{"id":71929,"date":"2021-04-26T13:38:03","date_gmt":"2021-04-26T18:38:03","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=71929"},"modified":"2026-03-28T11:01:53","modified_gmt":"2026-03-28T16:01:53","slug":"solving-a-system-of-equations-consisting-of-a-linear-equation-and-quadratic-equations","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/solving-a-system-of-equations-consisting-of-a-linear-equation-and-quadratic-equations\/","title":{"rendered":"Solving a System of Equations Consisting of a Linear Equation and Quadratic Equations"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_XEkSCu-3bPA\" 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_XEkSCu-3bPA\" data-source-videoID=\"XEkSCu-3bPA\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Solving a System of Equations Consisting of a Linear Equation and Quadratic Equations Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Solving a System of Equations Consisting of a Linear Equation and Quadratic Equations\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_XEkSCu-3bPA:hover {cursor:pointer;} img#videoThumbnailImage_XEkSCu-3bPA {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/07\/updated-solving-a-system-of-equations-consisting-of-a-linear-equation-and-quadratic-equations-64c3df6f6ca39.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_XEkSCu-3bPA\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_XEkSCu-3bPA\" 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\",\"Solving a System of Equations Consisting of a Linear Equation and Quadratic Equations\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_XEkSCu-3bPA\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_XEkSCu-3bPA\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_XEkSCu-3bPA\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction GdY_Function() {\n  var x = document.getElementById(\"GdY\");\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=\"GdY_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=\"GdY\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Linear_Equations_vs_Quadratic_Equations\" class=\"smooth-scroll\">Linear Equations vs. Quadratic Equations<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Solving_Systems_of_Equations\" class=\"smooth-scroll\">Solving Systems of Equations<\/a>\n<ul><\/li>\n<li class=\"toc-h3\"><a href=\"#Example_1\" class=\"smooth-scroll\">Example #1<\/a><\/li>\n<li class=\"toc-h3\"><a href=\"#Example_2\" class=\"smooth-scroll\">Example #2<\/a><\/li>\n<\/ul>\n<\/li>\n<li class=\"toc-h2\"><a href=\"#System_of_Equation_Practice_Questions\" class=\"smooth-scroll\">System of Equation 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=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this video on solving systems of equations with a linear and a quadratic equation!<\/p>\n<p>In this video, we will take a look at two different ways to approach these problems: graphically and algebraically. <\/p>\n<p>Let\u2019s get started!<\/p>\n<h2><span id=\"Linear_Equations_vs_Quadratic_Equations\" class=\"m-toc-anchor\"><\/span>Linear Equations vs. Quadratic Equations<\/h2>\n<p>\nBefore we get into how to solve these systems, let\u2019s review the differences between linear and quadratic equations.<\/p>\n<p><strong>Linear equations<\/strong> are equations that graph as a line. They are equations that can be written as \\(y=ax+b\\). Notice that neither \\(y\\) nor \\(x\\) is to a power or a root. <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/02\/linear-e@300-scaled.webp\" alt=\"\" width=\"\" height=\"\" class=\"aligncenter size-full wp-image-215971\"  role=\"img\" \/><\/p>\n<p><strong>Quadratic equations<\/strong> are equations of the form \\(y=ax^2+bx+c\\), and they graph as a parabola. Notice that the \\(y\\)-term is still not to a power or a root, but we do have an \\(x\\)-term that is squared.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/02\/quadratic-e@300-scaled.webp\" alt=\"\" width=\"\" height=\"\" class=\"aligncenter size-full wp-image-215971\"  role=\"img\" \/><\/p>\n<h2><span id=\"Solving_Systems_of_Equations\" class=\"m-toc-anchor\"><\/span>Solving Systems of Equations<\/h2>\n<p>\nWhen we solve systems of equations, we are taking two or more equations and finding the point, or points, where they intersect. When we have a linear equation and a quadratic equation, we will have zero, one, or two points of intersection.<\/p>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nThe first solving method we are going to look at is solving graphically. To do this, we simply graph our equations, either with a calculator or by hand, and see where the graphs intersect.<\/p>\n<p>Let\u2019s take a look at the equations:<\/p>\n<div class=\"examplesentence\">\\(y=2x+8\\) and \\(y=x^2\u2013x\u201320\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf we graph these two equations next to each other, we get something that looks like this:  <\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/02\/image_343.png\" alt=\"\" width=\"\" height=\"\" class=\"aligncenter size-full wp-image-215971\"  role=\"img\" style=\"box-shadow: 1.5px 1.5px 3px gray;\"  \/><\/p>\n<p>We can see from our graph that the two equations intersect at the points negative 4,0) and (7,22). (-4,0) and (7,22).<\/p>\n<p>But what if we aren\u2019t able to graph our equations? Well, that\u2019s where algebra comes in. The other way to solve systems like this is very similar to the substitution method of solving regular systems of linear equations.<\/p>\n<p>To find at what points our equations are equal to each other, we can substitute our \\(y\\)-values for each other. Looking back at our equations from earlier, we can see that since \\(y=2x+8\\), we can replace the \\(y\\) in \\(y=x^2\u2013x\u201320\\) with \\(2x+8\\), or in other words, set our equations equal to one another.<\/p>\n<p>When we write it out, that will look like this:<\/p>\n<div class=\"examplesentence\">\\(2x+8=x^2\u2013x\u201320\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn order to solve a quadratic equation for \\(x\\), we need to get all of our terms on one side and zero on the other. To do this, we will subtract \\(8\\) and \\(2x\\) from both sides.<\/p>\n<p>That gives us:<\/p>\n<div class=\"examplesentence\">\\(0=x^2\u20133x\u201328\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe can then factor this equation to get:<\/p>\n<div class=\"examplesentence\">\\(0=(x+4)(x\u20137)\\)<br \/>\n\\(x=-4, 7\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis tells us that our zeroes for this quadratic equation are -4 and 7. These are our values for \\(x\\) at the two points where our graphs intersect.<\/p>\n<p>To find our \\(y\\)-values for each of these points, we simply want to plug in our \\(x\\)-values to either equation and solve for \\(y\\). I\u2019m going to plug them into the linear equation because it will require fewer steps to get \\(y\\).<\/p>\n<div class=\"examplesentence\">\\(y=2(-4)+8=-8+8=0\\)<br \/>\n\\(y=2(7)+8=14+8=22\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo our points of intersection for this graph are \\((-4,0)\\) and \\((7,22)\\), which is what we found by graphing them.<\/p>\n<p>If we wanted one more way to check our answers, we can plug them back into our original equations and make sure our answers match up.<\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nLet\u2019s try one more example. Solve this system graphically and then algebraically:<\/p>\n<div class=\"examplesentence\">\\(12x\u2013y=1\\)<br \/>\n\\(y=2x^2\u20133x\u201328\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn order to graph our linear equation, we want to first rearrange it so that y is by itself on one side. We can do this by adding y and subtracting 1 from both sides.<\/p>\n<p>When we do this, it gives us:<\/p>\n<div class=\"examplesentence\">\\(y=12x\u20131\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we are able to graph both equations, and our graph will look like this:<\/p>\n<p>We can see that our graphs intersect at the points \\((-\\frac{3}{2},-19)\\) and \\((9,107)\\).<\/p>\n<p>Now let\u2019s check and see if we get the same answers using algebra.<\/p>\n<p>If we use our rearranged linear equation, we are able to set our two equations equal to each other and solve.<\/p>\n<div class=\"examplesentence\">\\(12x\u20131=2x^2\u20133x\u201328\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe can subtract 12x and add 1 to both sides to get all of our terms on one side and zero on the other.<\/p>\n<div class=\"examplesentence\">\\(0=2x^2\u201315x\u201327\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we want to factor our equation to get our x-values.<\/p>\n<div class=\"examplesentence\">\\(0=(2x+3)(x\u20139)\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf we set each equation individually to equal zero, we get:<\/p>\n<div class=\"examplesentence\">\\(2x+3=0\\)<br \/>\n\\(x\u20139=0\\)<\/div>\n<p>\n&nbsp;<br \/>\nOn this side, to solve for \\(x\\), we subtract 3 from both sides and then divide by 2. So \\(x\\) equals negative three-halves.<\/p>\n<p>On this side, we just simply add 9 to both sides.<\/p>\n<div class=\"examplesentence\">\\(x=9\\)<\/div>\n<p>\n&nbsp;<br \/>\nThese are our \\(x\\)-values for our two points of intersection. Now we want to find our \\(y\\)-values.<\/p>\n<p>So, we&#8217;re going to take our original equation, \\(y= 12x &#8211; 1\\), and plug in our \\(x\\)-value of negative three-halves. And then we&#8217;re gonna plug in our value of \\(x =9\\).<\/p>\n<div class=\"examplesentence\">\\(y=12(-\\frac{3}{2})\u20131=-18\u20131=-19\\)<br \/>\n\\(y=12(9)\u20131=108\u20131=107\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur two points of intersection are \\((-\\frac{3}{2},-19)\\) and \\((9,107)\\). This is just like we found when graphing!<\/p>\n<p>There\u2019s one important thing to point out before we go. In both of these examples, our equations factored nicely to give us our \\(x\\)-values. This will not always be the case. Sometimes you will not be able to factor and will need to use the quadratic equation to find your \\(x\\)-values. After you find your \\(x\\)-values, follow the same steps to find your points of intersection.<\/p>\n<p>I hope this video on solving systems with a linear and a quadratic function 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=\"System_of_Equation_Practice_Questions\" class=\"m-toc-anchor\"><\/span>System of Equation 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 \/>\nThe equations \\(y=x^2+2x+3\\) and \\(2y-2x=10\\) are graphed below. How many solutions are there for this system of equations?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\" wp-image-111138 aligncenter\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-1.png\" alt=\"parabola and linear line intersecting at (1,6) on a graph\" width=\"423\" height=\"413\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-1.png 1746w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-1-300x292.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-1-1024x998.png 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-1-768x749.png 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-1-1536x1497.png 1536w\" sizes=\"auto, (max-width: 423px) 100vw, 423px\" \/><\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">3 solutions <\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-2\">2 solutions <\/div><div class=\"PQ\"  id=\"PQ-1-3\">1 solution<\/div><div class=\"PQ\"  id=\"PQ-1-4\">0 solutions <\/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>A linear equation and a quadratic equation will have either zero, one, or two points of intersection. Each point of intersection is a solution to the system of equations. In this case, we see that the graphs for these equations intersect exactly two times, at \\((-2,3)\\) and \\((1,6)\\). Therefore, there are two solutions, and Choice B is the correct answer. <\/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 \/>\nWhich graph could be used to identify the solutions to the following system of equations?<\/p>\n<div class=\"yellow-math-quote\">\\(y=x^2-5x+3\\)<br \/>\n\\(y=x-6\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-2.png\" alt=\"parabola and linear line that do not intersect\" width=\"376\" height=\"370\" \/><\/div><div class=\"PQ\"  id=\"PQ-2-2\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-3.png\" alt=\"parabola and linear line intersecting around (1.5,-4.5) on a graph\" width=\"374\" height=\"369\" \/><\/div><div class=\"PQ\"  id=\"PQ-2-3\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-4.png\" alt=\"parabola and linear line intersecting at (-0.5,5.5) on a graph\" width=\"375\" height=\"383\" \/><\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-4\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-5.png\" alt=\"parabola and linear line intersecting at (3,-3) on a graph\" width=\"371\" height=\"367\" \/><\/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>Choice D is the correct answer because it correctly graphs both equations as stated.<\/p>\n<p>Choice A incorrectly graphed the quadratic function as \\(y=x^2+5x+3\\), shifting the parabola over 5 units.<\/p>\n<p>Choice B incorrectly graphed the quadratic function as \\(y=-x^2-5x+3\\), which flips the parabola in the opposite direction.<\/p>\n<p>Choice C incorrectly graphed the linear equation as \\(y=x+6\\), which changed the \\(y\\)-intercept from \u22126 to 6.<\/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 \/>\nTwo bus routes in Sunnydale can be graphed using the equations \\(y=-x^2-3x+2\\) and \\(y=x+6\\). Solve this system of equations algebraically to find out if there are any overlapping points on these two routes. If there are, what are the coordinate points of intersection for these two bus routes?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-3-1\">There is one point of intersection at \\((-2,4)\\). <\/div><div class=\"PQ\"  id=\"PQ-3-2\">There are no points of intersection. <\/div><div class=\"PQ\"  id=\"PQ-3-3\">There are two points of intersection at \\((-1,5)\\) and \\((5,11)\\). <\/div><div class=\"PQ\"  id=\"PQ-3-4\">There is one point of intersection at \\((4,-2)\\). <\/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>To solve algebraically, set the two equations equal to one another.<\/p>\n<p style=\"text-align: center\">\\(-x^2-3x+2=x+6\\)<\/p>\n<p>To solve the equation for \\(x\\), move all terms to one side of the equation and set it equal to zero. To do so, move \\(x+6\\) to the left side of the equation. First, subtract \\(x\\) from both sides to eliminate the \\(x\\) on the right. Since \\(x-x=0\\), the \\(x\\)\u2019s on the right side of the equation cancel out. Since \\(-3x-x=-4x\\), rewrite the equation using \\(-4x\\) on the left side of the equation.<\/p>\n<p style=\"text-align: center\">\\(-x^2-4x+2=6\\)<\/p>\n<p>Next, move the 6 on the right side of the equation to the left side of the equation. Subtract 6 from both sides of the equation. Since \\(6-6=0\\), write 0 on the right side of the equation. Since \\(2-6=-4\\), rewrite the equation using \u22124 on the left side of the equation.<\/p>\n<p style=\"text-align: center\">\\(-x^2-4x-4=0\\)<\/p>\n<p>From here, simplify the equation by multiplying each term by \u22121. Doing so makes each term positive, and having positive terms makes the equation easier to work with.<\/p>\n<p style=\"text-align: center\">\\(x^2+4x+4=0\\)<\/p>\n<p>Rewrite \\(ax^2+bx+c=0\\) in factored form.<\/p>\n<p>To factor \\(x^2+4x+4=0\\), identify two numbers with a product of \\(c\\), which is 4, and a sum of \\(b\\), which is also 4. These two numbers are 2 and 2 (\\(2+2=4\\) and \\(2\\times 2=4\\)).<\/p>\n<p style=\"text-align: center\">\\((x+2)(x+2)=0\\)<\/p>\n<p>Next, separate \\((x+2)(x+2)=0\\) into two separate equations and solve for \\(x\\). Since both equations are identical, \\(x=-2\\) for both. This indicates that there is one solution to the system of equations.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(x+2=0\\)<br \/>\n\\(x+2-2=0-2\\)<br \/>\n\\(x=-2\\)<\/p>\n<p>To identify the \\(y\\)-value for this point, substitute \u22122 for \\(x\\) back into the original equations, \\(y=-x^2-3x+2\\) and \\(y=x+6\\).<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(y=-x^2-3x+2\\)<br \/>\n\\(y=-(-2)^2-3(-2)+2\\)<br \/>\n\\(y=-4+6+2\\)<br \/>\n\\(y=4\\)<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(y=x+6\\)<br \/>\n\\(y=-2+6\\)<br \/>\n\\(y=4\\)<\/p>\n<p>Since \\(x=-2\\) and \\(y=4\\), the point of intersection is \\((-2,4)\\).<\/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 \/>\nWhich ordered pair is a possible solution to the system of equations shown below? Use a graph to help you solve.<\/p>\n<div class=\"yellow-math-quote\">\\(y=x^2+4x-5\\)<br \/>\n\\(y=2x+3\\)\n<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">\\((2,-7)\\)<\/div><div class=\"PQ\"  id=\"PQ-4-2\">\\((-2,-7)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-3\">\\((-4,-5)\\)<\/div><div class=\"PQ\"  id=\"PQ-4-4\">\\((4,5)\\)<\/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>When graphed, the two equations intersect at two points, \\((-4,-5)\\) and \\((2,7)\\).<\/p>\n<p>There are two solutions to this system of equations. Therefore, \\((-4,-5)\\) is one possible solution and Choice C is the correct answer.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/01\/Solving-Systems-with-Linear-and-Quadratic-Equations-SQ-6.png\" alt=\"parabola and linear line intersecting at (2,7) and (-4,-5) on a graph\" width=\"371\" height=\"367\" class=\"aligncenter\" \/><\/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 \/>\nStephanie is a city planner who is analyzing a map of street intersections to see where stoplights need to be installed. Specifically, she is looking at two streets that can be graphed using the equations \\(x+y=5\\) and \\(y+1=3x^2+2x\\). Solve this system of equations algebraically to find the coordinates for points of intersection on these two streets.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">\\((-1,-4)\\) and \\((2,-7)\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-2\">\\((1,4)\\) and \\((-2,7)\\)<\/div><div class=\"PQ\"  id=\"PQ-5-3\">\\((4,1)\\) and \\((7,-2)\\)<\/div><div class=\"PQ\"  id=\"PQ-5-4\">There are no points of intersection. <\/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>Start by rewriting both equations in standard form, \\(y=mx+b\\) for the linear equation and \\(y=ax^2+bx+c\\) for the quadratic equation. In standard form, the two equations are \\(y=-x+5\\) and \\(y=3x^2+2x-1\\).<\/p>\n<p style=\"text-align: center; line-height: 35px\">\n\\(x+y=5\\)<br \/>\n\\(x-x+y=5-x\\)<br \/>\n\\(y=-x+5\\)\n<\/p>\n<p style=\"text-align: center; line-height: 35px\">\n\\(y+1=3x^2+2x\\)<br \/>\n\\(y+1-1=3x^2+2x-1\\)<br \/>\n\\(y=3x^2+2x-1\\)\n<\/p>\n<p>To solve algebraically, set the two equations equal to one another.<\/p>\n<p style=\"text-align: center\">\\(-x+5=3x^2+2x-1\\)<\/p>\n<p>To solve the equation for \\(x\\), move all terms to one side of the equation and set it equal to zero. To do so, move \\(-x+5\\) to the right side of the equation. First, move \\(x\\) by adding \\(x\\) to both sides of the equation. Since \\(-x+x=0\\), the \\(x\\)\u2019s on the left side of the equation cancel out. Since \\(2x+x=3x\\), rewrite the equation using \\(3x\\) on the right side of the equation.<\/p>\n<p style=\"text-align: center\">\\(5=3x^2+3x-1\\)<\/p>\n<p>Next, move the 5 on the left side of the equation to the right side of the equation. Subtract 5 from both sides of the equation. Since \\(5-5=0\\), write \\(0\\) on the left side of the equation. Since \\(-1-5=-6\\), rewrite the equation using \\(-6\\) on the right side of the equation.<\/p>\n<p style=\"text-align: center\">\\(0=3x^2+3x-6\\)<\/p>\n<p>Write the equation in standard form by flipping both sides of the equation and rewriting it as \\(3x^2+3x-6=0\\).<\/p>\n<p style=\"text-align: center\">\\(3x^2+3x-6=0\\)<\/p>\n<p>Rewrite \\(ax^2+bx+c=0\\) in factored form, \\((x+\\text{_})(x+\\text{_})=0\\). To factor \\(3x^2+3x-6=0\\), identify two numbers with a product of \\(c\\), which is \\(-6\\), and a sum of \\(b\\), which is \\(3\\). Since \\(a=3\\), the factored equation looks like \\((3x+\\text{_})(x+\\text{_})=0\\). \\(-3\\) and \\(2\\) are the two numbers that work for the factored equation. \\(-3\\times2=-6\\) and \\(-3x+2(3x)=3x\\).<\/p>\n<p style=\"text-align: center\">\\((3x-3)(x+2)=0\\)<\/p>\n<p>Next, separate \\((3x-3)(x+2)=0\\) into two separate equations and solve for \\(x\\). Start with \\(3x-3=0\\). Isolate the variable to find that \\(x=1\\).<\/p>\n<p style=\"text-align: center; line-height: 50px\">\n\\(3x-3=0\\)<br \/>\n\\(3x-3+3=0+3\\)<br \/>\n\\(3x=3\\)<br \/>\n\\(\\dfrac{3x}{3}=\\dfrac{3}{3}\\)<br \/>\n\\(x=1\\)\n<\/p>\n<p>Solve \\(x+2=0\\). Isolate the variable to find that \\(x=-2\\).<\/p>\n<p style=\"text-align: center; line-height: 35px\">\n\\(x+2=0\\)<br \/>\n\\(x+2-2=0-2\\)<br \/>\n\\(x=-2\\)\n<\/p>\n<p>To identify the \\(y\\)-value for each point, substitute each \\(x\\)-value back into the original equation, \\(x+y=5\\), and solve.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\n\\(x+y=5\\)<br \/>\n\\(+y=5\\)<br \/>\n\\(1-1+y=5-1\\)<br \/>\n\\(y=4\\)\n<\/p>\n<p style=\"text-align: center; line-height: 35px\">\n\\(x+y=5\\)<br \/>\n\\((-2)+y=5\\)<br \/>\n\\(-2+2-y=5+2\\)<br \/>\n\\(y=7\\)\n<\/p>\n<p>When \\(x=1,y=4\\). When \\(x=-2,y=7\\). Therefore, the two points of intersections are \\((1,4)\\) and \\((-2,7)\\). Therefore, B is the correct answer.<\/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":187649,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-71929","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-linear-equations-videos","7":"page_category-quadratics-videos","8":"page_category-systems-of-equations-and-their-solutions-videos","9":"page_type-video","10":"content_type-practice-questions","11":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/71929","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=71929"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/71929\/revisions"}],"predecessor-version":[{"id":280736,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/71929\/revisions\/280736"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/187649"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=71929"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}