{"id":694,"date":"2013-05-28T14:36:14","date_gmt":"2013-05-28T14:36:14","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=694"},"modified":"2026-03-28T11:01:17","modified_gmt":"2026-03-28T16:01:17","slug":"the-elimination-method","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/the-elimination-method\/","title":{"rendered":"What is the Elimination Method?"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_OQl4Z5QVs5M\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_OQl4Z5QVs5M\" data-source-videoID=\"OQl4Z5QVs5M\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"What is the Elimination Method? 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There are four different ways to approach these kinds of problems: graphing, substitution, elimination, and using a matrix. Today we are going to focus on how to use the elimination method to find the point where two lines cross.<\/p>\n<h2><span id=\"What_is_the_Elimination_Method\" class=\"m-toc-anchor\"><\/span>What is the Elimination Method?<\/h2>\n<p>\nThe elimination method involves exactly what it sounds like, eliminating one of our terms to solve for the other. If we are using standard linear equations, in the form \\(ax+by=c\\), we want to find a number or numbers to multiply one or both of our equations by to get rid of either the x-term or the y-term by adding or subtracting the two equations.<\/p>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nLet\u2019s look at an example so I can show you what I mean.<\/p>\n<p>Find the point of intersection for these two lines: \\(2x+3y=14\\) and \\(x\u20134y=18\\).<\/p>\n<p>First, we want to see if there is any way to eliminate one of our terms by simply adding or subtracting our equations. That is not possible for this example, so we are going to need to multiply one of our equations by some number so that we can add or subtract them and get rid of either the \\(x\\)-term or the \\(y\\)-term.<\/p>\n<p>For this problem, we can multiply the second equation by 2 and then subtract to eliminate our \\(x\\)-terms.<\/p>\n<p>So we\u2019re gonna multiply this equation by 2.<\/p>\n<div class=\"examplesentence\">\\(2(x-4y=18)\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich will give us: \\(2x-8y=36\\). And we\u2019re gonna subtract \\(2x+3y=14\\).<\/p>\n<div class=\"examplesentence\">\\(2x-2x=0\\), \\(-8y-3y=-11y\\), and \\(36-14=22\\).<br \/>\n\\(-11y=22\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf we divide by -11 on both sides, we get that \\(y=-2\\).<\/p>\n<p>Solving this new equation gives us that \\(y=-2\\). And to find our \\(x\\)-term, we choose either equation, and plug in this \\(y\\)-value to solve for \\(x\\).<\/p>\n<p>I\u2019m going to use the equation \\(x\u20134y=18\\).<\/p>\n<div class=\"examplesentence\">\\(x-4(-2)=18\\)<br \/>\n\\(x+8=18\\)<\/div>\n<p>\n&nbsp;<br \/>\nSubtract 8 from both sides.<\/p>\n<div class=\"examplesentence\">\\(x=10\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur \\(x\\)-value for this system is 10. Since we are solving for the point of intersection between these two functions, we want to write our final answer as a coordinate point. Our point of intersection for this example is \\((10, -2)\\).<\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nLet\u2019s look at another example.<\/p>\n<div class=\"examplesentence\">\\(3x+4y=6\\)<br \/>\n\\(x-y=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nFor this example, we can multiply our bottom equation by either 3 or 4 to eliminate one of our terms. I am going to multiply by 4.<\/p>\n<div class=\"examplesentence\">\\(4(x-y=2)\\)<br \/>\n\\(4x-4y=8\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, since I want my \\(y\\)-terms to cancel out, I am going to add my two equations and solve for \\(x\\).<\/p>\n<p>And then we\u2019ll add the two equations.<\/p>\n<div class=\"examplesentence\">\\(3x+4x=7x\\), \\(4y-4y=0\\), and \\(6+8=14\\).<br \/>\n\\(7x=14\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf we divide by 7 on both sides, we get that \\(x=2\\).<\/p>\n<p>Now that we know \\(x=2\\), we can choose either equation to plug this value into and solve for <em>y<\/em>. I am going to use \\(x\u2013y=2\\).<\/p>\n<div class=\"examplesentence\">\\(2-y=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nSubtract 2 from both sides.<\/p>\n<div class=\"examplesentence\">\\(-y=0\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd divide by -1 on both sides.<\/p>\n<div class=\"examplesentence\">\\(y=0\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow all we have to do is write our answer in the form of a coordinate point, \\((2, 0)\\).<\/p>\n<h3><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nI want to try one more example before we go.<\/p>\n<div class=\"examplesentence\">\\(2x-7y=19\\)<br \/>\n\\(3x-11y=4\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis example is a little bit different from our other ones because there isn\u2019t one number we can easily multiply an equation by to eliminate a term, not without using fractions. So for this example, we are going to multiply both equations by a number in order to cancel something out. <\/p>\n<p>If we multiply our first equation by 3 and our second equation by 2, we will get a \\(6x\\) term in both equations that we can then cancel out. Let\u2019s try it.<\/p>\n<div class=\"examplesentence\" style=\"padding-bottom: -1em;\">\n<table class=\"NBTable\" style=\"margin: auto; width: 85%;\">\n<tbody>\n<tr>\n<td>\\(3(2x-7y=19)\\)<\/td>\n<td>\\(2(3x-11y=4)\\)<\/td>\n<\/tr>\n<tr>\n<td>\\(6x-21y=27\\)<\/td>\n<td>\\(6x-22y=8\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>\n&nbsp;<br \/>\nNow we subtract our two equations so that we have \\(6x-6x=0\\), and then \\(-21y-(-22y)=y\\) (it\u2019s like adding that \\(22y\\)). Now we do \\(57-8=49\\).<\/p>\n<div class=\"examplesentence\">\\(y=49\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis gives us that \\(y=49\\). Now we\u2019re going to use this information to plug \\(y\\) into either equation and solve for \\(x\\). I\u2019m going to use the equation \\(2x\u20137y=19\\).<\/p>\n<div class=\"examplesentence\">\\(2x-7(49)=19\\)<br \/>\n\\(2x-343=19\\)<br \/>\n\\(2x=362\\)<br \/>\n\\(x=181\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo \\(x=181\\). Our final answer then is the point \\((181,49)\\).<\/p>\n<hr>\n<p>\nRemember, when using the elimination method to solve systems of equations, the whole point is to try and get rid of one of your terms by multiplying one of the equations by a constant and then adding or subtracting your equations.<\/p>\n<p>I hope this video was helpful. Thanks for watching, and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Elimination_Method_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Elimination Method Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nUse the elimination method to solve the following system of equations:<\/p>\n<div class=\"yellow-math-quote\">\\(3x-y=12\\)<br \/>\n\\(2x+y=13\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\((5,2)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\((4,3)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">\\((5,3)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\((6,7)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>If the equations are added together, the \\(y\\)-terms will be eliminated. Adding the two equations results in \\(5x=25\\).<\/p>\n<p>Now, solve for \\(x\\) by dividing both sides by 5:<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(5x \\div 5 =5\\)<br \/>\n\\(25 \\div 5 = 5\\)<\/p>\n<p>Now plug in 5 for \\(x\\) into one of the original equations. If the first equation is used, \\(3x-y=12\\) becomes \\(3(5)-y=12\\), which simplifies to \\(15-y=12\\).<\/p>\n<p>Subtract 15 from both sides, and then divide both sides by \u22121 to get \\(y=3\\). We know that \\(x=5\\) and \\(y=3\\). This means that the lines will intersect at the point \\((5,3)\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nDetermine the solution to the following system of equations using the elimination method:<\/p>\n<div class=\"yellow-math-quote\">\\(3x+y=10\\)<br \/>\n\\(-4x-2y=2\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\((12,-23)\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\((12,-23)\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\((11,-20)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-4\">\\((11,-23)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>The simplest way to use the elimination method to solve this system of equations is to multiply the first equation by 2 so that the \\(y\\)-values cancel out.<\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\(2(3x+y=10)\\)<br \/>\n\\(6x+2y=20\\)<\/p>\n<p>Notice how \\(+2y\\) and \\(-2y\\) will cancel out if the two equations are added. <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Elimination-Method-Addition-Problem-1.svg\" alt=\"Two linear equations, 6x + 2y = 20 and -4x - 2y = 2, are added together to yield 2x = 22.\" width=\"154.16\" height=\"74.62\" class=\"aligncenter size-full wp-image-287669\"  role=\"img\" \/><\/p>\n<p>Solve for \\(x\\) by dividing both sides by 2.<\/p>\n<p style=\"text-align: center; line-height: 50px;\">\\(\\dfrac{2x}{2}=\\dfrac{22}{2}\\)<br \/>\n\\(x=11\\)<\/p>\n<p>Now plug 11 in for \\(x\\) into either of the original equations. If we use the first original equation, \\(3x+y=10\\) becomes \\(3(11)+y=10\\).<\/p>\n<p style=\"text-align: center;\">\\(33+y=10\\)<\/p>\n<p>Subtract 33 from both sides.<\/p>\n<p style=\"text-align: center; line-height: 35px;\">\n\\(33-33+y=10-33\\)<br \/>\n\\(y=-23\\)<\/p>\n<p>We know that \\(x=11\\) and \\(y=-23\\). Therefore, the lines will intersect at the point \\((11,-23)\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve the system of equations using the elimination method.<\/p>\n<div class=\"yellow-math-quote\">\\(3x+2y=7\\)<br \/>\n\\(5x-3y=37\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\((5,4)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-2\">\\((5,-4)\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\((-5,4)\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\((-5,-4)\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>One approach is to multiply the first equation by 3 and the second equation by 2. This will create a situation where the \\(y\\)-variables cancel out when the equations are added.<\/p>\n<p>Let\u2019s try the first approach:<\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\(3x+2y=7\\) becomes \\(9x+6y=21\\)<br \/>\n\\(5x-3y=37\\) becomes \\(10x-6y=74\\)\n<\/p>\n<p>Now, add the two new equations.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Elimination-Method-Addition-Problem-2.svg\" alt=\"Two linear equations, 9x + 6y = 21 and 10x - 6y = 74, are added to yield 19x = 95.\" width=\"154.16\" height=\"74.62\" class=\"aligncenter size-full wp-image-287672\"  role=\"img\" \/><\/p>\n<p>Solve for \\(x\\) by dividing both sides by 19.<\/p>\n<p style=\"text-align: center; line-height: 50px\">\\(\\dfrac{19x}{19}=\\dfrac{95}{19}\\)<br \/>\n\\(x=5\\)<\/p>\n<p>Now plug in 5 for \\(x\\) in either equation. If we use the original first equation \\(3x+2y=7\\) becomes \\(3(5)+2y=7\\).<\/p>\n<p style=\"text-align: center;\">\\(15+2y=7\\)<\/p>\n<p>Subtract 15 from both sides, and then divide both sides by 2.<\/p>\n<p style=\"text-align: center; line-height: 45px;\">\n\\(15-15+2y=7-15\\)<br \/>\n\\(2y=-8\\)<br \/>\n\\(\\large{\\frac{2y}{2}}\\normalsize{=}\\large{\\frac{-8}{2}}\\)<br \/>\n\\(y=-4\\)<\/p>\n<p>The value of \\(x\\) is 5, and the value of \\(y\\) is \u22124. This means the lines will intersect at the point \\((5,-4)\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nWillow spends $188 at the pet store. She purchases 7 small dog toys and 11 large dog toys. Sybil spends $236 at the same pet store. She purchases 13 small toys and 11 large toys. Determine the price of a small toy and a large toy. <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">Small toys cost $10 and large toys cost $12.<\/div><div class=\"PQ\"  id=\"PQ-4-2\">Small toys cost $8 and large toys cost $10.<\/div><div class=\"PQ\"  id=\"PQ-4-3\">Small toys cost $6 and large toys cost $14.<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-4\">Small toys cost $8 and large toys cost $12. <\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>If we let \\(x\\) equal the cost of small dog toys and \\(y\\) equal the cost of large dog toys, then we can set up the following equation to describe Willow\u2019s spending:<\/p>\n<p style=\"text-align: center\">\\(7x+11y=188\\)<\/p>\n<p>Similarly, we can set up an equation to describe Sybil\u2019s spending:<\/p>\n<p style=\"text-align: center\">\\(13x+11y=236\\)<\/;p><\/p>\n<p>The two equations can be set up as a system of equations, and then the elimination method can be used to determine the value of \\(x\\) and \\(y\\). If the second equation is subtracted from the first equation, the \\(y\\)-values will cancel out.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Elimination-Method-Addition-Problem-3.2.svg\" alt=\"A system of equations: 7x + 11y = 188 and 13x + 11y = 236, with subtraction shown resulting in -6x = -48.\" width=\"180.4\" height=\"77.9\" class=\"aligncenter size-full wp-image-287678\"  role=\"img\" \/><\/p>\n<p>Divide both sides by \u22126.<\/p>\n<p style=\"text-align: center; line-height: 50px;\">\n\\(\\dfrac{-6x}{-6}=\\dfrac{-48}{-6}\\)<br \/>\n\\(x=8\\)<\/p>\n<p>Now plug \\(x=8\\) into one of the original equations. If the first equation is used, \\(7x+11y=188\\) becomes \\(7(8)+11y=188\\).<\/p>\n<p style=\"text-align: center;\">\\(56+11y=188\\)<\/p>\n<p>Subtract 56 from both sides, and then divide both sides by 11.<\/p>\n<p style=\"text-align: center; line-height: 45px;\">\\(56-56+11y=188-56\\)<br \/>\n\\(11y=132\\)<br \/>\n\\(\\large{\\frac{11y}{11}}\\normalsize{=}\\large{\\frac{132}{11}}\\)<br \/>\n\\(y=12\\)<\/p>\n<p>Since \\(x\\) is the cost of small dog toys and \\(y\\) is the cost of large dog toys, this means that the small dog toys cost $8 and the large dog toys cost $12.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nA track and field athlete is practicing for an upcoming track meet. On Monday the athlete practices the long jump 6 times and the triple jump 9 times. This takes the athlete 39 minutes. On Tuesday the athlete practices the long jump 4 times and the triple jump 4 times. This takes the athlete 20 minutes. How long does it take the athlete to practice each event one time? <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">It takes the athlete 1 minute to practice the long jump and 3 minutes to practice the triple jump. <\/div><div class=\"PQ\"  id=\"PQ-5-2\">It takes the athlete 1 minutes to practice the long jump and 4 minutes to practice the triple jump. <\/div><div class=\"PQ\"  id=\"PQ-5-3\">It takes the athlete 4 minutes to practice the long jump and 4 minutes to practice the triple jump. <\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-4\">It takes the athlete 2 minutes to practice the long jump and 3 minutes to practice the triple jump.<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>We will let \\(x\\) equal the number of minutes it takes to practice the long jump and \\(y\\) equal the number of minutes it takes to practice the triple jump.<\/p>\n<ol style=\"list-style-type: none; margin-left: 1.2em\">\n<li style=\"margin-bottom: 1em\"><span style=\"font-weight: 600\">Equation #1:<\/span> On Monday, the athlete practices the long jump 6 times and the triple jump 9 times. This takes the athlete 39 minutes: \\(6x+9y=39\\).<\/li>\n<li><span style=\"font-weight: 600\">Equation #2:<\/span> On Tuesday, the athlete practices the long jump 4 times and the triple jump 4 times. This takes the athlete 20 minutes: \\(4x+4y=20\\).<\/li>\n<\/ol>\n<p>If we want the \\(x\\)-terms to cancel, then we can multiply the first equation by 2, and multiply the second equation by 3.<\/p>\n<p style=\"text-align: center; line-height: 35px;\">\nEquation #1: \\(2(6x+9y=39)\\) becomes \\(12x+18y=78\\)<br \/>\nEquation #2: \\(3(4x+4y=20)\\) becomes \\(12x+12y=60\\)\n<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/02\/Elimination-Method-Addition-Problem-4.svg\" alt=\"Two linear equations, 12x + 18y = 78 and 12x + 12y = 60, are subtracted to yield 6y = 18.\" width=\"165.64\" height=\"84.46\" class=\"aligncenter size-full wp-image-287681\"  role=\"img\" \/><\/p>\n<p>Divide both sides by 6.<\/p>\n<p style=\"text-align: center; line-height: 50px;\">\n\\(\\dfrac{6y}{6}=\\dfrac{18}{6}\\)<br \/>\n\\(y=3\\)<\/p>\n<p>Now plug \\(y=3\\) into one of the original equations. If we use the first equation, \\(6x+9y=39\\) becomes \\(6x+9(3)=39\\). <\/p>\n<p style=\"text-align: center;\">\\(6x+27=39\\)<\/p>\n<p>Subtract 27 from both sides and then divide both sides by 6. <\/p>\n<p style=\"text-align: center; line-height: 50px;\">\n\\(6x+27-27=39-27\\)<br \/>\n\\(6x=12\\)<br \/>\n\\(\\dfrac{6x}{6}=\\dfrac{12}{6}\\)<br \/>\n\\(x=2\\)<\/p>\n<p>We know \\(x=2\\), which means the athlete spends 2 minutes on the long jump. We also know \\(y=3\\), which means the athlete spends 3 minutes on the triple jump.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/algebra-i\/\">Return to Algebra I Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Algebra I Videos<\/p>\n","protected":false},"author":1,"featured_media":187037,"parent":0,"menu_order":39,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-694","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-math-advertising-group","7":"page_category-systems-of-equations-and-their-solutions-videos","8":"page_type-video","9":"content_type-practice-questions","10":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/694","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=694"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/694\/revisions"}],"predecessor-version":[{"id":247594,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/694\/revisions\/247594"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/187037"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=694"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}