{"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":"2018-07-17T19:52:39","modified_gmt":"2018-07-17T19:52:39","slug":"read-solve-systems-of-equations-by-graphing","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-suffolkccc-intermediatealgebra\/chapter\/read-solve-systems-of-equations-by-graphing\/","title":{"raw":"Solve Systems in Two Variables","rendered":"Solve 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>Recognize when a system is inconsistent from algebraic results<\/li>\r\n \t<li>Find the break even point for a cost and revenue system<\/li>\r\n \t<li>Write a system of equations based on attendance and cost data<\/li>\r\n \t<li>Solve a system using the elimination\u00a0method<\/li>\r\n \t<li>Solve a system using the elimination\u00a0method when multiplication is required<\/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) \\mathop = \\limits^?15\\text{ }\\text{True}\\hfill \\\\ 3\\left(4\\right)-\\left(7\\right) \\mathop= \\limits^?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)\\mathop = \\limits^{?} 8\\hfill &amp; \\hfill \\\\ \\text{ }8=8\\hfill &amp; \\text{True}\\hfill \\\\ 2\\left(5\\right)-9 \\mathop = \\limits^{?} \\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)\\mathop = \\limits^?-8\\hfill &amp; \\hfill \\\\ \\text{ }-8=-8\\hfill &amp; \\text{True}\\hfill \\\\ \\text{ }\\left(-3\\right)-\\left(-2\\right)\\mathop = \\limits^?-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\nConsider a\u00a0skateboard manufacturer\u2019s <strong>revenue function, <\/strong>this\u00a0is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[\/latex], where [latex]x=[\/latex] quantity and [latex]p=[\/latex] price. The revenue function is shown in orange in the graph below.\r\n\r\nThe <strong>cost function<\/strong> is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The <em>x<\/em>-axis represents the quantity of skateboards produced and sold in hundreds of units. The <em>y<\/em>-axis represents either cost or revenue in hundreds of dollars.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222649\/CNX_Precalc_Figure_09_01_0092.jpg\" alt=\"A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis. A line representing cost and a line representing revenue cross at the point (7,33), which is marked break-even. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"488\" height=\"347\" \/>\r\n\r\nThe point at which the two lines intersect is called the <strong>break-even point<\/strong>. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.\r\n\r\nThe shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGiven the cost function [latex]C\\left(x\\right)=0.85x+35,000[\/latex] and the revenue function [latex]R\\left(x\\right)=1.55x[\/latex], find the break-even point.\r\n\r\n[reveal-answer q=\"506309\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"506309\"]\r\n\r\nWrite the system of equations using [latex]y[\/latex] to replace function notation.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ y=0.85x+35,000\\end{array}\\hfill \\\\ y=1.55x\\hfill \\end{array}[\/latex]<\/p>\r\nSubstitute the expression [latex]0.85x+35,000[\/latex] from the first equation into the second equation and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0.85x+35,000=1.55x\\\\ 35,000=0.7x\\\\ 50,000=x\\end{array}[\/latex]<\/p>\r\nThen, we substitute [latex]x=50,000[\/latex] into either the cost function or the revenue function.\r\n<p style=\"text-align: center\">[latex]1.55\\left(50,000\\right)=77,500[\/latex]<\/p>\r\nThe break-even point is [latex]\\left(50,000,77,500\\right)[\/latex].\r\n\r\nThe cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222651\/CNX_Precalc_Figure_09_01_0102.jpg\" alt=\"A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"487\" height=\"390\" \/>\r\n\r\nThe company will make a profit after 50,000 units are produced.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will show how to write a system of linear equations given attendance and ticket cost data. We will then find the number of tickets purchased based on our system.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?\r\n\r\n[reveal-answer q=\"871151\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"871151\"]\r\n\r\nLet [latex]c=[\/latex] the number of children and [latex]a=[\/latex] the number of adults in attendance.\r\n\r\nThe total number of people is [latex]2,000[\/latex]. We can use this to write an equation for the number of people at the circus that day\r\n<p style=\"text-align: center\">[latex]c+a=2,000[\/latex]<\/p>\r\nThe revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[\/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[\/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.\r\n<p style=\"text-align: center\">[latex]25c+50a=70,000[\/latex]<\/p>\r\nWe now have a system of linear equations in two variables.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}c+a=2,000\\\\ 25c+50a=70,000\\end{array}[\/latex]<\/p>\r\nIn the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[\/latex] or [latex]a[\/latex]. We will solve for [latex]a[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}c+a=2,000\\\\ a=2,000-c\\end{array}[\/latex]<\/p>\r\nSubstitute the expression [latex]2,000-c[\/latex] in the second equation for [latex]a[\/latex] and solve for [latex]c[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l} 25c+50\\left(2,000-c\\right)=70,000\\hfill \\\\ 25c+100,000 - 50c=70,000\\hfill \\\\ \\text{ }-25c=-30,000\\hfill \\\\ \\text{ }c=1,200\\hfill \\end{array}[\/latex]<\/p>\r\nSubstitute [latex]c=1,200[\/latex] into the first equation to solve for [latex]a[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}1,200+a=2,000\\hfill \\\\ \\text{ }\\text{}a=800\\hfill \\end{array}[\/latex]<\/p>\r\nWe find that [latex]1,200[\/latex] children and [latex]800[\/latex] adults bought tickets to the circus that day.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our last video example we show how to set up a system of linear equations that represents the total cost for admission to a museum.\r\n\r\nhttps:\/\/youtu.be\/euh9ksWrq0A\r\n<h2>Elimination<\/h2>\r\nA third method of <strong>solving systems of linear equations<\/strong> is the <strong>elimination\u00a0method<\/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 elimination.\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]\\begin{array}{l}x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"522070\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522070\"]\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<div style=\"text-align: center\">[latex]\\frac{\\begin{array}{l}\\hfill \\\\ x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}}{\\text{}\\text{}\\text{}\\text{}\\text{}3y=2}[\/latex]<\/div>\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=\\frac{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+\\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]<\/p>\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<p style=\"text-align: center\">[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]<\/p>\r\nWe gain an important perspective on systems of equations by looking at the graphical representation. In the graph below, you will see 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 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 systems of linear equations.\r\n\r\nhttps:\/\/youtu.be\/M4IEmwcqR3c\r\n\r\nSometimes we have to do a couple of steps of algebra before we can eliminate a variable from a system solve it. 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 <strong>elimination method.<\/strong>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}3x+5y=-11\\hfill \\\\ \\hfill \\\\ x - 2y=11\\hfill \\end{array}[\/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 <em>x<\/em>-terms will add to zero.\r\n<div style=\"text-align: center\">[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]<\/div>\r\nNow, let\u2019s add them.\r\n<p style=\"text-align: center\">[latex]\\begin{array}\\ \\hfill 3x+5y=\u221211 \\\\ \\hfill \u22123x+6y=\u221233 \\\\ \\text{_____________} \\\\ \\hfill 11y=\u221244 \\\\ \\hfill y=\u22124 \\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{array}{c}3x+5y=-11\\\\ 3x+5\\left(-4\\right)=-11\\\\ 3x - 20=-11\\\\ 3x=9\\\\ x=3\\end{array}[\/latex]<\/p>\r\nOur solution is the ordered pair [latex]\\left(3,-4\\right)[\/latex]. Check the solution in the original second equation.\r\n<div style=\"text-align: center\">[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]<\/div>\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 in two variables by elimination.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{c}2x+3y=-16\\\\ 5x - 10y=30\\end{array}[\/latex]<\/div>\r\n[reveal-answer q=\"245990\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"245990\"]\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<div style=\"text-align: center\">[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]<\/div>\r\nThen, we add the two equations together.\r\n<p style=\"text-align: center\">[latex]\\begin{array}\\ \u221210x\u221215y=80 \\\\ 10x\u221220y=60 \\\\ \\text{______________} \\\\ \\text{ }\u221235y=140 \\\\ y=\u22124 \\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}{c}2x+3\\left(-4\\right)=-16\\\\ 2x - 12=-16\\\\ 2x=-4\\\\ x=-2\\end{array}[\/latex]<\/p>\r\nThe solution is [latex]\\left(-2,-4\\right)[\/latex]. Check it in the other 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\nBelow is a summary of the general steps for using the elimination method to solve a system of equations.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a system of equations, solve using the elimination\u00a0method.<\/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\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 in two variables by elimination.\r\n<p style=\"text-align: center\">[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>\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\r\n<p style=\"text-align: center\">[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>\r\nNow multiply the second equation by [latex]-1[\/latex] so that we can eliminate the <em>x<\/em>-variable.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}-1\\left(2x-y\\right)=-1\\left(4\\right)\\hfill \\\\ \\text{ }-2x+y=-4\\hfill \\end{array}[\/latex]<\/p>\r\nAdd the two equations to eliminate the <em>x<\/em>-variable and solve the resulting equation.\r\n<p style=\"text-align: center\">[latex]\\begin{array}\\ \\hfill 2x+y=18 \\\\ \\hfill\u22122x+y=\u22124 \\\\ \\text{_____________} \\\\ \\hfill 2y=14 \\\\ \\hfill y=7 \\end{array}[\/latex]<\/p>\r\nSubstitute [latex]y=7[\/latex] into the first equation.\r\n<p style=\"text-align: center\">[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>\r\nThe solution is [latex]\\left(\\frac{11}{2},7\\right)[\/latex]. Check it in the other equation.\r\n<p style=\"text-align: center\">[latex]\\begin{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>\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\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 elimination, the resulting equation will be an identity, such as [latex]0=0[\/latex]. The 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\nFind a solution to the system of equations using the <strong>elimination 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=\"10390\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"10390\"]\r\n\r\nWith the elimination 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\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\nSee 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 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><\/h2>","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>Recognize when a system is inconsistent from algebraic results<\/li>\n<li>Find the break even point for a cost and revenue system<\/li>\n<li>Write a system of equations based on attendance and cost data<\/li>\n<li>Solve a system using the elimination\u00a0method<\/li>\n<li>Solve a system using the elimination\u00a0method when multiplication is required<\/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) \\mathop = \\limits^?15\\text{ }\\text{True}\\hfill \\\\ 3\\left(4\\right)-\\left(7\\right) \\mathop= \\limits^?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)\\mathop = \\limits^{?} 8\\hfill & \\hfill \\\\ \\text{ }8=8\\hfill & \\text{True}\\hfill \\\\ 2\\left(5\\right)-9 \\mathop = \\limits^{?} \\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)\\mathop = \\limits^?-8\\hfill & \\hfill \\\\ \\text{ }-8=-8\\hfill & \\text{True}\\hfill \\\\ \\text{ }\\left(-3\\right)-\\left(-2\\right)\\mathop = \\limits^?-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<p>Consider a\u00a0skateboard manufacturer\u2019s <strong>revenue function, <\/strong>this\u00a0is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[\/latex], where [latex]x=[\/latex] quantity and [latex]p=[\/latex] price. The revenue function is shown in orange in the graph below.<\/p>\n<p>The <strong>cost function<\/strong> is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The <em>x<\/em>-axis represents the quantity of skateboards produced and sold in hundreds of units. The <em>y<\/em>-axis represents either cost or revenue in hundreds of dollars.<\/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\/01222649\/CNX_Precalc_Figure_09_01_0092.jpg\" alt=\"A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis. A line representing cost and a line representing revenue cross at the point (7,33), which is marked break-even. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"488\" height=\"347\" \/><\/p>\n<p>The point at which the two lines intersect is called the <strong>break-even point<\/strong>. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.<\/p>\n<p>The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Given the cost function [latex]C\\left(x\\right)=0.85x+35,000[\/latex] and the revenue function [latex]R\\left(x\\right)=1.55x[\/latex], find the break-even point.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q506309\">Show Solution<\/span><\/p>\n<div id=\"q506309\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the system of equations using [latex]y[\/latex] to replace function notation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\begin{array}{l}\\\\ y=0.85x+35,000\\end{array}\\hfill \\\\ y=1.55x\\hfill \\end{array}[\/latex]<\/p>\n<p>Substitute the expression [latex]0.85x+35,000[\/latex] from the first equation into the second equation and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0.85x+35,000=1.55x\\\\ 35,000=0.7x\\\\ 50,000=x\\end{array}[\/latex]<\/p>\n<p>Then, we substitute [latex]x=50,000[\/latex] into either the cost function or the revenue function.<\/p>\n<p style=\"text-align: center\">[latex]1.55\\left(50,000\\right)=77,500[\/latex]<\/p>\n<p>The break-even point is [latex]\\left(50,000,77,500\\right)[\/latex].<\/p>\n<p>The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.<\/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\/01222651\/CNX_Precalc_Figure_09_01_0102.jpg\" alt=\"A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.\" width=\"487\" height=\"390\" \/><\/p>\n<p>The company will make a profit after 50,000 units are produced.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will show how to write a system of linear equations given attendance and ticket cost data. We will then find the number of tickets purchased based on our system.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q871151\">Show Solution<\/span><\/p>\n<div id=\"q871151\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]c=[\/latex] the number of children and [latex]a=[\/latex] the number of adults in attendance.<\/p>\n<p>The total number of people is [latex]2,000[\/latex]. We can use this to write an equation for the number of people at the circus that day<\/p>\n<p style=\"text-align: center\">[latex]c+a=2,000[\/latex]<\/p>\n<p>The revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[\/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[\/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.<\/p>\n<p style=\"text-align: center\">[latex]25c+50a=70,000[\/latex]<\/p>\n<p>We now have a system of linear equations in two variables.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}c+a=2,000\\\\ 25c+50a=70,000\\end{array}[\/latex]<\/p>\n<p>In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[\/latex] or [latex]a[\/latex]. We will solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}c+a=2,000\\\\ a=2,000-c\\end{array}[\/latex]<\/p>\n<p>Substitute the expression [latex]2,000-c[\/latex] in the second equation for [latex]a[\/latex] and solve for [latex]c[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l} 25c+50\\left(2,000-c\\right)=70,000\\hfill \\\\ 25c+100,000 - 50c=70,000\\hfill \\\\ \\text{ }-25c=-30,000\\hfill \\\\ \\text{ }c=1,200\\hfill \\end{array}[\/latex]<\/p>\n<p>Substitute [latex]c=1,200[\/latex] into the first equation to solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}1,200+a=2,000\\hfill \\\\ \\text{ }\\text{}a=800\\hfill \\end{array}[\/latex]<\/p>\n<p>We find that [latex]1,200[\/latex] children and [latex]800[\/latex] adults bought tickets to the circus that day.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our last video example we show how to set up a system of linear equations that represents the total cost for admission to a museum.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Ex: Solve an Application Problem Using a System of Linear Equations (09x-43)\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/euh9ksWrq0A?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Elimination<\/h2>\n<p>A third method of <strong>solving systems of linear equations<\/strong> is the <strong>elimination\u00a0method<\/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 elimination.<\/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]\\begin{array}{l}x+2y=-1\\hfill \\\\ -x+y=3\\hfill \\end{array}[\/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>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<div style=\"text-align: center\">[latex]\\frac{\\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>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=\\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 style=\"text-align: center\">[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]<\/p>\n<p>The solution to this system is [latex]\\left(-\\frac{7}{3},\\frac{2}{3}\\right)[\/latex].<\/p>\n<p>Check the solution in the first equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{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<p>We gain an important perspective on systems of equations by looking at the graphical representation. In the graph below, you will see 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\" 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 systems of linear equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" 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>Sometimes we have to do a couple of steps of algebra before we can eliminate a variable from a system solve it. 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 <strong>elimination method.<\/strong><\/p>\n<p style=\"text-align: center\">[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=\"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 <em>x<\/em>-terms will add to zero.<\/p>\n<div style=\"text-align: center\">[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]<\/div>\n<p>Now, let\u2019s add them.<\/p>\n<p style=\"text-align: center\">[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 style=\"text-align: center\">[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<div style=\"text-align: center\">[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]<\/div>\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-9\" 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 in two variables by elimination.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{c}2x+3y=-16\\\\ 5x - 10y=30\\end{array}[\/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>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<div style=\"text-align: center\">[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]<\/div>\n<p>Then, we add the two equations together.<\/p>\n<p style=\"text-align: center\">[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 style=\"text-align: center\">[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<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>Below is a summary of the general steps for using the elimination method to solve a system of equations.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a system of equations, solve using the elimination\u00a0method.<\/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<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 in two variables by elimination.<\/p>\n<p style=\"text-align: center\">[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=\"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<\/p>\n<p style=\"text-align: center\">[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 style=\"text-align: center\">[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 style=\"text-align: center\">[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 style=\"text-align: center\">[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 style=\"text-align: center\">[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><\/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-10\" 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>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 elimination, the resulting equation will be an identity, such as [latex]0=0[\/latex]. 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>Find a solution to the system of equations using the <strong>elimination 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=\"q10390\">Show Solution<\/span><\/p>\n<div id=\"q10390\" class=\"hidden-answer\" style=\"display: none\">\n<p>With the elimination 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<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>See 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\" 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-11\" 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-12\" 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><\/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 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