{"id":1062,"date":"2018-03-07T23:44:57","date_gmt":"2018-03-07T23:44:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/?post_type=chapter&#038;p=1062"},"modified":"2022-11-22T18:51:37","modified_gmt":"2022-11-22T18:51:37","slug":"math-review-for-preparatory-chemistry","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/math-review-for-preparatory-chemistry\/","title":{"raw":"1.3 Math Review for Preparatory Chemistry","rendered":"1.3 Math Review for Preparatory Chemistry"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Table of Contents<\/h3>\r\nThis page contains several aids that can help you with essential mathematics required for success in Chemistry.\r\n<ul>\r\n \t<li>Place Value in the Decimal System<\/li>\r\n \t<li>Positive and Negative Numbers: Addition and Subtraction<\/li>\r\n \t<li>Positive and Negative Numbers: Multiplication and Division<\/li>\r\n \t<li>Algebra<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 id=\"Exponential\">Place Value in the Decimal System<\/h2>\r\nRecall that in the decimal system, each place in the number has a name and a multiplier value.\u00a0 For instance, to the left of the decimal point are the ones place, the tens place, the hundreds place, and so on.\u00a0 To the right of the decimal point are the tenths place, the hundredths place, the thousandths place, etc.\u00a0 The value of a digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) depends on its place in the number.\u00a0 For instance, a 3 in the ones place is worth three, but in the tens place it is worth thirty and in the hundredths place is worth three-hundredths.\r\n<div class=\"textbox examples\">\r\n<h3>Determining place value in the Decimal system<\/h3>\r\nIdentify the place value for each of the digits in the number 401.725.\r\n<p style=\"padding-left: 30px;\">a. 4<\/p>\r\n<p style=\"padding-left: 30px;\">b. 0<\/p>\r\n<p style=\"padding-left: 30px;\">c. 1<\/p>\r\n<p style=\"padding-left: 30px;\">d. 7<\/p>\r\n<p style=\"padding-left: 30px;\">e. 2<\/p>\r\n<p style=\"padding-left: 30px;\">f. 5<\/p>\r\n[reveal-answer q=\"120223\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"120223\"]\r\n<p style=\"padding-left: 30px;\">a. hundreds place<\/p>\r\n<p style=\"padding-left: 30px;\">b. tens place<\/p>\r\n<p style=\"padding-left: 30px;\">c. ones place<\/p>\r\n<p style=\"padding-left: 30px;\">d. tenths place<\/p>\r\n<p style=\"padding-left: 30px;\">e. hundredths place<\/p>\r\n<p style=\"padding-left: 30px;\">f. thousandths place<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 id=\"Exponential\">Positive and Negative Numbers: Addition and Subtraction<\/h2>\r\nOne way to think about positive and negative numbers is to use a number line like the one shown below.\u00a0 This may help you recall and understand the rules for adding, subtracting, multiplying, and dividing numbers to get answers with the correct absolute values and signs.\r\n\r\n<img class=\"aligncenter wp-image-1185 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2018\/03\/16170932\/Number-line-Math-Review.png\" alt=\"\" width=\"798\" height=\"95\" \/>\r\n\r\nAddition of a positive number means moving to the right on the number line, and subtraction of a positive number is moving to the left.\u00a0\u00a0 When the number added or subtracted has a negative sign, it means moving in the opposite direction.\u00a0 Use the number line to confirm the rules for adding and subtracting numbers:\r\n<ol>\r\n \t<li>For addition, the order of the numbers doesn\u2019t matter. For subtraction, the order does matter.<\/li>\r\n \t<li>The sum of two positive numbers is the sum of their absolute values, and the sign of the answer is positive.<\/li>\r\n \t<li>The sum of two negative numbers is the sum of their absolute values, and the sign of the answer is negative.<\/li>\r\n \t<li>The sum of a positive number and a negative number is the difference between their absolute values, and the sign of the answer is the same as the sign of the number with the larger absolute value.<\/li>\r\n \t<li>Subtracting a negative number is equivalent to adding.<\/li>\r\n<\/ol>\r\n<div class=\"textbox examples\">\r\n<h3>Addition and Subtraction of Numbers<\/h3>\r\nGive the answers to the following:\r\n<p style=\"padding-left: 30px;\">a. -13 + 23 = ____<\/p>\r\n<p style=\"padding-left: 30px;\">b. -8 - (-4) = ____<\/p>\r\n<p style=\"padding-left: 30px;\">c. 9 + (-12) = ____<\/p>\r\n<p style=\"padding-left: 30px;\">d. -12 + (-9) = ____<\/p>\r\n[reveal-answer q=\"180223\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"180223\"]\r\n<p style=\"padding-left: 30px;\">a. 10<\/p>\r\n<p style=\"padding-left: 30px;\">b. -4<\/p>\r\n<p style=\"padding-left: 30px;\">c. -3<\/p>\r\n<p style=\"padding-left: 30px;\">d. -21<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 id=\"Exponential\">Positive and Negative Numbers: Multiplication and Division<\/h2>\r\nMultiplication is repeated addition, so the order of numbers doesn\u2019t matter.\u00a0 It can be helpful to think of the negative sign as meaning \u201cthe opposite of\u201d when considering the rules for multiplying numbers that have signs.\r\n<ol>\r\n \t<li>Multiplying (or dividing) two positive or two negative numbers: the answer is positive.<\/li>\r\n \t<li>Multiplying (or dividing) one positive and one negative number: the answer is negative.<\/li>\r\n<\/ol>\r\nNote that these rules also apply to division, since dividing by a given number is the same as <em>multiplying<\/em> by its reciprocal.\u00a0 However, the order of numbers <em>does<\/em> matter for division.\u00a0 The rules for order of operations are:\r\n<ol>\r\n \t<li>PEMDAS (parentheses, exponents, multiplication, division, addition, subtraction)<\/li>\r\n \t<li>left to right<\/li>\r\n<\/ol>\r\n<div class=\"textbox examples\">\r\n<h3>Multiplication and Division of Numbers<\/h3>\r\nGive the answers to the following:\r\n<p style=\"padding-left: 30px;\">a. -13 \u00f7 23 = ____<\/p>\r\n<p style=\"padding-left: 30px;\">b. -8 \u00d7 (-4) = ____<\/p>\r\n<p style=\"padding-left: 30px;\">c. 9 \u00d7 (-12) = ____<\/p>\r\n<p style=\"padding-left: 30px;\">d. -12 \u00f7 (-9) = ____<\/p>\r\n<p style=\"padding-left: 30px;\">e. 50.2 \u00f7 (13.2 \u00d7 1.20) = ____<\/p>\r\n<p style=\"padding-left: 30px;\">f. 9.1 \u00d7 (1000 \u00f7 100) =<\/p>\r\n[reveal-answer q=\"170223\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"170223\"]\r\n<p style=\"padding-left: 30px;\">a. -0.57<\/p>\r\n<p style=\"padding-left: 30px;\">b. 32<\/p>\r\n<p style=\"padding-left: 30px;\">c. -108<\/p>\r\n<p style=\"padding-left: 30px;\">d. 1.33<\/p>\r\n<p style=\"padding-left: 30px;\">e. 3.17<\/p>\r\n<p style=\"padding-left: 30px;\">f. 91<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 id=\"Exponential\">Algebra<\/h2>\r\nMany chemical phenomena can be described by equations relating several different variables.\u00a0 If the values of all but one of the variables are known, the tools of algebra may be used to figure out the value of the remaining variable.\r\n\r\nThe tools of algebra are based on balance: a mathematical equality remains an equality as long as the same operations are performed on both sides of the equation.\u00a0 The goal is always to isolate the unknown (x) <strong>in the numerator<\/strong> of one side of the equation.\u00a0 The tools include:\r\n<ol>\r\n \t<li><strong>C<\/strong>ross-multiplying to eliminate fractions.<\/li>\r\n \t<li>Distributing a term that\u2019s outside of parentheses by applying it to each term inside the parentheses.<\/li>\r\n \t<li>Collecting like terms. This means bringing together parts of the equation that have the same variable or no variable.<\/li>\r\n \t<li>Adding or subtracting the same thing on both sides of the equation.<\/li>\r\n \t<li>Multiplying or dividing each side of the equation by the same thing.<\/li>\r\n \t<li>Taking the reciprocal of both sides of the equation.<\/li>\r\n<\/ol>\r\nBelow are some examples of solving for an unknown variable using the tools of algebra:\r\n<div class=\"textbox examples\">\r\n<h3>Algebra Example 1<\/h3>\r\nSolve x in the following equation:\r\n\r\n[latex]\\large 12x = 148[\/latex]\r\n\r\n[reveal-answer q=\"17022322\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"17022322\"]\r\n\r\nTo isolate x, divide both sides by 12:\r\n\r\n[latex]\\large \\frac{\\cancel{12}x}{\\cancel{12}} = \\frac{148}{12}[\/latex]\r\n\r\n[latex]\\large x = 12.33[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Algebra Example 2<\/h3>\r\nSolve x in the following equation:\r\n\r\n[latex]\\large \\frac{x}{9} = 54[\/latex]\r\n\r\n[reveal-answer q=\"17022323\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"17022323\"]\r\n\r\nTo isolate x, multiply both sides by 9:\r\n\r\n[latex]\\large \\cancel{9}\\times\\frac{x}{\\cancel{9}} = 54 \\times 9[\/latex]\r\n\r\n[latex]\\large x = 486[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Algebra Example 3<\/h3>\r\nSolve x in the following equation:\r\n\r\n[latex]\\large 3x + 5 = 23[\/latex]\r\n\r\n[reveal-answer q=\"17022324\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"17022324\"]\r\n\r\nFirst isolate the x term on one side by subtracting 5 from both sides:\r\n\r\n[latex]\\large (3x + \\cancel{5}) - \\cancel{5} = 23 - 5[\/latex]\r\n\r\nGiving\r\n\r\n[latex]\\large 3x = 18[\/latex]\r\n\r\nNext, solve for x by dividing both sides by 3\r\n\r\n[latex]\\large \\frac{\\cancel{3}x}{\\cancel{3}} = \\frac{18}{3}[\/latex]\r\n\r\n[latex]\\large x = 6[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Algebra Example 4<\/h3>\r\nSolve x in the following equation:\r\n\r\n[latex]\\large \\frac{x}{5} + 55 = 3x + 12[\/latex]\r\n\r\n[reveal-answer q=\"17022325\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"17022325\"]\r\n\r\nFirst subtract 55 from both sides of the equation\r\n\r\n[latex]\\large (\\frac{x}{5} + \\cancel{55}) - \\cancel{55} = 3x + 12 - 55[\/latex]\r\n\r\nGiving\r\n\r\n[latex]\\large \\frac{x}{5} = 3x - 43[\/latex]\r\n\r\nNext, multiple both sides by 5\r\n\r\n[latex]\\large \\cancel{5} \\times\\frac{x}{\\cancel{5}} = (3x - 43)\\times 5[\/latex]\r\n\r\n[latex]\\large x = 15x - 215[\/latex]\r\n\r\nNext, subtract 15x from both sides of the equation to isolate the x term\r\n\r\n[latex]\\large x - 15x = (\\cancel{15x} - 215)- \\cancel{15x}[\/latex]\r\n\r\nGiving\r\n\r\n[latex]\\large -14x = -215[\/latex]\r\n\r\nFinally, divide both sides by -14 to solve for x\r\n\r\n[latex]\\large \\frac{\\cancel{-14}x}{\\cancel{-14}} = \\frac{-215}{-14}[\/latex]\r\n\r\n[latex]\\large x = 15.36[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Table of Contents<\/h3>\n<p>This page contains several aids that can help you with essential mathematics required for success in Chemistry.<\/p>\n<ul>\n<li>Place Value in the Decimal System<\/li>\n<li>Positive and Negative Numbers: Addition and Subtraction<\/li>\n<li>Positive and Negative Numbers: Multiplication and Division<\/li>\n<li>Algebra<\/li>\n<\/ul>\n<\/div>\n<h2 id=\"Exponential\">Place Value in the Decimal System<\/h2>\n<p>Recall that in the decimal system, each place in the number has a name and a multiplier value.\u00a0 For instance, to the left of the decimal point are the ones place, the tens place, the hundreds place, and so on.\u00a0 To the right of the decimal point are the tenths place, the hundredths place, the thousandths place, etc.\u00a0 The value of a digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) depends on its place in the number.\u00a0 For instance, a 3 in the ones place is worth three, but in the tens place it is worth thirty and in the hundredths place is worth three-hundredths.<\/p>\n<div class=\"textbox examples\">\n<h3>Determining place value in the Decimal system<\/h3>\n<p>Identify the place value for each of the digits in the number 401.725.<\/p>\n<p style=\"padding-left: 30px;\">a. 4<\/p>\n<p style=\"padding-left: 30px;\">b. 0<\/p>\n<p style=\"padding-left: 30px;\">c. 1<\/p>\n<p style=\"padding-left: 30px;\">d. 7<\/p>\n<p style=\"padding-left: 30px;\">e. 2<\/p>\n<p style=\"padding-left: 30px;\">f. 5<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q120223\">Show Solution<\/span><\/p>\n<div id=\"q120223\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"padding-left: 30px;\">a. hundreds place<\/p>\n<p style=\"padding-left: 30px;\">b. tens place<\/p>\n<p style=\"padding-left: 30px;\">c. ones place<\/p>\n<p style=\"padding-left: 30px;\">d. tenths place<\/p>\n<p style=\"padding-left: 30px;\">e. hundredths place<\/p>\n<p style=\"padding-left: 30px;\">f. thousandths place<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 id=\"Exponential\">Positive and Negative Numbers: Addition and Subtraction<\/h2>\n<p>One way to think about positive and negative numbers is to use a number line like the one shown below.\u00a0 This may help you recall and understand the rules for adding, subtracting, multiplying, and dividing numbers to get answers with the correct absolute values and signs.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1185 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2835\/2018\/03\/16170932\/Number-line-Math-Review.png\" alt=\"\" width=\"798\" height=\"95\" \/><\/p>\n<p>Addition of a positive number means moving to the right on the number line, and subtraction of a positive number is moving to the left.\u00a0\u00a0 When the number added or subtracted has a negative sign, it means moving in the opposite direction.\u00a0 Use the number line to confirm the rules for adding and subtracting numbers:<\/p>\n<ol>\n<li>For addition, the order of the numbers doesn\u2019t matter. For subtraction, the order does matter.<\/li>\n<li>The sum of two positive numbers is the sum of their absolute values, and the sign of the answer is positive.<\/li>\n<li>The sum of two negative numbers is the sum of their absolute values, and the sign of the answer is negative.<\/li>\n<li>The sum of a positive number and a negative number is the difference between their absolute values, and the sign of the answer is the same as the sign of the number with the larger absolute value.<\/li>\n<li>Subtracting a negative number is equivalent to adding.<\/li>\n<\/ol>\n<div class=\"textbox examples\">\n<h3>Addition and Subtraction of Numbers<\/h3>\n<p>Give the answers to the following:<\/p>\n<p style=\"padding-left: 30px;\">a. -13 + 23 = ____<\/p>\n<p style=\"padding-left: 30px;\">b. -8 &#8211; (-4) = ____<\/p>\n<p style=\"padding-left: 30px;\">c. 9 + (-12) = ____<\/p>\n<p style=\"padding-left: 30px;\">d. -12 + (-9) = ____<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q180223\">Show Solution<\/span><\/p>\n<div id=\"q180223\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"padding-left: 30px;\">a. 10<\/p>\n<p style=\"padding-left: 30px;\">b. -4<\/p>\n<p style=\"padding-left: 30px;\">c. -3<\/p>\n<p style=\"padding-left: 30px;\">d. -21<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 id=\"Exponential\">Positive and Negative Numbers: Multiplication and Division<\/h2>\n<p>Multiplication is repeated addition, so the order of numbers doesn\u2019t matter.\u00a0 It can be helpful to think of the negative sign as meaning \u201cthe opposite of\u201d when considering the rules for multiplying numbers that have signs.<\/p>\n<ol>\n<li>Multiplying (or dividing) two positive or two negative numbers: the answer is positive.<\/li>\n<li>Multiplying (or dividing) one positive and one negative number: the answer is negative.<\/li>\n<\/ol>\n<p>Note that these rules also apply to division, since dividing by a given number is the same as <em>multiplying<\/em> by its reciprocal.\u00a0 However, the order of numbers <em>does<\/em> matter for division.\u00a0 The rules for order of operations are:<\/p>\n<ol>\n<li>PEMDAS (parentheses, exponents, multiplication, division, addition, subtraction)<\/li>\n<li>left to right<\/li>\n<\/ol>\n<div class=\"textbox examples\">\n<h3>Multiplication and Division of Numbers<\/h3>\n<p>Give the answers to the following:<\/p>\n<p style=\"padding-left: 30px;\">a. -13 \u00f7 23 = ____<\/p>\n<p style=\"padding-left: 30px;\">b. -8 \u00d7 (-4) = ____<\/p>\n<p style=\"padding-left: 30px;\">c. 9 \u00d7 (-12) = ____<\/p>\n<p style=\"padding-left: 30px;\">d. -12 \u00f7 (-9) = ____<\/p>\n<p style=\"padding-left: 30px;\">e. 50.2 \u00f7 (13.2 \u00d7 1.20) = ____<\/p>\n<p style=\"padding-left: 30px;\">f. 9.1 \u00d7 (1000 \u00f7 100) =<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q170223\">Show Solution<\/span><\/p>\n<div id=\"q170223\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"padding-left: 30px;\">a. -0.57<\/p>\n<p style=\"padding-left: 30px;\">b. 32<\/p>\n<p style=\"padding-left: 30px;\">c. -108<\/p>\n<p style=\"padding-left: 30px;\">d. 1.33<\/p>\n<p style=\"padding-left: 30px;\">e. 3.17<\/p>\n<p style=\"padding-left: 30px;\">f. 91<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 id=\"Exponential\">Algebra<\/h2>\n<p>Many chemical phenomena can be described by equations relating several different variables.\u00a0 If the values of all but one of the variables are known, the tools of algebra may be used to figure out the value of the remaining variable.<\/p>\n<p>The tools of algebra are based on balance: a mathematical equality remains an equality as long as the same operations are performed on both sides of the equation.\u00a0 The goal is always to isolate the unknown (x) <strong>in the numerator<\/strong> of one side of the equation.\u00a0 The tools include:<\/p>\n<ol>\n<li><strong>C<\/strong>ross-multiplying to eliminate fractions.<\/li>\n<li>Distributing a term that\u2019s outside of parentheses by applying it to each term inside the parentheses.<\/li>\n<li>Collecting like terms. This means bringing together parts of the equation that have the same variable or no variable.<\/li>\n<li>Adding or subtracting the same thing on both sides of the equation.<\/li>\n<li>Multiplying or dividing each side of the equation by the same thing.<\/li>\n<li>Taking the reciprocal of both sides of the equation.<\/li>\n<\/ol>\n<p>Below are some examples of solving for an unknown variable using the tools of algebra:<\/p>\n<div class=\"textbox examples\">\n<h3>Algebra Example 1<\/h3>\n<p>Solve x in the following equation:<\/p>\n<p>[latex]\\large 12x = 148[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q17022322\">Show Solution<\/span><\/p>\n<div id=\"q17022322\" class=\"hidden-answer\" style=\"display: none\">\n<p>To isolate x, divide both sides by 12:<\/p>\n<p>[latex]\\large \\frac{\\cancel{12}x}{\\cancel{12}} = \\frac{148}{12}[\/latex]<\/p>\n<p>[latex]\\large x = 12.33[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Algebra Example 2<\/h3>\n<p>Solve x in the following equation:<\/p>\n<p>[latex]\\large \\frac{x}{9} = 54[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q17022323\">Show Solution<\/span><\/p>\n<div id=\"q17022323\" class=\"hidden-answer\" style=\"display: none\">\n<p>To isolate x, multiply both sides by 9:<\/p>\n<p>[latex]\\large \\cancel{9}\\times\\frac{x}{\\cancel{9}} = 54 \\times 9[\/latex]<\/p>\n<p>[latex]\\large x = 486[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Algebra Example 3<\/h3>\n<p>Solve x in the following equation:<\/p>\n<p>[latex]\\large 3x + 5 = 23[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q17022324\">Show Solution<\/span><\/p>\n<div id=\"q17022324\" class=\"hidden-answer\" style=\"display: none\">\n<p>First isolate the x term on one side by subtracting 5 from both sides:<\/p>\n<p>[latex]\\large (3x + \\cancel{5}) - \\cancel{5} = 23 - 5[\/latex]<\/p>\n<p>Giving<\/p>\n<p>[latex]\\large 3x = 18[\/latex]<\/p>\n<p>Next, solve for x by dividing both sides by 3<\/p>\n<p>[latex]\\large \\frac{\\cancel{3}x}{\\cancel{3}} = \\frac{18}{3}[\/latex]<\/p>\n<p>[latex]\\large x = 6[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Algebra Example 4<\/h3>\n<p>Solve x in the following equation:<\/p>\n<p>[latex]\\large \\frac{x}{5} + 55 = 3x + 12[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q17022325\">Show Solution<\/span><\/p>\n<div id=\"q17022325\" class=\"hidden-answer\" style=\"display: none\">\n<p>First subtract 55 from both sides of the equation<\/p>\n<p>[latex]\\large (\\frac{x}{5} + \\cancel{55}) - \\cancel{55} = 3x + 12 - 55[\/latex]<\/p>\n<p>Giving<\/p>\n<p>[latex]\\large \\frac{x}{5} = 3x - 43[\/latex]<\/p>\n<p>Next, multiple both sides by 5<\/p>\n<p>[latex]\\large \\cancel{5} \\times\\frac{x}{\\cancel{5}} = (3x - 43)\\times 5[\/latex]<\/p>\n<p>[latex]\\large x = 15x - 215[\/latex]<\/p>\n<p>Next, subtract 15x from both sides of the equation to isolate the x term<\/p>\n<p>[latex]\\large x - 15x = (\\cancel{15x} - 215)- \\cancel{15x}[\/latex]<\/p>\n<p>Giving<\/p>\n<p>[latex]\\large -14x = -215[\/latex]<\/p>\n<p>Finally, divide both sides by -14 to solve for x<\/p>\n<p>[latex]\\large \\frac{\\cancel{-14}x}{\\cancel{-14}} = \\frac{-215}{-14}[\/latex]<\/p>\n<p>[latex]\\large x = 15.36[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1062\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li><strong>Authored by<\/strong>: Ryan Clemens. <strong>Provided by<\/strong>: Monroe Community College, Rochester,  NY. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":6181,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"\",\"author\":\"Ryan Clemens\",\"organization\":\"Monroe Community College, Rochester,  NY\",\"url\":\"\",\"project\":\"\",\"license\":\"cc0\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1062","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1062","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/users\/6181"}],"version-history":[{"count":23,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1062\/revisions"}],"predecessor-version":[{"id":1828,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1062\/revisions\/1828"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/parts\/21"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapters\/1062\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/media?parent=1062"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1062"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/contributor?post=1062"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/wp-json\/wp\/v2\/license?post=1062"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}