{"id":113747,"date":"2022-02-15T12:06:48","date_gmt":"2022-02-15T18:06:48","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=113747"},"modified":"2026-05-04T14:28:58","modified_gmt":"2026-05-04T19:28:58","slug":"solving-two-step-equations","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/solving-two-step-equations\/","title":{"rendered":"Solving Two-Step Equations"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_ugr6zOj0Xbk\" 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_ugr6zOj0Xbk\" data-source-videoID=\"ugr6zOj0Xbk\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Solving Two-Step Equations Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Solving Two-Step Equations\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_ugr6zOj0Xbk:hover {cursor:pointer;} img#videoThumbnailImage_ugr6zOj0Xbk {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/2173-thumbnail-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_ugr6zOj0Xbk\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_ugr6zOj0Xbk\" 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 Two-Step Equations\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_ugr6zOj0Xbk\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_ugr6zOj0Xbk\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_ugr6zOj0Xbk\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction 06B_Function() {\n  var x = document.getElementById(\"06B\");\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=\"06B_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=\"06B\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Example_1\" class=\"smooth-scroll\">Example 1<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Example_2\" class=\"smooth-scroll\">Example 2<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Example_3\" class=\"smooth-scroll\">Example 3<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#TwoStep_Equation_Practice_Questions\" class=\"smooth-scroll\">Two-Step 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>Hello! Today we are going to take a look at solving two-step equations.<\/p>\n<p>Let&#8217;s start with an example:<\/p>\n<h2><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example 1<\/h2>\n<div class=\"examplesentence\">\\(3x-4=11\\)<\/div>\n<p>\n&nbsp;<br \/>\nTo solve, we need to find the value of \\(x\\) that makes this statement true. We do this by getting \\(x\\) by itself on one side of the equation.<\/p>\n<p>When solving two-step equations, always undo the addition or subtraction first and then undo the multiplication or division.<\/p>\n<p>So, for this problem, we&#8217;re going to start by undoing the subtraction by adding 4 to both sides.<\/p>\n<div class=\"examplesentence\">\\(3x-4+4=11+4\\)<\/div>\n<p>\n&nbsp;<br \/>\nThat means these cancel out, and we&#8217;re left with \\(3x\\) on the left side, and \\(11+4=15\\).<\/p>\n<div class=\"examplesentence\">\\(3x=15\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, we undo the 3 being multiplied by \\(x\\) by dividing both sides by 3. Remember, any time a number is next to a letter, it means they are being multiplied. So this shows us multiplication. So, to undo that, we simply divide both sides by 3.<\/p>\n<div class=\"examplesentence\">\\(\\dfrac{3x}{3}=\\dfrac{15}{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur 3s cancel out and we&#8217;re left with \\(x\\) on the left side, and \\(15\\div 3=5\\).<\/p>\n<div class=\"examplesentence\">\\(x=5\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd that\u2019s our answer! We can check it by plugging it back into our original equation and seeing if it is a true statement. So let&#8217;s do that.<\/p>\n<div class=\"examplesentence\">\\(3(5)-4=11\\)<br \/>\n\\(15-4=11\\)<br \/>\n\\(11=11\\)<\/div>\n<p>\n&nbsp;<br \/>\n\\(11=11\\) is a true statement, so that verifies that \\(x=5\\) is the correct answer.<\/p>\n<h2><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example 2<\/h2>\n<p>\nLet\u2019s take a look at another problem.<\/p>\n<div class=\"examplesentence\">\\(\\dfrac{x}{4}+5=7\\)<\/div>\n<p>\n&nbsp;<br \/>\nRemember, we first want to undo the addition. So, to undo addition by 5, we&#8217;re going to subtract 5 from both sides.<\/p>\n<div class=\"examplesentence\">\\(\\dfrac{x}{4}+5-5=7-5\\)<\/div>\n<p>\n&nbsp;<br \/>\nThat gives us \\(\\frac{x}{4}\\) (or \\(x\\div 4\\)), these (5s) cancel out, remember, and then \\(7-5=2\\).<\/p>\n<div class=\"examplesentence\">\\(\\dfrac{x}{4}=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, now we&#8217;re left with division: \\(x\\) is being divided by 4. So, to undo this, we simply multiply both sides by 4.<\/p>\n<div class=\"examplesentence\">\\(4\\times \\dfrac{x}{4}=24\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur 4s cancel out and we&#8217;re left with \\(x\\) on the left side, and \\(2\\times 4=8\\).<\/p>\n<div class=\"examplesentence\">\\(x=8\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo our answer is \\(x=8\\).<\/p>\n<h2><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example 3<\/h2>\n<p>\nLet&#8217;s do one more example together before we go.<\/p>\n<div class=\"examplesentence\">\\(-3x+20=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, first we&#8217;re going to undo this addition by subtracting 20 from both sides.<\/p>\n<div class=\"examplesentence\">\\(-3x+20-20=2-20\\)<\/div>\n<p>\n&nbsp;<br \/>\nThat leaves us with:<\/p>\n<div class=\"examplesentence\">\\(-3x=-18\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then we need to get rid of this \u22123 being multiplied by \\(x\\), so we divide both sides by \u22123. <\/p>\n<div class=\"examplesentence\">\\(\\dfrac{-3x}{-3}=\\dfrac{-18}{-3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis leaves us with:<\/p>\n<div class=\"examplesentence\">\\(x=6\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo \\(x=6\\) is our answer. And there you have it!<\/p>\n<p>I hope that this video helped you better understand how to solve two-step equations. Thanks for watching, and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"TwoStep_Equation_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Two-Step 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 \/>\nSolve the following equation for \\(x\\):<\/p>\n<div class=\"yellow-math-quote\">\\(4x-5=19\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(x=14\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\(x=18\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">\\(x=6\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\(x=20\\)<\/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 the equation for \\(x\\), we want to undo any operations around x in order to isolate \\(x\\) on the left side of the equation. When solving two-step equations, always undo the addition or subtraction operation first, then undo the multiplication or division operation. <\/p>\n<p>Five is being subtracted from \\(4x\\). The opposite operation is addition. We need to add 5 to both sides of the equation.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(4x-5+5=19+5\\)<br \/>\n\\(4x=24\\)<\/p>\n<p>The number next to \\(x\\) implies we are multiplying it by \\(x\\), so we need to divide \\(4x\\) by 4 to undo the multiplication operation. Also, to keep our equation balanced, we need to divide by 4 on the right side of the equation.<\/p>\n<p style=\"text-align: center; line-height: 50px\">\\(\\dfrac{4x}{4}=\\dfrac{24}{4}\\)<br \/>\n\\(x=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 following equation for \\(x\\):<\/p>\n<div class=\"yellow-math-quote\">\\(\\large{\\frac{x}{3}}\\normalsize{+7=12}\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-2-1\">\\(x=15\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\(x=36\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\(x=5\\)<\/div><div class=\"PQ\"  id=\"PQ-2-4\">\\(x=57\\)<\/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>To solve the equation for \\(x\\), we want to undo any operations around \\(x\\) in order to isolate \\(x\\) on the left side of the equation. When solving two-step equations, always undo the addition or subtraction operation first, then undo the multiplication or division operation. <\/p>\n<p>In the equation, 7 is being added to \\(\\frac{x}{3}\\). The opposite operation is subtraction, so we need to subtract 7 from both sides of the equation.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(\\large{\\frac{x}{3}}\\normalsize{+7-7=12-7}\\)<br \/>\n\\(\\large{\\frac{x}{3}}\\normalsize{=5}\\)<\/p>\n<p>The fraction bar on the left side of the equation means we are dividing \\(x\\) by 3, so we need to multiply \\(\\frac{x}{3}\\) by 3 to undo the division operation. Also, to keep our equation balanced, we need to multiply 3 on the right side of the equation.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(3\\times\\large{\\frac{x}{3}}\\normalsize{=3\\times5}\\)<br \/>\n\\(x=15\\)<\/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 following equation for \\(x\\):<\/p>\n<div class=\"yellow-math-quote\">\\(-6x+27=3\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\(x=-24\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\(x=-26\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\(x=5\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-4\">\\(x=4\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>To solve the equation for \\(x\\), we want to undo any operations around \\(x\\) in order to isolate \\(x\\) on the left side of the equation. When solving two-step equations, always undo the addition or subtraction operation first, then undo the multiplication or division operation. <\/p>\n<p>In the equation, 27 is being added to \\(-6x\\). The opposite operation is subtraction, so we need to subtract 27 from both sides of the equation.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(-6x+27-27=3-27\\)<br \/>\n\\(-6x=-24\\)<\/p>\n<p>The number next to \\(x\\) implies we are multiplying it by \\(x\\), so we need to divide \\(-6x\\) by \u22126 to undo the multiplication operation. Also, to keep our equation balanced, we need to divide by \u22126 on the right-hand side of the equation to solve for \\(x\\).<\/p>\n<p style=\"text-align: center; line-height: 50px\">\\(\\dfrac{-6x}{-6}=\\frac{-24}{-6}\\)<br \/>\n\\(x=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 \/>\nYou want to save money to buy a new coat that costs $100. You have a part-time job that pays you $8 an hour after all deductions are taken out. If you have already saved $28 to buy the coat, how many more hours do you need to work to save enough money to buy the coat?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">8 hours<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-2\">9 hours<\/div><div class=\"PQ\"  id=\"PQ-4-3\">12 hours<\/div><div class=\"PQ\"  id=\"PQ-4-4\">10 hours<\/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 the variable \\(x\\) represent the number of hours you work. Multiplying $8 by the unknown number of hours you need to work gives you the amount of money you make from working.<\/p>\n<p>You have already saved $28, so we can write the expression \\(8x+28\\) to represent the total amount of money you will have saved after working \\(x\\) hours. You need to save $100 to buy the new coat, so we can write the equation below to find the number of hours you need to work.<\/p>\n<p style=\"text-align: center;\">\\(8x+28=100\\)<\/p>\n<p>To solve the equation for \\(x\\), we want to undo any operations around \\(x\\) in order to isolate \\(x\\) on the left-hand side of the equation. When solving two-step equations, always undo the addition or subtraction operation first, then undo the multiplication or division operation second. <\/p>\n<p>In the equation, 28 is being added to \\(8x\\). The opposite operation is subtraction, so we need to subtract 28 from both sides of the equation.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(8x+28-28=100-28\\)<br \/>\n\\(8x=72\\)<\/p>\n<p>The number next to \\(x\\) implies we are multiplying it by \\(x\\), so we need to divide 8x by 8 to undo the multiplication operation. Also, to keep our equation balanced, we need to divide by 8 on the right-hand side of the equation as well.<\/p>\n<p style=\"text-align: center; line-height: 50px\">\\(\\dfrac{8x}{8}=\\dfrac{72}{8}\\)<br \/>\n\\(x=9\\)<\/p>\n<p>Therefore, you need to work 9 hours to have enough money to buy the new coat.<\/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 \/>\nYou have counted ten less than one-half of the number of marbles that are in a bag. If you have counted 64 marbles, how many marbles are in the bag?<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">74 marbles<\/div><div class=\"PQ\"  id=\"PQ-5-2\">36 marbles<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-3\">148 marbles<\/div><div class=\"PQ\"  id=\"PQ-5-4\">108 marbles<\/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 the variable \\(x\\) represent the number of marbles in the bag. Subtracting 10 marbles from one-half of the number of marbles that are in the bag gives us the number of marbles that have been counted. The expression \\(\\frac{x}{2}-10\\) represents this statement. Since you have counted 64 marbles, we can write the equation below to find the number of marbles that are in the bag.<\/p>\n<p style=\"text-align: center;\">\\(\\large{\\frac{x}{2}}\\normalsize{-10=64}\\)<\/p>\n<p>To solve the equation for \\(x\\), we want to undo any operations around \\(x\\) in order to isolate \\(x\\) on the left side of the equation. When solving two-step equations, always undo the addition or subtraction operation first, then undo the multiplication or division operation second. <\/p>\n<p>In the equation, 10 is being subtracted from \\(\\frac{x}{2}\\). The opposite operation is addition, so we need to add 10 to both sides of the equation.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(\\large{\\frac{x}{2}}\\normalsize{-10+10=64+10}\\)<br \/>\n\\(\\large{\\frac{x}{2}}\\normalsize{=74}\\)<\/p>\n<p>The fraction bar on the left side of the equation means we are dividing \\(x\\) by 2, so we need to multiply \\(\\frac{x}{2}\\) by 2 to undo the division operation. To keep our equation balanced, we need to multiply 2 on the right side of the equation.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(2\\times \\large{\\frac{x}{2}}\\normalsize{=2\\times74}\\)<br \/>\n\\(x=148\\)<\/p>\n<p>Therefore, there are 148 marbles in the bag.<\/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":22,"featured_media":113750,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-113747","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-practice-question-videos","7":"page_type-video","8":"content_type-practice-questions","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/113747","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\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=113747"}],"version-history":[{"count":5,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/113747\/revisions"}],"predecessor-version":[{"id":293114,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/113747\/revisions\/293114"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/113750"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=113747"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}