\\ \\hfill y=7 \\end{array}[\/latex]\r\n\r\nSubstitute [latex]y=7[\/latex] into the first equation.\r\n\r\n[latex]\\begin{array}{l}2x+\\left(7\\right)=18\\hfill \\\\ \\text{ }2x=11\\hfill \\\\ \\text{ }x=\\frac{11}{2}\\hfill \\\\ \\text{ }=7.5\\hfill \\end{array}[\/latex]\r\n\r\nThe solution is [latex]\\left(\\frac{11}{2},7\\right)[\/latex]. Check it in the other equation.\r\n\r\n[latex]\\begin{array}{c}\\frac{x}{2}-\\frac{y}{4}=1\\\\ \\frac{\\frac{11}{2}}{2}-\\frac{7}{4}=1\\\\ \\frac{11}{4}-\\frac{7}{4}=1\\\\ \\frac{4}{4}=1\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\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\r\n[latex]\\begin{array}{c}2x+3y=8\\\\ 3x+5y=10\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"326265\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"326265\"][latex]\\left(10,-4\\right)[\/latex][\/hidden-answer]\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 Objectives<\/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>[latex]\\begin{array}{l}\\text{ }-x+y=-5\\hfill \\\\ \\text{ }2x - 5y=1\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q786744\">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>[latex]\\begin{array}{l}-x+y=-5\\hfill \\\\ \\text{ }y=x - 5\\hfill \\end{array}[\/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>[latex]\\begin{array}{l}\\text{ }2x - 5y=1\\hfill \\\\ 2x - 5\\left(x - 5\\right)=1\\hfill \\\\ \\text{ }2x - 5x+25=1\\hfill \\\\ \\text{ }-3x=-24\\hfill \\\\ \\text{ }x=8\\hfill \\end{array}[\/latex]<\/p>\n<p>Now, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].<\/p>\n<p>[latex]\\begin{array}{l}-\\left(8\\right)+y=-5\\hfill \\\\ \\text{ }y=3\\hfill \\end{array}[\/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>[latex]\\begin{array}{llll}-x+y=-5\\hfill & \\hfill & \\hfill & \\hfill \\\\ -\\left(8\\right)+\\left(3\\right)=-5\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\\\ 2x - 5y=1\\hfill & \\hfill & \\hfill & \\hfill \\\\ 2\\left(8\\right)-5\\left(3\\right)=1\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>You can use Desmos to help you solve a\u00a0system of equations by substitution. We will use the following system to show you how:<\/p>\n<p>[latex]\\begin{array}{l}x=y+3\\hfill \\\\ 4=3x - 2y\\hfill \\end{array}[\/latex]<\/p>\n<p>First, solve both equations for y:<\/p>\n<p>[latex]\\begin{array}{l}y=x-3\\hfill \\\\ y=\\frac{3}{2}x - 2\\hfill \\end{array}[\/latex]<\/p>\n<p>Now enter [latex]x-3=\\frac{3}{2}x - 2[\/latex] into Desmos. You will see that Desmos 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<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q783260\">Solution<\/span><\/p>\n<div id=\"q783260\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(-2,-5\\right)[\/latex]<\/div>\n<\/div>\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>&nbsp;<\/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>[latex]\\begin{array}{l}x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924657\">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>[latex]\\frac{\\begin{array}{l}\\hfill \\\\ x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}}{\\text{}\\text{}\\text{}\\text{}\\text{}3y=2}[\/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>[latex]\\begin{array}{l}3y=2\\hfill \\\\ \\text{ }y=\\frac{2}{3}\\hfill \\end{array}[\/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>[latex]\\begin{array}{l}\\text{ }-x+y=3\\hfill \\\\ \\text{ }-x+\\frac{2}{3}=3\\hfill \\\\ \\text{ }-x=3-\\frac{2}{3}\\hfill \\\\ \\text{ }-x=\\frac{7}{3}\\hfill \\\\ \\text{ }x=-\\frac{7}{3}\\hfill \\end{array}[\/latex]<br \/>\nThe 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>[latex]\\begin{array}{llll}\\text{ }x+2y=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }\\left(-\\frac{7}{3}\\right)+2\\left(\\frac{2}{3}\\right)=\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-\\frac{7}{3}+\\frac{4}{3}=\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-\\frac{3}{3}=\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-1=-1\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/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\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\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>[latex]\\begin{array}{l}3x+5y=-11\\hfill \\\\ \\hfill \\\\ x - 2y=11\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q883001\">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>[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill & \\hfill & \\hfill & \\hfill \\\\ -3\\left(x - 2y\\right)=-3\\left(11\\right)\\hfill & \\hfill & \\hfill & \\text{Multiply both sides by }-3.\\hfill \\\\ \\text{ }-3x+6y=-33\\hfill & \\hfill & \\hfill & \\text{Use the distributive property}.\\hfill \\end{array}[\/latex]<br \/>\nNow, let\u2019s add them.<\/p>\n<p>[latex]\\begin{array}\\ \\hfill 3x+5y=\u221211 \\\\ \\hfill \u22123x+6y=\u221233 \\\\ \\text{_____________} \\\\ \\hfill 11y=\u221244 \\\\ \\hfill y=\u22124 \\end{array}[\/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>[latex]\\begin{array}{c}3x+5y=-11\\\\ 3x+5\\left(-4\\right)=-11\\\\ 3x - 20=-11\\\\ 3x=9\\\\ x=3\\end{array}[\/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>[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\left(3\\right)-2\\left(-4\\right)=3+8\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }=11\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/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\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\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>[latex]\\begin{array}{c}2x - 7y=2\\\\ 3x+y=-20\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q609174\">Solution<\/span><\/p>\n<div id=\"q609174\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(-6,-2\\right)[\/latex]<\/div>\n<\/div>\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>[latex]\\begin{array}{c}2x+3y=-16\\\\ 5x - 10y=30\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q114755\">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>[latex]\\begin{array}{l} -5\\left(2x+3y\\right)=-5\\left(-16\\right)\\hfill \\\\ \\text{ }-10x - 15y=80\\hfill \\\\ \\text{ }2\\left(5x - 10y\\right)=2\\left(30\\right)\\hfill \\\\ \\text{ }10x - 20y=60\\hfill \\end{array}[\/latex]<\/p>\n<p>Then, we add the two equations together.<\/p>\n<p>[latex]\\begin{array}\\ \u221210x\u221215y=80 \\\\ 10x\u221220y=60 \\\\ \\text{______________} \\\\ \\text{ }\u221235y=140 \\\\ y=\u22124 \\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=-4[\/latex] into the original first equation.<\/p>\n<p>[latex]\\begin{array}{c}2x+3\\left(-4\\right)=-16\\\\ 2x - 12=-16\\\\ 2x=-4\\\\ x=-2\\end{array}[\/latex]<\/p>\n<p>The solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the other equation.<\/p>\n<p>[latex]\\begin{array}{r}\\hfill \\text{ }5x - 10y=30\\\\ \\hfill 5\\left(-2\\right)-10\\left(-4\\right)=30\\\\ \\hfill \\text{ }-10+40=30\\\\ \\hfill \\text{ }30=30\\end{array}[\/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\" data-media-type=\"image\/jpg\" \/><\/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>[latex]\\begin{array}{l}\\frac{x}{3}+\\frac{y}{6}=3\\hfill \\\\ \\frac{x}{2}-\\frac{y}{4}=\\text{ }1\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q359287\">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>[latex]\\begin{array}{l}6\\left(\\frac{x}{3}+\\frac{y}{6}\\right)=6\\left(3\\right)\\hfill \\\\ \\text{ }2x+y=18\\hfill \\\\ 4\\left(\\frac{x}{2}-\\frac{y}{4}\\right)=4\\left(1\\right)\\hfill \\\\ \\text{ }2x-y=4\\hfill \\end{array}[\/latex]<\/p>\n<p>Now multiply the second equation by [latex]-1[\/latex] so that we can eliminate the <em>x<\/em>-variable.<\/p>\n<p>[latex]\\begin{array}{l}-1\\left(2x-y\\right)=-1\\left(4\\right)\\hfill \\\\ \\text{ }-2x+y=-4\\hfill \\end{array}[\/latex]<\/p>\n<p>Add the two equations to eliminate the <em>x<\/em>-variable and solve the resulting equation.<\/p>\n<p>[latex]\\begin{array}\\ \\hfill 2x+y=18 \\\\ \\hfill\u22122x+y=\u22124 \\\\ \\text{_____________} \\\\ \\hfill 2y=14 \\\\ \\hfill y=7 \\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=7[\/latex] into the first equation.<\/p>\n<p>[latex]\\begin{array}{l}2x+\\left(7\\right)=18\\hfill \\\\ \\text{ }2x=11\\hfill \\\\ \\text{ }x=\\frac{11}{2}\\hfill \\\\ \\text{ }=7.5\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution is [latex]\\left(\\frac{11}{2},7\\right)[\/latex]. Check it in the other equation.<\/p>\n<p>[latex]\\begin{array}{c}\\frac{x}{2}-\\frac{y}{4}=1\\\\ \\frac{\\frac{11}{2}}{2}-\\frac{7}{4}=1\\\\ \\frac{11}{4}-\\frac{7}{4}=1\\\\ \\frac{4}{4}=1\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\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>[latex]\\begin{array}{c}2x+3y=8\\\\ 3x+5y=10\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q326265\">Solution<\/span><\/p>\n<div id=\"q326265\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(10,-4\\right)[\/latex]<\/div>\n<\/div>\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-1\" 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-3388\">\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>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><\/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":21,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Solving Systems of Equations using Elimination\",\"author\":\"James Sousa 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