{"id":849,"date":"2016-02-15T19:26:28","date_gmt":"2016-02-15T19:26:28","guid":{"rendered":"https:\/\/courses.candelalearning.com\/nrocarithmetic\/?post_type=chapter&#038;p=849"},"modified":"2018-01-03T23:56:02","modified_gmt":"2018-01-03T23:56:02","slug":"3-2-1-the-substitution-method","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/beginalgebra\/chapter\/3-2-1-the-substitution-method\/","title":{"raw":"Algebraic Methods for Solving Systems","rendered":"Algebraic Methods for Solving Systems"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the substitution method\r\n<ul>\r\n \t<li>Solve a system of equations using the substitution method.<\/li>\r\n \t<li>Recognize systems of equations that have no solution or an infinite number of solutions<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Use the elimination method without multiplication\r\n<ul>\r\n \t<li>Solve a system of equations when no multiplication is necessary to eliminate a variable<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Use the elimination method with multiplication\r\n<ul>\r\n \t<li>Use multiplication in combination with\u00a0the elimination method to solve a system of linear equations<\/li>\r\n \t<li>Recognize when the solution to a system of linear equations implies there are an infinite number of solutions<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Solve a system of equations using the substitution method<\/h2>\r\nIn the last couple sections, we verified that ordered pairs were solutions to systems, and we used graphs to classify how many solutions a system of two linear equations had. What if we are not given a point of intersection, or it is not obvious from a graph? Can we still find a solution to the system? Of course you can, using algebra!\r\n\r\nIn this section we will learn the substitution\u00a0method for finding a solution to a system of linear equations in two variables. We have used substitution in different ways throughout this course, for example when we were using the formulas for the area of a triangle and simple interest. We substituted values that we knew into the formula to solve for values that we did not know. \u00a0The idea is similar when applied to solving systems, there are just a few different steps in the process. You will first solve for one variable, and then substitute that expression into the other equation. Let's start with an example to see what this means.\r\n\r\n&nbsp;\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFind the value of <i>x<\/i> for this system.\r\n\r\nEquation A: [latex]4x+3y=\u221214[\/latex]\r\n\r\nEquation B: [latex]y=2[\/latex]\r\n\r\n[reveal-answer q=\"478211\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"478211\"]The problem asks to solve for <i>x<\/i>. Equation B gives you the value of <i>y<\/i>, [latex]y=2[\/latex], so you can substitute 2 into Equation A for y.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+3y=\u221214\\\\y=2\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<i>\u00a0<\/i><\/p>\r\nSubstitute\u00a0[latex]y=2[\/latex] into\u00a0Equation A.\r\n<p style=\"text-align: center;\">[latex]4x+3\\left(2\\right)=\u221214[\/latex]<\/p>\r\nSimplify and solve the equation for <i>x.<\/i>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+6=\u221214\\\\4x=\u221220\\\\x=\u22125\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=\u22125[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou can substitute a value for a variable even if it is an expression. Here\u2019s an example.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>x <\/i>and <i>y<\/i>.\r\n\r\nEquation A: [latex]y+x=3[\/latex]\r\n\r\nEquation B: [latex]x=y+5[\/latex]\r\n\r\n[reveal-answer q=\"300993\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"300993\"]The goal of the substitution method is to rewrite one of the equations in terms of a single variable. Equation B tells us that [latex]x=y+5[\/latex], so it makes sense to substitute [latex]y+5[\/latex] into Equation A for <i>x<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y+x=3\\\\x=y+5\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]y+5[\/latex] into Equation A for <i>x<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}y+x=3\\\\y+\\left(y+5\\right)=3\\end{array}[\/latex]<i><\/i><\/p>\r\nSimplify and solve the equation for <i>y.<\/i>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2y+5=\\,\\,\\,\\,3\\\\\\underline{\u22125\\,\\,\\,\\,\\,\u22125}\\\\2y=\u22122\\\\y=\u22121\\end{array}[\/latex]<\/p>\r\nNow find <i>x <\/i>by\u00a0substituting this value for <i>y<\/i> into either equation and solve for <i>x<\/i>. We will use Equation A here.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}y+x=3\\\\\u22121+x=3\\\\\\underline{+1\\,\\,\\,\\,\\,\\,\\,\\,\\,+1}\\\\x=4\\end{array}[\/latex]<\/p>\r\nFinally, check the solution\u00a0[latex]x=4[\/latex], [latex]y=\u22121[\/latex]\u00a0by substituting these values into each of the original equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}y+x=3\\\\\u22121+4=3\\\\3=3\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=y+5\\\\4=\u22121+5\\\\4=4\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=4[\/latex] and [latex]y=\u22121[\/latex]\r\n\r\nThe solution is [latex](4,\u22121)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nRemember, a solution to a system of equations must be a solution to each of the equations within the system. The ordered pair [latex](4,\u22121)[\/latex] does work for both equations, so you know that it is a solution to the system as well.\r\n\r\nLet\u2019s look at another example whose substitution involves the distributive property.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>x <\/i>and <i>y<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y = 3x + 6\\\\\u22122x + 4y = 4\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"240040\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"240040\"]Choose an equation to use for the substitution.\r\n\r\nThe first equation tells you how to express <i>y<\/i> in terms of <i>x<\/i>, so it makes sense to substitute 3<i>x<\/i> + 6 into the second equation for <i>y<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=3x+6\\\\\u22122x+4y=4\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]3x+6[\/latex] for<i> y<\/i> into the second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+4y=4\\\\\u22122x+4\\left(3x+6\\right)=4\\end{array}[\/latex]<i><\/i><\/p>\r\nSimplify and solve the equation for <i>x.<\/i>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+12x+24=4\\,\\,\\,\\,\\,\\,\\,\\\\10x+24=4\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{\u221224\\,\\,\u221224\\,\\,\\,\\,}\\\\10x=\u221220\\\\x=\u22122\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nTo find <i>y<\/i>, substitute this value for <i>x<\/i> back into one of the original equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=3x+6\\\\y=3\\left(\u22122\\right)+6\\\\y =\u22126+6\\\\y=0\\end{array}[\/latex]<\/p>\r\nCheck the solution [latex]x=\u22122[\/latex], [latex]y=0[\/latex]\u00a0by substituting them into each of the original equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=3x+6\\\\0=3\\left(\u22122\\right)+6\\\\0=\u22126+6\\\\0=0\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+4y=4\\\\\u22122\\left(\u22122\\right)+4\\left(0\\right)=4\\\\4+0=4\\\\4=4\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=\u22122[\/latex] and [latex]y=0[\/latex]\r\n\r\nThe solution is (\u22122, 0).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the examples above, one of the equations was already given to us in terms of the variable <i>x<\/i> or <i>y<\/i>. This allowed us to quickly substitute that value into the other equation and solve for one of the unknowns.\r\n\r\nSometimes you may have to rewrite one of the equations in terms of one of the variables first before you can substitute. In\u00a0the example below, you will first need to isolate one of the variables before you can substitute it into the other equation.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>x <\/i>and <i>y<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+3y=22\\\\3x+y=19\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"344538\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"344538\"]Choose an equation to use for the substitution. The second equation,\r\n\r\n[latex]3x+y=19[\/latex], can easily be rewritten in terms of <i>y<\/i>, so it makes sense to start there.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}2x+3y=22\\\\3x+y=19\\end{array}[\/latex]<\/p>\r\nRewrite [latex]3x+y=19[\/latex]\u00a0in terms of <i>y<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}3x+y=19\\\\y=19\u20133x\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]19\u20133x[\/latex] for <i>y<\/i> in the other equation.\r\n<p style=\"text-align: center;\"><i><\/i>[latex]\\begin{array}{r}2x+3y=22\\\\2x+3(19\u20133x)=22\\end{array}[\/latex]<i><\/i><\/p>\r\nSimplify and solve the equation for <i>x.<\/i>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+57\u20139x=22\\,\\,\\,\\,\\\\\u22127x+57=22\\,\\,\\,\\,\\\\\u22127x=\u221235\\\\x=5\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]x=5[\/latex] back into one of the original equations to solve for <i>y.<\/i>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}3x+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\3\\left(5\\right)+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\15+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\y=19\u221215\\\\y=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nCheck both solutions by substituting them into each of the original equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+3y=22\\\\2(5)+3\\left(4\\right)=22\\\\10+12=22\\\\22=22\\\\\\text{TRUE}\\\\\\\\3x+y=19\\\\3\\left(5\\right)+4= 19\\\\19=19\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x=5[\/latex] and [latex]y=4[\/latex]\r\n\r\nThe solution is (5, 4).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will be given an example of solving a systems 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. \u00a0It is really a matter of preference because sometimes solving for a variable will result in having to work with fractions. \u00a0As you become more experienced with algebra, you will be able to anticipate what choices will lead to more desirable outcomes.\r\n<h2 id=\"title2\">Recognize systems of equations that have no solution or an infinite number of solutions<\/h2>\r\nWhen we learned methods for solving linear equations in one variable, we found that some equations didn't have any solutions, and others had an infinite number of solutions. \u00a0We saw this behavior again\u00a0when we started describing solutions to systems of equations in two variables.\r\n\r\nRecall this example from Module 1 for solving linear equations in one variable:\r\n<p style=\"text-align: center;\">Solve for <i>x<\/i>.\u00a0[latex]12+2x\u20138=7x+5\u20135x[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{l}12+2x-8=7x+5-5x\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x+4=2x+5\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x+4=2x+5\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-2x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2x\\,\\,\\,\\,\\,\\,\\,\\,}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4= \\,5\\end{array}[\/latex]<\/p>\r\nThis false statement implies there are <strong>no solutions<\/strong> to this equation. \u00a0In the same way, you may see an outcome like this when you use the substitution method to find a solution to a system of linear equations in two variables. In the next example, you will see an example of a system of two equations that does not have a solution.\r\n\r\n&nbsp;\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>x<\/i> and <i>y<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=5x+4\\\\10x\u22122y=4\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"787022\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"787022\"]Since the first equation is [latex]y=5x+4[\/latex], you can substitute [latex]5x+4[\/latex] in for <i>y<\/i> in the second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}y=5x+4\\\\10x\u22122y=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\10x\u20132\\left(5x+4\\right)=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nExpand the expression on the left.\r\n<p style=\"text-align: center;\">[latex]10x\u201310x\u20138=4[\/latex]<\/p>\r\nCombine like terms on the left side of equation.\r\n<p style=\"text-align: center;\">[latex]10x\u201310x=0[\/latex], so you are left with [latex]\u22128=4[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}0\u20138=4\\\\\u22128=4\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe statement [latex]\u22128=4[\/latex] is false, so there is no solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nYou get the false statement [latex]\u22128=4[\/latex]. What does this mean? The graph of this system sheds some light on what is happening.\r\n\r\n<img class=\"aligncenter size-full wp-image-2994\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/20183038\/Screen-Shot-2016-04-20-at-11.30.22-AM.png\" alt=\"Two parallel lines. One line is y=5x+4, and the other line is 10x-2y=4.\" width=\"459\" height=\"458\" \/>\r\n\r\nThe lines are parallel, they never intersect and there is no solution to this system of linear equations. Note that the result [latex]\u22128=4[\/latex] is <b>not<\/b> a solution. It is simply a false statement and it indicates that there is <b>no<\/b> solution.\r\n\r\nWe have also seen linear equations in one variable and systems of equations in two variables that have an infinite number of solutions. \u00a0In the next example, you will see what happens when you apply the substitution method to a system with an infinite number of solutions.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for x and y.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,y=\u22120.5x\\\\9y=\u22124.5x\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[reveal-answer q=\"683508\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"683508\"]<\/p>\r\nSubstituting \u22120.5<i>x<\/i> for <i>y<\/i> in the second equation, you find the following:\r\n<div align=\"center\">\r\n\r\n[latex]\\begin{array}{r}9y=\u22124.5x\\\\9(\u22120.5x)=\u22124.5\\,\\,\\,\\\\\u22124.5x=\u22124.5x\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nThis time you get a true statement: [latex]\u22124.5x=\u22124.5x[\/latex]. But what does this type of answer mean? Again, graphing can help you make sense of this system.\r\n\r\n<img class=\"aligncenter size-full wp-image-2993\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/20182902\/Screen-Shot-2016-04-20-at-11.28.43-AM.png\" alt=\"Two overlapping lines that represent the same line. One line is 9y=-4.5x, and the other line is y=-0.5x.\" width=\"458\" height=\"460\" \/>\r\n\r\nThis system consists of two equations that both represent the same line; the two lines are collinear. Every point along the line will be a solution to the system, and that\u2019s why the substitution method yields a true statement. In this case, there are an infinite number of solutions.\r\n\r\nIn the following video you will see an example of solving a system that has an infinite number of solutions.\r\n\r\nhttps:\/\/youtu.be\/Pcqb109yK5Q\r\n\r\nIn the following video you will see an example of solving a system of equations that has no solutions.\r\n\r\nhttps:\/\/youtu.be\/kTtKfh5gFUc\r\n<h2>Solve a system of equations using the elimination\u00a0method<\/h2>\r\nThe <b>elimination method<\/b> for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an equation to eliminate one of the variable terms. \u00a0In this method, you may or may not need to multiply the terms in one equation by a number first. \u00a0We will first look at examples where no multiplication is necessary to use the elimination method. \u00a0In the next section you will see examples using multiplication after you are familiar with the idea of the elimination method.\r\n\r\nIt is easier to show rather than tell with this method, so let's dive right into some examples.\r\n\r\nIf you add the two equations,\r\n\r\n[latex]x\u2013y=\u22126[\/latex] and [latex]x+y=8[\/latex] together, watch what happens.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,x-y=\\,-6\\\\\\underline{+\\,x+y=\\,\\,\\,8}\\\\\\,2x+0\\,=\\,\\,\\,\\,2\\end{array}[\/latex]<\/p>\r\nYou have eliminated the <i>y<\/i> term, and this equation can be solved using the methods for solving equations with one variable.\r\n\r\nLet\u2019s see how this system is solved using the elimination method.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse elimination to solve the system.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x\u2013y=\u22126\\\\x+y=\\,\\,\\,\\,8\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"403819\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"403819\"]Add the equations.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{r}x-y=\\,\\,-6\\\\+\\underline{\\,\\,x+y=\\,\\,\\,\\,\\,8}\\\\\\,\\,\\,\\,\\,\\,2x\\,\\,\\,\\,\\,=\\,\\,\\,\\,\\,\\,2\\end{array}[\/latex]<\/p>\r\nSolve for <i>x<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x=2\\\\x=1\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]x=1[\/latex] into one of the original equations and solve for <i>y<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+y=8\\\\1+y=8\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=8\u20131\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=7\\end{array}[\/latex]<\/p>\r\nBe sure to check your answer in both equations!\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x\u2013y=\u22126\\\\1\u20137=\u22126\\\\\u22126=\u22126\\\\\\text{TRUE}\\\\\\\\x+y=8\\\\1+7=8\\\\8=8\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\nThe answers check.\r\n<h4>Answer<\/h4>\r\nThe solution is (1, 7).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nUnfortunately not all systems work out this easily. How about a system like [latex]2x+y=12[\/latex] and [latex]\u22123x+y=2[\/latex]. If you add these two equations together, no variables are eliminated.\r\n<p style=\"text-align: center;\">[latex] \\displaystyle \\begin{array}{l}\\,\\,\\,\\,2x+y=12\\\\\\underline{-3x+y=\\,\\,\\,2}\\\\-x+2y=14\\end{array}[\/latex]<\/p>\r\nBut you want to eliminate a variable. So let\u2019s add the opposite of one of the equations to the other equation. This means multiply every term in one of the equations by -1, so that the sign of every terms is opposite.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,2x+\\,\\,y\\,=12\\rightarrow2x+y=12\\rightarrow2x+y=12\\\\\u22123x+\\,\\,y\\,=2\\rightarrow\u2212\\left(\u22123x+y\\right)=\u2212(2)\\rightarrow3x\u2013y=\u22122\\\\\\,\\,\\,\\,5x+0y=10\\end{array}[\/latex]<\/p>\r\nYou have eliminated the <i>y<\/i> variable, and the problem can now be solved.\r\n\r\nThe following video describe a similar problem where you can eliminate one variable by adding the two equations together.\r\n\r\nhttps:\/\/youtu.be\/M4IEmwcqR3c\r\n<div class=\"textbox shaded\"><img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"63\" height=\"56\" \/>Caution! \u00a0When you add the opposite of one entire equation to another, make sure to change the sign of EVERY term on both sides of the equation. This is a very common\u00a0mistake to make.<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse elimination to solve the system.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=12\\\\\u22123x+y=2\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"702178\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"702178\"]You can eliminate the <i>y<\/i>-variable if you add the opposite of one of the equations to the other equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=12\\\\\u22123x+y=2\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nRewrite the second equation as its opposite.\r\n\r\nAdd.\u00a0Solve for <i>x<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=12\\,\\\\3x\u2013y=\u22122\\\\5x=10\\,\\\\x=2\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]y=2[\/latex] into one of the original equations and solve for <i>y<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(2\\right)+y=12\\\\4+y=12\\\\y=8\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nBe sure to check your answer in both equations!\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=12\\\\2\\left(2\\right)+8=12\\\\4+8=12\\\\12=12\\\\\\text{TRUE}\\\\\\\\\u22123x+y=2\\\\\u22123\\left(2\\right)+8=2\\\\\u22126+8=2\\\\2=2\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\nThe answers check.\r\n<h4>Answer<\/h4>\r\nThe solution is (2, 8).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following are two more examples showing how to solve linear systems of equations using elimination.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse elimination to solve the system.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\2x+5y=\\,25\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"438400\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"438400\"]Notice the coefficients of each variable in each equation. If you add these two equations, the <i>x<\/i> term will be eliminated since [latex]\u22122x+2x=0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\2x+5y=\\,25\\end{array}[\/latex]<\/p>\r\nAdd and solve for <i>y<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\2x+5y=25\\,\\\\8y=24\\,\\\\y=3\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]y=3[\/latex]<i> <\/i>into one of the original equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+5y=25\\\\2x+5\\left(3\\right)=25\\\\2x+15=25\\\\2x=10\\\\x=5\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nCheck solutions.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\\u22122\\left(5\\right)+3\\left(3\\right)=\u22121\\\\\u221210+9=\u22121\\\\\u22121=\u22121\\\\\\text{TRUE}\\\\\\\\2x+5y=25\\\\2\\left(5\\right)+5\\left(3\\right)=25\\\\10+15=25\\\\25=25\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\nThe answers check.\r\n<h4>Answer<\/h4>\r\nThe solution is (5, 3).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse elimination to solve for <i>x <\/i>and <i>y.<\/i>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+2y=14\\\\5x+2y=16\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"776093\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"776093\"]Notice the coefficients of each variable in each equation. You will need to add the opposite of one of the equations to eliminate the variable <i>y<\/i>, as [latex]2y+2y=4y[\/latex], but [latex]2y+\\left(\u22122y\\right)=0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+2y=14\\\\5x+2y=16\\end{array}[\/latex]<\/p>\r\n\u00a0Change one of the equations to its opposite, add and solve for <i>x<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+2y=14\\,\\,\\,\\,\\\\\u22125x\u20132y=\u221216\\\\\u2212x=\u22122\\,\\,\\,\\\\x=2\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]x=2[\/latex] into one of the original equations and solve for <i>y<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+2y=14\\\\4\\left(2\\right)+2y=14\\\\8+2y=14\\\\2y=6\\,\\,\\,\\\\y=3\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe solution is (2, 3).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nGo ahead and check this last example\u2014substitute (2, 3) into both equations. You get two true statements: 14=14 and 16=16!\r\n\r\nNotice that you could have used the opposite of the first equation rather than the second equation and gotten the same result.\r\n<h2 id=\"video1\" class=\"no-indent\" style=\"text-align: left;\">Recognize systems that have no solution or an infinite number of solutions<\/h2>\r\nJust as with the substitution method, the elimination method will sometimes eliminate <i>both v<\/i>ariables, and you end up with either a true statement or a false statement. Recall that a false statement means that there is no solution.\r\n\r\nLet\u2019s look at an example.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>x <\/i>and <i>y.<\/i>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}-x\u2013y=-4\\\\x+y=2\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"101540\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"101540\"]Add the equations to eliminate the\u00a0<em>x<\/em>-term.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}-x\u2013y=-4\\\\\\underline{x+y=2\\,\\,\\,}\\\\0=\u22122\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThere is no solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nGraphing these lines shows that they are parallel lines and as such do not share any point in common, verifying that there is no solution.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064435\/image040-1.jpg\" alt=\"Two parallel lines. One line is -x-y=-4. The other line is x+y=2.\" width=\"346\" height=\"345\" \/>\r\n\r\nIf both variables are eliminated and you are left with a true statement, this indicates that there are an infinite number of ordered pairs that satisfy both of the equations. In fact, the equations are the same line.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>x <\/i>and <i>y<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x+y=2\\,\\,\\,\\,\\\\-x\u2212y=-2\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"328100\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"328100\"]Add the equations to eliminate the\u00a0<i>x<\/i>-term.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x+y=2\\,\\,\\,\\,\\\\\\underline{-x\u2212y=-2}\\\\0=0\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThere are an infinite number of solutions.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nGraphing these two equations will help to illustrate what is happening.\r\n\r\n<b><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064437\/image041-1.jpg\" alt=\"Two overlapping lines. One is -x-y=-2, and the other is x+y=2.\" width=\"346\" height=\"345\" \/><\/b>\r\n<h2 id=\"video3\" class=\"no-indent\" style=\"text-align: left;\"><\/h2>\r\nIn the following video, a system of equations which has no solutions is solved using the method of elimination.\r\n\r\nhttps:\/\/youtu.be\/z5_ACYtzW98\r\n<h2 id=\"title2\">Solve a system of equations when multiplication is necessary to eliminate a variable<\/h2>\r\nMany times adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]\r\n\r\nIf you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let\u2019s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will allow you to eliminate\u00a0the same variable in the other equation.\r\n\r\nWe do this with multiplication.\u00a0\u00a0Notice that the first equation contains the term 4<i>y<\/i>, and the second equation contains the term <i>y<\/i>. If you multiply the second equation by \u22124, when you add both equations the <i>y<\/i> variables will add up to 0.\r\n\r\nThe following example takes you through all the steps to find a solution to this system.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>x<\/i> and <i>y<\/i>.\r\n\r\n<strong>Equation A:<\/strong> [latex]3x+4y=52[\/latex]\r\n\r\n<strong>Equation B:<\/strong> [latex]5x+y=30[\/latex]\r\n\r\n[reveal-answer q=\"815377\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"815377\"]Look for terms that can be eliminated. The equations do not have any <i>x<\/i> or <i>y<\/i> terms with the same coefficients.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]\r\n\r\nMultiply the second equation by [latex]\u22124[\/latex] so they do have the same coefficient.\r\n\r\n[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,3x+4y=52\\\\\u22124\\left(5x+y\\right)=\u22124\\left(30\\right)\\end{array}[\/latex]\r\n\r\nRewrite the system and add the equations.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\,\\,\\,\\,\\,\\,\\,\\\\\u221220x\u20134y=\u2212120\\end{array}[\/latex]\r\n\r\nSolve for <i>x<\/i>.\r\n\r\n[latex]\\begin{array}{l}\u221217x=-68\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x=4\\end{array}[\/latex]\r\n\r\nSubstitute [latex]x=4[\/latex] into one of the original equations to find <i>y<\/i>.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\3\\left(4\\right)+4y=52\\\\12+4y=52\\\\4y=40\\\\y=10\\end{array}[\/latex]\r\n\r\nCheck your answer.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\3\\left(4\\right)+4\\left(10\\right)=52\\\\12+40=52\\\\52=52\\\\\\text{TRUE}\\\\\\\\5x+y=30\\\\5\\left(4\\right)+10=30\\\\20+10=30\\\\30=30\\\\\\text{TRUE}\\end{array}[\/latex]\r\n\r\nThe answers check.\r\n<h4>Answer<\/h4>\r\nThe solution is (4, 10).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\"><img class=\" wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"55\" height=\"49\" \/>Caution! \u00a0When you\u00a0use multiplication to eliminate a variable, you must multiply EACH term in the equation by the number you choose. \u00a0Forgetting to multiply every term is a common mistake.<\/div>\r\nThere are other ways to solve this system. Instead of multiplying one equation in order to eliminate a variable when the equations were added, you could have multiplied <i>both<\/i> equations by different numbers.\r\n\r\nLet\u2019s remove the variable <i>x <\/i>this time. Multiply Equation A by 5 and Equation B by [latex]\u22123[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <em>x<\/em> and <em>y<\/em>.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"40585\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"40585\"]Look for terms that can be eliminated. The equations do not have any <i>x<\/i> or <i>y<\/i> terms with the same coefficient.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]\r\n\r\nIn order to use the elimination method, you have to create variables that have the same coefficient\u2014then you can eliminate them. Multiply the top equation by 5.\r\n\r\n[latex]\\begin{array}{r}5\\left(3x+4y\\right)=5\\left(52\\right)\\\\5x+y =30\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\15x+20y=260\\,\\,\\,\\,\\,\\,\\\\5x+y=30\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]\r\n\r\nNow multiply the bottom equation by \u22123.\r\n\r\n[latex]\\begin{array}{r}15x+20y=260\\,\\,\\,\\,\\,\\,\\,\\,\\\\-3(5x+y)=\u22123(30)\\\\15x+20y=260\\,\\,\\,\\,\\,\\,\\,\\,\\\\\u221215x\u20133y=\u221290\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]\r\n\r\nNext add the equations, and solve for <i>y<\/i>.\r\n\r\n[latex]\\begin{array}{r}15x+20y=260\\\\\u221215x\u20133y=\\,\u201390\\\\17y=170\\\\y=\\,\\,\\,10\\end{array}[\/latex]\r\n\r\nSubstitute [latex]y=10[\/latex] into one of the original equations to find <i>x<\/i>.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\3x+4\\left(10\\right)=52\\\\3x+40=52\\\\3x=12\\\\x=4\\,\\,\\,\\end{array}[\/latex]\r\n\r\nYou arrive at the same solution as before.\r\n<h4>Answer<\/h4>\r\nThe solution is (4, 10).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThese equations were multiplied by 5 and [latex]\u22123[\/latex] respectively, because that gave you terms that would add up to 0. Be sure to multiply all of the terms of the equation.\r\n\r\nIn the following video, you will see an example of using the elimination method for solving a system of equations.\r\n\r\nhttps:\/\/youtu.be\/_liDhKops2w\r\n\r\nIt is possible to use the elimination method with multiplication and get a result that indicates no solutions or infinitely many solutions, just as with the other methods we have learned for finding solutions to systems. In the following example, you will see a system that has infinitely many solutions.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>x<\/i> and <i>y<\/i>.\r\n\r\n<strong>Equation A:<\/strong> [latex]x-3y=-2[\/latex]\r\n\r\n<strong>Equation B:<\/strong> [latex]-2x+6y=4[\/latex]\r\n\r\n[reveal-answer q=\"815327\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"815327\"]Look for terms that can be eliminated. The equations do not have any <i>x<\/i> or <i>y<\/i> terms with the same coefficients.\r\n\r\n[latex]\\begin{array}{r}x-3y=-2\\\\-2x+6y=4\\end{array}[\/latex]\r\n\r\nMultiply the first equation by [latex]2[\/latex] so the x terms of\u00a0cancel out.\r\n\r\n[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\left(x-3y\\right)=2\\left(-2\\right)\\\\-2x+6y=4\\end{array}[\/latex]\r\n\r\nRewrite the system and add the equations.\r\n\r\n[latex]\\begin{array}{r}2x-6y=-4\\\\-2x+6y=4\\\\0x + 0y=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,0=0\\end{array}[\/latex]\r\n\r\nDoes this kind of solution\u00a0look familiar? \u00a0This represents a solution of all real numbers for linear equations, and it represents the same thing when you get this outcome with systems. If we solve both of these equations for y, you will see that they are the same equation.\r\n\r\nSolve equation A for y:\r\n\r\n[latex]\\begin{array}{r}x-3y=-2\\\\-3y=-x-2\\\\y=\\frac{1}{3}x+\\frac{2}{3}\\end{array}[\/latex]\r\n\r\nSolve equation B for y:\r\n\r\n[latex]\\begin{array}-2x+6y=4\\\\6y=2x+4\\\\y=\\frac{2}{6}x+\\frac{4}{6}\\end{array}[\/latex]\r\n\r\nReduce fractions by dividing the numerator and denominator of both fractions by 2:\r\n\r\n[latex]y=\\frac{1}{3}+\\frac{2}{3}[\/latex]\r\n\r\nBoth equations are the same when written in slope intercept form, and therefore the solution set for the system is all real numbers.\r\n<h4>Answer<\/h4>\r\nThe solution is: x and y can be all real numbers.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, the elimination method is used to solve a system of equations. Notice that one of the equations needs to be multiplied by a\u00a0negative one first. \u00a0Additionally, this system has an infinite number of solutions.\r\n\r\nhttps:\/\/youtu.be\/NRxh9Q16Ulk\r\n<h2>Summary<\/h2>\r\nThe substitution method is one way of solving systems of equations. To use the substitution method, use one equation to find an expression for one of the variables in terms of the other variable. Then substitute that expression in place of that variable in the second equation. You can then solve this equation as it will now have only one variable. Solving using the substitution method will yield one of three results: a single value for each variable within the system (indicating one solution), an untrue statement (indicating no solutions), or a true statement (indicating an infinite number of solutions).\r\n\r\nCombining equations is a powerful tool for solving a system of equations. Adding or subtracting two equations in order to eliminate a common variable is called the elimination (or addition) method. Once one variable is eliminated, it becomes much easier to solve for the other one.\r\n\r\nMultiplication can be used to set up matching terms in equations before they are combined to aid in finding a solution to a system. When using the multiplication method, it is important to multiply all the terms on both sides of the equation\u2014not just the one term you are trying to eliminate.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the substitution method\n<ul>\n<li>Solve a system of equations using the substitution method.<\/li>\n<li>Recognize systems of equations that have no solution or an infinite number of solutions<\/li>\n<\/ul>\n<\/li>\n<li>Use the elimination method without multiplication\n<ul>\n<li>Solve a system of equations when no multiplication is necessary to eliminate a variable<\/li>\n<\/ul>\n<\/li>\n<li>Use the elimination method with multiplication\n<ul>\n<li>Use multiplication in combination with\u00a0the elimination method to solve a system of linear equations<\/li>\n<li>Recognize when the solution to a system of linear equations implies there are an infinite number of solutions<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h2>Solve a system of equations using the substitution method<\/h2>\n<p>In the last couple sections, we verified that ordered pairs were solutions to systems, and we used graphs to classify how many solutions a system of two linear equations had. What if we are not given a point of intersection, or it is not obvious from a graph? Can we still find a solution to the system? Of course you can, using algebra!<\/p>\n<p>In this section we will learn the substitution\u00a0method for finding a solution to a system of linear equations in two variables. We have used substitution in different ways throughout this course, for example when we were using the formulas for the area of a triangle and simple interest. We substituted values that we knew into the formula to solve for values that we did not know. \u00a0The idea is similar when applied to solving systems, there are just a few different steps in the process. You will first solve for one variable, and then substitute that expression into the other equation. Let&#8217;s start with an example to see what this means.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Find the value of <i>x<\/i> for this system.<\/p>\n<p>Equation A: [latex]4x+3y=\u221214[\/latex]<\/p>\n<p>Equation B: [latex]y=2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q478211\">Show Solution<\/span><\/p>\n<div id=\"q478211\" class=\"hidden-answer\" style=\"display: none\">The problem asks to solve for <i>x<\/i>. Equation B gives you the value of <i>y<\/i>, [latex]y=2[\/latex], so you can substitute 2 into Equation A for y.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+3y=\u221214\\\\y=2\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<i>\u00a0<\/i><\/p>\n<p>Substitute\u00a0[latex]y=2[\/latex] into\u00a0Equation A.<\/p>\n<p style=\"text-align: center;\">[latex]4x+3\\left(2\\right)=\u221214[\/latex]<\/p>\n<p>Simplify and solve the equation for <i>x.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+6=\u221214\\\\4x=\u221220\\\\x=\u22125\\,\\,\\,\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=\u22125[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>You can substitute a value for a variable even if it is an expression. Here\u2019s an example.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>x <\/i>and <i>y<\/i>.<\/p>\n<p>Equation A: [latex]y+x=3[\/latex]<\/p>\n<p>Equation B: [latex]x=y+5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q300993\">Show Solution<\/span><\/p>\n<div id=\"q300993\" class=\"hidden-answer\" style=\"display: none\">The goal of the substitution method is to rewrite one of the equations in terms of a single variable. Equation B tells us that [latex]x=y+5[\/latex], so it makes sense to substitute [latex]y+5[\/latex] into Equation A for <i>x<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y+x=3\\\\x=y+5\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y+5[\/latex] into Equation A for <i>x<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}y+x=3\\\\y+\\left(y+5\\right)=3\\end{array}[\/latex]<i><\/i><\/p>\n<p>Simplify and solve the equation for <i>y.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2y+5=\\,\\,\\,\\,3\\\\\\underline{\u22125\\,\\,\\,\\,\\,\u22125}\\\\2y=\u22122\\\\y=\u22121\\end{array}[\/latex]<\/p>\n<p>Now find <i>x <\/i>by\u00a0substituting this value for <i>y<\/i> into either equation and solve for <i>x<\/i>. We will use Equation A here.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}y+x=3\\\\\u22121+x=3\\\\\\underline{+1\\,\\,\\,\\,\\,\\,\\,\\,\\,+1}\\\\x=4\\end{array}[\/latex]<\/p>\n<p>Finally, check the solution\u00a0[latex]x=4[\/latex], [latex]y=\u22121[\/latex]\u00a0by substituting these values into each of the original equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}y+x=3\\\\\u22121+4=3\\\\3=3\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=y+5\\\\4=\u22121+5\\\\4=4\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=4[\/latex] and [latex]y=\u22121[\/latex]<\/p>\n<p>The solution is [latex](4,\u22121)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Remember, a solution to a system of equations must be a solution to each of the equations within the system. The ordered pair [latex](4,\u22121)[\/latex] does work for both equations, so you know that it is a solution to the system as well.<\/p>\n<p>Let\u2019s look at another example whose substitution involves the distributive property.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>x <\/i>and <i>y<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y = 3x + 6\\\\\u22122x + 4y = 4\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q240040\">Show Solution<\/span><\/p>\n<div id=\"q240040\" class=\"hidden-answer\" style=\"display: none\">Choose an equation to use for the substitution.<\/p>\n<p>The first equation tells you how to express <i>y<\/i> in terms of <i>x<\/i>, so it makes sense to substitute 3<i>x<\/i> + 6 into the second equation for <i>y<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=3x+6\\\\\u22122x+4y=4\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]3x+6[\/latex] for<i> y<\/i> into the second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+4y=4\\\\\u22122x+4\\left(3x+6\\right)=4\\end{array}[\/latex]<i><\/i><\/p>\n<p>Simplify and solve the equation for <i>x.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+12x+24=4\\,\\,\\,\\,\\,\\,\\,\\\\10x+24=4\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{\u221224\\,\\,\u221224\\,\\,\\,\\,}\\\\10x=\u221220\\\\x=\u22122\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>To find <i>y<\/i>, substitute this value for <i>x<\/i> back into one of the original equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=3x+6\\\\y=3\\left(\u22122\\right)+6\\\\y =\u22126+6\\\\y=0\\end{array}[\/latex]<\/p>\n<p>Check the solution [latex]x=\u22122[\/latex], [latex]y=0[\/latex]\u00a0by substituting them into each of the original equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=3x+6\\\\0=3\\left(\u22122\\right)+6\\\\0=\u22126+6\\\\0=0\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+4y=4\\\\\u22122\\left(\u22122\\right)+4\\left(0\\right)=4\\\\4+0=4\\\\4=4\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=\u22122[\/latex] and [latex]y=0[\/latex]<\/p>\n<p>The solution is (\u22122, 0).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the examples above, one of the equations was already given to us in terms of the variable <i>x<\/i> or <i>y<\/i>. This allowed us to quickly substitute that value into the other equation and solve for one of the unknowns.<\/p>\n<p>Sometimes you may have to rewrite one of the equations in terms of one of the variables first before you can substitute. In\u00a0the example below, you will first need to isolate one of the variables before you can substitute it into the other equation.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>x <\/i>and <i>y<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+3y=22\\\\3x+y=19\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q344538\">Show Solution<\/span><\/p>\n<div id=\"q344538\" class=\"hidden-answer\" style=\"display: none\">Choose an equation to use for the substitution. The second equation,<\/p>\n<p>[latex]3x+y=19[\/latex], can easily be rewritten in terms of <i>y<\/i>, so it makes sense to start there.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}2x+3y=22\\\\3x+y=19\\end{array}[\/latex]<\/p>\n<p>Rewrite [latex]3x+y=19[\/latex]\u00a0in terms of <i>y<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}3x+y=19\\\\y=19\u20133x\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]19\u20133x[\/latex] for <i>y<\/i> in the other equation.<\/p>\n<p style=\"text-align: center;\"><i><\/i>[latex]\\begin{array}{r}2x+3y=22\\\\2x+3(19\u20133x)=22\\end{array}[\/latex]<i><\/i><\/p>\n<p>Simplify and solve the equation for <i>x.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+57\u20139x=22\\,\\,\\,\\,\\\\\u22127x+57=22\\,\\,\\,\\,\\\\\u22127x=\u221235\\\\x=5\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]x=5[\/latex] back into one of the original equations to solve for <i>y.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}3x+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\3\\left(5\\right)+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\15+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\y=19\u221215\\\\y=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Check both solutions by substituting them into each of the original equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+3y=22\\\\2(5)+3\\left(4\\right)=22\\\\10+12=22\\\\22=22\\\\\\text{TRUE}\\\\\\\\3x+y=19\\\\3\\left(5\\right)+4= 19\\\\19=19\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x=5[\/latex] and [latex]y=4[\/latex]<\/p>\n<p>The solution is (5, 4).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will be given an example of solving a systems 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. \u00a0It is really a matter of preference because sometimes solving for a variable will result in having to work with fractions. \u00a0As you become more experienced with algebra, you will be able to anticipate what choices will lead to more desirable outcomes.<\/p>\n<h2 id=\"title2\">Recognize systems of equations that have no solution or an infinite number of solutions<\/h2>\n<p>When we learned methods for solving linear equations in one variable, we found that some equations didn&#8217;t have any solutions, and others had an infinite number of solutions. \u00a0We saw this behavior again\u00a0when we started describing solutions to systems of equations in two variables.<\/p>\n<p>Recall this example from Module 1 for solving linear equations in one variable:<\/p>\n<p style=\"text-align: center;\">Solve for <i>x<\/i>.\u00a0[latex]12+2x\u20138=7x+5\u20135x[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}12+2x-8=7x+5-5x\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x+4=2x+5\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2x+4=2x+5\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-2x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2x\\,\\,\\,\\,\\,\\,\\,\\,}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4= \\,5\\end{array}[\/latex]<\/p>\n<p>This false statement implies there are <strong>no solutions<\/strong> to this equation. \u00a0In the same way, you may see an outcome like this when you use the substitution method to find a solution to a system of linear equations in two variables. In the next example, you will see an example of a system of two equations that does not have a solution.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>x<\/i> and <i>y<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y=5x+4\\\\10x\u22122y=4\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q787022\">Show Solution<\/span><\/p>\n<div id=\"q787022\" class=\"hidden-answer\" style=\"display: none\">Since the first equation is [latex]y=5x+4[\/latex], you can substitute [latex]5x+4[\/latex] in for <i>y<\/i> in the second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}y=5x+4\\\\10x\u22122y=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\10x\u20132\\left(5x+4\\right)=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Expand the expression on the left.<\/p>\n<p style=\"text-align: center;\">[latex]10x\u201310x\u20138=4[\/latex]<\/p>\n<p>Combine like terms on the left side of equation.<\/p>\n<p style=\"text-align: center;\">[latex]10x\u201310x=0[\/latex], so you are left with [latex]\u22128=4[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}0\u20138=4\\\\\u22128=4\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The statement [latex]\u22128=4[\/latex] is false, so there is no solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>You get the false statement [latex]\u22128=4[\/latex]. What does this mean? The graph of this system sheds some light on what is happening.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2994\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/20183038\/Screen-Shot-2016-04-20-at-11.30.22-AM.png\" alt=\"Two parallel lines. One line is y=5x+4, and the other line is 10x-2y=4.\" width=\"459\" height=\"458\" \/><\/p>\n<p>The lines are parallel, they never intersect and there is no solution to this system of linear equations. Note that the result [latex]\u22128=4[\/latex] is <b>not<\/b> a solution. It is simply a false statement and it indicates that there is <b>no<\/b> solution.<\/p>\n<p>We have also seen linear equations in one variable and systems of equations in two variables that have an infinite number of solutions. \u00a0In the next example, you will see what happens when you apply the substitution method to a system with an infinite number of solutions.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for x and y.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,y=\u22120.5x\\\\9y=\u22124.5x\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q683508\">Show Solution<\/span><\/p>\n<div id=\"q683508\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substituting \u22120.5<i>x<\/i> for <i>y<\/i> in the second equation, you find the following:<\/p>\n<div style=\"margin: auto;\">\n<p>[latex]\\begin{array}{r}9y=\u22124.5x\\\\9(\u22120.5x)=\u22124.5\\,\\,\\,\\\\\u22124.5x=\u22124.5x\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>This time you get a true statement: [latex]\u22124.5x=\u22124.5x[\/latex]. But what does this type of answer mean? Again, graphing can help you make sense of this system.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2993\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/04\/20182902\/Screen-Shot-2016-04-20-at-11.28.43-AM.png\" alt=\"Two overlapping lines that represent the same line. One line is 9y=-4.5x, and the other line is y=-0.5x.\" width=\"458\" height=\"460\" \/><\/p>\n<p>This system consists of two equations that both represent the same line; the two lines are collinear. Every point along the line will be a solution to the system, and that\u2019s why the substitution method yields a true statement. In this case, there are an infinite number of solutions.<\/p>\n<p>In the following video you will see an example of solving a system that has an infinite number of solutions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Solve a System of Equations Using Substitution - Infinite Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Pcqb109yK5Q?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the following video you will see an example of solving a system of equations that has no solutions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  Solve a System of Equations Using Substitution - No Solution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/kTtKfh5gFUc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Solve a system of equations using the elimination\u00a0method<\/h2>\n<p>The <b>elimination method<\/b> for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an equation to eliminate one of the variable terms. \u00a0In this method, you may or may not need to multiply the terms in one equation by a number first. \u00a0We will first look at examples where no multiplication is necessary to use the elimination method. \u00a0In the next section you will see examples using multiplication after you are familiar with the idea of the elimination method.<\/p>\n<p>It is easier to show rather than tell with this method, so let&#8217;s dive right into some examples.<\/p>\n<p>If you add the two equations,<\/p>\n<p>[latex]x\u2013y=\u22126[\/latex] and [latex]x+y=8[\/latex] together, watch what happens.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,x-y=\\,-6\\\\\\underline{+\\,x+y=\\,\\,\\,8}\\\\\\,2x+0\\,=\\,\\,\\,\\,2\\end{array}[\/latex]<\/p>\n<p>You have eliminated the <i>y<\/i> term, and this equation can be solved using the methods for solving equations with one variable.<\/p>\n<p>Let\u2019s see how this system is solved using the elimination method.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use elimination to solve the system.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x\u2013y=\u22126\\\\x+y=\\,\\,\\,\\,8\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q403819\">Show Solution<\/span><\/p>\n<div id=\"q403819\" class=\"hidden-answer\" style=\"display: none\">Add the equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{r}x-y=\\,\\,-6\\\\+\\underline{\\,\\,x+y=\\,\\,\\,\\,\\,8}\\\\\\,\\,\\,\\,\\,\\,2x\\,\\,\\,\\,\\,=\\,\\,\\,\\,\\,\\,2\\end{array}[\/latex]<\/p>\n<p>Solve for <i>x<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x=2\\\\x=1\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]x=1[\/latex] into one of the original equations and solve for <i>y<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+y=8\\\\1+y=8\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=8\u20131\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=7\\end{array}[\/latex]<\/p>\n<p>Be sure to check your answer in both equations!<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x\u2013y=\u22126\\\\1\u20137=\u22126\\\\\u22126=\u22126\\\\\\text{TRUE}\\\\\\\\x+y=8\\\\1+7=8\\\\8=8\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>The answers check.<\/p>\n<h4>Answer<\/h4>\n<p>The solution is (1, 7).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Unfortunately not all systems work out this easily. How about a system like [latex]2x+y=12[\/latex] and [latex]\u22123x+y=2[\/latex]. If you add these two equations together, no variables are eliminated.<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle \\begin{array}{l}\\,\\,\\,\\,2x+y=12\\\\\\underline{-3x+y=\\,\\,\\,2}\\\\-x+2y=14\\end{array}[\/latex]<\/p>\n<p>But you want to eliminate a variable. So let\u2019s add the opposite of one of the equations to the other equation. This means multiply every term in one of the equations by -1, so that the sign of every terms is opposite.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,2x+\\,\\,y\\,=12\\rightarrow2x+y=12\\rightarrow2x+y=12\\\\\u22123x+\\,\\,y\\,=2\\rightarrow\u2212\\left(\u22123x+y\\right)=\u2212(2)\\rightarrow3x\u2013y=\u22122\\\\\\,\\,\\,\\,5x+0y=10\\end{array}[\/latex]<\/p>\n<p>You have eliminated the <i>y<\/i> variable, and the problem can now be solved.<\/p>\n<p>The following video describe a similar problem where you can eliminate one variable by adding the two equations together.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" 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 shaded\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"63\" height=\"56\" \/>Caution! \u00a0When you add the opposite of one entire equation to another, make sure to change the sign of EVERY term on both sides of the equation. This is a very common\u00a0mistake to make.<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use elimination to solve the system.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=12\\\\\u22123x+y=2\\,\\,\\,\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q702178\">Show Solution<\/span><\/p>\n<div id=\"q702178\" class=\"hidden-answer\" style=\"display: none\">You can eliminate the <i>y<\/i>-variable if you add the opposite of one of the equations to the other equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=12\\\\\u22123x+y=2\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Rewrite the second equation as its opposite.<\/p>\n<p>Add.\u00a0Solve for <i>x<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=12\\,\\\\3x\u2013y=\u22122\\\\5x=10\\,\\\\x=2\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=2[\/latex] into one of the original equations and solve for <i>y<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(2\\right)+y=12\\\\4+y=12\\\\y=8\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Be sure to check your answer in both equations!<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+y=12\\\\2\\left(2\\right)+8=12\\\\4+8=12\\\\12=12\\\\\\text{TRUE}\\\\\\\\\u22123x+y=2\\\\\u22123\\left(2\\right)+8=2\\\\\u22126+8=2\\\\2=2\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>The answers check.<\/p>\n<h4>Answer<\/h4>\n<p>The solution is (2, 8).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following are two more examples showing how to solve linear systems of equations using elimination.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use elimination to solve the system.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\2x+5y=\\,25\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q438400\">Show Solution<\/span><\/p>\n<div id=\"q438400\" class=\"hidden-answer\" style=\"display: none\">Notice the coefficients of each variable in each equation. If you add these two equations, the <i>x<\/i> term will be eliminated since [latex]\u22122x+2x=0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\2x+5y=\\,25\\end{array}[\/latex]<\/p>\n<p>Add and solve for <i>y<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\2x+5y=25\\,\\\\8y=24\\,\\\\y=3\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=3[\/latex]<i> <\/i>into one of the original equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+5y=25\\\\2x+5\\left(3\\right)=25\\\\2x+15=25\\\\2x=10\\\\x=5\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Check solutions.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\\u22122\\left(5\\right)+3\\left(3\\right)=\u22121\\\\\u221210+9=\u22121\\\\\u22121=\u22121\\\\\\text{TRUE}\\\\\\\\2x+5y=25\\\\2\\left(5\\right)+5\\left(3\\right)=25\\\\10+15=25\\\\25=25\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>The answers check.<\/p>\n<h4>Answer<\/h4>\n<p>The solution is (5, 3).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use elimination to solve for <i>x <\/i>and <i>y.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+2y=14\\\\5x+2y=16\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q776093\">Show Solution<\/span><\/p>\n<div id=\"q776093\" class=\"hidden-answer\" style=\"display: none\">Notice the coefficients of each variable in each equation. You will need to add the opposite of one of the equations to eliminate the variable <i>y<\/i>, as [latex]2y+2y=4y[\/latex], but [latex]2y+\\left(\u22122y\\right)=0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+2y=14\\\\5x+2y=16\\end{array}[\/latex]<\/p>\n<p>\u00a0Change one of the equations to its opposite, add and solve for <i>x<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+2y=14\\,\\,\\,\\,\\\\\u22125x\u20132y=\u221216\\\\\u2212x=\u22122\\,\\,\\,\\\\x=2\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]x=2[\/latex] into one of the original equations and solve for <i>y<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4x+2y=14\\\\4\\left(2\\right)+2y=14\\\\8+2y=14\\\\2y=6\\,\\,\\,\\\\y=3\\,\\,\\,\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The solution is (2, 3).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Go ahead and check this last example\u2014substitute (2, 3) into both equations. You get two true statements: 14=14 and 16=16!<\/p>\n<p>Notice that you could have used the opposite of the first equation rather than the second equation and gotten the same result.<\/p>\n<h2 id=\"video1\" class=\"no-indent\" style=\"text-align: left;\">Recognize systems that have no solution or an infinite number of solutions<\/h2>\n<p>Just as with the substitution method, the elimination method will sometimes eliminate <i>both v<\/i>ariables, and you end up with either a true statement or a false statement. Recall that a false statement means that there is no solution.<\/p>\n<p>Let\u2019s look at an example.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>x <\/i>and <i>y.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}-x\u2013y=-4\\\\x+y=2\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q101540\">Show Solution<\/span><\/p>\n<div id=\"q101540\" class=\"hidden-answer\" style=\"display: none\">Add the equations to eliminate the\u00a0<em>x<\/em>-term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}-x\u2013y=-4\\\\\\underline{x+y=2\\,\\,\\,}\\\\0=\u22122\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>There is no solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Graphing these lines shows that they are parallel lines and as such do not share any point in common, verifying that there is no solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064435\/image040-1.jpg\" alt=\"Two parallel lines. One line is -x-y=-4. The other line is x+y=2.\" width=\"346\" height=\"345\" \/><\/p>\n<p>If both variables are eliminated and you are left with a true statement, this indicates that there are an infinite number of ordered pairs that satisfy both of the equations. In fact, the equations are the same line.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>x <\/i>and <i>y<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x+y=2\\,\\,\\,\\,\\\\-x\u2212y=-2\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q328100\">Show Solution<\/span><\/p>\n<div id=\"q328100\" class=\"hidden-answer\" style=\"display: none\">Add the equations to eliminate the\u00a0<i>x<\/i>-term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x+y=2\\,\\,\\,\\,\\\\\\underline{-x\u2212y=-2}\\\\0=0\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>There are an infinite number of solutions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Graphing these two equations will help to illustrate what is happening.<\/p>\n<p><b><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064437\/image041-1.jpg\" alt=\"Two overlapping lines. One is -x-y=-2, and the other is x+y=2.\" width=\"346\" height=\"345\" \/><\/b><\/p>\n<h2 id=\"video3\" class=\"no-indent\" style=\"text-align: left;\"><\/h2>\n<p>In the following video, a system of equations which has no solutions is solved using the method of elimination.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex:  System of Equations Using Elimination (No Solution)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/z5_ACYtzW98?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 id=\"title2\">Solve a system of equations when multiplication is necessary to eliminate a variable<\/h2>\n<p>Many times adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\n<p>If you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let\u2019s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will allow you to eliminate\u00a0the same variable in the other equation.<\/p>\n<p>We do this with multiplication.\u00a0\u00a0Notice that the first equation contains the term 4<i>y<\/i>, and the second equation contains the term <i>y<\/i>. If you multiply the second equation by \u22124, when you add both equations the <i>y<\/i> variables will add up to 0.<\/p>\n<p>The following example takes you through all the steps to find a solution to this system.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>x<\/i> and <i>y<\/i>.<\/p>\n<p><strong>Equation A:<\/strong> [latex]3x+4y=52[\/latex]<\/p>\n<p><strong>Equation B:<\/strong> [latex]5x+y=30[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q815377\">Show Solution<\/span><\/p>\n<div id=\"q815377\" class=\"hidden-answer\" style=\"display: none\">Look for terms that can be eliminated. The equations do not have any <i>x<\/i> or <i>y<\/i> terms with the same coefficients.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\n<p>Multiply the second equation by [latex]\u22124[\/latex] so they do have the same coefficient.<\/p>\n<p>[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,3x+4y=52\\\\\u22124\\left(5x+y\\right)=\u22124\\left(30\\right)\\end{array}[\/latex]<\/p>\n<p>Rewrite the system and add the equations.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\,\\,\\,\\,\\,\\,\\,\\\\\u221220x\u20134y=\u2212120\\end{array}[\/latex]<\/p>\n<p>Solve for <i>x<\/i>.<\/p>\n<p>[latex]\\begin{array}{l}\u221217x=-68\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x=4\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]x=4[\/latex] into one of the original equations to find <i>y<\/i>.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\3\\left(4\\right)+4y=52\\\\12+4y=52\\\\4y=40\\\\y=10\\end{array}[\/latex]<\/p>\n<p>Check your answer.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\3\\left(4\\right)+4\\left(10\\right)=52\\\\12+40=52\\\\52=52\\\\\\text{TRUE}\\\\\\\\5x+y=30\\\\5\\left(4\\right)+10=30\\\\20+10=30\\\\30=30\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>The answers check.<\/p>\n<h4>Answer<\/h4>\n<p>The solution is (4, 10).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"55\" height=\"49\" \/>Caution! \u00a0When you\u00a0use multiplication to eliminate a variable, you must multiply EACH term in the equation by the number you choose. \u00a0Forgetting to multiply every term is a common mistake.<\/div>\n<p>There are other ways to solve this system. Instead of multiplying one equation in order to eliminate a variable when the equations were added, you could have multiplied <i>both<\/i> equations by different numbers.<\/p>\n<p>Let\u2019s remove the variable <i>x <\/i>this time. Multiply Equation A by 5 and Equation B by [latex]\u22123[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <em>x<\/em> and <em>y<\/em>.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q40585\">Show Solution<\/span><\/p>\n<div id=\"q40585\" class=\"hidden-answer\" style=\"display: none\">Look for terms that can be eliminated. The equations do not have any <i>x<\/i> or <i>y<\/i> terms with the same coefficient.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\n<p>In order to use the elimination method, you have to create variables that have the same coefficient\u2014then you can eliminate them. Multiply the top equation by 5.<\/p>\n<p>[latex]\\begin{array}{r}5\\left(3x+4y\\right)=5\\left(52\\right)\\\\5x+y =30\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\15x+20y=260\\,\\,\\,\\,\\,\\,\\\\5x+y=30\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Now multiply the bottom equation by \u22123.<\/p>\n<p>[latex]\\begin{array}{r}15x+20y=260\\,\\,\\,\\,\\,\\,\\,\\,\\\\-3(5x+y)=\u22123(30)\\\\15x+20y=260\\,\\,\\,\\,\\,\\,\\,\\,\\\\\u221215x\u20133y=\u221290\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Next add the equations, and solve for <i>y<\/i>.<\/p>\n<p>[latex]\\begin{array}{r}15x+20y=260\\\\\u221215x\u20133y=\\,\u201390\\\\17y=170\\\\y=\\,\\,\\,10\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=10[\/latex] into one of the original equations to find <i>x<\/i>.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\3x+4\\left(10\\right)=52\\\\3x+40=52\\\\3x=12\\\\x=4\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>You arrive at the same solution as before.<\/p>\n<h4>Answer<\/h4>\n<p>The solution is (4, 10).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>These equations were multiplied by 5 and [latex]\u22123[\/latex] respectively, because that gave you terms that would add up to 0. Be sure to multiply all of the terms of the equation.<\/p>\n<p>In the following video, you will see an example of using the elimination method for solving a system of equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" 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<p>It is possible to use the elimination method with multiplication and get a result that indicates no solutions or infinitely many solutions, just as with the other methods we have learned for finding solutions to systems. In the following example, you will see a system that has infinitely many solutions.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>x<\/i> and <i>y<\/i>.<\/p>\n<p><strong>Equation A:<\/strong> [latex]x-3y=-2[\/latex]<\/p>\n<p><strong>Equation B:<\/strong> [latex]-2x+6y=4[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q815327\">Show Solution<\/span><\/p>\n<div id=\"q815327\" class=\"hidden-answer\" style=\"display: none\">Look for terms that can be eliminated. The equations do not have any <i>x<\/i> or <i>y<\/i> terms with the same coefficients.<\/p>\n<p>[latex]\\begin{array}{r}x-3y=-2\\\\-2x+6y=4\\end{array}[\/latex]<\/p>\n<p>Multiply the first equation by [latex]2[\/latex] so the x terms of\u00a0cancel out.<\/p>\n<p>[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\left(x-3y\\right)=2\\left(-2\\right)\\\\-2x+6y=4\\end{array}[\/latex]<\/p>\n<p>Rewrite the system and add the equations.<\/p>\n<p>[latex]\\begin{array}{r}2x-6y=-4\\\\-2x+6y=4\\\\0x + 0y=0\\\\\\,\\,\\,\\,\\,\\,\\,\\,0=0\\end{array}[\/latex]<\/p>\n<p>Does this kind of solution\u00a0look familiar? \u00a0This represents a solution of all real numbers for linear equations, and it represents the same thing when you get this outcome with systems. If we solve both of these equations for y, you will see that they are the same equation.<\/p>\n<p>Solve equation A for y:<\/p>\n<p>[latex]\\begin{array}{r}x-3y=-2\\\\-3y=-x-2\\\\y=\\frac{1}{3}x+\\frac{2}{3}\\end{array}[\/latex]<\/p>\n<p>Solve equation B for y:<\/p>\n<p>[latex]\\begin{array}-2x+6y=4\\\\6y=2x+4\\\\y=\\frac{2}{6}x+\\frac{4}{6}\\end{array}[\/latex]<\/p>\n<p>Reduce fractions by dividing the numerator and denominator of both fractions by 2:<\/p>\n<p>[latex]y=\\frac{1}{3}+\\frac{2}{3}[\/latex]<\/p>\n<p>Both equations are the same when written in slope intercept form, and therefore the solution set for the system is all real numbers.<\/p>\n<h4>Answer<\/h4>\n<p>The solution is: x and y can be all real numbers.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, the elimination method is used to solve a system of equations. Notice that one of the equations needs to be multiplied by a\u00a0negative one first. \u00a0Additionally, this system has an infinite number of solutions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Ex:  System of Equations Using Elimination (Infinite Solutions)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/NRxh9Q16Ulk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>The substitution method is one way of solving systems of equations. To use the substitution method, use one equation to find an expression for one of the variables in terms of the other variable. Then substitute that expression in place of that variable in the second equation. You can then solve this equation as it will now have only one variable. Solving using the substitution method will yield one of three results: a single value for each variable within the system (indicating one solution), an untrue statement (indicating no solutions), or a true statement (indicating an infinite number of solutions).<\/p>\n<p>Combining equations is a powerful tool for solving a system of equations. Adding or subtracting two equations in order to eliminate a common variable is called the elimination (or addition) method. Once one variable is eliminated, it becomes much easier to solve for the other one.<\/p>\n<p>Multiplication can be used to set up matching terms in equations before they are combined to aid in finding a solution to a system. When using the multiplication method, it is important to multiply all the terms on both sides of the equation\u2014not just the one term you are trying to eliminate.<\/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-849\">\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>Ex 2: Solve a System of Equations Using Substitution. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/MIXL35YRzRw\">https:\/\/youtu.be\/MIXL35YRzRw<\/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 Substitution - Infinite Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Pcqb109yK5Q\">https:\/\/youtu.be\/Pcqb109yK5Q<\/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 Substitution - No Solution. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/kTtKfh5gFUc\">https:\/\/youtu.be\/kTtKfh5gFUc<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em><\/li><li>Unit 13: Graphing, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 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: System of Equations Using Elimination (Infinite Solutions) . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/NRxh9Q16Ulk\">https:\/\/youtu.be\/NRxh9Q16Ulk<\/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: System of Equations Using Elimination (Infinite Solutions). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/NRxh9Q16Ulk\">https:\/\/youtu.be\/NRxh9Q16Ulk<\/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) for Lumen Learning. <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><\/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":20,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex 2: Solve a System of Equations Using Substitution\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/MIXL35YRzRw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Solve a System of Equations Using Substitution - 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