{"id":2075,"date":"2016-07-01T22:36:57","date_gmt":"2016-07-01T22:36:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2075"},"modified":"2018-05-17T02:00:41","modified_gmt":"2018-05-17T02:00:41","slug":"read-solve-systems-of-equations-by-graphing","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/read-solve-systems-of-equations-by-graphing\/","title":{"raw":"Linear Systems in Two Variables","rendered":"Linear Systems in Two Variables"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Define and classify solutions to systems of linear equations\r\n<ul>\r\n \t<li>Recognize\u00a0consistent and inconsistent, dependent and independent systems of linear equations<\/li>\r\n \t<li>Determine whether an ordered pair is a solution to a system of linear equations<\/li>\r\n \t<li>Solve a system of linear equations by graphing<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Methods for solving\u00a0systems\r\n<ul>\r\n \t<li>Use substitution to solve a system algebraically<\/li>\r\n \t<li>Solve a system using the addition\u00a0method<\/li>\r\n \t<li>Recognize when a system is inconsistent from algebraic results<\/li>\r\n \t<li>Recognize when a system is dependent using algebraic results<\/li>\r\n<\/ul>\r\n<\/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 the unique 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. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.\r\n\r\nIn this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{c}2x+y=\\text{ }15\\\\ 3x-y=\\text{ }5\\end{array}[\/latex]<\/div>\r\nThe <em>solution<\/em> to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{l}2\\left(4\\right)+\\left(7\\right)=15\\text{ }\\text{True}\\hfill \\\\ 3\\left(4\\right)-\\left(7\\right)=5\\text{ }\\text{True}\\hfill \\end{array}[\/latex]<\/div>\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, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a <strong>dependent system<\/strong> if the equations have the same slope and the same <em>y<\/em>-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.\r\n\r\nAnother type of system of linear equations is an <strong>inconsistent system<\/strong>, which is one in which the equations represent two parallel lines. The lines have the same slope and different <em>y-<\/em>intercepts. There are no points common to both lines; hence, there is no solution to the system.\r\n<div class=\"textbox\">\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<div>\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. Notice that the two lines are parallel and will never intersect.<\/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\n<\/div>\r\nBelow are graphical representations of each type of 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\r\nThe independent and dependent systems are also consistent because they both have at least one solution.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution.<\/h3>\r\n<ol>\r\n \t<li>Substitute the ordered pair into each equation in the system.<\/li>\r\n \t<li>Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nDetermine whether the ordered pair [latex]\\left(5,1\\right)[\/latex] is a solution to the given system of equations.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{l}x+3y=8\\hfill \\\\ 2x - 9=y\\hfill \\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"322824\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"322824\"]\r\n\r\nSubstitute the ordered pair [latex]\\left(5,1\\right)[\/latex] into both equations.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{ll}\\left(5\\right)+3\\left(1\\right)=8\\hfill &amp; \\hfill \\\\ \\text{ }8=8\\hfill &amp; \\text{True}\\hfill \\\\ 2\\left(5\\right)-9=\\left(1\\right)\\hfill &amp; \\hfill \\\\ \\text{ }\\text{1=1}\\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]<\/p>\r\nThe ordered pair [latex]\\left(5,1\\right)[\/latex] satisfies both equations, so it is the solution to the system.\r\n\r\nWe can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222634\/CNX_Precalc_Figure_09_01_0032.jpg\" alt=\"A graph of two lines running through the point five, one. The first line's equation is x plus 3y equals 8. The second line's equation is 2x minus 9 equals y.\" width=\"487\" height=\"365\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we will show another example of how to verify whether an ordered pair is a solution to a system of equations.\r\n\r\nhttps:\/\/youtu.be\/2IxgKgjX00k\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{array}{c}2x+y=-8\\\\ x-y=-1\\end{array}[\/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{array}{c}2x+y=-8\\\\ y=-2x - 8\\end{array}[\/latex]<\/p>\r\nSolve the second equation for [latex]y[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}x-y=-1\\\\ y=x+1\\end{array}[\/latex]<\/p>\r\nGraph both equations on the same set of axes as in the figure 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 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\nGraphing can be used if the system is inconsistent or dependent.\u00a0In both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.\r\n\r\nIn the following video we show another example of how to identify whether a graphed system has a solution, and identify what type of solution is represented.\r\n\r\nhttps:\/\/youtu.be\/ZolxtOjcEQY\r\n\r\nIn our last video we show how to solve a system of equations by first graphing the lines, then identifying the type of solution the system has.\r\n\r\nhttps:\/\/youtu.be\/Lv832rXAQ5k\r\n<h2>Substitution<\/h2>\r\nSolving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a <strong>system of linear equations<\/strong> that are more precise than graphing. One such method is solving a system of equations by the <strong>substitution method<\/strong>, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of two equations in two variables, solve using the substitution method.<\/h3>\r\n<ol>\r\n \t<li>Solve one of the two equations for one of the variables in terms of the other.<\/li>\r\n \t<li>Substitute the expression for this variable into the second equation, then solve for the remaining variable.<\/li>\r\n \t<li>Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.<\/li>\r\n \t<li>Check the solution in both equations.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve the following system of equations by substitution.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }-x+y=-5\\hfill \\\\ \\text{ }2x - 5y=1\\hfill \\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"748381\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"748381\"]\r\n\r\nFirst, we will solve the first equation for [latex]y[\/latex].\r\n<div style=\"text-align: center\">[latex]\\begin{array}{l}-x+y=-5\\hfill \\\\ \\text{ }y=x - 5\\hfill \\end{array}[\/latex]<\/div>\r\nNow we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }2x - 5y=1\\hfill \\\\ 2x - 5\\left(x - 5\\right)=1\\hfill \\\\ \\text{ }2x - 5x+25=1\\hfill \\\\ \\text{ }-3x=-24\\hfill \\\\ \\text{ }x=8\\hfill \\end{array}[\/latex]<\/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}{l}-\\left(8\\right)+y=-5\\hfill \\\\ \\text{ }y=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 1 or [latex]\u20131[\/latex] so that we do not have to deal with fractions.\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. It is really a matter of preference because sometimes solving for a variable will result in having to work with fractions. As you become more experienced with algebra, you will be able to anticipate what choices will lead to more desirable outcomes.\r\n\r\nRecall that an <strong>inconsistent system<\/strong> consists of parallel lines that have the same slope but different <em>y<\/em>-intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as [latex]12=0[\/latex].\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]\\begin{array}{l}\\text{ }x=9 - 2y\\hfill \\\\ x+2y=13\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"86393\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"86393\"]\r\n\r\nWe can approach this problem in two ways. Because one equation is already solved for <em>x<\/em>, the most obvious step is to use substitution.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}x+2y=13\\hfill \\\\ \\left(9 - 2y\\right)+2y=13\\hfill \\\\ 9+0y=13\\hfill \\\\ 9=13\\hfill \\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\nThe second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }x=9 - 2y\\hfill \\\\ 2y=-x+9\\hfill \\\\ \\text{ }y=-\\frac{1}{2}x+\\frac{9}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nWe then convert the second equation expressed to slope-intercept form.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}x+2y=13\\hfill \\\\ \\text{ }2y=-x+13\\hfill \\\\ \\text{ }y=-\\frac{1}{2}x+\\frac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nComparing the equations, we see that they have the same slope but different <em>y<\/em>-intercepts. Therefore, the lines are parallel and do not intersect.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ y=-\\frac{1}{2}x+\\frac{9}{2}\\end{array}\\hfill \\\\ y=-\\frac{1}{2}x+\\frac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nWriting the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.\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<h4>Answer<\/h4>\r\nThere is no solution to this system of linear equations.\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\nIn our next video we show that a system can have an infinite number of solutions.\r\n\r\nhttps:\/\/youtu.be\/Pcqb109yK5Q\r\n\r\n\r\n<h3>Solving Systems of Equations in Two Variables by the Addition Method<\/h3>\r\nA third method of <strong>solving systems of linear equations<\/strong> is the <strong>addition method,\u00a0<\/strong>this method is also called the\u00a0<strong>elimination method<\/strong>. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of equations, solve using the addition method.<\/h3>\r\n<ol>\r\n \t<li>Write both equations with <em>x<\/em>- and <em>y<\/em>-variables on the left side of the equal sign and constants on the right.<\/li>\r\n \t<li>Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.<\/li>\r\n \t<li>Solve the resulting equation for the remaining variable.<\/li>\r\n \t<li>Substitute that value into one of the original equations and solve for the second variable.<\/li>\r\n \t<li>Check the solution by substituting the values into the other equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a System by the Addition Method<\/h3>\r\nSolve the given system of equations by addition.\r\n\r\n[latex]\\begin{array}{l}x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"924657\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"924657\"]\r\n\r\nBoth equations are already set equal to a constant. Notice that the coefficient of [latex]x[\/latex] in the second equation, \u20131, is the opposite of the coefficient of [latex]x[\/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.\r\n\r\n[latex]\\frac{\\begin{array}{l}\\hfill \\\\ x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}}{\\text{}\\text{}\\text{}\\text{}\\text{}3y=2}[\/latex]\r\n\r\nNow that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].\r\n\r\n[latex]\\begin{array}{l}3y=2\\hfill \\\\ \\text{ }y=\\frac{2}{3}\\hfill \\end{array}[\/latex]\r\n\r\nThen, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex].\r\n\r\n[latex]\\begin{array}{l}\\text{ }-x+y=3\\hfill \\\\ \\text{ }-x+\\frac{2}{3}=3\\hfill \\\\ \\text{ }-x=3-\\frac{2}{3}\\hfill \\\\ \\text{ }-x=\\frac{7}{3}\\hfill \\\\ \\text{ }x=-\\frac{7}{3}\\hfill \\end{array}[\/latex]\r\nThe solution to this system is [latex]\\left(-\\frac{7}{3},\\frac{2}{3}\\right)[\/latex].\r\n\r\nCheck the solution in the first equation.\r\n\r\n[latex]\\begin{array}{llll}\\text{ }x+2y=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }\\left(-\\frac{7}{3}\\right)+2\\left(\\frac{2}{3}\\right)=\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-\\frac{7}{3}+\\frac{4}{3}=\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-\\frac{3}{3}=\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-1=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]\r\n<h4>Analysis of the Solution<\/h4>\r\nWe gain an important perspective on systems of equations by looking at the graphical representation. See the graph below\u00a0to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183607\/CNX_Precalc_Figure_09_01_0042.jpg\" alt=\"A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.\" width=\"487\" height=\"291\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try IT<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115130&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Addition Method When Multiplication of One Equation Is Required<\/h3>\r\nSolve the given system of equations by the <strong>addition method<\/strong>.\r\n\r\n[latex]\\begin{array}{l}3x+5y=-11\\hfill \\\\ \\hfill \\\\ x - 2y=11\\hfill \\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"883001\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"883001\"]\r\n\r\nAdding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[\/latex] in it and the second equation has [latex]x[\/latex]. So if we multiply the second equation by [latex]-3,\\text{}[\/latex] the <em>x<\/em>-terms will add to zero.\r\n\r\n[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ -3\\left(x - 2y\\right)=-3\\left(11\\right)\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Multiply both sides by }-3.\\hfill \\\\ \\text{ }-3x+6y=-33\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Use the distributive property}.\\hfill \\end{array}[\/latex]\r\nNow, let\u2019s add them.\r\n\r\n[latex]\\begin{array}\\ \\hfill 3x+5y=\u221211 \\\\ \\hfill \u22123x+6y=\u221233 \\\\ \\text{_____________} \\\\ \\hfill 11y=\u221244 \\\\ \\hfill y=\u22124 \\end{array}[\/latex]\r\n\r\nFor the last step, we substitute [latex]y=-4[\/latex] into one of the original equations and solve for [latex]x[\/latex].\r\n\r\n[latex]\\begin{array}{c}3x+5y=-11\\\\ 3x+5\\left(-4\\right)=-11\\\\ 3x - 20=-11\\\\ 3x=9\\\\ x=3\\end{array}[\/latex]\r\n\r\nOur solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.\r\n\r\n[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\left(3\\right)-2\\left(-4\\right)=3+8\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183609\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the system of equations by addition.\r\n\r\n[latex]\\begin{array}{c}2x - 7y=2\\\\ 3x+y=-20\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"609174\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"609174\"][latex]\\left(-6,-2\\right)[\/latex][\/hidden-answer]\r\n<iframe id=\"mom15\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115120&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Addition Method When Multiplication of Both Equations Is Required<\/h3>\r\nSolve the given system of equations in two variables by addition.\r\n\r\n[latex]\\begin{array}{c}2x+3y=-16\\\\ 5x - 10y=30\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"114755\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"114755\"]\r\n\r\nOne equation has [latex]2x[\/latex] and the other has [latex]5x[\/latex]. The least common multiple is [latex]10x[\/latex] so we will have to multiply both equations by a constant in order to eliminate one variable. Let\u2019s eliminate [latex]x[\/latex] by multiplying the first equation by [latex]-5[\/latex] and the second equation by [latex]2[\/latex].\r\n\r\n[latex]\\begin{array}{l} -5\\left(2x+3y\\right)=-5\\left(-16\\right)\\hfill \\\\ \\text{ }-10x - 15y=80\\hfill \\\\ \\text{ }2\\left(5x - 10y\\right)=2\\left(30\\right)\\hfill \\\\ \\text{ }10x - 20y=60\\hfill \\end{array}[\/latex]\r\n\r\nThen, we add the two equations together.\r\n\r\n[latex]\\begin{array}\\ \u221210x\u221215y=80 \\\\ 10x\u221220y=60 \\\\ \\text{______________} \\\\ \\text{ }\u221235y=140 \\\\ y=\u22124 \\end{array}[\/latex]\r\n\r\nSubstitute [latex]y=-4[\/latex] into the original first equation.\r\n\r\n[latex]\\begin{array}{c}2x+3\\left(-4\\right)=-16\\\\ 2x - 12=-16\\\\ 2x=-4\\\\ x=-2\\end{array}[\/latex]\r\n\r\nThe solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the other equation.\r\n\r\n[latex]\\begin{array}{r}\\hfill \\text{ }5x - 10y=30\\\\ \\hfill 5\\left(-2\\right)-10\\left(-4\\right)=30\\\\ \\hfill \\text{ }-10+40=30\\\\ \\hfill \\text{ }30=30\\end{array}[\/latex]\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183611\/CNX_Precalc_Figure_09_01_0062.jpg\" alt=\"A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.\" width=\"487\" height=\"366\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Addition Method in Systems of Equations Containing Fractions<\/h3>\r\nSolve the given system of equations in two variables by addition.\r\n\r\n[latex]\\begin{array}{l}\\frac{x}{3}+\\frac{y}{6}=3\\hfill \\\\ \\frac{x}{2}-\\frac{y}{4}=\\text{ }1\\hfill \\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"359287\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"359287\"]\r\n\r\nFirst clear each equation of fractions by multiplying both sides of the equation by the least common denominator.\r\n\r\n[latex]\\begin{array}{l}6\\left(\\frac{x}{3}+\\frac{y}{6}\\right)=6\\left(3\\right)\\hfill \\\\ \\text{ }2x+y=18\\hfill \\\\ 4\\left(\\frac{x}{2}-\\frac{y}{4}\\right)=4\\left(1\\right)\\hfill \\\\ \\text{ }2x-y=4\\hfill \\end{array}[\/latex]\r\n\r\nNow multiply the second equation by [latex]-1[\/latex] so that we can eliminate the <em>x<\/em>-variable.\r\n\r\n[latex]\\begin{array}{l}-1\\left(2x-y\\right)=-1\\left(4\\right)\\hfill \\\\ \\text{ }-2x+y=-4\\hfill \\end{array}[\/latex]\r\n\r\nAdd the two equations to eliminate the <em>x<\/em>-variable and solve the resulting equation.\r\n\r\n[latex]\\begin{array}\\ \\hfill 2x+y=18 \\\\ \\hfill\u22122x+y=\u22124 \\\\ \\text{_____________} \\\\ \\hfill 2y=14 \\\\ \\hfill y=7 \\end{array}[\/latex]\r\n\r\nSubstitute [latex]y=7[\/latex] into the first equation.\r\n\r\n[latex]\\begin{array}{l}2x+\\left(7\\right)=18\\hfill \\\\ \\text{ }2x=11\\hfill \\\\ \\text{ }x=\\frac{11}{2}\\hfill \\\\ \\text{ }=7.5\\hfill \\end{array}[\/latex]\r\n\r\nThe solution is [latex]\\left(\\frac{11}{2},7\\right)[\/latex]. Check it in the other equation.\r\n\r\n[latex]\\begin{array}{c}\\frac{x}{2}-\\frac{y}{4}=1\\\\ \\frac{\\frac{11}{2}}{2}-\\frac{7}{4}=1\\\\ \\frac{11}{4}-\\frac{7}{4}=1\\\\ \\frac{4}{4}=1\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the system of equations by addition.\r\n\r\n[latex]\\begin{array}{c}2x+3y=8\\\\ 3x+5y=10\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"326265\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"326265\"][latex]\\left(10,-4\\right)[\/latex][\/hidden-answer]\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115110&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\nin the following video we present more examples of how to use the addition (elimination) method to solve a system of two linear equations.\r\n\r\nhttps:\/\/youtu.be\/ova8GSmPV4o\r\n<h2>Classify Solutions to Systems<\/h2>\r\nNow that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an <strong>inconsistent system<\/strong> consists of parallel lines that have the same slope but different [latex]y[\/latex] -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as [latex]12=0[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Inconsistent System of Equations<\/h3>\r\nSolve the following system of equations.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }x=9 - 2y\\hfill \\\\ x+2y=13\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"888134\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"888134\"]\r\n\r\nWe can approach this problem in two ways. Because one equation is already solved for [latex]x[\/latex], the most obvious step is to use substitution.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}x+2y=13\\hfill \\\\ \\left(9 - 2y\\right)+2y=13\\hfill \\\\ 9+0y=13\\hfill \\\\ 9=13\\hfill \\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\nThe second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }x=9 - 2y\\hfill \\\\ 2y=-x+9\\hfill \\\\ \\text{ }y=-\\frac{1}{2}x+\\frac{9}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nWe then convert the second equation expressed to slope-intercept form.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}x+2y=13\\hfill \\\\ \\text{ }2y=-x+13\\hfill \\\\ \\text{ }y=-\\frac{1}{2}x+\\frac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nComparing the equations, we see that they have the same slope but different <em>y<\/em>-intercepts. Therefore, the lines are parallel and do not intersect.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ y=-\\frac{1}{2}x+\\frac{9}{2}\\end{array}\\hfill \\\\ y=-\\frac{1}{2}x+\\frac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWriting the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183613\/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\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the following system of equations in two variables.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}2y - 2x=2\\\\ 2y - 2x=6\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"681787\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"681787\"]No solution. It is an inconsistent system.[\/hidden-answer]\r\n<iframe id=\"mom11\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29699&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Expressing the Solution of a System of Dependent Equations Containing Two Variables<\/h3>\r\nRecall that a <strong>dependent system<\/strong> of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as [latex]0=0[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Solution to a Dependent System of Linear Equations<\/h3>\r\nFind a solution to the system of equations using the <strong>addition method<\/strong>.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}x+3y=2\\\\ 3x+9y=6\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"390828\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"390828\"]\r\n\r\nWith the addition method, we want to eliminate one of the variables by adding the equations. In this case, let\u2019s 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}{l}\\text{ }x+3y=2\\hfill \\\\ \\left(-3\\right)\\left(x+3y\\right)=\\left(-3\\right)\\left(2\\right)\\hfill \\\\ \\text{ }-3x - 9y=-6\\hfill \\end{array}[\/latex]<\/p>\r\nNow add the equations.\r\n<p style=\"text-align: center\">[latex]\\begin{array} \\hfill\u22123x\u22129y=\u22126 \\\\ \\hfill+3x+9y=6 \\\\ \\hfill \\text{_____________} \\\\ \\hfill 0=0 \\end{array}[\/latex]<\/p>\r\nWe can see that there will be an infinite number of solutions that satisfy both equations.\r\n<h4>Analysis of the Solution<\/h4>\r\nIf we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let\u2019s look at what happens when we convert the system to slope-intercept form.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }x+3y=2\\hfill \\\\ \\text{ }3y=-x+2\\hfill \\\\ \\text{ }y=-\\frac{1}{3}x+\\frac{2}{3}\\hfill \\\\ 3x+9y=6\\hfill \\\\ \\text{ }9y=-3x+6\\hfill \\\\ \\text{ }y=-\\frac{3}{9}x+\\frac{6}{9}\\hfill \\\\ \\text{ }y=-\\frac{1}{3}x+\\frac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nLook at the graph below. Notice the results are the same. The general solution to the system is [latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183615\/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\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Writing the general solution<\/h3>\r\nIn the previous example, we presented an analysis of the solution to the following system of equations:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}x+3y=2\\\\ 3x+9y=6\\end{array}[\/latex]<\/p>\r\nAfter a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as\u00a0[latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex]. Why would we write the solution this way? In some ways, this representation tells us a lot. \u00a0It tells us that x can be anything, x is x. \u00a0It also tells us that y is going to depend on x, just like when we write a function rule. \u00a0In this case, depending on what you put in for x, y will be defined in terms of x as [latex]-\\frac{1}{3}x+\\frac{2}{3}[\/latex].\r\n\r\nIn other words, there are infinitely many (x,y) pairs that will satisfy this system of equations, and they all fall on the line\u00a0[latex]y=-\\frac{1}{3}x+\\frac{2}{3}[\/latex].\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the following system of equations in two variables.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ \\text{ }\\text{}\\text{}y - 2x=5\\end{array}\\hfill \\\\ -3y+6x=-15\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"218404\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"218404\"]The system is dependent so there are infinite solutions of the form [latex]\\left(x,2x+5\\right)[\/latex].[\/hidden-answer]\r\n<iframe id=\"mom17\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15665&amp;theme=oea&amp;iframe_resize_id=mom17\" width=\"100%\" height=\"300\"><\/iframe>\r\n<\/div>\r\nIn the following video we show another example of solving a system that is dependent using elimination (addition).\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><\/h2>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Define and classify solutions to systems of linear equations\n<ul>\n<li>Recognize\u00a0consistent and inconsistent, dependent and independent systems of linear equations<\/li>\n<li>Determine whether an ordered pair is a solution to a system of linear equations<\/li>\n<li>Solve a system of linear equations by graphing<\/li>\n<\/ul>\n<\/li>\n<li>Methods for solving\u00a0systems\n<ul>\n<li>Use substitution to solve a system algebraically<\/li>\n<li>Solve a system using the addition\u00a0method<\/li>\n<li>Recognize when a system is inconsistent from algebraic results<\/li>\n<li>Recognize when a system is dependent using algebraic results<\/li>\n<\/ul>\n<\/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 the unique 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. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution.<\/p>\n<p>In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{c}2x+y=\\text{ }15\\\\ 3x-y=\\text{ }5\\end{array}[\/latex]<\/div>\n<p>The <em>solution<\/em> to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{l}2\\left(4\\right)+\\left(7\\right)=15\\text{ }\\text{True}\\hfill \\\\ 3\\left(4\\right)-\\left(7\\right)=5\\text{ }\\text{True}\\hfill \\end{array}[\/latex]<\/div>\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, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a <strong>dependent system<\/strong> if the equations have the same slope and the same <em>y<\/em>-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions.<\/p>\n<p>Another type of system of linear equations is an <strong>inconsistent system<\/strong>, which is one in which the equations represent two parallel lines. The lines have the same slope and different <em>y-<\/em>intercepts. There are no points common to both lines; hence, there is no solution to the system.<\/p>\n<div class=\"textbox\">\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<div>\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. Notice that the two lines are parallel and will never intersect.<\/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<\/div>\n<p>Below are graphical representations of each type of 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<p>The independent and dependent systems are also consistent because they both have at least one solution.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution.<\/h3>\n<ol>\n<li>Substitute the ordered pair into each equation in the system.<\/li>\n<li>Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Determine whether the ordered pair [latex]\\left(5,1\\right)[\/latex] is a solution to the given system of equations.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{l}x+3y=8\\hfill \\\\ 2x - 9=y\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q322824\">Show Solution<\/span><\/p>\n<div id=\"q322824\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the ordered pair [latex]\\left(5,1\\right)[\/latex] into both equations.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{ll}\\left(5\\right)+3\\left(1\\right)=8\\hfill & \\hfill \\\\ \\text{ }8=8\\hfill & \\text{True}\\hfill \\\\ 2\\left(5\\right)-9=\\left(1\\right)\\hfill & \\hfill \\\\ \\text{ }\\text{1=1}\\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/p>\n<p>The ordered pair [latex]\\left(5,1\\right)[\/latex] satisfies both equations, so it is the solution to the system.<\/p>\n<p>We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines.<\/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\/01222634\/CNX_Precalc_Figure_09_01_0032.jpg\" alt=\"A graph of two lines running through the point five, one. The first line's equation is x plus 3y equals 8. The second line's equation is 2x minus 9 equals y.\" width=\"487\" height=\"365\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we will show another example of how to verify whether an ordered pair is a solution to a system of equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Determine if an Ordered Pair is a Solution to a System of Linear Equations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/2IxgKgjX00k?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\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{array}{c}2x+y=-8\\\\ x-y=-1\\end{array}[\/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{array}{c}2x+y=-8\\\\ y=-2x - 8\\end{array}[\/latex]<\/p>\n<p>Solve the second equation for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}x-y=-1\\\\ y=x+1\\end{array}[\/latex]<\/p>\n<p>Graph both equations on the same set of axes as in the figure 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 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<p>Graphing can be used if the system is inconsistent or dependent.\u00a0In both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.<\/p>\n<p>In the following video we show another example of how to identify whether a graphed system has a solution, and identify what type of solution is represented.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Determine the Number of Solutions to a System of Linear Equations From a Graph\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ZolxtOjcEQY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In our last video we show how to solve a system of equations by first graphing the lines, then identifying the type of solution the system has.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 2:  Solve a System of Equations by Graphing\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Lv832rXAQ5k?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Substitution<\/h2>\n<p>Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a <strong>system of linear equations<\/strong> that are more precise than graphing. One such method is solving a system of equations by the <strong>substitution method<\/strong>, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a system of two equations in two variables, solve using the substitution method.<\/h3>\n<ol>\n<li>Solve one of the two equations for one of the variables in terms of the other.<\/li>\n<li>Substitute the expression for this variable into the second equation, then solve for the remaining variable.<\/li>\n<li>Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.<\/li>\n<li>Check the solution in both equations.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve the following system of equations by substitution.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }-x+y=-5\\hfill \\\\ \\text{ }2x - 5y=1\\hfill \\end{array}[\/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>First, we will solve the first equation for [latex]y[\/latex].<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{l}-x+y=-5\\hfill \\\\ \\text{ }y=x - 5\\hfill \\end{array}[\/latex]<\/div>\n<p>Now we can substitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }2x - 5y=1\\hfill \\\\ 2x - 5\\left(x - 5\\right)=1\\hfill \\\\ \\text{ }2x - 5x+25=1\\hfill \\\\ \\text{ }-3x=-24\\hfill \\\\ \\text{ }x=8\\hfill \\end{array}[\/latex]<\/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}{l}-\\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 1 or [latex]\u20131[\/latex] so that we do not have to deal with fractions.<\/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-4\" title=\"Ex 2:  Solve a System of Equations Using Substitution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/MIXL35YRzRw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>If you had chosen the other equation to start with in the previous example, you would still be able to find the same solution. It is really a matter of preference because sometimes solving for a variable will result in having to work with fractions. As you become more experienced with algebra, you will be able to anticipate what choices will lead to more desirable outcomes.<\/p>\n<p>Recall that an <strong>inconsistent system<\/strong> consists of parallel lines that have the same slope but different <em>y<\/em>-intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as [latex]12=0[\/latex].<\/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]\\begin{array}{l}\\text{ }x=9 - 2y\\hfill \\\\ x+2y=13\\hfill \\end{array}[\/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>We can approach this problem in two ways. Because one equation is already solved for <em>x<\/em>, the most obvious step is to use substitution.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}x+2y=13\\hfill \\\\ \\left(9 - 2y\\right)+2y=13\\hfill \\\\ 9+0y=13\\hfill \\\\ 9=13\\hfill \\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>The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }x=9 - 2y\\hfill \\\\ 2y=-x+9\\hfill \\\\ \\text{ }y=-\\frac{1}{2}x+\\frac{9}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>We then convert the second equation expressed to slope-intercept form.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}x+2y=13\\hfill \\\\ \\text{ }2y=-x+13\\hfill \\\\ \\text{ }y=-\\frac{1}{2}x+\\frac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>Comparing the equations, we see that they have the same slope but different <em>y<\/em>-intercepts. Therefore, the lines are parallel and do not intersect.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ y=-\\frac{1}{2}x+\\frac{9}{2}\\end{array}\\hfill \\\\ y=-\\frac{1}{2}x+\\frac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.<\/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<h4>Answer<\/h4>\n<p>There is no solution to this system of linear equations.<\/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-5\" 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>In our next video we show that a system can have an infinite number of solutions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" 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<h3>Solving Systems of Equations in Two Variables by the Addition Method<\/h3>\n<p>A third method of <strong>solving systems of linear equations<\/strong> is the <strong>addition method,\u00a0<\/strong>this method is also called the\u00a0<strong>elimination method<\/strong>. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, solve using the addition method.<\/h3>\n<ol>\n<li>Write both equations with <em>x<\/em>&#8211; and <em>y<\/em>-variables on the left side of the equal sign and constants on the right.<\/li>\n<li>Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.<\/li>\n<li>Solve the resulting equation for the remaining variable.<\/li>\n<li>Substitute that value into one of the original equations and solve for the second variable.<\/li>\n<li>Check the solution by substituting the values into the other equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a System by the Addition Method<\/h3>\n<p>Solve the given system of equations by addition.<\/p>\n<p>[latex]\\begin{array}{l}x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924657\">Solution<\/span><\/p>\n<div id=\"q924657\" class=\"hidden-answer\" style=\"display: none\">\n<p>Both equations are already set equal to a constant. Notice that the coefficient of [latex]x[\/latex] in the second equation, \u20131, is the opposite of the coefficient of [latex]x[\/latex] in the first equation, 1. We can add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.<\/p>\n<p>[latex]\\frac{\\begin{array}{l}\\hfill \\\\ x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}}{\\text{}\\text{}\\text{}\\text{}\\text{}3y=2}[\/latex]<\/p>\n<p>Now that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex].<\/p>\n<p>[latex]\\begin{array}{l}3y=2\\hfill \\\\ \\text{ }y=\\frac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>Then, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex].<\/p>\n<p>[latex]\\begin{array}{l}\\text{ }-x+y=3\\hfill \\\\ \\text{ }-x+\\frac{2}{3}=3\\hfill \\\\ \\text{ }-x=3-\\frac{2}{3}\\hfill \\\\ \\text{ }-x=\\frac{7}{3}\\hfill \\\\ \\text{ }x=-\\frac{7}{3}\\hfill \\end{array}[\/latex]<br \/>\nThe solution to this system is [latex]\\left(-\\frac{7}{3},\\frac{2}{3}\\right)[\/latex].<\/p>\n<p>Check the solution in the first equation.<\/p>\n<p>[latex]\\begin{array}{llll}\\text{ }x+2y=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }\\left(-\\frac{7}{3}\\right)+2\\left(\\frac{2}{3}\\right)=\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-\\frac{7}{3}+\\frac{4}{3}=\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-\\frac{3}{3}=\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-1=-1\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We gain an important perspective on systems of equations by looking at the graphical representation. See the graph below\u00a0to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183607\/CNX_Precalc_Figure_09_01_0042.jpg\" alt=\"A graph of two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y=negative 1. The second line's equation is negative x + y equals 3.\" width=\"487\" height=\"291\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115130&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Addition Method When Multiplication of One Equation Is Required<\/h3>\n<p>Solve the given system of equations by the <strong>addition method<\/strong>.<\/p>\n<p>[latex]\\begin{array}{l}3x+5y=-11\\hfill \\\\ \\hfill \\\\ x - 2y=11\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q883001\">Solution<\/span><\/p>\n<div id=\"q883001\" class=\"hidden-answer\" style=\"display: none\">\n<p>Adding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[\/latex] in it and the second equation has [latex]x[\/latex]. So if we multiply the second equation by [latex]-3,\\text{}[\/latex] the <em>x<\/em>-terms will add to zero.<\/p>\n<p>[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill & \\hfill & \\hfill & \\hfill \\\\ -3\\left(x - 2y\\right)=-3\\left(11\\right)\\hfill & \\hfill & \\hfill & \\text{Multiply both sides by }-3.\\hfill \\\\ \\text{ }-3x+6y=-33\\hfill & \\hfill & \\hfill & \\text{Use the distributive property}.\\hfill \\end{array}[\/latex]<br \/>\nNow, let\u2019s add them.<\/p>\n<p>[latex]\\begin{array}\\ \\hfill 3x+5y=\u221211 \\\\ \\hfill \u22123x+6y=\u221233 \\\\ \\text{_____________} \\\\ \\hfill 11y=\u221244 \\\\ \\hfill y=\u22124 \\end{array}[\/latex]<\/p>\n<p>For the last step, we substitute [latex]y=-4[\/latex] into one of the original equations and solve for [latex]x[\/latex].<\/p>\n<p>[latex]\\begin{array}{c}3x+5y=-11\\\\ 3x+5\\left(-4\\right)=-11\\\\ 3x - 20=-11\\\\ 3x=9\\\\ x=3\\end{array}[\/latex]<\/p>\n<p>Our solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.<\/p>\n<p>[latex]\\begin{array}{llll}\\text{ }x - 2y=11\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\left(3\\right)-2\\left(-4\\right)=3+8\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }=11\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183609\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the system of equations by addition.<\/p>\n<p>[latex]\\begin{array}{c}2x - 7y=2\\\\ 3x+y=-20\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q609174\">Solution<\/span><\/p>\n<div id=\"q609174\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(-6,-2\\right)[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom15\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115120&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Addition Method When Multiplication of Both Equations Is Required<\/h3>\n<p>Solve the given system of equations in two variables by addition.<\/p>\n<p>[latex]\\begin{array}{c}2x+3y=-16\\\\ 5x - 10y=30\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q114755\">Solution<\/span><\/p>\n<div id=\"q114755\" class=\"hidden-answer\" style=\"display: none\">\n<p>One equation has [latex]2x[\/latex] and the other has [latex]5x[\/latex]. The least common multiple is [latex]10x[\/latex] so we will have to multiply both equations by a constant in order to eliminate one variable. Let\u2019s eliminate [latex]x[\/latex] by multiplying the first equation by [latex]-5[\/latex] and the second equation by [latex]2[\/latex].<\/p>\n<p>[latex]\\begin{array}{l} -5\\left(2x+3y\\right)=-5\\left(-16\\right)\\hfill \\\\ \\text{ }-10x - 15y=80\\hfill \\\\ \\text{ }2\\left(5x - 10y\\right)=2\\left(30\\right)\\hfill \\\\ \\text{ }10x - 20y=60\\hfill \\end{array}[\/latex]<\/p>\n<p>Then, we add the two equations together.<\/p>\n<p>[latex]\\begin{array}\\ \u221210x\u221215y=80 \\\\ 10x\u221220y=60 \\\\ \\text{______________} \\\\ \\text{ }\u221235y=140 \\\\ y=\u22124 \\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=-4[\/latex] into the original first equation.<\/p>\n<p>[latex]\\begin{array}{c}2x+3\\left(-4\\right)=-16\\\\ 2x - 12=-16\\\\ 2x=-4\\\\ x=-2\\end{array}[\/latex]<\/p>\n<p>The solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the other equation.<\/p>\n<p>[latex]\\begin{array}{r}\\hfill \\text{ }5x - 10y=30\\\\ \\hfill 5\\left(-2\\right)-10\\left(-4\\right)=30\\\\ \\hfill \\text{ }-10+40=30\\\\ \\hfill \\text{ }30=30\\end{array}[\/latex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183611\/CNX_Precalc_Figure_09_01_0062.jpg\" alt=\"A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.\" width=\"487\" height=\"366\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Addition Method in Systems of Equations Containing Fractions<\/h3>\n<p>Solve the given system of equations in two variables by addition.<\/p>\n<p>[latex]\\begin{array}{l}\\frac{x}{3}+\\frac{y}{6}=3\\hfill \\\\ \\frac{x}{2}-\\frac{y}{4}=\\text{ }1\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q359287\">Solution<\/span><\/p>\n<div id=\"q359287\" class=\"hidden-answer\" style=\"display: none\">\n<p>First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.<\/p>\n<p>[latex]\\begin{array}{l}6\\left(\\frac{x}{3}+\\frac{y}{6}\\right)=6\\left(3\\right)\\hfill \\\\ \\text{ }2x+y=18\\hfill \\\\ 4\\left(\\frac{x}{2}-\\frac{y}{4}\\right)=4\\left(1\\right)\\hfill \\\\ \\text{ }2x-y=4\\hfill \\end{array}[\/latex]<\/p>\n<p>Now multiply the second equation by [latex]-1[\/latex] so that we can eliminate the <em>x<\/em>-variable.<\/p>\n<p>[latex]\\begin{array}{l}-1\\left(2x-y\\right)=-1\\left(4\\right)\\hfill \\\\ \\text{ }-2x+y=-4\\hfill \\end{array}[\/latex]<\/p>\n<p>Add the two equations to eliminate the <em>x<\/em>-variable and solve the resulting equation.<\/p>\n<p>[latex]\\begin{array}\\ \\hfill 2x+y=18 \\\\ \\hfill\u22122x+y=\u22124 \\\\ \\text{_____________} \\\\ \\hfill 2y=14 \\\\ \\hfill y=7 \\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=7[\/latex] into the first equation.<\/p>\n<p>[latex]\\begin{array}{l}2x+\\left(7\\right)=18\\hfill \\\\ \\text{ }2x=11\\hfill \\\\ \\text{ }x=\\frac{11}{2}\\hfill \\\\ \\text{ }=7.5\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution is [latex]\\left(\\frac{11}{2},7\\right)[\/latex]. Check it in the other equation.<\/p>\n<p>[latex]\\begin{array}{c}\\frac{x}{2}-\\frac{y}{4}=1\\\\ \\frac{\\frac{11}{2}}{2}-\\frac{7}{4}=1\\\\ \\frac{11}{4}-\\frac{7}{4}=1\\\\ \\frac{4}{4}=1\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the system of equations by addition.<\/p>\n<p>[latex]\\begin{array}{c}2x+3y=8\\\\ 3x+5y=10\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q326265\">Solution<\/span><\/p>\n<div id=\"q326265\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(10,-4\\right)[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115110&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>in the following video we present more examples of how to use the addition (elimination) method to solve a system of two linear equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Solving Systems of Equations using Elimination\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ova8GSmPV4o?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Classify Solutions to Systems<\/h2>\n<p>Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an <strong>inconsistent system<\/strong> consists of parallel lines that have the same slope but different [latex]y[\/latex] -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as [latex]12=0[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Inconsistent System of Equations<\/h3>\n<p>Solve the following system of equations.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }x=9 - 2y\\hfill \\\\ x+2y=13\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q888134\">Solution<\/span><\/p>\n<div id=\"q888134\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can approach this problem in two ways. Because one equation is already solved for [latex]x[\/latex], the most obvious step is to use substitution.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}x+2y=13\\hfill \\\\ \\left(9 - 2y\\right)+2y=13\\hfill \\\\ 9+0y=13\\hfill \\\\ 9=13\\hfill \\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>The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }x=9 - 2y\\hfill \\\\ 2y=-x+9\\hfill \\\\ \\text{ }y=-\\frac{1}{2}x+\\frac{9}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>We then convert the second equation expressed to slope-intercept form.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}x+2y=13\\hfill \\\\ \\text{ }2y=-x+13\\hfill \\\\ \\text{ }y=-\\frac{1}{2}x+\\frac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>Comparing the equations, we see that they have the same slope but different <em>y<\/em>-intercepts. Therefore, the lines are parallel and do not intersect.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ y=-\\frac{1}{2}x+\\frac{9}{2}\\end{array}\\hfill \\\\ y=-\\frac{1}{2}x+\\frac{13}{2}\\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown 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\/896\/2016\/11\/03183613\/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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the following system of equations in two variables.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}2y - 2x=2\\\\ 2y - 2x=6\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q681787\">Solution<\/span><\/p>\n<div id=\"q681787\" class=\"hidden-answer\" style=\"display: none\">No solution. It is an inconsistent system.<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom11\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29699&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<h3>Expressing the Solution of a System of Dependent Equations Containing Two Variables<\/h3>\n<p>Recall that a <strong>dependent system<\/strong> of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as [latex]0=0[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Solution to a Dependent System of Linear Equations<\/h3>\n<p>Find a solution to the system of equations using the <strong>addition method<\/strong>.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}x+3y=2\\\\ 3x+9y=6\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q390828\">Solution<\/span><\/p>\n<div id=\"q390828\" class=\"hidden-answer\" style=\"display: none\">\n<p>With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let\u2019s 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}{l}\\text{ }x+3y=2\\hfill \\\\ \\left(-3\\right)\\left(x+3y\\right)=\\left(-3\\right)\\left(2\\right)\\hfill \\\\ \\text{ }-3x - 9y=-6\\hfill \\end{array}[\/latex]<\/p>\n<p>Now add the equations.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array} \\hfill\u22123x\u22129y=\u22126 \\\\ \\hfill+3x+9y=6 \\\\ \\hfill \\text{_____________} \\\\ \\hfill 0=0 \\end{array}[\/latex]<\/p>\n<p>We can see that there will be an infinite number of solutions that satisfy both equations.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let\u2019s look at what happens when we convert the system to slope-intercept form.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }x+3y=2\\hfill \\\\ \\text{ }3y=-x+2\\hfill \\\\ \\text{ }y=-\\frac{1}{3}x+\\frac{2}{3}\\hfill \\\\ 3x+9y=6\\hfill \\\\ \\text{ }9y=-3x+6\\hfill \\\\ \\text{ }y=-\\frac{3}{9}x+\\frac{6}{9}\\hfill \\\\ \\text{ }y=-\\frac{1}{3}x+\\frac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>Look at the graph below. Notice the results are the same. The general solution to the system is [latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03183615\/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\" \/><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Writing the general solution<\/h3>\n<p>In the previous example, we presented an analysis of the solution to the following system of equations:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}x+3y=2\\\\ 3x+9y=6\\end{array}[\/latex]<\/p>\n<p>After a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as\u00a0[latex]\\left(x, -\\frac{1}{3}x+\\frac{2}{3}\\right)[\/latex]. Why would we write the solution this way? In some ways, this representation tells us a lot. \u00a0It tells us that x can be anything, x is x. \u00a0It also tells us that y is going to depend on x, just like when we write a function rule. \u00a0In this case, depending on what you put in for x, y will be defined in terms of x as [latex]-\\frac{1}{3}x+\\frac{2}{3}[\/latex].<\/p>\n<p>In other words, there are infinitely many (x,y) pairs that will satisfy this system of equations, and they all fall on the line\u00a0[latex]y=-\\frac{1}{3}x+\\frac{2}{3}[\/latex].<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the following system of equations in two variables.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ \\text{ }\\text{}\\text{}y - 2x=5\\end{array}\\hfill \\\\ -3y+6x=-15\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q218404\">Solution<\/span><\/p>\n<div id=\"q218404\" class=\"hidden-answer\" style=\"display: none\">The system is dependent so there are infinite solutions of the form [latex]\\left(x,2x+5\\right)[\/latex].<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom17\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15665&amp;theme=oea&amp;iframe_resize_id=mom17\" width=\"100%\" height=\"300\"><\/iframe>\n<\/div>\n<p>In the following video we show another example of solving a system that is dependent using elimination (addition).<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" 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 &#8211; it has no solutions which means the lines the equations represent are parallel to each other.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" 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><\/h2>\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-2075\">\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>Determine if an Ordered Pair is a Solution to a System of Linear Equations. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/2IxgKgjX00k\">https:\/\/youtu.be\/2IxgKgjX00k<\/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>Determine the Number of Solutions to a System of Linear Equations From a Graph. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ZolxtOjcEQY\">https:\/\/youtu.be\/ZolxtOjcEQY<\/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><\/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 by Graphing. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Lv832rXAQ5k\">https:\/\/youtu.be\/Lv832rXAQ5k<\/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 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 and Education. <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\/about\/pdm\">Public Domain: No Known Copyright<\/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: Solve a System of Equations Using Eliminations (Fractions). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/s3S64b1DrtQ\">https:\/\/youtu.be\/s3S64b1DrtQ<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/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 (No Solution). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/z5_ACYtzW98\">https:\/\/youtu.be\/z5_ACYtzW98<\/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>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al... <strong>Provided by<\/strong>: Open Stax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free:  http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 29699. <strong>Authored by<\/strong>: McClure, Caren. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 115164, 115120, 115110. <strong>Authored by<\/strong>: Shabazian, Roy. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, 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