{"id":119,"date":"2023-11-08T16:09:58","date_gmt":"2023-11-08T16:09:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/solving-systems-of-equations-by-substitution\/"},"modified":"2024-08-13T05:00:41","modified_gmt":"2024-08-13T05:00:41","slug":"2-r-solving-linear-systems-of-equations-by-substitution-and-elimination","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/2-r-solving-linear-systems-of-equations-by-substitution-and-elimination\/","title":{"raw":"2.R Solving Linear Systems of Equations by Substitution and Elimination","rendered":"2.R Solving Linear Systems of Equations by Substitution and Elimination"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Investigate types of solutions to systems of equations with two variables graphically.<\/li>\r\n \t<li>Solve linear systems of equations with two variables by substitution and elimination.<\/li>\r\n \t<li>Identify inconsistent or dependent systems of equations containing two variables.<\/li>\r\n<\/ul>\r\n<\/div>\r\nA <strong>system of linear equations<\/strong> consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find a solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions.\r\n\r\nFor example, consider the following system of linear equations in two variables.\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}2x+y=15\\\\3x-y=5\\end{cases}[\/latex]<\/p>\r\nA <strong>solution<\/strong>\u00a0to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair\u00a0 [latex](4, 7)[\/latex] is the solution to the system of linear equations. We can verify the solution by substituting the values [latex]\\color{green}{x=4},\\color{blue}{y=7}[\/latex] into each equation to see if the ordered pair satisfies both equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2\\color{green}{\\left(4\\right)}+\\color{blue}{\\left(7\\right)}=15\\text{ }\\text{True}\\hfill \\\\ 3\\color{green}{\\left(4\\right)}-\\color{blue}{\\left(7\\right)}=5\\text{ }\\text{True}\\hfill \\end{array}[\/latex]<\/p>\r\nIn addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A <strong>consistent system<\/strong> of equations has at least one solution. A consistent system is considered to be an <strong>independent system<\/strong> if it has a single solution like the example we just explored. A consistent system is considered to be a <strong>dependent system<\/strong> if the equations are equivalent. In other words, every solution to one equation will also be a solution to the other equation. Thus, there are an infinite number of solutions.\r\n\r\nAnother type of system of linear equations is an <strong>inconsistent system\u00a0<\/strong>which is one where the equations cannot both be true for a single ordered pair.\u00a0Hence, there is no solution to the system.\u00a0 An easy example is\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}x+y=3\\\\x+y=4\\end{cases}[\/latex]<\/p>\r\nClearly, the expression [latex]x+y[\/latex] cannot possibly have two different values at once.\r\n<div class=\"textbox shaded\">\r\n<h3>A General Note: Types of Linear Systems<\/h3>\r\nThere are three types of systems of linear equations in two variables and three types of solutions.\r\n\r\nBelow are graphical representations of each type of system.\u00a0 It is important to understand that where the two lines cross represents an ordered pair that satisfies both equations, thus is a solution to the system.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222630\/CNX_Precalc_Figure_09_01_002n2.jpg\" alt=\"Graphs of an independent system, an inconsistent system, and a dependent system. The independent system has two lines which cross at the point seven-fifths, negative eleven fifths. The inconsistent system shows two parallel lines. The dependent system shows a single line running through the points negative one, negative two and one, two.\" width=\"945\" height=\"479\" \/>\r\n<ul>\r\n \t<li>An <strong>independent system<\/strong> has exactly one solution pair [latex]\\left(x,y\\right)[\/latex]. The point where the two lines intersect is the only solution.<\/li>\r\n \t<li>An <strong>inconsistent system<\/strong> has no solution. The two lines are parallel and will never intersect.\u00a0 They have the same slope but different [latex]y[\/latex]-intercepts.<\/li>\r\n \t<li>A <strong>dependent system<\/strong> has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.<\/li>\r\n<\/ul>\r\nThe independent and dependent systems are also consistent because they both have at least one solution.\r\n\r\n<\/div>\r\n<h2>Solving Systems of Equations by Graphing<\/h2>\r\nThere are multiple methods of solving systems of linear equations. For a <strong>system of linear equations<\/strong> in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the following system of equations by graphing. Identify the type of system.\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}2x&amp;+\\ \\ y=-8\\\\ x&amp;-\\ \\ y=-1\\end{cases}[\/latex]<\/p>\r\n[reveal-answer q=\"336799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"336799\"]\r\n\r\nSolve the first equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+y&amp;=-8\\\\ y&amp;=-2x - 8\\end{align}[\/latex]<\/p>\r\nSolve the second equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x-y&amp;=-1\\\\ -y&amp;=-x-1\\\\ y&amp;=x+1\\end{align}[\/latex]<\/p>\r\nGraph both equations on the same set of axes as seen below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222636\/CNX_Precalc_Figure_09_01_0122.jpg\" alt=\"A graph of two lines running through the point negative 3, negative 2. The first line's equation is y equals minus 2x minus 8. The second line's equation is y equals x+1.\" width=\"487\" height=\"316\" \/>\r\n\r\nThe lines appear to intersect at the point [latex]\\left(-3,-2\\right)[\/latex]. We can check to make sure that this is the solution to the system by substituting the ordered pair into both original equations\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}2\\left(-3\\right)+\\left(-2\\right)=-8\\hfill &amp; \\hfill \\\\ \\text{ }-8=-8\\hfill &amp; \\text{True}\\hfill \\\\ \\text{ }\\left(-3\\right)-\\left(-2\\right)=-1\\hfill &amp; \\hfill \\\\ \\text{ }-1=-1\\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solution to the system is the ordered pair [latex]\\left(-3,-2\\right)[\/latex], so the system is independent.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Substitution Method<\/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 algebraic methods of solving a <strong>system of linear equations<\/strong> that can always find an exact solution.\r\n\r\nConsider the system\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}y=2x\\\\x+y=12\\end{cases}[\/latex]<\/p>\r\nSince [latex]y[\/latex] and [latex]2x[\/latex] are the same quantity by the first equation, they can be used interchangeably in the problem from this point forward. Thus we can replace the\u00a0[latex]\\color{green}{y}[\/latex] in the second equation with [latex]\\color{green}{2x}.[\/latex] This is called a <strong>substitution<\/strong> and is the basis of the <strong>substitution method<\/strong>. In the second equation now,\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+\\color{green}{(2x)}&amp;=12\\\\3x&amp;=12\\\\x&amp;=4\\end{align}[\/latex]<\/p>\r\nOnce we know that [latex]x=4,[\/latex] use this value in the first equation to obtain\u00a0[latex]y=2(4)=8.[\/latex] The system has solution [latex](4,8).[\/latex]\r\n<div class=\"textbox shaded\">\r\n<h3>Solving by Substitution<\/h3>\r\n<ol>\r\n \t<li>Solve one of the two equations for a variable.<\/li>\r\n \t<li>Substitute the expression for this variable into the second equation, and then solve for the remaining variable.<\/li>\r\n \t<li>Use the value of the second variable to find the value of the first.<\/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<\/h3>\r\nSolve the following system of equations by substitution.\r\n<div style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{llll}-x&amp;+&amp;y&amp;=-5\\\\ 2x&amp;-&amp;5y&amp;=1\\\\ \\end{array}\\right. [\/latex]<\/div>\r\n[reveal-answer q=\"748381\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"748381\"]\r\n\r\nWe will solve the first equation for [latex]y[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}-x+y&amp;=-5\\\\ y&amp;=x - 5\\end{align}[\/latex]<\/div>\r\nNow, we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.\u00a0 Notice that this results in an equation where [latex]x[\/latex] is the only variable.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}2x - 5y&amp;=1\\hfill \\\\ 2x - 5\\color{green}{\\left(x - 5\\right)}&amp;=1\\hfill \\\\ 2x - 5x+25&amp;=1\\hfill \\\\ -3x&amp;=-24\\hfill \\\\ \\text{ }x&amp;=8\\hfill \\end{array}[\/latex]<\/div>\r\nNow, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}-\\left(8\\right)+y&amp;=-5\\hfill \\\\ \\text{ }y&amp;=3\\hfill \\end{array}[\/latex]<\/div>\r\nOur solution is [latex]\\left(8,3\\right)[\/latex].\r\n\r\nCheck the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{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]<\/p>\r\nThe substitution method can be used to solve any linear system in two variables, but the method works best if one of the equations contains a coefficient of [latex]1[\/latex] or [latex]\u20131[\/latex] so that we do not have to deal with fractions.\u00a0 This is why we chose to solve for\u00a0[latex]y[\/latex] in the first step.\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 system of two equations using the substitution method.\r\n\r\nhttps:\/\/youtu.be\/MIXL35YRzRw\r\n\r\nIf you had chosen the other equation to start with in the previous example, you would still be able to find the same solution. It is really a matter of preference, because sometimes solving for a variable will result in having to work with fractions. Substitution works best if one of the variables is easily solved for.\r\n\r\nHere is an example of what it looks like to solve an\u00a0<strong>inconsistent system<\/strong>\u00a0using substitution. We can tell it is inconsistent when we arrive at a line where there are no variables left and the numerical statement is false.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the following system of equations.\r\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{llll}x&amp;=&amp;9 &amp;- \\ \\ 2y\\\\ x&amp;+&amp;2y&amp;=\\ 13 \\end{array}\\right. [\/latex]<\/p>\r\n[reveal-answer q=\"86393\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"86393\"]\r\n\r\nSince the first equation is already solved for [latex]x,[\/latex] we can substitute the expression [latex]9-2y[\/latex] for [latex]x[\/latex] in the second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}x+2y&amp;=13\\\\ \\color{green}{\\left(9 - 2y\\right)}+2y&amp;=13 \\\\ 9+0y&amp;=13 \\\\ 9&amp;=13 \\end{array}[\/latex]<\/p>\r\nClearly, this statement is a contradiction because [latex]9\\ne 13[\/latex]. Therefore, the system has no solution.\r\n\r\nHere are the two equations solved to slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ y=-\\dfrac{1}{2}x+\\dfrac{9}{2}\\end{array}\\hfill \\\\ y=-\\dfrac{1}{2}x+\\dfrac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nWe can see they have the same slope and different [latex]y[\/latex]-intercepts. Therefore, the lines are parallel and do not intersect. This confirms visually that the system is inconsistent.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222645\/CNX_Precalc_Figure_09_01_0072.jpg\" alt=\"A graph of two parallel lines. The first line's equation is y equals negative one-half x plus 13 over two. The second line's equation is y equals negative one-half x plus 9 over two.\" width=\"488\" height=\"297\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video, we show another example of using substitution to solve a system that has no solution.\r\n\r\nhttps:\/\/youtu.be\/kTtKfh5gFUc\r\n\r\nThe next video illustrates what happens if solving a dependent system by substitution. This case is similar to the \"no solution\" case, but instead of obtaining a false statement like\u00a0[latex]12=0[\/latex] while solving, you obtain a true statement like\u00a0[latex]12=12[\/latex].\r\n\r\nhttps:\/\/youtu.be\/Pcqb109yK5Q\r\n\r\nWhen a system is dependent, we will write our answer in set-builder notation. For example, suppose a dependent system has equation [latex]5x-2y=10.[\/latex] In set builder notation, we write the solution as [latex]\\left\\{(x,y)\\; |\\; 5x-2y=10\\right\\}.[\/latex] Remember that this indicates there are infinitely many solutions since there are infinitely many ordered pairs that satisfy the equation.\r\n\r\nThere are two things to note here. First, the left side of the notation dictates that solutions to the system have the form of ordered pairs [latex](x,y).[\/latex] Second, we do not need to list the other equation that was in the system at all, since in a dependent system the two equations are equivalent to each other.\r\n<h2>The Elimination Method<\/h2>\r\nA third method of solving systems of linear equations is the <strong>elimination\u00a0method<\/strong>. In this method, we add two equations to each other with the goal of eliminating one variable entirely. This takes advantage of the addition principle:\u00a0 If [latex]a=b[\/latex], then [latex]a+c=b+c[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{l}x+y=\\text{ }5\\\\x-y=\\text{ }11\\end{array} \\right.[\/latex]<\/p>\r\nConsider the above system. The first equation asserts that\u00a0[latex]x+y[\/latex] and\u00a0[latex]5[\/latex] are the same quantity, so we can add\u00a0[latex]\\color{green}{(x+y)}[\/latex] to the left side of the second equation, and\u00a0[latex]\\color{blue}{5}[\/latex] to the right side.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}x-y+\\color{green}{(x+y)}&amp;=11+\\color{blue}{(5)}\\\\2x&amp;=\\text{ }16\\end{array}[\/latex]<\/p>\r\nNotice since the [latex]y[\/latex] terms had opposite coefficients they canceled, leaving only [latex]x[\/latex] to solve for. The rest of the steps are similar to the substitution process. We divide to obtain [latex]x=8,[\/latex] then use this value in equation one to obtain [latex]8+y=5,[\/latex] or [latex]y=-3.[\/latex] The solution is the ordered pair [latex](8, -3).[\/latex]\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the given system of equations by the elimination method.\r\n<div style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}x+2y&amp;=-1\\hfill \\\\ -x+y&amp;=3\\hfill \\end{array}\\right.[\/latex]<\/div>\r\n[reveal-answer q=\"522070\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522070\"]\r\n\r\nFirst make sure the equations are lined up in columns of like terms and like symbols. Notice that the coefficient of [latex]x[\/latex] in the second equation,\u00a0[latex]\u20131[\/latex], is the opposite of the coefficient of [latex]x[\/latex] in the first equation,\u00a0[latex]1[\/latex]. We can add the two equations to eliminate [latex]x[\/latex]. We will typically show this step with the equations added vertically as follows.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{\\begin{array}{l}\\hfill \\\\ \\:\\:x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}}{\\text{}\\text{}\\text{}\\text{}\\text{}\\:\\:\\:\\:\\:\\:3y=2}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNow that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3y=2\\hfill \\\\ \\text{ }y=\\dfrac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThen, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }-x+y=3\\hfill \\\\ \\text{ }-x+\\dfrac{2}{3}=3\\hfill \\\\ \\text{ }-x=3-\\dfrac{2}{3}\\hfill \\\\ \\text{ }-x=\\dfrac{7}{3}\\hfill \\\\ \\text{ }\\:\\:\\:\\:\\:x=-\\dfrac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solution to this system is [latex]\\left(-\\dfrac{7}{3},\\dfrac{2}{3}\\right)[\/latex].\r\n\r\nCheck the solution in the first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&amp;=-1 \\\\ \\left(-\\dfrac{7}{3}\\right)+2\\left(\\dfrac{2}{3}\\right)&amp;=-1\\\\ -\\dfrac{7}{3}+\\dfrac{4}{3}&amp;=-1\\\\ -\\dfrac{3}{3}&amp;=-1\\\\ -1&amp;=-1&amp;&amp; \\text{True}\\end{align}[\/latex]<\/p>\r\nYou can check yourself that the solution also satisfies the second equation.\u00a0 The graph confirms that there is exactly one solution and there is no need to look for other solutions.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222638\/CNX_Precalc_Figure_09_01_0042.jpg\" alt=\"A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.\" width=\"487\" height=\"291\" \/>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will see another example of how to use the method of elimination to solve a system of linear equations.\r\n\r\nhttps:\/\/youtu.be\/M4IEmwcqR3c\r\n\r\nIf neither the [latex]x[\/latex] coefficients nor the [latex]y[\/latex] coefficients are opposites to begin with, we must first multiply one or both equations by a constant. In the next example, you will see a technique where we multiply one of the equations in the system by a number that will allow us to eliminate one of the variables.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the given system of equations by the elimination method<strong>.<\/strong>\r\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}3x+5y&amp;=-11\\hfill \\\\ x - 2y&amp;=11\\hfill \\end{array} \\right.[\/latex]<\/p>\r\n[reveal-answer q=\"843118\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843118\"]\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 [latex]x[\/latex] terms will cancel. Make sure to multiply all three terms in the second equation by [latex]-3.[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}3x+5y&amp;=-11&amp;\\xrightarrow{\\phantom{\\cdot \\; -3}} &amp; 3x+5y &amp;=-11\\\\\r\nx-2y&amp;=11&amp;\\xrightarrow{\\cdot \\; -3} &amp; -3x+6y&amp;=-33 \\\\\r\n&amp; &amp; &amp; \\text{_______}&amp;\\text{______} \\\\\r\n&amp; &amp; &amp; 11y&amp;=-44 \\\\\r\n&amp; &amp; &amp; y &amp;=-4 \\\\ \\end{array}[\/latex]<\/p>\r\nFor the last step, we substitute [latex]y=-4[\/latex] into one of the original equations and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3x+5y&amp;=-11\\\\ 3x+5\\left(-4\\right)&amp;=-11\\\\ 3x - 20&amp;=-11\\\\ 3x&amp;=9\\\\ x&amp;=3\\end{align}[\/latex]<\/p>\r\nOur solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation (you should also verify that it satisfies the first equation).\r\n<div style=\"text-align: center;\">[latex]\\begin{align}x - 2y&amp;=11\\\\ \\left(3\\right)-2\\left(-4\\right)&amp;=11\\\\ 11&amp;=11 &amp;&amp; \\text{True}\\hfill \\end{align}[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222640\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nBelow is another video example of using the elimination method to solve a system of linear equations.\r\n\r\nhttps:\/\/youtu.be\/_liDhKops2w\r\n\r\nIn the next example, we will see that sometimes both equations need to be multiplied by different numbers in order for one variable to be eliminated.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the given system of equations by elimination.\r\n<div style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}2x+3y&amp;=-16\\\\ 5x - 10y&amp;=30\\end{array}\\right.[\/latex]<\/div>\r\n[reveal-answer q=\"245990\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"245990\"]\r\n\r\nLet's focus on eliminating [latex]x.[\/latex] The Least Common Multiple (LCM) of the [latex]x[\/latex] coefficients of [latex]2[\/latex] and [latex]5[\/latex] is [latex]10,[\/latex] so multiply\u00a0the first equation by [latex]-5[\/latex] and the second equation by [latex]2.[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}2x+3y&amp;=-16&amp;\\xrightarrow{\\cdot \\; -5} &amp; -10x-15y &amp;=80\\\\\r\n5x-10y&amp;=30&amp;\\xrightarrow{\\cdot \\; 2} &amp; 10x-20y&amp;=60 \\\\\r\n&amp; &amp; &amp; \\text{_______}&amp;\\text{______} \\\\\r\n&amp; &amp; &amp; -35y&amp;=140 \\\\\r\n&amp; &amp; &amp; y &amp;=-4 \\\\ \\end{array}[\/latex]<\/p>\r\nSubstitute [latex]y=-4[\/latex] into the original first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}2x+3\\left(-4\\right)&amp;=-16\\\\ 2x - 12&amp;=-16\\\\ 2x&amp;=-4\\\\ x&amp;=-2\\end{array}[\/latex]<\/p>\r\nThe solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the second original equation.\r\n<div style=\"text-align: center;\">[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]<\/div>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222643\/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\" \/> [\/hidden-answer]\r\n\r\n<\/div>\r\nDid you notice in the last example that an alternative approach could have been to start by dividing equation two by [latex]5[\/latex]?\r\n\r\nIn the next\u00a0example, we will show how to solve a system with fractions. As with single linear equations, the easiest way to solve is to clear the fractions first with the least common denominator.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the given system of equations by elimination.\r\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}\\dfrac{x}{3}+\\dfrac{y}{6}&amp;=\\text{ }3\\hfill \\\\ \\dfrac{x}{2}-\\dfrac{y}{4}&amp;=\\text{ }1\\hfill \\end{array}\\right.[\/latex]<\/p>\r\n[reveal-answer q=\"288325\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"288325\"]\r\n\r\nFirst clear each equation of fractions by multiplying both sides of the equation by the least common denominator. After this step, solve the resulting system in the same way as the other examples.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}\\dfrac{x}{3}+\\dfrac{y}{6}&amp;=3&amp;\\xrightarrow{\\cdot \\; 6} &amp; 2x+y&amp;=18\\\\\r\n\\dfrac{x}{2}-\\dfrac{y}{4}&amp;=1&amp;\\xrightarrow{\\cdot \\; 4} &amp; 2x-y&amp;=4 \\\\\r\n&amp; &amp; &amp; \\text{_______}&amp;\\text{______} \\\\\r\n&amp; &amp; &amp; 4x&amp;=22 \\\\\r\n&amp; &amp; &amp; x &amp;=\\dfrac{22}{4}=\\dfrac{11}{2} \\\\ \\end{array}[\/latex]<\/p>\r\nSubstitute [latex]x=\\dfrac{11}{2}[\/latex] into the modified version of the first equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}2\\left(\\dfrac{11}{2}\\right)+y&amp;=18\\hfill \\\\ \\text{ }11+y&amp;=18\\hfill \\\\ \\text{ }y&amp;=7\\hfill \\end{array}[\/latex]<\/p>\r\nThe solution is [latex]\\left(\\dfrac{11}{2},7\\right)[\/latex]. Check it in the modified second equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x-y=4\\\\ 2\\left(\\dfrac{11}{2}\\right)-7=4\\\\ 11-7=4 \\\\ 4=4\\end{array}[\/latex]<\/p>\r\n\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\nIn the following video, you will find one more example of using the elimination method to solve a system; this one has coefficients that are fractions.\r\n\r\nhttps:\/\/youtu.be\/s3S64b1DrtQ\r\n\r\nThe last example includes two equations that represent the same line and are therefore dependent.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the given system of equations by elimination.\r\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}x+3y&amp;=\\text{ }2\\\\ 3x+9y&amp;=\\text{ }6\\end{array}\\right.[\/latex]<\/p>\r\n[reveal-answer q=\"10390\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"10390\"]\r\n\r\nIn this case, focus on eliminating [latex]x[\/latex]. If we multiply both sides of the first equation by [latex]-3[\/latex], then we will be able to eliminate the [latex]x[\/latex] -variable.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}x+3y&amp;=2&amp;\\xrightarrow{\\cdot \\; -3} &amp; -3x-9y&amp;=-6\\\\\r\n3x+9y&amp;=6&amp;\\xrightarrow{\\phantom{\\cdot \\; -3}} &amp; 3x+9y&amp;=6 \\\\\r\n&amp; &amp; &amp; \\text{_______}&amp;\\text{______} \\\\\r\n&amp; &amp; &amp; 0&amp;=0 \\\\ \\end{array}[\/latex]<\/p>\r\nSince we have an identity with all variables simultaneously eliminated, an infinite number of solutions satisfy both equations.\u00a0The general solution to the system is [latex]\\left\\{(x,y)\\; | \\; x+3y=2\\right\\}[\/latex].\r\n\r\nLook at what happens when we convert the system to slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x+3y&amp;=2 &amp;&amp; \\color{blue}{\\textsf{Equation 1}} \\\\ 3y&amp;=-x+2 \\\\ y&amp;=-\\dfrac{1}{3}x+\\dfrac{2}{3} \\\\ &amp; \\\\ 3x+9y&amp;=6&amp;&amp; \\color{blue}{\\textsf{Equation 2}} \\\\ 9y&amp;=-3x+6 \\\\ y&amp;=-\\dfrac{3}{9}x+\\dfrac{6}{9}\\\\ y&amp;=-\\dfrac{1}{3}x+\\dfrac{2}{3}\\end{align}[\/latex]<\/p>\r\nWe confirm that our lines are the same in slope-intercept form so they are equivalent equations, which means they have the same solution set and the same graph.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222647\/CNX_Precalc_Figure_09_01_0082.jpg\" alt=\"A graph of two lines that overlap each other. The first line's equation is x+3y=2. The second line's equation is 3x-9y=6.\" width=\"487\" height=\"366\" \/>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of solving a system that is dependent using elimination.\r\n\r\nhttps:\/\/youtu.be\/NRxh9Q16Ulk\r\n\r\nIn our last video example, we present a system that is inconsistent; it has no solutions which means the lines the equations represent are parallel to each other.\r\n\r\nhttps:\/\/youtu.be\/z5_ACYtzW98\r\n<h2>Summary<\/h2>\r\nSystems of equations are typically solved algebraically by either the substitution or the elimination method. Substitution is preferred if one variable is already solved for, or if one of the variables can be easily solved for without introducing fractions. Elimination is preferred if substitution would be difficult, or if a variable would cancel upon adding the equations as they are written.\r\n\r\nDependent systems are represented graphically by two copies of the same line and have infinite solutions. While solving, a system is dependent if you reach an equation with no variables and a true statement. Inconsistent systems are represented graphically by two parallel lines and have no solution. While solving, a system is inconsistent if you reach an equation with no variables and a false statement.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Investigate types of solutions to systems of equations with two variables graphically.<\/li>\n<li>Solve linear systems of equations with two variables by substitution and elimination.<\/li>\n<li>Identify inconsistent or dependent systems of equations containing two variables.<\/li>\n<\/ul>\n<\/div>\n<p>A <strong>system of linear equations<\/strong> consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find a solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions.<\/p>\n<p>For example, consider the following system of linear equations in two variables.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}2x+y=15\\\\3x-y=5\\end{cases}[\/latex]<\/p>\n<p>A <strong>solution<\/strong>\u00a0to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair\u00a0 [latex](4, 7)[\/latex] is the solution to the system of linear equations. We can verify the solution by substituting the values [latex]\\color{green}{x=4},\\color{blue}{y=7}[\/latex] into each equation to see if the ordered pair satisfies both equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2\\color{green}{\\left(4\\right)}+\\color{blue}{\\left(7\\right)}=15\\text{ }\\text{True}\\hfill \\\\ 3\\color{green}{\\left(4\\right)}-\\color{blue}{\\left(7\\right)}=5\\text{ }\\text{True}\\hfill \\end{array}[\/latex]<\/p>\n<p>In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A <strong>consistent system<\/strong> of equations has at least one solution. A consistent system is considered to be an <strong>independent system<\/strong> if it has a single solution like the example we just explored. A consistent system is considered to be a <strong>dependent system<\/strong> if the equations are equivalent. In other words, every solution to one equation will also be a solution to the other equation. Thus, there are an infinite number of solutions.<\/p>\n<p>Another type of system of linear equations is an <strong>inconsistent system\u00a0<\/strong>which is one where the equations cannot both be true for a single ordered pair.\u00a0Hence, there is no solution to the system.\u00a0 An easy example is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}x+y=3\\\\x+y=4\\end{cases}[\/latex]<\/p>\n<p>Clearly, the expression [latex]x+y[\/latex] cannot possibly have two different values at once.<\/p>\n<div class=\"textbox shaded\">\n<h3>A General Note: Types of Linear Systems<\/h3>\n<p>There are three types of systems of linear equations in two variables and three types of solutions.<\/p>\n<p>Below are graphical representations of each type of system.\u00a0 It is important to understand that where the two lines cross represents an ordered pair that satisfies both equations, thus is a solution to the system.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222630\/CNX_Precalc_Figure_09_01_002n2.jpg\" alt=\"Graphs of an independent system, an inconsistent system, and a dependent system. The independent system has two lines which cross at the point seven-fifths, negative eleven fifths. The inconsistent system shows two parallel lines. The dependent system shows a single line running through the points negative one, negative two and one, two.\" width=\"945\" height=\"479\" \/><\/p>\n<ul>\n<li>An <strong>independent system<\/strong> has exactly one solution pair [latex]\\left(x,y\\right)[\/latex]. The point where the two lines intersect is the only solution.<\/li>\n<li>An <strong>inconsistent system<\/strong> has no solution. The two lines are parallel and will never intersect.\u00a0 They have the same slope but different [latex]y[\/latex]-intercepts.<\/li>\n<li>A <strong>dependent system<\/strong> has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.<\/li>\n<\/ul>\n<p>The independent and dependent systems are also consistent because they both have at least one solution.<\/p>\n<\/div>\n<h2>Solving Systems of Equations by Graphing<\/h2>\n<p>There are multiple methods of solving systems of linear equations. For a <strong>system of linear equations<\/strong> in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the following system of equations by graphing. Identify the type of system.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}2x&+\\ \\ y=-8\\\\ x&-\\ \\ y=-1\\end{cases}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q336799\">Show Solution<\/span><\/p>\n<div id=\"q336799\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solve the first equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x+y&=-8\\\\ y&=-2x - 8\\end{align}[\/latex]<\/p>\n<p>Solve the second equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x-y&=-1\\\\ -y&=-x-1\\\\ y&=x+1\\end{align}[\/latex]<\/p>\n<p>Graph both equations on the same set of axes as seen below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222636\/CNX_Precalc_Figure_09_01_0122.jpg\" alt=\"A graph of two lines running through the point negative 3, negative 2. The first line's equation is y equals minus 2x minus 8. The second line's equation is y equals x+1.\" width=\"487\" height=\"316\" \/><\/p>\n<p>The lines appear to intersect at the point [latex]\\left(-3,-2\\right)[\/latex]. We can check to make sure that this is the solution to the system by substituting the ordered pair into both original equations<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}2\\left(-3\\right)+\\left(-2\\right)=-8\\hfill & \\hfill \\\\ \\text{ }-8=-8\\hfill & \\text{True}\\hfill \\\\ \\text{ }\\left(-3\\right)-\\left(-2\\right)=-1\\hfill & \\hfill \\\\ \\text{ }-1=-1\\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution to the system is the ordered pair [latex]\\left(-3,-2\\right)[\/latex], so the system is independent.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>The Substitution Method<\/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 algebraic methods of solving a <strong>system of linear equations<\/strong> that can always find an exact solution.<\/p>\n<p>Consider the system<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}y=2x\\\\x+y=12\\end{cases}[\/latex]<\/p>\n<p>Since [latex]y[\/latex] and [latex]2x[\/latex] are the same quantity by the first equation, they can be used interchangeably in the problem from this point forward. Thus we can replace the\u00a0[latex]\\color{green}{y}[\/latex] in the second equation with [latex]\\color{green}{2x}.[\/latex] This is called a <strong>substitution<\/strong> and is the basis of the <strong>substitution method<\/strong>. In the second equation now,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+\\color{green}{(2x)}&=12\\\\3x&=12\\\\x&=4\\end{align}[\/latex]<\/p>\n<p>Once we know that [latex]x=4,[\/latex] use this value in the first equation to obtain\u00a0[latex]y=2(4)=8.[\/latex] The system has solution [latex](4,8).[\/latex]<\/p>\n<div class=\"textbox shaded\">\n<h3>Solving by Substitution<\/h3>\n<ol>\n<li>Solve one of the two equations for a variable.<\/li>\n<li>Substitute the expression for this variable into the second equation, and then solve for the remaining variable.<\/li>\n<li>Use the value of the second variable to find the value of the first.<\/li>\n<li>Check the solution in both equations.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the following system of equations by substitution.<\/p>\n<div style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{llll}-x&+&y&=-5\\\\ 2x&-&5y&=1\\\\ \\end{array}\\right.[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q748381\">Show Solution<\/span><\/p>\n<div id=\"q748381\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will solve the first equation for [latex]y[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}-x+y&=-5\\\\ y&=x - 5\\end{align}[\/latex]<\/div>\n<p>Now, we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.\u00a0 Notice that this results in an equation where [latex]x[\/latex] is the only variable.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}2x - 5y&=1\\hfill \\\\ 2x - 5\\color{green}{\\left(x - 5\\right)}&=1\\hfill \\\\ 2x - 5x+25&=1\\hfill \\\\ -3x&=-24\\hfill \\\\ \\text{ }x&=8\\hfill \\end{array}[\/latex]<\/div>\n<p>Now, we substitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}-\\left(8\\right)+y&=-5\\hfill \\\\ \\text{ }y&=3\\hfill \\end{array}[\/latex]<\/div>\n<p>Our solution is [latex]\\left(8,3\\right)[\/latex].<\/p>\n<p>Check the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{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<p>The substitution method can be used to solve any linear system in two variables, but the method works best if one of the equations contains a coefficient of [latex]1[\/latex] or [latex]\u20131[\/latex] so that we do not have to deal with fractions.\u00a0 This is why we chose to solve for\u00a0[latex]y[\/latex] in the first step.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will be given an example of solving a system of two equations using the substitution method.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 2:  Solve a System of Equations Using Substitution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/MIXL35YRzRw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>If you had chosen the other equation to start with in the previous example, you would still be able to find the same solution. It is really a matter of preference, because sometimes solving for a variable will result in having to work with fractions. Substitution works best if one of the variables is easily solved for.<\/p>\n<p>Here is an example of what it looks like to solve an\u00a0<strong>inconsistent system<\/strong>\u00a0using substitution. We can tell it is inconsistent when we arrive at a line where there are no variables left and the numerical statement is false.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the following system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{llll}x&=&9 &- \\ \\ 2y\\\\ x&+&2y&=\\ 13 \\end{array}\\right.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q86393\">Show Solution<\/span><\/p>\n<div id=\"q86393\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the first equation is already solved for [latex]x,[\/latex] we can substitute the expression [latex]9-2y[\/latex] for [latex]x[\/latex] in the second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}x+2y&=13\\\\ \\color{green}{\\left(9 - 2y\\right)}+2y&=13 \\\\ 9+0y&=13 \\\\ 9&=13 \\end{array}[\/latex]<\/p>\n<p>Clearly, this statement is a contradiction because [latex]9\\ne 13[\/latex]. Therefore, the system has no solution.<\/p>\n<p>Here are the two equations solved to slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ y=-\\dfrac{1}{2}x+\\dfrac{9}{2}\\end{array}\\hfill \\\\ y=-\\dfrac{1}{2}x+\\dfrac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>We can see they have the same slope and different [latex]y[\/latex]-intercepts. Therefore, the lines are parallel and do not intersect. This confirms visually that the system is inconsistent.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222645\/CNX_Precalc_Figure_09_01_0072.jpg\" alt=\"A graph of two parallel lines. The first line's equation is y equals negative one-half x plus 13 over two. The second line's equation is y equals negative one-half x plus 9 over two.\" width=\"488\" height=\"297\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next video, we show another example of using substitution to solve a system that has no solution.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" 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<p>The next video illustrates what happens if solving a dependent system by substitution. This case is similar to the &#8220;no solution&#8221; case, but instead of obtaining a false statement like\u00a0[latex]12=0[\/latex] while solving, you obtain a true statement like\u00a0[latex]12=12[\/latex].<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" 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>When a system is dependent, we will write our answer in set-builder notation. For example, suppose a dependent system has equation [latex]5x-2y=10.[\/latex] In set builder notation, we write the solution as [latex]\\left\\{(x,y)\\; |\\; 5x-2y=10\\right\\}.[\/latex] Remember that this indicates there are infinitely many solutions since there are infinitely many ordered pairs that satisfy the equation.<\/p>\n<p>There are two things to note here. First, the left side of the notation dictates that solutions to the system have the form of ordered pairs [latex](x,y).[\/latex] Second, we do not need to list the other equation that was in the system at all, since in a dependent system the two equations are equivalent to each other.<\/p>\n<h2>The Elimination Method<\/h2>\n<p>A third method of solving systems of linear equations is the <strong>elimination\u00a0method<\/strong>. In this method, we add two equations to each other with the goal of eliminating one variable entirely. This takes advantage of the addition principle:\u00a0 If [latex]a=b[\/latex], then [latex]a+c=b+c[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{l}x+y=\\text{ }5\\\\x-y=\\text{ }11\\end{array} \\right.[\/latex]<\/p>\n<p>Consider the above system. The first equation asserts that\u00a0[latex]x+y[\/latex] and\u00a0[latex]5[\/latex] are the same quantity, so we can add\u00a0[latex]\\color{green}{(x+y)}[\/latex] to the left side of the second equation, and\u00a0[latex]\\color{blue}{5}[\/latex] to the right side.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}x-y+\\color{green}{(x+y)}&=11+\\color{blue}{(5)}\\\\2x&=\\text{ }16\\end{array}[\/latex]<\/p>\n<p>Notice since the [latex]y[\/latex] terms had opposite coefficients they canceled, leaving only [latex]x[\/latex] to solve for. The rest of the steps are similar to the substitution process. We divide to obtain [latex]x=8,[\/latex] then use this value in equation one to obtain [latex]8+y=5,[\/latex] or [latex]y=-3.[\/latex] The solution is the ordered pair [latex](8, -3).[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the given system of equations by the elimination method.<\/p>\n<div style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}x+2y&=-1\\hfill \\\\ -x+y&=3\\hfill \\end{array}\\right.[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q522070\">Show Solution<\/span><\/p>\n<div id=\"q522070\" class=\"hidden-answer\" style=\"display: none\">\n<p>First make sure the equations are lined up in columns of like terms and like symbols. Notice that the coefficient of [latex]x[\/latex] in the second equation,\u00a0[latex]\u20131[\/latex], is the opposite of the coefficient of [latex]x[\/latex] in the first equation,\u00a0[latex]1[\/latex]. We can add the two equations to eliminate [latex]x[\/latex]. We will typically show this step with the equations added vertically as follows.<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{\\begin{array}{l}\\hfill \\\\ \\:\\:x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}}{\\text{}\\text{}\\text{}\\text{}\\text{}\\:\\:\\:\\:\\:\\:3y=2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Now that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3y=2\\hfill \\\\ \\text{ }y=\\dfrac{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 style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }-x+y=3\\hfill \\\\ \\text{ }-x+\\dfrac{2}{3}=3\\hfill \\\\ \\text{ }-x=3-\\dfrac{2}{3}\\hfill \\\\ \\text{ }-x=\\dfrac{7}{3}\\hfill \\\\ \\text{ }\\:\\:\\:\\:\\:x=-\\dfrac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution to this system is [latex]\\left(-\\dfrac{7}{3},\\dfrac{2}{3}\\right)[\/latex].<\/p>\n<p>Check the solution in the first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+2y&=-1 \\\\ \\left(-\\dfrac{7}{3}\\right)+2\\left(\\dfrac{2}{3}\\right)&=-1\\\\ -\\dfrac{7}{3}+\\dfrac{4}{3}&=-1\\\\ -\\dfrac{3}{3}&=-1\\\\ -1&=-1&& \\text{True}\\end{align}[\/latex]<\/p>\n<p>You can check yourself that the solution also satisfies the second equation.\u00a0 The graph confirms that there is exactly one solution and there is no need to look for other solutions.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222638\/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\" \/>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will see another example of how to use the method of elimination to solve a system of linear equations.<\/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<p>If neither the [latex]x[\/latex] coefficients nor the [latex]y[\/latex] coefficients are opposites to begin with, we must first multiply one or both equations by a constant. In the next example, you will see a technique where we multiply one of the equations in the system by a number that will allow us to eliminate one of the variables.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the given system of equations by the elimination method<strong>.<\/strong><\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}3x+5y&=-11\\hfill \\\\ x - 2y&=11\\hfill \\end{array} \\right.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q843118\">Show Solution<\/span><\/p>\n<div id=\"q843118\" 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 [latex]x[\/latex] terms will cancel. Make sure to multiply all three terms in the second equation by [latex]-3.[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}3x+5y&=-11&\\xrightarrow{\\phantom{\\cdot \\; -3}} & 3x+5y &=-11\\\\  x-2y&=11&\\xrightarrow{\\cdot \\; -3} & -3x+6y&=-33 \\\\  & & & \\text{_______}&\\text{______} \\\\  & & & 11y&=-44 \\\\  & & & y &=-4 \\\\ \\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 style=\"text-align: center;\">[latex]\\begin{align}3x+5y&=-11\\\\ 3x+5\\left(-4\\right)&=-11\\\\ 3x - 20&=-11\\\\ 3x&=9\\\\ x&=3\\end{align}[\/latex]<\/p>\n<p>Our solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation (you should also verify that it satisfies the first equation).<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}x - 2y&=11\\\\ \\left(3\\right)-2\\left(-4\\right)&=11\\\\ 11&=11 && \\text{True}\\hfill \\end{align}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222640\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Below is another video example of using the elimination method to solve a system of linear equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" 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>In the next example, we will see that sometimes both equations need to be multiplied by different numbers in order for one variable to be eliminated.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the given system of equations by elimination.<\/p>\n<div style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}2x+3y&=-16\\\\ 5x - 10y&=30\\end{array}\\right.[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q245990\">Show Solution<\/span><\/p>\n<div id=\"q245990\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let&#8217;s focus on eliminating [latex]x.[\/latex] The Least Common Multiple (LCM) of the [latex]x[\/latex] coefficients of [latex]2[\/latex] and [latex]5[\/latex] is [latex]10,[\/latex] so multiply\u00a0the first equation by [latex]-5[\/latex] and the second equation by [latex]2.[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}2x+3y&=-16&\\xrightarrow{\\cdot \\; -5} & -10x-15y &=80\\\\  5x-10y&=30&\\xrightarrow{\\cdot \\; 2} & 10x-20y&=60 \\\\  & & & \\text{_______}&\\text{______} \\\\  & & & -35y&=140 \\\\  & & & y &=-4 \\\\ \\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=-4[\/latex] into the original first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}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 second original equation.<\/p>\n<div style=\"text-align: center;\">[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]<\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222643\/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\" \/> <\/div>\n<\/div>\n<\/div>\n<p>Did you notice in the last example that an alternative approach could have been to start by dividing equation two by [latex]5[\/latex]?<\/p>\n<p>In the next\u00a0example, we will show how to solve a system with fractions. As with single linear equations, the easiest way to solve is to clear the fractions first with the least common denominator.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the given system of equations by elimination.<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}\\dfrac{x}{3}+\\dfrac{y}{6}&=\\text{ }3\\hfill \\\\ \\dfrac{x}{2}-\\dfrac{y}{4}&=\\text{ }1\\hfill \\end{array}\\right.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q288325\">Show Solution<\/span><\/p>\n<div id=\"q288325\" 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. After this step, solve the resulting system in the same way as the other examples.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}\\dfrac{x}{3}+\\dfrac{y}{6}&=3&\\xrightarrow{\\cdot \\; 6} & 2x+y&=18\\\\  \\dfrac{x}{2}-\\dfrac{y}{4}&=1&\\xrightarrow{\\cdot \\; 4} & 2x-y&=4 \\\\  & & & \\text{_______}&\\text{______} \\\\  & & & 4x&=22 \\\\  & & & x &=\\dfrac{22}{4}=\\dfrac{11}{2} \\\\ \\end{array}[\/latex]<\/p>\n<p>Substitute [latex]x=\\dfrac{11}{2}[\/latex] into the modified version of the first equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}2\\left(\\dfrac{11}{2}\\right)+y&=18\\hfill \\\\ \\text{ }11+y&=18\\hfill \\\\ \\text{ }y&=7\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution is [latex]\\left(\\dfrac{11}{2},7\\right)[\/latex]. Check it in the modified second equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x-y=4\\\\ 2\\left(\\dfrac{11}{2}\\right)-7=4\\\\ 11-7=4 \\\\ 4=4\\end{array}[\/latex]<\/p>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will find one more example of using the elimination method to solve a system; this one has coefficients that are fractions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Ex: Solve a System of Equations Using Eliminations (Fractions)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/s3S64b1DrtQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The last example includes two equations that represent the same line and are therefore dependent.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the given system of equations by elimination.<\/p>\n<p style=\"text-align: center;\">[latex]\\left\\{ \\begin{array}{rl}x+3y&=\\text{ }2\\\\ 3x+9y&=\\text{ }6\\end{array}\\right.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q10390\">Show Solution<\/span><\/p>\n<div id=\"q10390\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this case, focus on eliminating [latex]x[\/latex]. If we multiply both sides of the first equation by [latex]-3[\/latex], then we will be able to eliminate the [latex]x[\/latex] -variable.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rllrl}x+3y&=2&\\xrightarrow{\\cdot \\; -3} & -3x-9y&=-6\\\\  3x+9y&=6&\\xrightarrow{\\phantom{\\cdot \\; -3}} & 3x+9y&=6 \\\\  & & & \\text{_______}&\\text{______} \\\\  & & & 0&=0 \\\\ \\end{array}[\/latex]<\/p>\n<p>Since we have an identity with all variables simultaneously eliminated, an infinite number of solutions satisfy both equations.\u00a0The general solution to the system is [latex]\\left\\{(x,y)\\; | \\; x+3y=2\\right\\}[\/latex].<\/p>\n<p>Look at what happens when we convert the system to slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x+3y&=2 && \\color{blue}{\\textsf{Equation 1}} \\\\ 3y&=-x+2 \\\\ y&=-\\dfrac{1}{3}x+\\dfrac{2}{3} \\\\ & \\\\ 3x+9y&=6&& \\color{blue}{\\textsf{Equation 2}} \\\\ 9y&=-3x+6 \\\\ y&=-\\dfrac{3}{9}x+\\dfrac{6}{9}\\\\ y&=-\\dfrac{1}{3}x+\\dfrac{2}{3}\\end{align}[\/latex]<\/p>\n<p>We confirm that our lines are the same in slope-intercept form so they are equivalent equations, which means they have the same solution set and the same graph.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222647\/CNX_Precalc_Figure_09_01_0082.jpg\" alt=\"A graph of two lines that overlap each other. The first line's equation is x+3y=2. The second line's equation is 3x-9y=6.\" width=\"487\" height=\"366\" \/>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show another example of solving a system that is dependent using elimination.<\/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<p>In our last video example, we present a system that is inconsistent; it has no solutions which means the lines the equations represent are parallel to each other.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" 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>Summary<\/h2>\n<p>Systems of equations are typically solved algebraically by either the substitution or the elimination method. Substitution is preferred if one variable is already solved for, or if one of the variables can be easily solved for without introducing fractions. Elimination is preferred if substitution would be difficult, or if a variable would cancel upon adding the equations as they are written.<\/p>\n<p>Dependent systems are represented graphically by two copies of the same line and have infinite solutions. While solving, a system is dependent if you reach an equation with no variables and a true statement. Inconsistent systems are represented graphically by two parallel lines and have no solution. While solving, a system is inconsistent if you reach an equation with no variables and a false statement.<\/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-119\">\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>Ex 2: Solve a System of Equations Using Substitution. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <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>Unit 14: Systems of Equations and Inequalities, 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\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/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) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/kTtKfh5gFUc\">https:\/\/youtu.be\/kTtKfh5gFUc<\/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) . <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 an Application Problem Using a System of Linear Equations (09x-43). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/euh9ksWrq0A\">https:\/\/youtu.be\/euh9ksWrq0A<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/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":395986,"menu_order":1,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Solve a System of Equations Using Substitution\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/MIXL35YRzRw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Solve a System of Equations Using Substitution - 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