{"id":692,"date":"2013-05-28T14:35:22","date_gmt":"2013-05-28T14:35:22","guid":{"rendered":"http:\/\/www.mometrix.com\/academy\/?page_id=692"},"modified":"2026-03-26T09:30:31","modified_gmt":"2026-03-26T14:30:31","slug":"systems-of-equations","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/systems-of-equations\/","title":{"rendered":"How to Solve Systems of Equations"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_cZh_FhY-Ors\" 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_cZh_FhY-Ors\" data-source-videoID=\"cZh_FhY-Ors\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"How to Solve Systems of Equations Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"How to Solve Systems of Equations\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_cZh_FhY-Ors:hover {cursor:pointer;} img#videoThumbnailImage_cZh_FhY-Ors {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/184-solving-systems-of-equations-substitution-elimination-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_cZh_FhY-Ors\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_cZh_FhY-Ors\" 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\",\"How to Solve Systems of Equations\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_cZh_FhY-Ors\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_cZh_FhY-Ors\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_cZh_FhY-Ors\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction MAb_Function() {\n  var x = document.getElementById(\"MAb\");\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=\"MAb_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=\"MAb\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Solving_Systems_of_Equations\" class=\"smooth-scroll\">Solving Systems of Equations<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Equations_as_Scenarios\" class=\"smooth-scroll\">Equations as Scenarios<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Substitution\" class=\"smooth-scroll\">Substitution<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Elimination\" class=\"smooth-scroll\">Elimination<\/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=\"#System_of_Equations_Practice_Problems\" class=\"smooth-scroll\">System of Equations Practice Problems<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"overview\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"overview\">Overview<\/label><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=\"overview-spoiler\">\n<p>An <strong>equation<\/strong> is a statement that two mathematical <strong>expressions <\/strong>are equal. Expressions represent quantities and can be strictly numerical or contain some combination of numerical terms and terms containing variables. For instance, \\(6 + 3\\) is a numerical expression, and \\(2x \u2013 7\\) is an expression that contains a variable.<\/p>\n<p>Equations are like sentences, and they can be evaluated as true or false. For example, \\(2 \u00d7 5 = 10\\) is a true statement because the quantity \\(2\\times 5\\) is equal to the quantity 10. On the other hand, \\(25 = 4\\times 5\\) is a false statement, because the quantity 25 is not equal to the quantity \\(4\\times 5\\).<\/p>\n<p>When equations contain variables, the job of solving is to discover the value(s) of the variables that make the equation true. For example, \\(x = 10\\) is only true when \\(x\\) has the value of 10, while \\(2x = 10\\) is only true when \\(x\\) has the value of 5.<\/p>\n<p>Typically, equations are approached from the standpoint of \u201cbalancing,\u201d \u201cperforming inverse operations,\u201d and \u201cisolating the variable.\u201d<\/p>\n<h2><span id=\"Solving_Systems_of_Equations\" class=\"m-toc-anchor\"><\/span>Solving Systems of Equations<\/h2>\n<p>Here\u2019s an example of how the process of solving equations might be approached.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-53663\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations1.png\" sizes=\"auto, (max-width: 871px) 100vw, 871px\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations1.png 871w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations1-300x284.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations1-768x727.png 768w\" alt=\"Infographic showing how to solve 3 b plus 6 equals 24\" width=\"871\" height=\"825\" \/><\/p>\n<h2><span id=\"Equations_as_Scenarios\" class=\"m-toc-anchor\"><\/span>Equations as Scenarios<\/h2>\n<p>Expressions often represent quantities that are related in a certain way. Some common examples are\\(l\\), \\(x\\), and \\(w\\), which often represent the area of a rectangle, and \\(\\sqrt{(x_2 \u2013 x_1)^2 + (y_2 \u2013 y_1)^2} \\), which often represents the distance between two points on the coordinate plane. Sometimes, we need to create our own equations based on given information.<\/p>\n<p><strong>Example:<\/strong> Max is Ruby\u2019s little brother. Ruby is half of three times Max\u2019s age, and the sum of their ages is 25 years. How old is each sibling?<\/p>\n<p>Let the expression \\(a\\) represent Max\u2019s age. Then Ruby\u2019s age is represented by the expression \\(\\frac{3a}{2}\\) and the equation we need to solve is \\(a+ \\frac{3a}{2} = 25\\). In other words, Max\u2019s age plus Ruby\u2019s age is 25.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-53667\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations2.png\" sizes=\"auto, (max-width: 901px) 100vw, 901px\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations2.png 901w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations2-300x295.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations2-768x754.png 768w\" alt=\"Infographic showing how equations can be scenarios\" width=\"901\" height=\"885\" \/><\/p>\n<p>Sometimes, there are multiple ways to visualize an equation. Sometimes, the order we use depends on how we see the terms. For example, let\u2019s look at a couple of different ways we can visualize and solve \\(\\frac{3x}{4} = 30 \\):<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-53697\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations3.png\" sizes=\"auto, (max-width: 943px) 100vw, 943px\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations3.png 943w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations3-300x117.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations3-768x300.png 768w\" alt=\"Infograph showing how to divide 3 x divided by 4\" width=\"943\" height=\"368\" \/><\/p>\n<p>Visualizing the problem this way will lead you down two different paths to solving it:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-53665\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations4.png\" sizes=\"auto, (max-width: 939px) 100vw, 939px\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations4.png 939w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations4-287x300.png 287w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations4-768x802.png 768w\" alt=\"Infographic showing how to solve 3 x over 4 equals 30\" width=\"939\" height=\"981\" \/><\/p>\n<p>Finally, let\u2019s look at a couple of different ways we can visualize and solve \\(\\frac{3}{4x} = 30:\\)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-large wp-image-53664\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations5-853x1024.png\" sizes=\"auto, (max-width: 853px) 100vw, 853px\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations5-853x1024.png 853w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations5-250x300.png 250w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations5-768x922.png 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2019\/08\/Systems-of-Equations5.png 947w\" alt=\"Infographic showing how to solve 3 x over 4 equals 30\" width=\"853\" height=\"1024\" \/><\/p>\n<\/div>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hey guys! Welcome to this video over systems of equations.<\/p>\n<p>A system of equations is a group of two or more <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/linear-equations\/\">equations<\/a>, and each of the equations within the group have at least one unknown variable.<\/p>\n<p>When given a system of equations, the goal is to find the value for each of the unknown variables.<\/p>\n<p>In this video, we will discuss two tools to help you solve for the unknown variables: substitution, and elimination.<\/p>\n<h2><span id=\"Substitution\" class=\"m-toc-anchor\"><\/span>Substitution<\/h2>\n<p>\nSubstitution is a way of solving the system by getting rid of all of the variables, except for one, then solving for that equation.<\/p>\n<p>The best time to use substitution is when you have a variable that has a coefficient of 1 or -1. The reason why is, because if it has a coefficient of 1 or -1, then you don\u2019t need to undo multiplication or division, you just need to undo addition or subtraction in order to isolate a variable.<\/p>\n<p>There are three steps that you need to follow in order to be able to solve the system using substitution.<\/p>\n<div class=\"transcriptcallout\" style=\"text-align: left;\">\n<ol style=\"margin-bottom: 0em; margin-left: 2em;\">\n<li style=\"margin-bottom: 10px;\">Solve for your \\(x\\) or \\(y\\) in one equation.<\/li>\n<li style=\"margin-bottom: 10px;\">Plug your \\(x\\) or \\(y\\) in that you solved for into the other equation, then solve.<\/li>\n<li>Use the number that you get when you solve to solve for the other variable, or variables, depending on how many equations that you have.<\/li>\n<\/ol>\n<\/div>\n<p>\n&nbsp;<br \/>\nSo let\u2019s get started!<\/p>\n<h3><span id=\"Example\" class=\"m-toc-anchor\"><\/span>Example<\/h3>\n<div class=\"examplesentence\">\\(4x+3y=2\\)<br \/>\n\\(x-9y=-19\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn this example, we can see that our second equation has a variable with 1 as the coefficient. So, that lets us know to solve for \\(x\\). Now, you can actually solve for any of the variables. It will just always be easier to solve for one that has a coefficient of 1 or -1.<\/p>\n<p>Now, to do this we just add \\(9y\\) to both sides.<\/p>\n<div class=\"examplesentence\">\\(x= 9y -19\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, we\u2019ve done what our first step tells us to do and we solved for one of our variables. Step two now tells us to plug in the variable that we have solved for into our other equation, then solve. So we&#8217;re going to plug our \\(x\\) into \\(4x+3y = 2\\). Because we know that \\(x=9y -19\\), we&#8217;ll plug this into our \\(x\\)-value here and then solve. <\/p>\n<div class=\"examplesentence\">\\(4(9y-19)+3y=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow that we have it down to one variable, we are able to solve for the value of that variable, so in this case, \\(y\\).<\/p>\n<p>Let\u2019s rewrite this by multiply our 4 by everything inside our parentheses. <\/p>\n<div class=\"examplesentence\">\\(36y &#8211; 76 + 3y = 2\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we can add our 76 to both sides. You can do this multiple ways. If you wanted to add your \\(y\\)s together first, then do that. I\u2019m just doing what is easier in my mind.<\/p>\n<p>Once I add my 76 to both sides and add my \\(y\\)s together, I get \\(39y= 78\\). <\/p>\n<p>Now, I divide both sides by 39.<\/p>\n<div class=\"examplesentence\">\\(y=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nUse the number that you get when you solve to solve for the other variable, or variables.<\/p>\n<p>So now all we have to do is take our \\(y\\)-value and plug it into one of our original equations. It doesn&#8217;t matter which one! You can plug it into the first or the second, but I am going to plug it into the second equation, just because.<\/p>\n<p>So that gives you:<\/p>\n<div class=\"examplesentence\">\\(x -9(2) = -19\\)<br \/>\n\\(x &#8211; 18 = -19\\)<br \/>\n\\(x=-1\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd we\u2019re done! We have found the value of both of our variables using substitution.<\/p>\n<p>Now, let\u2019s take a look at how to solve for a system using elimination.<\/p>\n<h2><span id=\"Elimination\" class=\"m-toc-anchor\"><\/span>Elimination<\/h2>\n<p>\nThe reason this tool is called elimination is because you add together the two equations in order to eliminate one of your variables.<\/p>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example 1<\/h3>\n<div class=\"examplesentence\">\\(3x  &#8211;  4y  = -27\\)<br \/>\n\\(7x  + 4y  = 57\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe can tell just by looking at it, that our \\(y\\)s will cancel out. So once we add the two equations together. We have \\(10x = 30\\).<\/p>\n<p>And we can divide both sides by 10: \\(x=3\\)<\/p>\n<p>We can take our \\(x\\)-value and plug it into either of our original equations. I\u2019ll plug it into our first one here. So, we have:<\/p>\n<div class=\"examplesentence\">\\(3(3) &#8211; 4y = -27\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, and then I\u2019ll rewrite this:<\/p>\n<div class=\"examplesentence\">\\(9 &#8211; 4y = -27\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, we need to move our 9 to the other side by subtracting 9.<\/p>\n<div class=\"examplesentence\">\\(-4y=-36\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we divide both sides by our -4 here.<\/p>\n<div class=\"examplesentence\">\\(y=9\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo now we&#8217;ve solved and found both of our variables. That was a good example to learn how elimination works, but it may not always be that our terms cancel so easily.<\/p>\n<p>Like in this example:<\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example 2<\/h3>\n<div class=\"examplesentence\">\\(9x &#8211; 3y = -57\\)<br \/>\n\\(2x + 6y = 34\\)<\/div>\n<p>\n&nbsp;<br \/>\nNone of our terms cancel right off the bat, so we&#8217;ll need to do a little manipulation in order to get them to do what we want them to do.<\/p>\n<p>So, what could we do to get our terms to cancel? Well, there are a couple of things, but the easiest, as it appears to me, is to multiply our first equation by 2. That will allow us to cancel our \\(y\\)s, because we will be adding a \\(-6\\) and a \\(+6\\). So, let\u2019s do that.<\/p>\n<p>We have this equation (\\(9x &#8211; 3y = -57\\)) all being multiplied by 2, so let\u2019s rewrite that and we get:<\/p>\n<div class=\"examplesentence\">\\(2(9x &#8211; 3y = -57)\\)<br \/>\n\\(18x &#8211; 6y = -114\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, we need to take this and add it to our other equation so we can get our \\(y\\)s to cancel out. So that would give us \\(20x = -80\\).<\/p>\n<p>Now, we need to divide both of our sides by 20, which gives us \\(x=-4\\).<\/p>\n<p>So, now that we know that \\(x=-4\\), we can plug in our \\(-4\\) where our \\(x\\)s are in either one of the original equations.<\/p>\n<p>I\u2019ll plug it into the second one. So we have:<\/p>\n<div class=\"examplesentence\">\\(2(-4) + 6y = 34\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd I\u2019ll simplify this and rewrite it.<\/p>\n<div class=\"examplesentence\">\\(-8 + 6y =34\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow I&#8217;ll move and my 8 to both sides: \\(6y=42\\).<\/p>\n<p>Now we divide both sides by 6 here: \\(y=7\\)<\/p>\n<p>So our final answer to this problem is this:<\/p>\n<div class=\"examplesentence\">\\(x=-4 \\text{ and } y=7\\)<\/div>\n<p>\n&nbsp;<br \/>\nI hope that this video on how to solve a system of equations using substitution and elimination has been helpful.<\/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;\">How do you solve systems of equations? <\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>A system of equations refers to a set or collection of equations that share the same variables. The goal of solving systems of equations is to identify the location where the lines intersect when the equations are graphed. The \\((x,y)\\) ordered pair of this intersection point is considered the solution of the system.<\/p>\n<p>There are generally four methods for solving systems of equations: substitution, elimination, graphing, and matrices. The method that you select depends on the structure of the equations. For example, if the equations are already in slope-intercept form \\((y=mx+b)\\), then the graphing method is a convenient option. <\/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;\">What are the four methods for solving systems of equations? <\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>When solving systems of equations, you have a few options as far as which method you choose. The method that you select will depend on the structure of the equations.<\/p>\n<p>Selecting the strategy that is most convenient and efficient is of course the goal. The more you work with systems, the more trained your eye will become in selecting the best method.<\/p>\n<p>The four methods to choose from include substitution, elimination, graphing, and matrices. Each approach is quite different, but every strategy will ultimately arrive at the same goal, which is to locate the \\((x,y)\\) ordered pair where the lines intersect. <\/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;\">Why do we use systems of equations? <\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>Systems of equations have many useful real-world applications. For example, whenever you are given two unknown values, as well as information that connects the two unknown values, a system of equations can be set up in order to solve for the two unknowns.<\/p>\n<p>Let\u2019s say the admission fee to an amusement park is $20 for children and $30 for adults. Let\u2019s also say that 215 people entered the park today, and that the total admission fees collected today was $3,500. We can solve for the number of children and adult tickets sold by setting up a system of equations.<\/p>\n<p>We have two unknown values, and we have enough information connecting the two values to set up our equations. To solve this problem, we first need to select a strategy (substitution, elimination, graphing, or matrices).<\/p>\n<p>Solving systems of equations is a skill that has many valuable applications. This is why the process of solving systems is considered a major branch of algebra.<\/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;\">What is the difference between an equation and a system of equations? <\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>A linear equation can be graphed by plugging in values for \\(x\\) and then calculating the corresponding \\(y\\)-values. These \\(x\\)&#8211; and \\(y\\)-coordinates can be graphed and will eventually form a line.<\/p>\n<p>When there is more than one equation, and they share the same variables, the equations are called a \u201csystem\u201d.<\/p>\n<p>An example of a system of equations would be:<\/p>\n<p style=\"text-align: center\">\\(y=2x+5\\)<br \/>\\(3y=4x-8\\)<\/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;\">What careers use systems of equations? <\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>The exciting thing about learning how to solve systems of equations is that the skill can be applied to almost any career! Anytime you are faced with a scenario where you have two unknown values, and enough information to compare the two values, setting up and solving a system of equations will allow you to identify the amount of each value.<\/p>\n<p>This is often useful in the field of finance, business, and sales. Specifically careers that involve calculations with costs, revenue, and profit. <\/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 are systems of equations used in real life? <\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>This skill has applications in many areas of our daily lives. For example, we can use a system of equations to determine how many calories we burn using different machines at the gym. If we use the rowing machine for 30 minutes and the treadmill for 20 minutes, and we burn a total of 430 calories, this can be set up as equation #1: \\(30r+20w=430\\). If we go to the gym the following day and use the rowing machine for 50 minutes and the treadmill for 10 minutes, burning a total of 600 calories, this can be set up as equation #2: \\(50r+10w=600\\).<\/p>\n<p>We have two unknown values (how many calories we burn per minute on each machine), and we have enough information comparing both of the values. This means that we can solve the problem by setting up and solving a system of equations.<\/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 solve systems of equations by substitution? <\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>Solving systems of equations by substitution follows three basic steps.<\/p>\n<ol>\n<li>Solve one equation for one of the variables.<\/li>\n<li>Substitute this expression into the other equation, and solve for the missing variable.<\/li>\n<li>Substitute this answer into one of the equations in order to solve for the other variable.<\/ul>\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 elimination in algebra?<\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>Systems of equations can be solved using elimination. This method follows three basic steps. <\/p>\n<ol>\n<li style=\"margin-bottom: 12px\">Manipulate the equations so that one variable will cancel out when the equations are added or subtracted.<\/li>\n<li style=\"margin-bottom: 12px\">Add or subtract the equations so that one variable is eliminated. Solve for the remaining variable.<\/li>\n<li>Plug the solved variable back into one of the original equations in order to solve for the other variable.<\/li>\n<\/ol>\n<p>For example, the following system can be solved by elimination.<\/p>\n<p style=\"text-align: center\">\\(3x-4y=-5\\)<br \/>\\(5x+8y=-1\\)<\/p>\n<h4 style=\"margin-bottom: 0.2em\">Step 1<\/h4>\n<p>In this case, we can multiply the first equation by 2 so that the \\(y\\)-variables will cancel out when the equations are added. In this case, \\(3x-4y=-5\\) becomes \\(6x-8y=-10\\).<\/p>\n<h4 style=\"margin-bottom: 0.2em\">Step 2<\/h4>\n<p> Add or subtract the equations so that one variable is eliminated. Solve for the remaining variable.<\/p>\n<p style=\"text-align: center\">\\(6x-8y=-10\\)<br \/>\\(5x+8y=-1\\)<\/p>\n<p>When the equations are added, the result is \\(11x=-11\\). Now we can solve for \\(x\\) by dividing both sides of the equation by 11. <\/p>\n<p style=\"text-align: center\">\\(x=-1\\)<\/p>\n<h4 style=\"margin-bottom: 0.2em\">Step 3<\/h4>\n<p>Let\u2019s plug \u20131 in for \\(x\\) in the first equation. <\/p>\n<p style=\"text-align: center\">\\(3x-4y=-5\\)<\/p>\n<p> The equation now becomes \\(3(-1)-4y=-5\\), which simplifies to \\(y=\\frac{1}{2}\\).<\/p>\n<p>This means \\(x=-1\\) and \\(y=\\frac{1}{2}\\), which means that the solution to the system of equations is the ordered pair \\((-1, \\frac{1}{2})\\).<\/p>\n<p>If the two original equations were graphed, this ordered pair is where the two lines would intersect.<\/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;\">Can you subtract systems of equations? <\/p>\n<\/p><\/div>\n<div class=\"a_item\">\n<h4 class=\"letter text_bold\">A<\/h4>\n<p>When solving a system of equations using the elimination method, the equations can be added or <em>subtracted<\/em> in order to eliminate a variable. For example, if you notice that both equations have the same variable with the same sign, then elimination using subtraction is likely the most efficient method.<\/p>\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=\"System_of_Equations_Practice_Problems\" class=\"m-toc-anchor\"><\/span>System of Equations Practice Problems<\/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 substitution method to solve the following system of equations:<\/p>\n<div class=\"yellow-math-quote\">\\(y=3x+2\\)<br \/>\n\\(y=5x+6\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\((-1,-1)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\((1,11)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">\\((-2,-4)\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\((4,-2)\\)<\/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>To solve this system using the substitution method, we start by noting that both equations are already solved for \\(y\\):<\/p>\n<p style=\"text-align: center\">\\(y = 3x + 2\\)<br \/>\n\\(y = 5x + 6\\)<\/p>\n<p>Because both expressions equal \\(y\\), we can set them equal to each other. Substitute \\(3x + 2\\) for \\(y\\) in the second equation:<\/p>\n<p style=\"text-align: center\">\\(3x + 2 = 5x + 6\\)<\/p>\n<p>Next, solve this equation for \\(x\\). Subtract \\(5x\\) from both sides:<\/p>\n<p style=\"text-align: center\">\\(3x + 2 &#8211; 5x = 6\\)\n<\/p>\n<p>This simplifies to \\(-2x + 2 = 6\\). Now subtract 2 from both sides:<\/p>\n<p style=\"text-align: center\">\\(-2x = 4\\)<\/p>\n<p>Then, divide both sides by \u20132 to get \\(x = -2\\).<\/p>\n<p>With the value of \\(x\\) found, substitute \\(-2\\) back into either original equation to find \\(y\\). Using the first equation:<\/p>\n<p style=\"text-align: center\">\\(y = 3(-2) + 2\\)<br \/>\n\\(y = -6 + 2\\)<br \/>\n\\(y = -4\\)<\/p>\n<p>Therefore, the solution to the system is the ordered pair \\((-2,-4)\\).<\/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 \/>\nUse the elimination method to solve the following system of equations:<\/p>\n<div class=\"yellow-math-quote\">\\(2x-3y=-9\\)<br \/>\n\\(x+4y=23\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\((0,3)\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\((7,4)\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\((5,3)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-4\">\\((3,5)\\)<\/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>We can multiply the second equation by \u20132 so that the \\(x\\)-variables are eliminated when the two equations are combined. <\/p>\n<p style=\"text-align: center;\">\\(-2x+4y=23\\)<br \/>\n\\(-2x-8y=-46\\)<\/p>\n<p>Combine the two equations to eliminate \\(x\\). Then solve the resulting equation for \\(y\\).<\/p>\n<p style=\"text-align: center\">\\(2x &#8211; 3y = -9\\)<br \/>\n\\(-2x &#8211; 8y = -46\\)<\/p>\n<p style=\"text-align: center\">\\(-11y = -55\\)<\/p>\n<p>Divide both sides of the equation by \\(-11\\) and simplify to find \\(y = 5\\).<\/p>\n<p>Now we can substitute the \\(y\\)-value into either of the original equations to find the \\(x\\)-value. Substituting \\(y=5\\) into the second equation, we get:<\/p>\n<p style=\"text-align: center;\">\\(x+4(5)=23\\)<br \/>\n\\(x+20=23\\)<br \/>\n\\(x+20-20=23-20\\)<br \/>\n\\(x=3\\)<\/p>\n<p>The answer as an ordered pair is \\((3,5)\\).<\/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 \/>\nUse the elimination method to solve the following system of equations:<\/p>\n<div class=\"yellow-math-quote\">\\(x+\\frac{1}{3}y=7\\)<br \/>\n\\(3x-y=9\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\((6,5)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-2\">\\((5,6)\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\((8,-3)\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\((1,-6)\\)<\/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>We can multiply the first equation by \\(3\\) so that the \\(y\\)-variables are eliminated when the two equations are combined. After multiplying, the first equation becomes:<\/p>\n<p style=\"text-align: center\">\\(3(x + \\tfrac{1}{3}y) = 21\\)<br \/>\n\\(3x + y = 21\\)<\/p>\n<p>The second equation remains (\\(3x &#8211; y = 9\\)).<\/p>\n<p>Combine the two equations to eliminate \\(y\\), then solve the resulting equation for \\(x\\). Adding the equations gives:<\/p>\n<p style=\"text-align: center\">\\((3x + y) + (3x &#8211; y) = 21 + 9\\)<br \/>\n\\(6x = 30\\)<\/p>\n<p>Divide both sides of the equation by \\(6\\) and simplify to get \\(x = 5\\).<\/p>\n<p>Now we can substitute the \\(x\\)-value into either of the original equations to find the \\(y\\)-value. Substituting \\(x = 5\\) into the first equation, we get:<\/p>\n<p style=\"text-align: center\">\\(5 + \\tfrac{1}{3}y = 7\\)<\/p>\n<p>Subtract 5 from both sides of the equation:<\/p>\n<p style=\"text-align: center\">\\(\\tfrac{1}{3}y = 2\\)<\/p>\n<p>Multiply both sides by \\(3\\) to solve for \\(y\\), which results in \\(y = 6\\).<\/p>\n<p>The answer is the ordered pair \\((5,6)\\).<\/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 sum of two numbers is 75. Four times the smaller number equals the larger number. What are the two numbers?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">12 and 48<\/div><div class=\"PQ\"  id=\"PQ-4-2\">10 and 10<\/div><div class=\"PQ\"  id=\"PQ-4-3\">30 and 45<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-4\">15 and 60<\/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>Let \\(x\\) be the smaller number and \\(y\\) be the larger number. The sum of the two numbers is 75, so we can write the equation \\(x + y = 75\\).<\/p>\n<p>Since the larger number is four times the smaller number, we also have \\(y = 4x\\).<\/p>\n<p>We can solve this system of linear equations to find each number.<\/p>\n<p>Since the second equation is solved for \\(y\\), we will use the substitution method by substituting its value into the first equation. Replacing \\(y\\) with \\(4x\\) gives:<\/p>\n<p style=\"text-align: center\">\\(x + 4x = 75\\)<\/p>\n<p>Now, solve the equation for \\(x\\). Combine like terms on the left-hand side:<\/p>\n<p style=\"text-align: center\">\\(5x = 75\\)<\/p>\n<p>Divide both sides of the equation by 5:<\/p>\n<p style=\"text-align: center\">\\(x = 15\\)<\/p>\n<p>Next, substitute the value of \\(x\\) into either of the original equations to find the value of \\(y\\). Substituting \\(x = 15\\) into the first equation, we have:<\/p>\n<p style=\"text-align: center\">\\(15 + y = 75\\)<\/p>\n<p>Subtract 15 from both sides of the equation:<\/p>\n<p style=\"text-align: center\">\\(y = 60\\)<\/p>\n<p>Since \\(x = 15\\) and \\(y = 60\\), the two numbers are 15 and 60.<\/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 local fast-food restaurant sells hotdogs for $2.75 each and hamburgers for $4.50 each. During a given lunch rush, the restaurant sells a combined total of 225 hotdogs and hamburgers for a combined revenue of $881.25. How many hotdogs and how many hamburgers did the restaurant sell during the lunch rush?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">100 hot dogs and 125 hamburgers<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-2\">75 hot dogs and 150 hamburgers<\/div><div class=\"PQ\"  id=\"PQ-5-3\">150 hot dogs and 75 hamburgers<\/div><div class=\"PQ\"  id=\"PQ-5-4\">85 hot dogs and 140 hamburgers<\/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>Let \\(x\\) be the number of hot dogs sold and \\(y\\) be the number of hamburgers sold. The restaurant sold a combined total of 225 items, so we can write the equation \\(x + y = 225\\).<\/p>\n<p>The revenue generated by the hot dog sales is \\(2.75x\\), and the revenue generated by the hamburger sales is \\(4.50y\\). Since the combined revenue is $881.25, we have the equation \\(2.75x + 4.50y = 881.25\\).<\/p>\n<p>So, we have the following system of equations:<\/p>\n<p style=\"text-align: center\">\\(x + y = 225\\)<br \/>\n\\(2.75x + 4.50y = 881.25\\)<\/p>\n<p>We will use the elimination method by manipulating the equations so that one variable will cancel out when the equations are added or subtracted. We can multiply the first equation by \u20132.75 so that the \\(x\\)-variables are eliminated when the two equations are combined.<\/p>\n<p>After multiplying, the equation becomes:<\/p>\n<p style=\"text-align: center\">\\(-2.75x &#8211; 2.75y = -618.75\\)<\/p>\n<p>Now combine the two equations to eliminate \\(x\\), then solve the resulting equation for \\(y\\). Adding the equations gives:<\/p>\n<p style=\"text-align: center\">\\(-2.75x &#8211; 2.75y + 2.75x + 4.50y\\) \\(\\:= -618.75 + 881.25\\)<\/p>\n<p>This simplifies to \\(1.75y = 262.50\\).<\/p>\n<p>Divide both sides of the equation by \\(1.75\\) and simplify to get \\(y = 150\\).<\/p>\n<p>Now we can substitute the \\(y\\)-value into either of the original equations to find the \\(x\\)-value. Substituting \\(y = 150\\) into the first equation, we get:<\/p>\n<p style=\"text-align: center\">\\(x + 150 = 225\\)<\/p>\n<p>Subtract 150 from both sides of the equation to get \\(x = 75\\).<\/p>\n<p>So, the restaurant sold 75 hot dogs and 150 hamburgers.<\/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":91228,"parent":0,"menu_order":38,"comment_status":"closed","ping_status":"open","template":"","meta":{"footnotes":""},"class_list":{"0":"post-692","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\/692","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=692"}],"version-history":[{"count":7,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/692\/revisions"}],"predecessor-version":[{"id":91699,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/692\/revisions\/91699"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/91228"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=692"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}