{"id":701,"date":"2013-05-28T14:39:24","date_gmt":"2013-05-28T14:39:24","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=701"},"modified":"2026-03-25T11:28:29","modified_gmt":"2026-03-25T16:28:29","slug":"comparison-of-methods-for-solving-systems","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/comparison-of-methods-for-solving-systems\/","title":{"rendered":"Top 3 Methods for Solving Systems of Equations"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_SWe7R-ciFAs\" 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_SWe7R-ciFAs\" data-source-videoID=\"SWe7R-ciFAs\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Top 3 Methods for Solving Systems of Equations Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Top 3 Methods for Solving Systems of Equations\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_SWe7R-ciFAs:hover {cursor:pointer;} img#videoThumbnailImage_SWe7R-ciFAs {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/07\/updated-top-3-methods-for-solving-systems-of-equations-64bee3694b0dc.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_SWe7R-ciFAs\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_SWe7R-ciFAs\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Top 3 Methods for Solving Systems of Equations\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_SWe7R-ciFAs\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_SWe7R-ciFAs\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_SWe7R-ciFAs\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction acZ_Function() {\n  var x = document.getElementById(\"acZ\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"acZ_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"acZ\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Substitution,_Elimination,_and_Augmented_Matrix\" class=\"smooth-scroll\">Substitution, Elimination, and Augmented Matrix<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Elimination_Method_Example\" class=\"smooth-scroll\">Elimination Method Example<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Substitution_Method_Example\" class=\"smooth-scroll\">Substitution Method Example<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Augmented_Matrix_Example\" class=\"smooth-scroll\">Augmented Matrix Example<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Frequently_Asked_Questions\" class=\"smooth-scroll\">Frequently Asked Questions<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Systems_of_Equations_Practice_Questions\" class=\"smooth-scroll\">Systems of Equations Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"FAQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"FAQs\">FAQs<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hey, guys! Welcome to this video over comparing different methods for solving a system of equations.<\/p>\n<p>If you recall, a <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/systems-of-equations\/\">system of equations<\/a> is when you have more than one equation with unknown variables in a given problem. So, in order to solve that problem, you need to be able to find the value of all the variables in each equation.<\/p>\n<h2><span id=\"Substitution,_Elimination,_and_Augmented_Matrix\" class=\"m-toc-anchor\"><\/span>Substitution, Elimination, and Augmented Matrix<\/h2>\n<p>\nThere are three different ways that you could do this: the substitution method, elimination method, and using an augmented matrix.<\/p>\n<p>In this video, I\u2019m assuming that you already know how to perform each method, so I want to spend a lot of time explaining not how to do them but rather when to use each method.<\/p>\n<p>First, I will verbally tell you when to use each method, then I will write out three different examples, and we will decide together which method is most efficient for each system.<\/p>\n<h3><span id=\"When_to_Use_the_Substitution_Method\" class=\"m-toc-anchor\"><\/span>When to Use the Substitution Method<\/h3>\n<p>\nYou should use the substitution method when one of the <strong>variables<\/strong> in one of your equations has already been isolated (it has a coefficient of 1).<\/p>\n<h3><span id=\"When_to_Use_the_Elimination_Method\" class=\"m-toc-anchor\"><\/span>When to Use the Elimination Method<\/h3>\n<p>\nYou should use the elimination method when the same variables in all of the equations share the same coefficient, or when they share the same but negative coefficient.<\/p>\n<h3><span id=\"When_to_Use_an_Augmented_Matrix\" class=\"m-toc-anchor\"><\/span>When to Use an Augmented Matrix<\/h3>\n<p>\nYou would use an augmented matrix when the substitution and elimination method are either impractical or impossible altogether.<\/p>\n<p>Now, let\u2019s look at three different systems, and use what we\u2019ve just learned to think through which method is most useful for each system.<\/p>\n<div class=\"examplesentence\"><strong>EXAMPLE SYSTEM #1<\/strong><br \/>\n\\(5x &#8211; 58y = -883\\)<br \/>\n\\(-5x + 2y = -13\\)<\/div>\n<p>\n&nbsp;<\/p>\n<div class=\"examplesentence\"><strong>EXAMPLE SYSTEM #2<\/strong><br \/>\n\\(9x + 4y = 65\\)<br \/>\n\\(x &#8211; 18y = -2\\)<\/div>\n<p>\n&nbsp;<\/p>\n<div class=\"examplesentence\"><strong>EXAMPLE SYSTEM #3<\/strong><br \/>\n\\(2x + 7y &#8211; 3z = 47\\)<br \/>\n\\(x &#8211; 4y + 8z = -33\\)<br \/>\n\\(7x + 2y +10z = 11\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, what we will do is go through each system, decide which method would be most efficient, and then solve with that method.<\/p>\n<h2><span id=\"Elimination_Method_Example\" class=\"m-toc-anchor\"><\/span>Elimination Method Example<\/h2>\n<p>\nAlright, let\u2019s look at this first equation.<\/p>\n<div class=\"examplesentence\">\\(5x &#8211; 58y = -883\\)<br \/>\n\\(-5x + 2y = -13\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, thinking back to the explanation I gave on when to use each method, notice what I said about elimination: \u201cYou should use the elimination method when the same variables in all of the equations share the same coefficient, or when they share the same but negative coefficient.\u201d<\/p>\n<p>Well this exact thing is true in the case of this particular system. So, let\u2019s solve this system using <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/the-elimination-method\/\">elimination<\/a>.<\/p>\n<div class=\"examplesentence\">\\(-56y = -896\\)<br \/>\n\\(y = 16\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, we plug our \\(y\\)-variable back into one of the original equations. I\u2019ll plug it into the first.<\/p>\n<div class=\"examplesentence\">\\(5x &#8211; 58(16) = -883\\)<br \/>\n\\(5x &#8211; 928 = -883\\)<br \/>\n\\(5x = 45\\)<br \/>\n\\(x = 9\\)<\/div>\n<p>\n&nbsp;<br \/>\nGreat, so we\u2019ve solved this system using elimination, because our same two variables had the same coefficient or when they share the same but negative coefficient (like in our case).<\/p>\n<h2><span id=\"Substitution_Method_Example\" class=\"m-toc-anchor\"><\/span>Substitution Method Example<\/h2>\n<p>\nLet\u2019s move on to system #2.<\/p>\n<div class=\"examplesentence\">\\(9x + 4y = 65\\)<br \/>\n\\(x &#8211; 18y = -2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAlright, so again, let&#8217;s think back on what was said in our explanation on when to use each method. Recall what was said about substitution: \u201cYou should use the substitution method when one of the variables in one of your equations has already been isolated.\u201d<\/p>\n<p>Well, such is the case with this system. Our \\(x\\)-variable in our second equation has a coefficient of 1. So, let\u2019s solve this system using substitution.<\/p>\n<div class=\"examplesentence\">\\(9(18y &#8211; 2) + 4y = 65\\)<br \/>\n\\(162y &#8211; 18 + 4y = 65\\)<br \/>\n\\(166y = 83\\)<br \/>\n\\(y = \\frac{1}{2}\\)<br \/>\n\\(x = 18(\\frac{1}{2}) &#8211; 2\\)<br \/>\n\\(x = 7\\)<\/div>\n<p>\n&nbsp;<br \/>\nThat was very simple to solve using substitution. Remember, the signifier to help you know when to use it is if one of the equations has a variable that is already isolated.<\/p>\n<h2><span id=\"Augmented_Matrix_Example\" class=\"m-toc-anchor\"><\/span>Augmented Matrix Example<\/h2>\n<p>\nLet\u2019s look at our last system, system #3.<\/p>\n<div class=\"examplesentence\">\\(2x + 7y &#8211; 3z = 47\\)<br \/>\n\\(x &#8211; 4y + 8z = -33\\)<br \/>\n\\(7x + 2y +10z = 11\\)<\/div>\n<p>\n&nbsp;<br \/>\nRemember, what we said about when to use an augmented matrix. Well, right now is a good time. Using elimination or substitution for that matter would take a lot more work than would using an augmented matrix. <\/p>\n<p>So, let\u2019s set up our matrix and solve.<\/p>\n<div class=\"examplesentence\" style=\"font-size: 80%;\">\\(\\begin{bmatrix} \\left.\\begin{matrix} 2&#038; 7&#038; -3\\\\ 1&#038; -4&#038; 8\\\\ 7&#038; 2&#038; 10 \\end{matrix}\\right| &#038; \\begin{matrix} 47\\\\ -33\\\\ 11 \\end{matrix} \\end{bmatrix}\\)<br \/>\n\\(R_{1}\\leftrightarrow R_{2}\\)<br \/>\n<span style=\"line-height:140px\">\\(\\begin{bmatrix} \\left.\\begin{matrix} 1&#038; -4&#038; 8\\\\ 2&#038; 7&#038; -3\\\\ 7&#038; 2&#038; 10 \\end{matrix}\\right| &#038; \\begin{matrix} -33\\\\ 47\\\\ 11 \\end{matrix} \\end{bmatrix}\\)<br \/>\n<span style=\"line-height:140px\">\\(\\begin{matrix} \\\\ -2R_{1}+R_{2}=R_{2}\\\\ -7R_{1}+R_{3}=R_{3} \\end{matrix}\\begin{bmatrix} \\left.\\begin{matrix} 1&#038; -4&#038; 8\\\\ 0&#038; 15&#038; -19\\\\ 0&#038; 30&#038; -46 \\end{matrix}\\right| &#038; \\begin{matrix} -33\\\\ 113\\\\ 242 \\end{matrix} \\end{bmatrix}\\)<br \/>\n<span style=\"line-height:140px\">\\(\\begin{matrix} \\\\ \\frac{R_{2}}{15}=R_{2}\\\\ \\end{matrix}\\begin{bmatrix} \\left.\\begin{matrix} 1&#038; -4&#038; 8\\\\ 0&#038; 1&#038; \\frac{-19}{15}\\\\ 0&#038; 30&#038; -46 \\end{matrix}\\right| &#038; \\begin{matrix} -33\\\\ \\frac{113}{15}\\\\ 242 \\end{matrix} \\end{bmatrix}\\)<br \/>\n<span style=\"line-height:140px\">\\(\\begin{matrix} 4R_{2}+R_{1}=R_{1}\\\\ \\\\ -30R_{2}+R_{3}=R_{3} \\end{matrix}\\begin{bmatrix} \\left.\\begin{matrix} 1&#038; 0&#038; \\frac{44}{15}\\\\ 0&#038; 1&#038; \\frac{-19}{15}\\\\ 0&#038; 0&#038; -8 \\end{matrix}\\right| &#038; \\begin{matrix} \\frac{-43}{15}\\\\ \\frac{113}{15}\\\\ 16 \\end{matrix} \\end{bmatrix}\\)<br \/>\n<span style=\"line-height:140px\">\\(\\begin{matrix} \\\\ \\\\ \\frac{R_{3}}{-8}=R_{3} \\end{matrix}\\begin{bmatrix} \\left.\\begin{matrix} 1&#038; 0&#038; \\frac{44}{15}\\\\ 0&#038; 1&#038; \\frac{-19}{15}\\\\ 0&#038; 0&#038; 1 \\end{matrix}\\right| &#038; \\begin{matrix} \\frac{-43}{15}\\\\ \\frac{113}{15}\\\\ -2 \\end{matrix} \\end{bmatrix}\\)<br \/>\n<span style=\"line-height:140px\">\\(\\begin{matrix} -\\frac{44}{15}R_{3}+R_{1}=R_{1}\\\\ \\frac{19}{15}R_{3}+R_{2}=R_{2}\\\\ \\end{matrix}\\begin{bmatrix} \\left.\\begin{matrix} 1&#038; 0&#038; 0\\\\ 0&#038; 1&#038; 0\\\\ 0&#038; 0&#038; 1 \\end{matrix}\\right| &#038; \\begin{matrix} 3\\\\ 5\\\\ -2 \\end{matrix} \\end{bmatrix}\\begin{bmatrix} x\\\\ y\\\\ z \\end{bmatrix}\\)<\/p>\n<p style=\"margin: -2em;\">\\(x=3\\), \\(y=5\\), \\(z=-2\\)<\/p>\n<\/div>\n<p>\n&nbsp;<br \/>\nI hope that this video over the comparison of methods for solving systems was helpful for you.<\/p>\n<p>See you guys next time!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"FAQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Frequently_Asked_Questions\" class=\"m-toc-anchor\"><\/span>Frequently Asked Questions<\/h2>\n<div class=\"faq-list\">\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">What is a system of equations?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>Systems of equations are two or more equations that can be used to solve one another.<\/p>\n<p>For example: <\/p>\n<p style=\"text-align:center;\"> \\(\\begin{align*}3x+4 &#038;= y\\\\ 2x+3y &#038;= -9\\end{align*}\\)<\/p>\n<\/p><\/div>\n<\/p><\/div>\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">How do you do the substitution method?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>The substitution method of solving linear equations involves substituting one equation for a variable in the other equation, solving for one of the variables, and then using that variable and one of the original equations to solve for the other variable.<\/p>\n<div class=\"lightbulb-example-2\"><span class=\"lightbulb-icon\">\ud83d\udca1<\/span><span class=\"faq-example-question\">Example: Solve both \\(2x + 3y = 15\\) and <span style=\"white-space:nowrap\">\\(y = 2x + 1\\).<\/span><\/span><\/p>\n<hr style=\"padding: 0; margin-top: -0.2em; margin-bottom: 1.2em\">First, substitute \\(2x + 1\\) for \\(y\\): <\/p>\n<p style=\"text-align:center;\"> \\(2x + 3(2x + 1) = 15\\)<\/p>\n<p>Next, solve for \\(x\\):  <\/p>\n<p style=\"text-align: center\"> \\(2x + 3(2x + 1) = 15\\)<\/p>\n<p style=\"text-align: center\">\\(2x + 6x + 3 = 15\\)<\/p>\n<p style=\"text-align: center\">\\(8x + 3 = 15\\)<\/p>\n<p style=\"text-align: center\">\\(8x = 12\\)<\/p>\n<p style=\"text-align: center\">\\(x = 3\\)<\/p>\n<p>  Solve for \\(y\\) by substituting 3 for \\(x\\) in either original equation <\/p>\n<p style=\"text-align:center;\"> \\(y = 2(3) + 1 = 6 + 1 = 7\\)<\/p>\n<p style=\"margin-bottom: 0em\">The solution to these two equations is the point \\((3,7)\\).<\/p>\n<\/div>\n<\/p><\/div>\n<\/p><\/div>\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">How do you do the elimination method?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>To use the elimination method of solving systems of equations, manipulate one of the equations so it can be added to, or subtracted from, the other equation where one variable will cancel out. Then, solve for the other variable. Finally, use that variable to solve for the one that originally was eliminated.<\/p>\n<div class=\"lightbulb-example-2\"><span class=\"lightbulb-icon\">\ud83d\udca1<\/span><span class=\"faq-example-question\">Example: \\(\\begin{align*}3x-2y &#038;= 14\\\\ 6x-7y &#038;= 11\\end{align*}\\)<\/span><\/p>\n<hr style=\"padding: 0; margin-top: -0.2em; margin-bottom: 1.2em\">First, multiply the top equation by -2. <\/p>\n<p style=\"text-align:center;\"> \\(-2(3x-2y=14)\\)<br \/> \\(-6x+4y=-28\\)<\/p>\n<p> Then, add the two equations.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/01\/Solving-Systems-example-1.svg\" alt=\"A system of equations being solved by elimination: -6x + 4y = -28 and 6x + 7y = 11, resulting in -3y = -17.\" width=\"156\" height=\"84\" class=\"aligncenter size-full wp-image-286441\"  role=\"img\" \/><\/p>\n<p>Solve for \\(y\\) by dividing by -3 on both sides. <\/p>\n<p style=\"text-align:center;\"> \\(y=\\dfrac{17}{3}\\)<\/p>\n<p> Then, substitute \\(y\\) in either original equation to solve for \\(x\\). <\/p>\n<p style=\"text-align:center;\">\\(3x-2(\\frac{17}{3})=14\\)<\/p>\n<p style=\"text-align:center;\">\\(3x-\\frac{34}{3}=14\\)<\/p>\n<p style=\"text-align:center;\">\\(3x-\\frac{34}{3}=\\frac{42}{3}\\)<\/p>\n<p style=\"text-align:center;\">\\(3x=\\frac{76}{3}\\)<\/p>\n<p style=\"text-align:center;\">\\(x=\\frac{76}{9}\\)<\/p>\n<p> The solution to this system is the point (\\(\\frac{76}{9},\\frac{17}{3}\\)).<\/p><\/div>\n<\/p><\/div>\n<\/p><\/div>\n<div class=\"qa_wrap\">\n<div class=\"q_item text_bold\">\n<h4 class=\"letter\">Q<\/h4>\n<p style=\"line-height: unset;\">What is an augmented matrix?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>An augmented matrix is formed by appending the entries from one matrix onto the end of another.<\/p>\n<div class=\"lightbulb-example-2\"><span class=\"lightbulb-icon\">\ud83d\udca1<\/span><span class=\"faq-example-question\">Example: <\/p>\n<p style=\"text-align:center;\"> \\(M=\\begin{bmatrix}1&#038;2\\\\3&#038;4\\end{bmatrix}\\: \\: \\: I=\\begin{bmatrix}1&#038;0\\\\0&#038;1\\end{bmatrix}\\)<\/p>\n<p><\/span><\/p>\n<hr style=\"padding: 0; margin-top: -0.2em; margin-bottom: 1.2em\">Augmented Matrix:<\/p>\n<p style=\"text-align:center;\">\\(\\begin{bmatrix}1&#038;2&#038;| \\: \\:1&#038;0\\\\3&#038;4&#038;|\\: \\:0&#038;1\\end{bmatrix}\\)<\/p>\n<\/div>\n<\/p><\/div>\n<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Systems_of_Equations_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Systems of Equations 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 \/>\nSolve the system of equations by substitution:<\/p>\n<div class=\"yellow-math-quote\">\\(2x-y=12\\)<br \/>\n\\(x-y=3\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\((3,2)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\((6,9)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-3\">\\((12,3)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-4\">\\((9,6)\\)<\/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>Let\u2019s start by solving the second equation for \\(x\\).<\/p>\n<p>The equation \\(x\u2212y=3\\) becomes \\(x=3+y\\).<\/p>\n<p>Now that we have isolated \\(x\\), we can substitute this in for \u201c\\(x\\)\u201d in the other equation in order to solve for \\(y\\). <\/p>\n<p>This means \\(2x\u2212y=12\\) becomes \\(2(3+y)\u2212y=12\\).<\/p>\n<p>From here, we can solve for \\(y\\) because we are now only dealing with one variable. <\/p>\n<p>The equation \\(2(3+y)\u2212y=12\\) becomes \\(y=6\\).<\/p>\n<p>Now that we have solved for \\(y\\), we can plug this value into one of the original equations in order to solve for \\(x\\). Let\u2019s use the first original equation (\\(2x\u2212y=12\\)).<\/p>\n<p>The equation \\(2x\u2212y=12\\) becomes \\(2x\u22126=12\\) and when we isolate the variable \\(x\\), we end up with \\(x=9\\).<\/p>\n<p>Our solution is the ordered pair \\((9,6)\\).<\/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 \/>\nSolve the system of equations using substitution.<\/p>\n<div class=\"yellow-math-quote\">\\(2x+3y=12\\)<br \/>\n\\(x+y=5\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\((2,3)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-2\">\\((3,2)\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\((5,12)\\)<\/div><div class=\"PQ\"  id=\"PQ-2-4\">\\((12,5)\\)<\/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>Let\u2019s start by solving the second equation for \\(x\\). <\/p>\n<p>The equation \\(x+y=5\\) becomes \\(x=\u2212y+5\\).<\/p>\n<p>Now that we have isolated \\(x\\), we can substitute this in for \\(x\\) in the other equation in order to solve for \\(y\\).<\/p>\n<p>This means that \\(2x+3y=12\\) becomes \\(2(\u2212y+5)+3y=12\\).<\/p>\n<p>From here, we can solve for \\(y\\) because we are now only dealing with one variable.<\/p>\n<p>This means that \\(2(\u2212y+5)+3y=12\\) becomes \\(y=2\\).<\/p>\n<p>Now that we have solved for \\(y\\), we can plug this value into one of the original equations in order to solve for \\(x\\). Let\u2019s use the second original equation (\\(x+y=5\\)).<\/p>\n<p>This means that \\(x+y=5\\) becomes \\(x+(2)=5\\) and when we isolate the variable \\(x\\), we end up with \\(x=3\\). <\/p>\n<p>Our solution is the ordered pair \\((3,2)\\).<\/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 by elimination: <\/p>\n<div class=\"yellow-math-quote\">\\(2x+3y=15\\)<br \/>\n\\(x\u22123y=3\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\((1,6)\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\((3,5)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-3\">\\((6,1)\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\((5,3)\\)<\/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>Elimination is a useful strategy for this system of equations because we can see that the terms \\(3y\\) and \\(-3y\\) will cancel out.<\/p>\n<p>Let\u2019s begin the process by adding the two equations.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/01\/Solving-Systems-second-example.svg\" alt=\"A math equations on a white background.\" width=\"133\" height=\"81\" class=\"aligncenter size-full wp-image-286447\"  role=\"img\" \/><\/p>\n<p>Since \\(3y\\) and \\(-3y\\) cancel out, we are left with \\(3x=18\\), which simplifies to \\(x=6\\).<\/p>\n<p>From this point, we can simply plug in 6 for \\(x\\) in either equation in order to solve for \\(y\\).<\/p>\n<p>Let\u2019s use the second original equation: \\(x\u22123y=3\\).<\/p>\n<p>The equation \\(x\u22123y=3\\) becomes \\((6)\u22123y=3\\)<\/p>\n<p>From here, we can isolate the variable \\(y\\). <\/p>\n<p style=\"text-align: center\">\\(y=1\\)<\/p>\n<p>The solution is the ordered pair \\((6,1)\\). <\/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 \/>\nThe admission fee at an amusement park is $2.50 for children and $4.50 for adults. On Monday 2,000 people entered the amusement park and $8,000 was collected. How many children and how many adults went to the amusement park on Monday? <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">1,000 children and 1,000 adults <\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-2\">500 children and 1,500 adults <\/div><div class=\"PQ\"  id=\"PQ-4-3\">400 children and 1,600 adults <\/div><div class=\"PQ\"  id=\"PQ-4-4\">800 children and 1,200 adults <\/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>We can solve for the number of adults and children by setting up a system of equations.<\/p>\n<p>Let\u2019s set up two equations, one for the number of people and one for the cost. Let\u2019s have \\(a\\) represent adults and \\(c\\)represent children. <\/p>\n<p style=\"text-align:center;\">\\(a+c=2,000\\)<br \/>\n\\(4.5a+2.5c=8,000\\)<\/p>\n<p>From here, it appears that the substitution method would be most efficient because we have coefficients of 1. Let\u2019s solve the first equation for \\(a\\). Now we have \\(a=2,000\u2013c\\).<\/p>\n<p>We can plug in this value for \\(a\\) into the other equation.<\/p>\n<p style=\"text-align:center;\">\n\\(4.5(2,000-c)+2.5c=8,000\\)<\/p>\n<p>From here, we can isolate the variable \\(c\\).<\/p>\n<p style=\"text-align:center;\">\n\\(c=500\\)<\/p>\n<p>Now that we have solved for \\(c\\), we can plug 500 in for \\(c\\) in either of the original equations. Let\u2019s use the first equation. <\/p>\n<p>The equation \\(a+c=2,000\\) becomes \\(a+500=2,000\\), which means \\(a=1,500\\).<\/p>\n<p>The number of children is 500 and the number of adults is 1,500.<\/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 potter is selling bowls and cups at an art fair. This morning he sold 30 bowls and four cups and made a total of $1,040. Later in the afternoon, he sold eight bowls for a total of $256. Find the price per bowl and cup.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-5-1\">Bowl = $32, Cup = $20 <\/div><div class=\"PQ\"  id=\"PQ-5-2\">Bowl = $35, Cup = $22 <\/div><div class=\"PQ\"  id=\"PQ-5-3\">Bowl = $12, Cup = $30 <\/div><div class=\"PQ\"  id=\"PQ-5-4\">Bowl = $34, Cup = $24 <\/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 can solve for the number of cups and bowls by setting up a system of equations.<\/p>\n<p>Two equations that match the scenario would be \\(30B+4C=1,040\\) and \\(8B=256\\), where \\(B\\) represents bowls and \\(C\\) represents cups. <\/p>\n<p>Let\u2019s solve the second equation for \\(B\\).<\/p>\n<p>The equation \\(8B=256\\) becomes \\(B=32\\).<\/p>\n<p>From this point, we can substitute 32 into the other equation for \\(B\\).<\/p>\n<p>This means that \\(30B+4C=1,040\\) becomes \\(30(32)+4C=1,040\\).<\/p>\n<p>We can now solve for \\(C\\).<\/p>\n<p style=\"text-align: center\">\\(C=20\\)<\/p>\n<p>Bowls cost $32 and cups cost $20.<\/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":186233,"parent":0,"menu_order":42,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-701","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\/701","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=701"}],"version-history":[{"count":5,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/701\/revisions"}],"predecessor-version":[{"id":245080,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/701\/revisions\/245080"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/186233"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=701"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}