{"id":111805,"date":"2022-02-02T16:58:44","date_gmt":"2022-02-02T22:58:44","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=111805"},"modified":"2026-03-28T11:48:15","modified_gmt":"2026-03-28T16:48:15","slug":"solving-formulas-for-specific-variables","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/solving-formulas-for-specific-variables\/","title":{"rendered":"Solving Formulas for Specific Variables"},"content":{"rendered":"<h1>Solving Formulas for Specific Variables<\/h1>\n\n\t\t\t<div id=\"mmDeferVideoEncompass_e0jgtf1u9I0\" 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_e0jgtf1u9I0\" data-source-videoID=\"e0jgtf1u9I0\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Solving Formulas for Specific Variables Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Solving Formulas for Specific Variables\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_e0jgtf1u9I0:hover {cursor:pointer;} img#videoThumbnailImage_e0jgtf1u9I0 {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/2172-thumbnail-3.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_e0jgtf1u9I0\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_e0jgtf1u9I0\" 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 Formulas for Specific Variables\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_e0jgtf1u9I0\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_e0jgtf1u9I0\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_e0jgtf1u9I0\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\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! Welcome to this video on <strong>solving equations for specific variables<\/strong>. This is an algebraic topic that has a lot of practical applications in the science fields.<\/p>\n<p>For example, let\u2019s say we have Newton\u2019s second law of motion:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(F=ma\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>and instead of being given the mass and acceleration to find the force, we are given the force and the acceleration and are asked to find the mass. All we have to do is rearrange this formula to solve for the variable \\(m\\). So, since \\(m\\) is being multiplied by \\(a\\), we need to divide both sides by \\(a\\) to undo this multiplication.<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(\\frac{F}{a}=\\frac{ma}{a}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>So on the left side we&#8217;ll have \\(\\frac{F}{a}\\), and on the right side our \\(a\\)&#8217;s cancel out, and we&#8217;re left with \\(m\\).<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(\\frac{F}{a}=m\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>So this is our equation. Now we can easily plug in our values for force and acceleration and find our mass.<\/p>\n<p>What if we\u2019re given the equation:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(e=mc^{2}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>and we\u2019re asked to solve for \\(c\\)? Well, there&#8217;s multiplication between \\(m\\) and \\(c^{2}\\), and a square that we need to undo. When simplifying equations, we use the order of operations (PEMDAS). To rearrange equations, we need to do this backwards, so we need to first undo multiplication and then undo the square (or the exponent). So, we said we need to undo multiplication first, so we&#8217;re multiplying \\(m\\) by \\(c^{2}\\), so we need to do the opposite of multiplication, which is division, and divide both sides by \\(m\\).<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(\\frac{e}{m}=\\frac{mc^{2}}{m}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>So this will give us, \\(\\frac{e}{m}=c^{2}\\), because our \\(m\\)&#8217;s cancel out.<\/p>\n<p>Now we need to undo the square by square rooting both sides.<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(\\sqrt{\\frac{e}{m}}=\\sqrt{c^{2}}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>So if we take the square root of both sides, we can leave this left side as \\(\\sqrt{\\frac{e}{m}}\\), and if you have \\(\\sqrt{c^{2}}\\), it cancels out, which is what we were wanting. And we&#8217;re left with \\(c\\).<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(\\sqrt{\\frac{e}{m}}=c\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>So this is our answer!<\/p>\n<p>Let\u2019s try one more example. This one doesn&#8217;t have a science application like the other ones do, but it&#8217;s something you might see in your algebra class. <\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(3xy-4=2z\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp; <\/p>\n<p>And I want you to solve for the variable \\(y\\).<\/p>\n<p>So, remember, we need to do the reverse order of operations and we need to get rid of everything on this side, except for the \\(y\\), because that&#8217;s what we&#8217;re solving for. So we&#8217;re going to start by adding 4 to both sides.<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(3xy-4+4=2z+4\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp; <\/p>\n<p>These terms cancel out on the left and we&#8217;re left with:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(3xy=2z+4\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp; <\/p>\n<p>Now, since both 3 and \\(x\\) are being multiplied by \\(y\\), we can simply divide <em>both<\/em> sides of our equation by \\(3x\\). We can do the division of the 3 and the division of the \\(x\\) at the exact same time, and that&#8217;s what this will look like. When we do this though, make sure you divide the <em>entire<\/em> right side by \\(3x\\). If you only do the \\(2z\\) part or only the 4 part, you&#8217;ll get the wrong answer, so make sure you do the <em>entire<\/em> thing by \\(3x\\).<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(\\frac{3xy}{3x}=\\frac{2z+4}{3x}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp; <\/p>\n<p>So, over here our \\(3x\\) terms cancel out and we&#8217;re left with \\(y\\).<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(y=\\frac{2z+4}{3x}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp; <\/p>\n<p>And that\u2019s our answer!<\/p>\n<p>I hope this 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=\"Specific_Variable_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Specific Variable 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 equation of a line in slope-intercept form is \\(y=mx+b\\), where \\(m\\) is the slope of the line, \\(b\\) is the \\(y\\)-intercept of the line, and \\(x\\) and \\(y\\) are the coordinates of a point \\((x,y)\\) on the line.<\/p>\n<p>Solve the equation for \\(m\\).<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(m=\\large{\\frac{y}{b}}\\normalsize{-x}\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\(m=\\large{\\frac{y-x}{b}}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">\\(m=\\large{\\frac{y-b}{x}}\\)<\/div><div class=\"PQ\"  id=\"PQ-1-4\">\\(m=\\large{\\frac{y}{x}}\\normalsize{-b}\\)<\/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>We can rearrange the equation using algebraic methods such as adding, subtracting, multiplying, or dividing terms to both sides of the equation to solve for the specified variable. <\/p>\n<p>To isolate the variable \\(m\\), start by subtracting \\(b\\) from both sides.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(y-b=mx+b-b\\)<br \/>\n\\(y-b=mx\\)<\/p>\n<p>Then, divide by \\(x\\) on both sides.<\/p>\n<p style=\"text-align: center; line-height: 60px\">\\(\\dfrac{y-b}{x}=\\dfrac{mx}{x}\\)<br \/>\n\\(\\dfrac{y-b}{x}=m\\)<\/p>\n<p>Therefore, \\(m=\\large{\\frac{y-b}{x}}\\).<\/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 \/>\nThe equation for the area of a trapezoid is given below in terms of its height, \\(h\\), and two bases, \\(b_1\\) and \\(b_2\\). Solve the equation for \\(b_1\\).<\/p>\n<div class=\"yellow-math-quote\">\\(A=\\frac{1}{2}(b_1+b_2)h\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-2-1\">\\(b_1=\\large{\\frac{2A}{h}}\\normalsize{-b_2}\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\(b_1=\\large{\\frac{2A-b_2}{h}}\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\(b_1=\\large{\\frac{Ah}{2}}\\normalsize{-b_2}\\)<\/div><div class=\"PQ\"  id=\"PQ-2-4\">\\(b_1=\\large{\\frac{A}{2b_2}}\\normalsize{-h}\\)<\/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 rearrange the equation using algebraic methods such as adding, subtracting, multiplying, or dividing terms to both sides of its equation to solve for the specified variable.<\/p>\n<p>To solve the equation for \\(b_1\\), start by multiplying both sides by 2.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(2\\cdot A=2 \\cdot \\frac{1}{2}(b_1+b_2)h\\)<br \/>\n\\(2A=(b_1+b_2)h\\)<\/p>\n<p>Next, divide both sides by \\(h\\).<\/p>\n<p style=\"text-align: center; line-height: 60px\">\\(\\dfrac{2A}{h}=\\dfrac{(b_1+b_2)h}{h}\\)<br \/>\n\\(\\dfrac{2A}{h}=b_1+b_2\\)<\/p>\n<p>Finally, subtract \\(b_2\\) from both sides.<\/p>\n<p style=\"text-align: center; line-height: 60px\">\\(\\dfrac{2A}{h}-b_2=(b_1+b_2)-b_2\\)<br \/>\n\\(\\dfrac{2A}{h}-b_2=b_1\\)<\/p>\n<p>Therefore, \\(b_1=\\large{\\frac{2A}{h}}\\normalsize{-b_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 \/>\nThe equation for the volume of a right circular cone is given below in terms of its radius, \\(r\\), and its height, \\(h\\). Solve the equation for the cone\u2019s radius.<\/p>\n<div class=\"yellow-math-quote\">\\(V=\\frac{1}{3}\\pi r^2h\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\(r=\\large{\\frac{\\sqrt{\\pi V}}{3h}}\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\(r=\\large{\\frac{\\sqrt{3V}}{\\pi h}}\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\(r=\\sqrt{\\large{\\frac{\\pi V}{3h}}}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-4\">\\(r=\\sqrt{\\large{\\frac{3V}{\\pi h}}}\\)<\/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>Solve for the radius, \\(r\\), by first multiplying both sides by 3.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(3\\cdot V=3\\cdot\\frac{1}{3}\\pi r^2h\\)<br \/>\n\\(3V=\\pi r^2h\\)<\/p>\n<p>Then, divide both sides by \\(\\pi h\\).<\/p>\n<p style=\"text-align: center; line-height: 60px\">\\(\\dfrac{3V}{\\pi h}=\\dfrac{\\pi r^2h}{\\pi h}\\)<br \/>\n\\(\\dfrac{3V}{\\pi h}=r^2\\)<\/p>\n<p>Finally, take the square root of both sides.<\/p>\n<p style=\"text-align: center; line-height: 60px\">\\(\\sqrt{\\large{\\frac{3V}{\\pi h}}}=\\sqrt{r^2}\\)<br \/>\n\\(\\sqrt{\\large{\\frac{3V}{\\pi h}}}=r\\)<\/p>\n<p>Therefore, \\(r=\\sqrt{\\large{\\frac{3V}{\\pi h}}}\\).<\/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 temperature of an outdoor thermometer reads \u22124\u00b0C. The equation to find the temperature in degrees Celsius, \\(C\\), given the temperature in degrees Fahrenheit, \\(F\\), is given below. You want to know the temperature outside in degrees Fahrenheit. Which of the following equations can be used to directly find the temperature in degrees Fahrenheit?<\/p>\n<div class=\"yellow-math-quote\">\\(C=\\large{\\frac{5}{9}}\\normalsize{(F-32)}\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">\\(F=\\large{\\frac{9}{5}}\\normalsize{(C+32)}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-2\">\\(F=\\large{\\frac{9}{5}}\\normalsize{C+32}\\)<\/div><div class=\"PQ\"  id=\"PQ-4-3\">\\(F=\\large{\\frac{5}{9}}\\normalsize{(}\\large{\\frac{C}{F-2}}\\normalsize{)}\\)<\/div><div class=\"PQ\"  id=\"PQ-4-4\">\\(F=\\large{\\frac{5}{9}}\\normalsize{C+32}\\)<\/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>To find the temperature in degrees Fahrenheit in terms of the temperature in degrees Celsius, we need to solve for \\(F\\).<\/p>\n<p>Start by multiplying both sides by \\(\\frac{9}{5}\\).<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(\\large{\\frac{9}{5}}\\normalsize{\\cdot \\: C=\\,}\\large{\\frac{9}{5}}\\normalsize{\\, \\cdot\\, }\\large{\\frac{5}{9}}\\normalsize{(F-32)}\\)<br \/>\n\\(\\large{\\frac{9}{5}}\\normalsize{C=F-32}\\)<\/p>\n<p>From here, add 32 to both sides.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(\\large{\\frac{9}{5}}\\normalsize{C+32=F-32+32}\\)<br \/>\n\\(\\large{\\frac{9}{5}}\\normalsize{C+32=F}\\)<\/p>\n<p>Therefore, the formula for finding the temperature in degrees Fahrenheit when given the temperature in degrees Celsius is \\(F=\\frac{9}{5}C+32\\).<\/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 \/>\nThe cost, \\(C\\), in dollars, for a company to produce \\(x\\) textbooks is given by the equation below. Which of the following equations can be used to directly find the number of textbooks produced at a given cost?<\/p>\n<div class=\"yellow-math-quote\">\\(C=3(x+10)-12\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">\\(x=\\large{\\frac{C}{x+10}}\\normalsize{+12}\\)<\/div><div class=\"PQ\"  id=\"PQ-5-2\">\\(x=\\large{\\frac{C}{3}}\\normalsize{+4}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-3\">\\(x=\\large{\\frac{C+12}{3}}\\normalsize{-10}\\)<\/div><div class=\"PQ\"  id=\"PQ-5-4\">\\(x=\\large{\\frac{C}{3}}\\normalsize{+2}\\)<\/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>To find the number of textbooks in terms of their cost to produce, we need to solve for \\(x\\). Start by adding 12 to both sides of the equation.<\/p>\n<p style=\"text-align: center ;line-height: 35px\">\\(C+12=3(x+10)-12+12\\)<br \/>\n\\(C+12=3(x+10)\\)<\/p>\n<p>From here, divide both sides by 3.<\/p>\n<p style=\"text-align: center; line-height: 60px\">\\(\\dfrac{C+12}{3}=\\dfrac{3(x+10)}{3}\\)<br \/>\n\\(\\dfrac{C+12}{3}=x+10\\)<\/p>\n<p>Then, subtract 10 from both sides.<\/p>\n<p style=\"text-align: center; line-height: 60px\">\\(\\dfrac{C+12}{3}-10=x+10-10\\)<br \/>\n\\(\\dfrac{C+12}{3}-10=x\\)<\/p>\n<p>Therefore, the number of textbooks in terms of their cost to produce is \\(x=\\large{\\frac{C+12}{3}}\\normalsize{-10}\\).<\/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>Solving Formulas for Specific Variables Return to Algebra I Videos<\/p>\n","protected":false},"author":22,"featured_media":111808,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-111805","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\/111805","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=111805"}],"version-history":[{"count":5,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/111805\/revisions"}],"predecessor-version":[{"id":286189,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/111805\/revisions\/286189"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/111808"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=111805"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}