{"id":16570,"date":"2019-10-03T19:20:02","date_gmt":"2019-10-03T19:20:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-the-elimination-method-with-multiplication\/"},"modified":"2024-05-01T19:01:20","modified_gmt":"2024-05-01T19:01:20","slug":"read-the-elimination-method-with-multiplication","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-the-elimination-method-with-multiplication\/","title":{"raw":"The Elimination Method With Multiplication","rendered":"The Elimination Method With Multiplication"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the elimination method with multiplication<\/li>\r\n \t<li>Express the solution of a dependent system of equations containing two variables<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 id=\"title2\">Solve a system of equations when multiplication is necessary to eliminate a variable<\/h2>\r\nMany times adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]\r\n\r\nIf you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let\u2019s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will allow you to eliminate\u00a0the same variable in the other equation.\r\n\r\nWe do this with multiplication.\u00a0\u00a0Notice that the first equation contains the term [latex]4y[\/latex], and the second equation contains the term [latex]y[\/latex]. If you multiply the second equation by [latex]\u22124[\/latex], when you add both equations the y variables will add up to [latex]0[\/latex].\r\n\r\nThe following example takes you through all the steps to find a solution to this system.\r\n\r\n&nbsp;\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for [latex]x[\/latex] and [latex]y[\/latex].\r\n\r\n<strong>Equation A:<\/strong> [latex]3x+4y=52[\/latex]\r\n\r\n<strong>Equation B:<\/strong> [latex]5x+y=30[\/latex]\r\n\r\n[reveal-answer q=\"815377\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"815377\"]Look for terms that can be eliminated. The equations do not have any x or y terms with the same coefficients.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]\r\n\r\nMultiply the second equation by [latex]\u22124[\/latex] so they do have the same coefficient.\r\n\r\n[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,3x+4y=52\\\\\u22124\\left(5x+y\\right)=\u22124\\left(30\\right)\\end{array}[\/latex]\r\n\r\nRewrite the system and add the equations.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\,\\,\\,\\,\\,\\,\\,\\\\\u221220x\u20134y=\u2212120\\end{array}[\/latex]\r\n\r\nSolve for [latex]x[\/latex].\r\n\r\n[latex]\\begin{array}{l}\u221217x=-68\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x=4\\end{array}[\/latex]\r\n\r\nSubstitute [latex]x=4[\/latex] into one of the original equations to find y.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\3\\left(4\\right)+4y=52\\\\12+4y=52\\\\4y=40\\\\y=10\\end{array}[\/latex]\r\n\r\nCheck your answer.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\3\\left(4\\right)+4\\left(10\\right)=52\\\\12+40=52\\\\52=52\\\\\\text{TRUE}\\\\\\\\5x+y=30\\\\5\\left(4\\right)+10=30\\\\20+10=30\\\\30=30\\\\\\text{TRUE}\\end{array}[\/latex]\r\n\r\nThe answers check.\r\n<h4>Answer<\/h4>\r\nThe solution is [latex](4, 10)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\"><img class=\" wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"55\" height=\"49\" \/>Caution! \u00a0When you\u00a0use multiplication to eliminate a variable, you must multiply EACH term in the equation by the number you choose. \u00a0Forgetting to multiply every term is a common mistake.<\/div>\r\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 \\\\ 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 \\\\ -3x+6y=-33\\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Use the distributive property}.\\hfill \\end{array}[\/latex]<\/div>\r\nNow, let us 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)=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\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 in which we multiply one of the equations be a constant.\r\n\r\nhttps:\/\/youtu.be\/_liDhKops2w\r\n\r\nIt is worth demonstrating that there is more than one way to solve a system.\u00a0 Consider our first example. Instead of multiplying one equation in order to eliminate a variable when the equations were added, we could have multiplied both equations by different numbers.\r\n\r\nLet\u2019s remove the variable [latex]x[\/latex] this time. Multiply Equation A by [latex]5[\/latex] and Equation B by [latex]\u22123[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for [latex]x[\/latex] and [latex]y[\/latex].\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"40585\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"40585\"]Look for terms that can be eliminated. The equations do not have any x or y terms with the same coefficient.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]\r\n\r\nIn order to use the elimination method, you have to create variables that have the same coefficient\u2014then you can eliminate them. Multiply the top equation by [latex]5[\/latex].\r\n\r\n[latex]\\begin{array}{r}5\\left(3x+4y\\right)=5\\left(52\\right)\\\\5x+y =30\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\15x+20y=260\\,\\,\\,\\,\\,\\,\\\\5x+y=30\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]\r\n\r\nNow multiply the bottom equation by [latex]\u22123[\/latex].\r\n\r\n[latex]\\begin{array}{r}15x+20y=260\\,\\,\\,\\,\\,\\,\\,\\,\\\\-3(5x+y)=\u22123(30)\\\\15x+20y=260\\,\\,\\,\\,\\,\\,\\,\\,\\\\\u221215x\u20133y=\u221290\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]\r\n\r\nNext add the equations, and solve for [latex]y[\/latex].\r\n\r\n[latex]\\begin{array}{r}15x+20y=260\\\\\u221215x\u20133y=\\,\u201390\\\\17y=170\\\\y=\\,\\,\\,10\\end{array}[\/latex]\r\n\r\nSubstitute [latex]y=10[\/latex] into one of the original equations to find [latex]x[\/latex].\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\3x+4\\left(10\\right)=52\\\\3x+40=52\\\\3x=12\\\\x=4\\,\\,\\,\\end{array}[\/latex]\r\n\r\nYou arrive at the same solution as before.\r\n<h4>Answer<\/h4>\r\nThe solution is [latex](4, 10)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThese equations were multiplied by [latex]5[\/latex] and [latex]\u22123[\/latex] respectively, because that gave you terms that would add up to [latex]0[\/latex]. Be sure to multiply all of the terms of the equation.\r\n\r\nIn the next example, we will see that sometimes we have to multiply both numbers 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 second original equation.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill \\text{ }5x - 10y=30\\\\ \\hfill 5\\left(-2\\right)-10\\left(-4\\right)=30\\\\ \\hfill \\text{ }-10+40=30\\\\ \\hfill \\text{ }30=30\\end{array}[\/latex]<\/div>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222643\/CNX_Precalc_Figure_09_01_0062.jpg\" alt=\"A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.\" width=\"487\" height=\"366\" \/> [\/hidden-answer]\r\n\r\n<\/div>\r\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\u00a0<\/em>and <em>y<\/em>-variables on the left side of the equal sign and constants on the right.<\/li>\r\n \t<li>Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.<\/li>\r\n \t<li>Solve the resulting equation for the remaining variable.<\/li>\r\n \t<li>Substitute that value into one of the original equations and solve for the second variable.<\/li>\r\n \t<li>Check the solution by substituting the values into the other equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]38343[\/ohm_question]\r\n\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}\\dfrac{x}{3}+\\dfrac{y}{6}=3\\hfill \\\\ \\dfrac{x}{2}-\\dfrac{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(\\dfrac{x}{3}+\\dfrac{y}{6}\\right)=6\\left(3\\right)\\hfill \\\\ \\text{ }2x+y=18\\hfill \\\\ 4\\left(\\dfrac{x}{2}-\\dfrac{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=\\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>[\/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\nIt is possible to use the elimination method with multiplication and get a result that indicates no solutions or infinitely many solutions, just as with the other methods we have learned for finding solutions to systems.\u00a0 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.\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, 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 \\\\ \\hfill3x+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 slope-intercept form, we might know what the solution would look like before adding. 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=-\\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\nSee the graph below. Notice the results are the same. The general solution to the system is [latex]\\left(x, -\\dfrac{1}{3}x+\\dfrac{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, the elimination method is used to solve a system of equations. Notice that one of the equations needs to be multiplied by a\u00a0negative one first. \u00a0Additionally, this system has an infinite number of solutions.\r\n\r\nhttps:\/\/youtu.be\/NRxh9Q16Ulk\r\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<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]115192[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 class=\"no-indent\">Summary<\/h2>\r\n<p class=\"no-indent\">Multiplication can be used to set up matching terms in equations before they are combined to aid in finding a solution to a system. When using the multiplication method, it is important to multiply all the terms on both sides of the equation\u2014not just the one term you are trying to eliminate.<\/p>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the elimination method with multiplication<\/li>\n<li>Express the solution of a dependent system of equations containing two variables<\/li>\n<\/ul>\n<\/div>\n<h2 id=\"title2\">Solve a system of equations when multiplication is necessary to eliminate a variable<\/h2>\n<p>Many times adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\n<p>If you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let\u2019s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will allow you to eliminate\u00a0the same variable in the other equation.<\/p>\n<p>We do this with multiplication.\u00a0\u00a0Notice that the first equation contains the term [latex]4y[\/latex], and the second equation contains the term [latex]y[\/latex]. If you multiply the second equation by [latex]\u22124[\/latex], when you add both equations the y variables will add up to [latex]0[\/latex].<\/p>\n<p>The following example takes you through all the steps to find a solution to this system.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<p><strong>Equation A:<\/strong> [latex]3x+4y=52[\/latex]<\/p>\n<p><strong>Equation B:<\/strong> [latex]5x+y=30[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q815377\">Show Solution<\/span><\/p>\n<div id=\"q815377\" class=\"hidden-answer\" style=\"display: none\">Look for terms that can be eliminated. The equations do not have any x or y terms with the same coefficients.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\n<p>Multiply the second equation by [latex]\u22124[\/latex] so they do have the same coefficient.<\/p>\n<p>[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,3x+4y=52\\\\\u22124\\left(5x+y\\right)=\u22124\\left(30\\right)\\end{array}[\/latex]<\/p>\n<p>Rewrite the system and add the equations.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\,\\,\\,\\,\\,\\,\\,\\\\\u221220x\u20134y=\u2212120\\end{array}[\/latex]<\/p>\n<p>Solve for [latex]x[\/latex].<\/p>\n<p>[latex]\\begin{array}{l}\u221217x=-68\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,x=4\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]x=4[\/latex] into one of the original equations to find y.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\3\\left(4\\right)+4y=52\\\\12+4y=52\\\\4y=40\\\\y=10\\end{array}[\/latex]<\/p>\n<p>Check your answer.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\3\\left(4\\right)+4\\left(10\\right)=52\\\\12+40=52\\\\52=52\\\\\\text{TRUE}\\\\\\\\5x+y=30\\\\5\\left(4\\right)+10=30\\\\20+10=30\\\\30=30\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n<p>The answers check.<\/p>\n<h4>Answer<\/h4>\n<p>The solution is [latex](4, 10)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"55\" height=\"49\" \/>Caution! \u00a0When you\u00a0use multiplication to eliminate a variable, you must multiply EACH term in the equation by the number you choose. \u00a0Forgetting to multiply every term is a common mistake.<\/div>\n<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 \\\\ 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 \\\\ -3x+6y=-33\\hfill & \\hfill & \\hfill & \\text{Use the distributive property}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Now, let us 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)=11\\hfill & \\hfill & \\hfill & \\hfill \\\\ 11=11\\hfill & \\hfill & \\hfill & \\text{True}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222640\/CNX_Precalc_Figure_09_01_0052.jpg\" alt=\"A graph of two lines that cross at the point 3, negative 4. The first line's equation is 3x+5y=-11. The second line's equation is x-2y=11.\" width=\"487\" height=\"327\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Below is another video example of using the elimination method to solve a system of linear equations in which we multiply one of the equations be a constant.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 2:  Solve a System of Equations Using the Elimination Method\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/_liDhKops2w?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>It is worth demonstrating that there is more than one way to solve a system.\u00a0 Consider our first example. Instead of multiplying one equation in order to eliminate a variable when the equations were added, we could have multiplied both equations by different numbers.<\/p>\n<p>Let\u2019s remove the variable [latex]x[\/latex] this time. Multiply Equation A by [latex]5[\/latex] and Equation B by [latex]\u22123[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q40585\">Show Solution<\/span><\/p>\n<div id=\"q40585\" class=\"hidden-answer\" style=\"display: none\">Look for terms that can be eliminated. The equations do not have any x or y terms with the same coefficient.<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\n<p>In order to use the elimination method, you have to create variables that have the same coefficient\u2014then you can eliminate them. Multiply the top equation by [latex]5[\/latex].<\/p>\n<p>[latex]\\begin{array}{r}5\\left(3x+4y\\right)=5\\left(52\\right)\\\\5x+y =30\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\15x+20y=260\\,\\,\\,\\,\\,\\,\\\\5x+y=30\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Now multiply the bottom equation by [latex]\u22123[\/latex].<\/p>\n<p>[latex]\\begin{array}{r}15x+20y=260\\,\\,\\,\\,\\,\\,\\,\\,\\\\-3(5x+y)=\u22123(30)\\\\15x+20y=260\\,\\,\\,\\,\\,\\,\\,\\,\\\\\u221215x\u20133y=\u221290\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Next add the equations, and solve for [latex]y[\/latex].<\/p>\n<p>[latex]\\begin{array}{r}15x+20y=260\\\\\u221215x\u20133y=\\,\u201390\\\\17y=170\\\\y=\\,\\,\\,10\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]y=10[\/latex] into one of the original equations to find [latex]x[\/latex].<\/p>\n<p>[latex]\\begin{array}{r}3x+4y=52\\\\3x+4\\left(10\\right)=52\\\\3x+40=52\\\\3x=12\\\\x=4\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>You arrive at the same solution as before.<\/p>\n<h4>Answer<\/h4>\n<p>The solution is [latex](4, 10)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>These equations were multiplied by [latex]5[\/latex] and [latex]\u22123[\/latex] respectively, because that gave you terms that would add up to [latex]0[\/latex]. Be sure to multiply all of the terms of the equation.<\/p>\n<p>In the next example, we will see that sometimes we have to multiply both numbers 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 second original equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill \\text{ }5x - 10y=30\\\\ \\hfill 5\\left(-2\\right)-10\\left(-4\\right)=30\\\\ \\hfill \\text{ }-10+40=30\\\\ \\hfill \\text{ }30=30\\end{array}[\/latex]<\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/01222643\/CNX_Precalc_Figure_09_01_0062.jpg\" alt=\"A graph of two lines that cross the point -2,-4. The first line's equation is 2x+3y=-16. The second line's equation is 5x-10y=30.\" width=\"487\" height=\"366\" \/> <\/div>\n<\/div>\n<\/div>\n<p>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\u00a0<\/em>and <em>y<\/em>-variables on the left side of the equal sign and constants on the right.<\/li>\n<li>Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.<\/li>\n<li>Solve the resulting equation for the remaining variable.<\/li>\n<li>Substitute that value into one of the original equations and solve for the second variable.<\/li>\n<li>Check the solution by substituting the values into the other equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm38343\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=38343&theme=oea&iframe_resize_id=ohm38343&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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}\\dfrac{x}{3}+\\dfrac{y}{6}=3\\hfill \\\\ \\dfrac{x}{2}-\\dfrac{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(\\dfrac{x}{3}+\\dfrac{y}{6}\\right)=6\\left(3\\right)\\hfill \\\\ \\text{ }2x+y=18\\hfill \\\\ 4\\left(\\dfrac{x}{2}-\\dfrac{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=\\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<\/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-2\" 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>It is possible to use the elimination method with multiplication and get a result that indicates no solutions or infinitely many solutions, just as with the other methods we have learned for finding solutions to systems.\u00a0 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, 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 \\\\ \\hfill3x+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 slope-intercept form, we might know what the solution would look like before adding. 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=-\\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>See the graph below. Notice the results are the same. The general solution to the system is [latex]\\left(x, -\\dfrac{1}{3}x+\\dfrac{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, the elimination method is used to solve a system of equations. Notice that one of the equations needs to be multiplied by a\u00a0negative one first. \u00a0Additionally, this system has an infinite number of solutions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" 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><br \/>\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.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" 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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm115192\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=115192&theme=oea&iframe_resize_id=ohm115192&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 class=\"no-indent\">Summary<\/h2>\n<p class=\"no-indent\">Multiplication can be used to set up matching terms in equations before they are combined to aid in finding a solution to a system. When using the multiplication method, it is important to multiply all the terms on both sides of the equation\u2014not just the one term you are trying to eliminate.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-16570\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex: System of Equations Using Elimination (Infinite Solutions). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/NRxh9Q16Ulk\">https:\/\/youtu.be\/NRxh9Q16Ulk<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>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\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Solve a System of Equations Using the Elimination Method. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/_liDhKops2w\">https:\/\/youtu.be\/_liDhKops2w<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169554,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex: System of Equations Using Elimination (Infinite Solutions)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/NRxh9Q16Ulk\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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