{"id":1088,"date":"2022-03-12T22:08:33","date_gmt":"2022-03-12T22:08:33","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=1088"},"modified":"2025-11-15T18:14:26","modified_gmt":"2025-11-15T18:14:26","slug":"2-6-3-the-addition-method-for-solving-systems-of-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/2-6-3-the-addition-method-for-solving-systems-of-linear-equations\/","title":{"raw":"2.6.3: The Elimination Method","rendered":"2.6.3: The Elimination Method"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n<div class=\"wrapper\">\r\n<div id=\"wrap\">\r\n<div id=\"content\" role=\"main\">\r\n<div id=\"post-107\" class=\"standard post-107 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Outcomes<\/h1>\r\n<ul>\r\n \t<li>Solve a system of linear equations using the elimination method<\/li>\r\n \t<li>Write the general solution of a dependent system as an ordered pair in terms of one variable<\/li>\r\n<\/ul>\r\n<\/div>\r\nA third method of\u00a0solving systems of linear equations is the\u00a0<em><strong>elimination\u00a0method<\/strong><\/em>. The idea behind this method is to add the two equations in the system together while eliminating one of the variables. The result will be a new linear equation in one variable that we can solve.\u00a0 We get this new equation by adding equations where the same term in each equation has opposite coefficients, so that when they are added the sum is zero. <span style=\"font-size: 1rem; text-align: initial;\">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 through addition.<\/span>\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nSolve the given system of equations by elimination.\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}x+2y&amp;=-1 \\\\ -x+y&amp;=3 \\end{aligned}\\end{equation}[\/latex]<\/p>\r\n\r\n<h4>Solution<\/h4>\r\nBoth equations have the [latex]x[\/latex] and [latex]y[\/latex] terms on the same side of the equations. Notice that the coefficient of [latex]x[\/latex] in the second equation,\u00a0[latex]\u20131[\/latex], is the opposite of the coefficient of [latex]x[\/latex] in the first equation,\u00a0[latex]1[\/latex]. This means we can immediately add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}x+2y&amp;=-1\\\\-x+y&amp;=3\\\\ \\hline 3y&amp;=2\\end{align}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nNow that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3y=2\\hfill \\\\ \\text{ }y=\\dfrac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThen, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }-x+y=3\\hfill \\\\ \\text{ }-x+\\dfrac{2}{3}=3\\hfill \\\\ \\text{ }-x=3-\\dfrac{2}{3}\\hfill \\\\ \\text{ }-x=\\dfrac{7}{3}\\hfill \\\\ \\text{ }\\:\\:\\:\\:\\:x=-\\dfrac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solution to this system is [latex]\\left(-\\dfrac{7}{3},\\dfrac{2}{3}\\right)[\/latex].\r\n\r\nCheck the solution in the first equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{ }x+2y=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }\\left(-\\dfrac{7}{3}\\right)+2\\left(\\dfrac{2}{3}\\right)=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-\\dfrac{7}{3}+\\dfrac{4}{3}=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-\\dfrac{3}{3}=-1\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\text{ }-1=-1\\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, we can 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 class=\"alignnone\" 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=\"Two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y==1. The second line's equation is negative x + y=3.\" width=\"487\" height=\"291\" \/>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nNotice that we used the second equation to determine [latex]x[\/latex] after solving for [latex]y[\/latex], then we used the first equation to check our answer. We could have used the first equation to\u00a0determine [latex]x[\/latex] after solving for [latex]y[\/latex], then we would have used the second equation to check our answer. We use both equations as we know that the solution will work in the equation we use to determine the second variable, so we use the other one to check that the solution works in both equations.\r\n\r\nThe following video shows an example of how to use the method of elimination to solve a system of linear equations.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/M4IEmwcqR3c?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.3-1.docx\">Transcript-2.6.3-1<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nSolve the given system of equations by elimination.\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}3x+2y&amp;=6 \\\\ 3x-2y&amp;=6 \\end{aligned}\\end{equation}[\/latex]<\/p>\r\n[reveal-answer q=\"hjm239\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm239\"]\r\n\r\n[latex](x, y)=(2, 0)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nSometimes we have to do a couple of steps of algebra before we can eliminate a variable from a system and solve it. In the next example, we 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 examples\">\r\n<h3>Example 2<\/h3>\r\nSolve the given system of equations by the\u00a0elimination method.\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}3x+5y&amp;=-11\\hfill \\\\ x - 2y&amp;=11\\hfill \\end{aligned}[\/latex]<\/p>\r\n\r\n<h4>Solution<\/h4>\r\nAdding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[\/latex] in it and the second equation has [latex]x[\/latex]. So if we multiply the second equation by [latex]-3,\\text{}[\/latex] the [latex]x[\/latex]-terms will 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 \\\\ -3x+6y=-33\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Use the distributive property}.\\hfill \\end{array}[\/latex]<\/div>\r\nNow, add the equations:\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)=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ 11=11\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{True}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"wp-nocaption aligncenter\"><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=\"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\" \/><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nNext is another video example of using the elimination method to solve a system of linear equations.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/_liDhKops2w?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.3-2.docx\">Transcript-2.6.3-2<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nSolve the given system of equations by the\u00a0elimination method.\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}3x+4y&amp;=16\\hfill \\\\ x - 2y&amp;=2\\hfill \\end{aligned}[\/latex]<\/p>\r\n[reveal-answer q=\"hjm553\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm553\"][latex](x, y)=(4, 1)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\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 examples\">\r\n<h3>Example 3<\/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<h4>Solution<\/h4>\r\nIf we decide to eliminate [latex]x[\/latex], we should notice that one equation has [latex]2x[\/latex] and the other has [latex]5x[\/latex]. We need to make the coefficients opposites of one another by finding a number that 2 and 5 both divide into. The least common multiple of 2 and 5 is 10, so we will multiply both equations by a constant that will result in the coefficients of [latex]x[\/latex] being 10 and \u201310. We can achieve opposite coefficients of [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;\">\\[\r\n\\begin{array}{rl}\r\n\\textcolor{blue}{-5}(2x + 3y) &amp;= \\textcolor{blue}{-5} \\cdot (-16) \\\\\r\n-10x - 15y &amp;= 80 \\\\\r\n\\\\\r\n\\textcolor{green}{2}(5x - 10y) &amp;= \\textcolor{green}{2} \\cdot 30 \\\\\r\n10x - 20y &amp;= 60\r\n\\end{array}\r\n\\]<\/div>\r\nThen, we add the two equations together to eliminate [latex]x[\/latex]:\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 one of the original equations:\r\n\r\nEquation 1:\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 second original equation:\r\n\r\nEquation 2:\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<div><\/div>\r\n<img class=\"alignnone\" 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=\"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<\/div>\r\n<\/div>\r\nNotice that we checked our solution in the original equations, not the modified equations. This is to guarantee that we did not make an arithmetical mistake when we multiplied the original equations by a constant.\r\n\r\nHere 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>solve a system of linear equations using the elimination\u00a0method<\/h3>\r\n<ol>\r\n \t<li>Write both equations with [latex]x[\/latex]-<em>\u00a0<\/em>and [latex]y[\/latex]-terms on the left side of the equals sign and constants on the right.<\/li>\r\n \t<li>Write one equation above the other, lining up corresponding variables. Use multiplication as necessary so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation.<\/li>\r\n \t<li>Add the equations.<\/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 other variable.<\/li>\r\n \t<li>Check the solution by substituting the values into the other original equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nSolve the given system of equations by the\u00a0elimination method.\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}5x+-2y&amp;=20\\hfill \\\\ 3x + 5y&amp;=-19\\hfill \\end{aligned}[\/latex]<\/p>\r\n[reveal-answer q=\"hjm036\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm036\"][latex](x,y)=(2,-5)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next\u00a0example, we will show how to solve a system with fractions. The easiest way to solve this system is to clear the fractions first by multiplying the equation by the least common denominator.\r\n<div class=\"textbox examples\">\r\n<h3>Example 4<\/h3>\r\nSolve the given system of equations in two variables by elimination.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\dfrac{x}{3}+\\dfrac{y}{6}=3\\hfill \\\\ \\dfrac{x}{2}-\\dfrac{y}{4}=\\text{ }1\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<h4>Solution<\/h4>\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;\">\\[\r\n\\begin{array}{rl}\r\n\\textcolor{blue}{6} \\left(\\dfrac{x}{3}+\\dfrac{y}{6}\\right) &amp;= \\textcolor{blue}{6} \\cdot 3 \\\\\r\n2x + y &amp;= 18 \\\\\r\n\\\\\r\n\\textcolor{blue}{4} \\left(\\dfrac{x}{2}-\\dfrac{y}{4}\\right) &amp;= \\textcolor{blue}{4} \\cdot 1 \\\\\r\n2x - y &amp;= 4\r\n\\end{array}\r\n\\]<\/p>\r\nNow multiply the second equation by [latex]-1[\/latex] so that we can eliminate the [latex]x[\/latex]-term:\r\n<p style=\"text-align: center;\">\\[\r\n\\begin{array}{rl}\r\n\\textcolor{blue}{-1} (2x - y) &amp;= \\textcolor{blue}{-1} \\cdot 4 \\\\\r\n-2x + y &amp;= -4\r\n\\end{array}\r\n\\]<\/p>\r\nAdd the two equations to eliminate the [latex]x[\/latex]-terms 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, and solve for [latex]x[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+\\left(7\\right)=18\\hfill \\\\ \\text{ }2x=11\\hfill \\\\ \\text{ }x=\\dfrac{11}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nThe solution is [latex]\\left(\\dfrac{11}{2},7\\right)[\/latex]. Check it in the other equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x-y=4\\\\ 2(\\dfrac{11}{2})-7=4\\\\ 11-7=4 \\\\ 4=4\\end{array}[\/latex]<\/p>\r\n\r\n<div><\/div>\r\n<\/div>\r\nThe following video shows another example of using the elimination method to solve a system; this one has coefficients that are fractions.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/s3S64b1DrtQ?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.3-3.docx\">Transcript-2.6.3-3<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nSolve the given system of equations by the\u00a0elimination method.\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x+\\frac{1}{5}y&amp;=4\\hfill \\\\ \\frac{1}{2}x + y&amp;=11\\hfill \\end{aligned}[\/latex]<\/p>\r\n[reveal-answer q=\"hjm507\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm507\"][latex](x,y)=(2,10)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nThe elimination method can also identify <strong>dependent systems and inconsistent systems.\u00a0 <\/strong>If after adding to eliminate one of the terms, the resulting equation is an identity such as 0=0, the system is dependent, and if the resulting equation is a contradiction such as 5 = 8, the system is inconsistent.\r\n<div class=\"textbox examples\">\r\n<h3>EXAMPLE 5<\/h3>\r\nFind a solution to the system of equations using the\u00a0elimination method.\r\n.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x+3y=2\\\\ 3x+9y=6\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Solution<\/h4>\r\nWith the elimination method, we want to eliminate one of the variables by adding the equations.\r\n\r\nLet's 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 by addition:\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 \\\\ \\hfill3x+9y=6 \\\\ \\hfill \\text{_____________} \\\\ \\hfill 0=0 \\end{array}[\/latex]<\/p>\r\nWe end up with an identity which means that there will be an infinite number of solutions that satisfy both equations.\r\n<h4>Answer<\/h4>\r\nThere are an infinite number of solution that lie on the line [latex]x+3y=2[\/latex].\r\n\r\nNote: If we rewrite both equations in slope-intercept form, we end up with the same equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+3y=2\\hfill \\\\ \\text{ }3y=-x+2\\hfill \\\\ \\text{ }y=-\\dfrac{1}{3}x+\\dfrac{2}{3}\\hfill \\\\ 3x+9y=6\\hfill \\\\ \\text{ }9y=-3x+6\\hfill \\\\ \\text{ }y=-\\dfrac{3}{9}x+\\dfrac{6}{9}\\hfill \\\\ \\text{ }y=-\\dfrac{1}{3}x+\\dfrac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThe graph shows that both lines are the same.\r\n\r\n&nbsp;\r\n\r\n<img class=\"alignnone\" 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=\"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=\"309\" height=\"232\" \/>\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\nNotice that when we have a dependent system we write the set of all solutions as an ordered pair in term of one variable. In the example, we wrote [latex]y[\/latex] in terms of [latex]x[\/latex] and wrote the solution as\u00a0[latex]\\left(x, -\\dfrac{1}{3}x+\\dfrac{2}{3}\\right)[\/latex].\u00a0 We could also have written the solution in terms of [latex]y[\/latex]: [latex]\\left (-3y+2, y\\right )[\/latex].\r\n\r\nThe following video shows another example of solving a system that is dependent using elimination.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/NRxh9Q16Ulk?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.3-4.docx\">Transcript-2.6.3-4<\/a>\r\n\r\nThe last video example presents a system that is inconsistent; it has no solutions, which means the lines the equations represent are parallel to each other, and the system is inconsistent.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/z5_ACYtzW98?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.3-5.docx\">Transcript-2.6.3-5<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 5<\/h3>\r\nSolve the system of equations using the elimination method.\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\frac{5}{2}x-y&amp;=4\\\\-\\frac{5}{4}x+\\frac{1}{2}y&amp;=5\\end{aligned}[\/latex]<\/p>\r\n[reveal-answer q=\"hjm827\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm827\"]No solution.[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 6<\/h3>\r\nSolve the system of equations using the elimination method.\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\frac{2}{3}x-\\frac{1}{4}y&amp;=-1\\\\-6x+\\frac{9}{4}y&amp;=9\\end{aligned}[\/latex]<\/p>\r\n[reveal-answer q=\"hjm487\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm487\"]\r\n\r\nInfinite solutions on the line [latex]\\frac{2}{3}x-\\frac{1}{4}y=-1[\/latex]\r\n\r\n[latex] \\LARGE \\{\\normalsize\\left (x, \\frac{8}{3}x+4\\right )\\LARGE\\}[\/latex] or [latex]\\LARGE\\{\\normalsize\\left (\\frac{3}{8}y-\\frac{3}{2}, y\\right )\\LARGE \\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<div class=\"wrapper\">\n<div id=\"wrap\">\n<div id=\"content\" role=\"main\">\n<div id=\"post-107\" class=\"standard post-107 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div class=\"textbox learning-objectives\">\n<h1>Learning Outcomes<\/h1>\n<ul>\n<li>Solve a system of linear equations using the elimination method<\/li>\n<li>Write the general solution of a dependent system as an ordered pair in terms of one variable<\/li>\n<\/ul>\n<\/div>\n<p>A third method of\u00a0solving systems of linear equations is the\u00a0<em><strong>elimination\u00a0method<\/strong><\/em>. The idea behind this method is to add the two equations in the system together while eliminating one of the variables. The result will be a new linear equation in one variable that we can solve.\u00a0 We get this new equation by adding equations where the same term in each equation has opposite coefficients, so that when they are added the sum is zero. <span style=\"font-size: 1rem; text-align: initial;\">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 through addition.<\/span><\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Solve the given system of equations by elimination.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}x+2y&=-1 \\\\ -x+y&=3 \\end{aligned}\\end{equation}[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>Both equations have the [latex]x[\/latex] and [latex]y[\/latex] terms on the same side of the equations. Notice that the coefficient of [latex]x[\/latex] in the second equation,\u00a0[latex]\u20131[\/latex], is the opposite of the coefficient of [latex]x[\/latex] in the first equation,\u00a0[latex]1[\/latex]. This means we can immediately add the two equations to eliminate [latex]x[\/latex] without needing to multiply by a constant.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}x+2y&=-1\\\\-x+y&=3\\\\ \\hline 3y&=2\\end{align}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>Now that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3y=2\\hfill \\\\ \\text{ }y=\\dfrac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>Then, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }-x+y=3\\hfill \\\\ \\text{ }-x+\\dfrac{2}{3}=3\\hfill \\\\ \\text{ }-x=3-\\dfrac{2}{3}\\hfill \\\\ \\text{ }-x=\\dfrac{7}{3}\\hfill \\\\ \\text{ }\\:\\:\\:\\:\\:x=-\\dfrac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution to this system is [latex]\\left(-\\dfrac{7}{3},\\dfrac{2}{3}\\right)[\/latex].<\/p>\n<p>Check the solution in the first equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}\\text{ }x+2y=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }\\left(-\\dfrac{7}{3}\\right)+2\\left(\\dfrac{2}{3}\\right)=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-\\dfrac{7}{3}+\\dfrac{4}{3}=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-\\dfrac{3}{3}=-1\\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }-1=-1\\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, we can 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\" class=\"alignnone\" 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=\"Two lines that cross at the point negative seven-thirds, two-thirds. The first line's equation is x+2y==1. The second line's equation is negative x + y=3.\" width=\"487\" height=\"291\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Notice that we used the second equation to determine [latex]x[\/latex] after solving for [latex]y[\/latex], then we used the first equation to check our answer. We could have used the first equation to\u00a0determine [latex]x[\/latex] after solving for [latex]y[\/latex], then we would have used the second equation to check our answer. We use both equations as we know that the solution will work in the equation we use to determine the second variable, so we use the other one to check that the solution works in both equations.<\/p>\n<p>The following video shows an example of how to use the method of elimination to solve a system of linear equations.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/M4IEmwcqR3c?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.3-1.docx\">Transcript-2.6.3-1<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Solve the given system of equations by elimination.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}3x+2y&=6 \\\\ 3x-2y&=6 \\end{aligned}\\end{equation}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm239\">Show Answer<\/span><\/p>\n<div id=\"qhjm239\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](x, y)=(2, 0)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Sometimes we have to do a couple of steps of algebra before we can eliminate a variable from a system and solve it. In the next example, we 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 examples\">\n<h3>Example 2<\/h3>\n<p>Solve the given system of equations by the\u00a0elimination method.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}3x+5y&=-11\\hfill \\\\ x - 2y&=11\\hfill \\end{aligned}[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>Adding these equations as presented will not eliminate a variable. However, we see that the first equation has [latex]3x[\/latex] in it and the second equation has [latex]x[\/latex]. So if we multiply the second equation by [latex]-3,\\text{}[\/latex] the [latex]x[\/latex]-terms will 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 \\\\ -3x+6y=-33\\hfill & \\hfill & \\hfill & \\text{Use the distributive property}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Now, add the equations:<\/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)=11\\hfill & \\hfill & \\hfill & \\hfill \\\\ 11=11\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"wp-nocaption aligncenter\"><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=\"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\" \/><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Next is another video example of using the elimination method to solve a system of linear equations.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/_liDhKops2w?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.3-2.docx\">Transcript-2.6.3-2<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Solve the given system of equations by the\u00a0elimination method.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}3x+4y&=16\\hfill \\\\ x - 2y&=2\\hfill \\end{aligned}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm553\">Show Answer<\/span><\/p>\n<div id=\"qhjm553\" class=\"hidden-answer\" style=\"display: none\">[latex](x, y)=(4, 1)[\/latex]<\/div>\n<\/div>\n<\/div>\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 examples\">\n<h3>Example 3<\/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<h4>Solution<\/h4>\n<p>If we decide to eliminate [latex]x[\/latex], we should notice that one equation has [latex]2x[\/latex] and the other has [latex]5x[\/latex]. We need to make the coefficients opposites of one another by finding a number that 2 and 5 both divide into. The least common multiple of 2 and 5 is 10, so we will multiply both equations by a constant that will result in the coefficients of [latex]x[\/latex] being 10 and \u201310. We can achieve opposite coefficients of [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;\">\\[<br \/>\n\\begin{array}{rl}<br \/>\n\\textcolor{blue}{-5}(2x + 3y) &amp;= \\textcolor{blue}{-5} \\cdot (-16) \\\\<br \/>\n-10x &#8211; 15y &amp;= 80 \\\\<br \/>\n\\\\<br \/>\n\\textcolor{green}{2}(5x &#8211; 10y) &amp;= \\textcolor{green}{2} \\cdot 30 \\\\<br \/>\n10x &#8211; 20y &amp;= 60<br \/>\n\\end{array}<br \/>\n\\]<\/div>\n<p>Then, we add the two equations together to eliminate [latex]x[\/latex]:<\/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 one of the original equations:<\/p>\n<p>Equation 1:<\/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 second original equation:<\/p>\n<p>Equation 2:<\/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<div><\/div>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" 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=\"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<p>Notice that we checked our solution in the original equations, not the modified equations. This is to guarantee that we did not make an arithmetical mistake when we multiplied the original equations by a constant.<\/p>\n<p>Here 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>solve a system of linear equations using the elimination\u00a0method<\/h3>\n<ol>\n<li>Write both equations with [latex]x[\/latex]&#8211;<em>\u00a0<\/em>and [latex]y[\/latex]-terms on the left side of the equals sign and constants on the right.<\/li>\n<li>Write one equation above the other, lining up corresponding variables. Use multiplication as necessary so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation.<\/li>\n<li>Add the equations.<\/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 other variable.<\/li>\n<li>Check the solution by substituting the values into the other original equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Solve the given system of equations by the\u00a0elimination method.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}5x+-2y&=20\\hfill \\\\ 3x + 5y&=-19\\hfill \\end{aligned}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm036\">Show Answer<\/span><\/p>\n<div id=\"qhjm036\" class=\"hidden-answer\" style=\"display: none\">[latex](x,y)=(2,-5)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>In the next\u00a0example, we will show how to solve a system with fractions. The easiest way to solve this system is to clear the fractions first by multiplying the equation by the least common denominator.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4<\/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}\\dfrac{x}{3}+\\dfrac{y}{6}=3\\hfill \\\\ \\dfrac{x}{2}-\\dfrac{y}{4}=\\text{ }1\\hfill \\end{array}[\/latex]<\/p>\n<h4>Solution<\/h4>\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;\">\\[<br \/>\n\\begin{array}{rl}<br \/>\n\\textcolor{blue}{6} \\left(\\dfrac{x}{3}+\\dfrac{y}{6}\\right) &amp;= \\textcolor{blue}{6} \\cdot 3 \\\\<br \/>\n2x + y &amp;= 18 \\\\<br \/>\n\\\\<br \/>\n\\textcolor{blue}{4} \\left(\\dfrac{x}{2}-\\dfrac{y}{4}\\right) &amp;= \\textcolor{blue}{4} \\cdot 1 \\\\<br \/>\n2x &#8211; y &amp;= 4<br \/>\n\\end{array}<br \/>\n\\]<\/p>\n<p>Now multiply the second equation by [latex]-1[\/latex] so that we can eliminate the [latex]x[\/latex]-term:<\/p>\n<p style=\"text-align: center;\">\\[<br \/>\n\\begin{array}{rl}<br \/>\n\\textcolor{blue}{-1} (2x &#8211; y) &amp;= \\textcolor{blue}{-1} \\cdot 4 \\\\<br \/>\n-2x + y &amp;= -4<br \/>\n\\end{array}<br \/>\n\\]<\/p>\n<p>Add the two equations to eliminate the [latex]x[\/latex]-terms 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, and solve for [latex]x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x+\\left(7\\right)=18\\hfill \\\\ \\text{ }2x=11\\hfill \\\\ \\text{ }x=\\dfrac{11}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>The solution is [latex]\\left(\\dfrac{11}{2},7\\right)[\/latex]. Check it in the other equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2x-y=4\\\\ 2(\\dfrac{11}{2})-7=4\\\\ 11-7=4 \\\\ 4=4\\end{array}[\/latex]<\/p>\n<div><\/div>\n<\/div>\n<p>The following video shows another example of using the elimination method to solve a system; this one has coefficients that are fractions.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/s3S64b1DrtQ?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.3-3.docx\">Transcript-2.6.3-3<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Solve the given system of equations by the\u00a0elimination method.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}x+\\frac{1}{5}y&=4\\hfill \\\\ \\frac{1}{2}x + y&=11\\hfill \\end{aligned}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm507\">Show Answer<\/span><\/p>\n<div id=\"qhjm507\" class=\"hidden-answer\" style=\"display: none\">[latex](x,y)=(2,10)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>The elimination method can also identify <strong>dependent systems and inconsistent systems.\u00a0 <\/strong>If after adding to eliminate one of the terms, the resulting equation is an identity such as 0=0, the system is dependent, and if the resulting equation is a contradiction such as 5 = 8, the system is inconsistent.<\/p>\n<div class=\"textbox examples\">\n<h3>EXAMPLE 5<\/h3>\n<p>Find a solution to the system of equations using the\u00a0elimination method.<br \/>\n.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x+3y=2\\\\ 3x+9y=6\\end{array}[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>With the elimination method, we want to eliminate one of the variables by adding the equations.<\/p>\n<p>Let&#8217;s 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 by addition:<\/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 \\\\ \\hfill3x+9y=6 \\\\ \\hfill \\text{_____________} \\\\ \\hfill 0=0 \\end{array}[\/latex]<\/p>\n<p>We end up with an identity which means that there will be an infinite number of solutions that satisfy both equations.<\/p>\n<h4>Answer<\/h4>\n<p>There are an infinite number of solution that lie on the line [latex]x+3y=2[\/latex].<\/p>\n<p>Note: If we rewrite both equations in slope-intercept form, we end up with the same equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }x+3y=2\\hfill \\\\ \\text{ }3y=-x+2\\hfill \\\\ \\text{ }y=-\\dfrac{1}{3}x+\\dfrac{2}{3}\\hfill \\\\ 3x+9y=6\\hfill \\\\ \\text{ }9y=-3x+6\\hfill \\\\ \\text{ }y=-\\dfrac{3}{9}x+\\dfrac{6}{9}\\hfill \\\\ \\text{ }y=-\\dfrac{1}{3}x+\\dfrac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>The graph shows that both lines are the same.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" 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=\"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=\"309\" height=\"232\" \/><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<p>Notice that when we have a dependent system we write the set of all solutions as an ordered pair in term of one variable. In the example, we wrote [latex]y[\/latex] in terms of [latex]x[\/latex] and wrote the solution as\u00a0[latex]\\left(x, -\\dfrac{1}{3}x+\\dfrac{2}{3}\\right)[\/latex].\u00a0 We could also have written the solution in terms of [latex]y[\/latex]: [latex]\\left (-3y+2, y\\right )[\/latex].<\/p>\n<p>The following video shows another example of solving a system that is dependent using elimination.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/NRxh9Q16Ulk?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.3-4.docx\">Transcript-2.6.3-4<\/a><\/p>\n<p>The last video example presents a system that is inconsistent; it has no solutions, which means the lines the equations represent are parallel to each other, and the system is inconsistent.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/z5_ACYtzW98?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.6.3-5.docx\">Transcript-2.6.3-5<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 5<\/h3>\n<p>Solve the system of equations using the elimination method.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\frac{5}{2}x-y&=4\\\\-\\frac{5}{4}x+\\frac{1}{2}y&=5\\end{aligned}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm827\">Show Answer<\/span><\/p>\n<div id=\"qhjm827\" class=\"hidden-answer\" style=\"display: none\">No solution.<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 6<\/h3>\n<p>Solve the system of equations using the elimination method.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\frac{2}{3}x-\\frac{1}{4}y&=-1\\\\-6x+\\frac{9}{4}y&=9\\end{aligned}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm487\">Show Answer<\/span><\/p>\n<div id=\"qhjm487\" class=\"hidden-answer\" style=\"display: none\">\n<p>Infinite solutions on the line [latex]\\frac{2}{3}x-\\frac{1}{4}y=-1[\/latex]<\/p>\n<p>[latex]\\LARGE \\{\\normalsize\\left (x, \\frac{8}{3}x+4\\right )\\LARGE\\}[\/latex] or [latex]\\LARGE\\{\\normalsize\\left (\\frac{3}{8}y-\\frac{3}{2}, y\\right )\\LARGE \\}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1088\">\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>Adapted and revised. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Try It hjm487; hjm827; hjm507; hjm036; hjm553; hjm239. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <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 1: Solve a System of Equations Using the Elimination Method. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/M4IEmwcqR3c\">https:\/\/youtu.be\/M4IEmwcqR3c<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Solve a System of Equations Using the Elimination Method. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/_liDhKops2w\">https:\/\/youtu.be\/_liDhKops2w<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Solve a System of Equations Using Eliminations (Fractions). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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 <\/div>\n\t\t\t <\/section>","protected":false},"author":422608,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Solve a System of Equations Using the Elimination Method\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/M4IEmwcqR3c\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Solve a System of Equations Using the Elimination Method\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/_liDhKops2w\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Solve a System of Equations Using Eliminations (Fractions)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/s3S64b1DrtQ\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: System of Equations Using Elimination (Infinite Solutions)\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/NRxh9Q16Ulk\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: System of Equations Using Elimination (No Solution)\",\"author\":\"James Sousa (Mathispower4u.com) 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University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1088","chapter","type-chapter","status-publish","hentry"],"part":963,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1088","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/users\/422608"}],"version-history":[{"count":19,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1088\/revisions"}],"predecessor-version":[{"id":4654,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1088\/revisions\/4654"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/parts\/963"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/1088\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/media?parent=1088"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1088"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/contributor?post=1088"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/license?post=1088"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}