{"id":810,"date":"2021-09-20T00:07:29","date_gmt":"2021-09-20T00:07:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/?post_type=chapter&#038;p=810"},"modified":"2022-06-18T17:50:44","modified_gmt":"2022-06-18T17:50:44","slug":"4-2-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/4-2-2\/","title":{"raw":"4.2.2 Solving Linear Inequalities","rendered":"4.2.2 Solving Linear Inequalities"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3 style=\"text-align: center;\">Learning Outcomes<\/h3>\r\n<ul style=\"text-align: left;\">\r\n \t<li>Solve single-step linear inequalities<\/li>\r\n \t<li>Solve multi-step linear inequalities<\/li>\r\n \t<li>Write a solution set in interval notation<\/li>\r\n \t<li>Write solution set in set-builder notation<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key words<\/h3>\r\n<ul>\r\n \t<li><strong>Solution set<\/strong>: a set containing the solutions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 style=\"text-align: left;\">Properties of Inequality<\/h2>\r\nA linear inequality is similar to a linear equation in many ways, but do the properties of equality still hold? An inequality like [latex]2\\lt 5[\/latex] can be visualized as an unbalanced scale:\r\n<p style=\"text-align: start; font-size: 16px;\"><img class=\"aligncenter wp-image-1877 \" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5676\/2021\/09\/27185947\/Inequality-Scale-300x259.png\" alt=\"Picture of an unbalanced scale.\" width=\"206\" height=\"178\" \/><\/p>\r\n<p style=\"text-align: start; font-size: 16px;\">It seems logical that adding or subtracting the same amount from both sides of the scale will not change the imbalance.<\/p>\r\nConsider the simple inequality [latex]2 \\lt 5[\/latex]. Let's add [latex]4[\/latex] to each side and determine if the inequality still holds:\r\n\r\n[latex]\\begin{equation}\\begin{aligned}2 &amp; \\lt\\ 5 \\\\ 2 \\color{blue}{+4} &amp; \\lt 5\\color{blue}{+4} \\\\ 6 &amp; \\lt 9 \\end{aligned}\\end{equation}[\/latex]\r\n\r\nAdding the same number to both sides keeps the inequality in tact. Let's try subtracting [latex]3[\/latex] from both sides to see what happens:\r\n\r\n[latex]\\begin{equation}\\begin{aligned}2 &amp; \\lt\\ 5 \\\\ 2 \\color{blue}{-3} &amp; \\lt 5\\color{blue}{-3} \\\\ -1 &amp; \\lt 2 \\end{aligned}\\end{equation}[\/latex]\r\n\r\nSubtracting the same number to both sides keeps the inequality in tact.\r\n\r\nThe addition and subtraction property of equality applies to inequalities. But what of the multiplication and division properties of equality? Let's consider the inequality [latex]2\\lt 5[\/latex] again, and start by multiplying both sides by a positive number.\r\n\r\n[latex]\\begin{equation}\\begin{aligned}2 &amp; \\lt\\ 5 \\\\ 2 \\color{blue}{\\cdot 3} &amp; \\lt 5\\color{blue}{\\cdot 3} \\\\ 6 &amp; \\lt 15 \\end{aligned}\\end{equation}[\/latex]\r\n\r\nThe inequality still holds. But what about multiplying by a negative number?\r\n\r\n[latex]\\begin{equation}\\begin{aligned}2 &amp; \\lt\\ 5 \\\\ 2 \\color{blue}{\\cdot (-4)} &amp; \\lt 5\\color{blue}{\\cdot (-4)} \\\\ -8 &amp; \\lt -20 \\end{aligned}\\end{equation}[\/latex]\r\n\r\nThuis results in a false statement! To make the statement true, we must flip the [latex]\\lt[\/latex] sign to a\u00a0[latex]\\gt[\/latex] sign because\u00a0[latex]-8\\gt -20[\/latex]. The same thing happens if we try to divide by a negative number: we must flip the inequality sign.\r\n<div class=\"textbox shaded\">\r\n<h3>Properties of inequality<\/h3>\r\n<p style=\"text-align: center;\">For all real numbers [latex]a,\\,b,\\,c[\/latex], if [latex]a&lt;b[\/latex] then [latex]a+c\\lt b+c[\/latex].<\/p>\r\n<p style=\"text-align: center;\">Adding or subtracting the same term to both sides an inequality keeps the inequality true.<\/p>\r\n<p style=\"text-align: center;\">For all real numbers [latex]a,\\,b,\\,c[\/latex] with [latex]c\\gt 0[\/latex], if [latex]a&lt;b[\/latex] then [latex]a\\cdot c\\lt b\\cdot c[\/latex].<\/p>\r\n<p style=\"text-align: center;\">Multiplying or dividing by a <strong>positive term<\/strong> to both\u00a0sides an inequality keeps the inequality true.<\/p>\r\n<p style=\"text-align: center;\">For all real numbers [latex]a,\\,b,\\,c[\/latex] with [latex]c\\lt 0[\/latex], if [latex]a\\lt b[\/latex] then [latex]a\\cdot c\\gt b\\cdot c[\/latex].<\/p>\r\n<p style=\"text-align: center;\">To multiply or divide by a <strong>negative term<\/strong>\u00a0on both\u00a0sides of an inequality we must <strong>reverse<\/strong> the inequality sign.<\/p>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1rem; text-align: initial;\">Solving an linear inequality follows the same rules as solving a linear equation. In the first example, we use the multiplication and division property to isolate the variable.<\/span>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve.\r\n\r\n1. [latex]3x\\lt 6[\/latex]\u00a0 \u00a0Write your answer in set-builder notation.\r\n\r\n2. [latex]-2x - 1\\ge 5[\/latex]\u00a0 \u00a0Write your answer in interval notation.\r\n\r\n3. [latex]5-x\\geq 10[\/latex]\u00a0 \u00a0Graph your answer on a number line.\r\n<h4>Solution:<\/h4>\r\n1.\r\n[latex]\\begin{equation}\\begin{aligned} 3x &amp; &lt; 6 \\\\ \\color{blue}{\\frac{1}{3}}\\left( 3x \\right) &amp; &lt; \\left ( 6\\right )\\color{blue}{\\frac{1}{3}}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Multiply both sides by }\\frac{1}{3} \\text{ (or divide both sides by }3) \\\\ x &amp; \\lt 2 \\end{aligned}\\end{equation}[\/latex]\r\n\r\nAnswer: [latex]\\{\\;x\\;\\large |\\normalsize\\;\\;x \\lt 2,\\;x\\in\\mathbb{R}\\;\\}[\/latex]\r\n\r\n&nbsp;\r\n\r\n2.\r\n[latex]\\begin{equation}\\begin{aligned}-2x - 1 &amp; \\ge 5\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Add 1 to both sides} \\\\ -2x &amp; \\ge 6\\\\ \\color{blue}{\\left(-\\frac{1}{2}\\right)}\\left(-2x\\right) &amp; \\color{blue}{\\le} \\left(6\\right)\\color{blue}{\\left(-\\frac{1}{2}\\right)}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Multiply both sides by }-\\frac{1}{2} \\text{ and reverse the inequality sign} \\\\ x &amp; \\le -3\\end{aligned}\\end{equation}[\/latex]\r\n\r\nAnswer: [latex]x\\in \\left ( -\\infty,\\,-3\\right ][\/latex]\r\n\r\n&nbsp;\r\n\r\n3.\r\n[latex]\\begin{equation}\\begin{aligned}5-x &amp; \\geq10 \\\\ -x &amp; \\geq 5\\\\ \\color{blue}{\\left(-1\\right)}\\left(-x\\right) &amp; \\color{blue}{\\leq} \\left(5\\right)\\color{blue}{\\left(-1\\right)}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Multiply both sides by } -1\\text{ and reverse the sign}\\\\ x &amp; \\leq-5\\end{aligned}\\end{equation}[\/latex]\r\n\r\nAnswer:\u00a0<img class=\"aligncenter wp-image-1884 \" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5676\/2021\/09\/27235040\/Number-line-x%E2%89%A4-5.png\" alt=\"number line\" width=\"392\" height=\"79\" \/>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve:\r\n\r\n1. [latex]4x\\ge -8[\/latex]\u00a0 \u00a0Write your answer in set-builder notation.\r\n\r\n2. [latex]-5x\\gt 4[\/latex]\u00a0 \u00a0Write your answer in interval notation.\r\n\r\n3. [latex]-3x\\lt -9[\/latex]\u00a0Graph your answer on a number line.\r\n\r\n[reveal-answer q=\"hjm106\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm106\"]\r\n<ol>\r\n \t<li>[latex]\\{\\;x\\;\\large |\\normalsize\\;\\;x \\ge 2,\\;x\\in\\mathbb{R}\\;\\}[\/latex]<\/li>\r\n \t<li>[latex]x\\in\\left (-\\infty,\\,-\\frac{4}{5}\\right )[\/latex]<\/li>\r\n \t<li><img class=\"wp-image-1885 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5676\/2021\/09\/28000018\/x3-number-line.png\" alt=\"number line x&gt;3\" width=\"473\" height=\"96\" \/><\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn our next example, we will use the addition property to solve inequalities.\r\n<div class=\"textbox exercises\" style=\"text-align: left;\">\r\n<h3>Example<\/h3>\r\nSolve. Write the answer in interval notation.\r\n\r\n1. [latex]x - 15\\lt 4[\/latex]\r\n\r\n2. [latex]6\\ge x - 1[\/latex]\r\n\r\n3. [latex]x+7\\gt 9[\/latex]\r\n<h4>Solution:<\/h4>\r\n1.\r\n[latex]\\begin{equation}\\begin{aligned} x - 15 &amp; \\lt 4 \\\\ x - 15 \\color{blue}{+15} &amp; \\lt 4 \\color{blue}{+15}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Add 15 to both sides.}\\\\ x &amp; \\lt 19 \\end{aligned}\\end{equation}[\/latex]\r\n\r\nAnswer: [latex]x\\in\\left ( -\\infty,\\,19\\right )[\/latex]\r\n\r\n&nbsp;\r\n\r\n2.\r\n[latex]\\begin{equation}\\begin{aligned}6 &amp; \\geq x - 1 \\\\ 6\\color{blue}{+1} &amp; \\geq x - 1 \\color{blue}{+1} \\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Add 1 to both sides}. \\\\ 7 &amp; \\geq x \\end{aligned}\\end{equation}[\/latex]\r\n\r\nAnswer: [latex]x\\in\\left ( -\\infty,\\,7\\right ][\/latex]\r\n\r\n&nbsp;\r\n\r\n3.\r\n[latex]\\begin{equation}\\begin{aligned}x+7 &amp; \\gt 9\\\\ x+7 \\color{blue}{- 7} &amp; \\gt 9\\color{blue}{- 7}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Subtract 7 from both sides}.\\\\x &amp; \\gt 2\\end{aligned}\\end{equation}[\/latex]\r\n\r\nAnswer: [latex]x\\in\\left [ 2,\\,\\infty\\right )[\/latex]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve. Write the answer in interval notation.\r\n\r\n1. [latex]x-8\\lt -6[\/latex]\r\n\r\n2. [latex]x+7\\gt 7[\/latex]\r\n\r\n3. [latex]x+\\frac{2}{5}\\geq -\\frac{2}{5}[\/latex]\r\n\r\n[reveal-answer q=\"HJM540\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"HJM540\"]\r\n<ol>\r\n \t<li>[latex]x\\in\\left (-\\infty,\\,2\\right )[\/latex]<\/li>\r\n \t<li>[latex]x\\in\\left (0,\\,\\infty\\right )[\/latex]<\/li>\r\n \t<li>[latex]x\\in\\left [-\\frac{4}{5},\\,\\infty\\right )[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nThe following video shows examples of solving single-step inequalities using the multiplication and addition properties.\r\n\r\n[embed]https:\/\/youtu.be\/1Z22Xh66VFM[\/embed]\r\n\r\nThe following video shows examples of solving inequalities with the variable on the right side.\r\n\r\n[embed]https:\/\/youtu.be\/RBonYKvTCLU[\/embed]\r\n<h2 style=\"text-align: left;\">Solving Multi-Step Inequalities<\/h2>\r\nAs the previous examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations. We just have to remember to <em>reverse the sign if we multiply or divide by a negative term<\/em>. To isolate the variable and solve,\u00a0we combine like terms and perform operations with the multiplication and addition properties.\r\n<div class=\"textbox exercises\" style=\"text-align: left;\">\r\n<h3>Example<\/h3>\r\nSolve the inequality: [latex]13 - 7x\\ge 10x - 4[\/latex]. Write the solution in interval notation.\r\n\r\n&nbsp;\r\n<h4>Solution<\/h4>\r\nSolving this inequality is similar to solving an equation up until the last step.\r\n<div style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}13 - 7x &amp; \\ge 10x - 4 \\\\ 13 - 17x &amp; \\ge -4\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Move variable terms to one side of the inequality by adding }17x\\text{ to both sides}. \\\\-17x &amp; \\ge -17\\;\\;\\;\\;\\;\\;\\;\\;\\text{Isolate the variable term by subtracting }17\\text{ from both sides}. \\\\ x &amp; \\le 1\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Dividing both sides by -17 reverses the inequality}.\\end{aligned}\\end{equation}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThe solution set is given by the interval [latex]\\left(-\\infty ,1\\right][\/latex].\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]143594[\/ohm_question]\r\n\r\n<\/div>\r\nIn the next example, we solve an inequality that contains fractions. Notice how we need to reverse the inequality sign at the end because we multiply by a negative.\r\n<div class=\"textbox exercises\" style=\"text-align: left;\">\r\n<h3>Example<\/h3>\r\nSolve the following inequality and write the answer in interval notation: [latex]-\\frac{3}{4} x\\ge -\\frac{5}{8} +\\frac{2}{3} x[\/latex].\r\n\r\n&nbsp;\r\n<h4>Solution<\/h4>\r\n<div style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-\\frac{3}{4}x &amp; \\ge -\\frac{5}{8}+\\frac{2}{3}x \\\\ -\\frac{3}{4}x-\\frac{2}{3}x &amp; \\ge -\\frac{5}{8}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Make the left side the variable side}. \\\\ -\\frac{9}{12}x-\\frac{8}{12}x &amp; \\ge -\\frac{5}{8}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Build equivalent fractions with a common denominator}. \\\\ -\\frac{17}{12}x &amp; \\ge -\\frac{5}{8} \\\\ x &amp; \\le -\\frac{5}{8}\\left(-\\frac{12}{17}\\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Multiplying by a negative number reverses the inequality}.\\\\ x &amp; \\le \\frac{15}{34} \\end{aligned}\\end{equation}[\/latex]<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div>The solution set is the interval [latex]\\left(-\\infty ,\\frac{15}{34}\\right][\/latex]. This can also be written:\u00a0 [latex]x\\in\\left(-\\infty ,\\frac{15}{34}\\right][\/latex]<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\nInstead of working with the fractions in this last example, we could clear the fractions by multiplying both sides by a common denominator.\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nSolve the following inequality and write the answer in interval notation: [latex]-\\frac{3}{4} x\\ge -\\frac{5}{8} +\\frac{2}{3} x[\/latex].\r\n\r\n&nbsp;\r\n<h4>Solution<\/h4>\r\nThe least common multiple of 4, 8 and 3 is 24. We will multiply both sides of the inequality by 24. This requires applying the distributive property to the two terms on the right side of the inequality.\r\n<div style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}\\color{blue}{\\frac{24}{1}}\\left (\\frac{-3}{4}\\right ) x &amp; \\ge\\color{blue}{\\frac{24}{1}}\\left ( \\frac{-5}{8}\\right )+\\color{blue}{\\frac{24}{1}}\\left ( \\frac{2}{3}\\right ) x \\;\\;\\;\\;\\;\\text{Multiply by the least common multiple}\\\\ -18x &amp; \\ge-15+16x \\\\ -18x-16x &amp; \\ge -15\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Subtract }16x\\text{ from both sides.}\\\\ -34x &amp;\\ge -15\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Divide both sides by }-34\\text{ and reverse the sign.} \\\\ x &amp; \\le \\frac{15}{34}\\end{aligned}\\end{equation}[\/latex]<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div>The solution set is the interval [latex]\\left(-\\infty ,\\frac{15}{34}\\right][\/latex]. This can also be written:\u00a0 [latex]x\\in\\left(-\\infty ,\\frac{15}{34}\\right][\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve the following inequality and write the answer in interval notation: [latex]\\frac{1}{3} x\\le -\\frac{4}{5} +\\frac{3}{4} x[\/latex].\r\n\r\n[reveal-answer q=\"hjm260\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm260\"]\r\n\r\n[latex]x\\in\\left [\\frac{48}{25},\\,\\infty\\right )[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3 style=\"text-align: center;\">Learning Outcomes<\/h3>\n<ul style=\"text-align: left;\">\n<li>Solve single-step linear inequalities<\/li>\n<li>Solve multi-step linear inequalities<\/li>\n<li>Write a solution set in interval notation<\/li>\n<li>Write solution set in set-builder notation<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key words<\/h3>\n<ul>\n<li><strong>Solution set<\/strong>: a set containing the solutions<\/li>\n<\/ul>\n<\/div>\n<h2 style=\"text-align: left;\">Properties of Inequality<\/h2>\n<p>A linear inequality is similar to a linear equation in many ways, but do the properties of equality still hold? An inequality like [latex]2\\lt 5[\/latex] can be visualized as an unbalanced scale:<\/p>\n<p style=\"text-align: start; font-size: 16px;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1877\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5676\/2021\/09\/27185947\/Inequality-Scale-300x259.png\" alt=\"Picture of an unbalanced scale.\" width=\"206\" height=\"178\" \/><\/p>\n<p style=\"text-align: start; font-size: 16px;\">It seems logical that adding or subtracting the same amount from both sides of the scale will not change the imbalance.<\/p>\n<p>Consider the simple inequality [latex]2 \\lt 5[\/latex]. Let&#8217;s add [latex]4[\/latex] to each side and determine if the inequality still holds:<\/p>\n<p>[latex]\\begin{equation}\\begin{aligned}2 & \\lt\\ 5 \\\\ 2 \\color{blue}{+4} & \\lt 5\\color{blue}{+4} \\\\ 6 & \\lt 9 \\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Adding the same number to both sides keeps the inequality in tact. Let&#8217;s try subtracting [latex]3[\/latex] from both sides to see what happens:<\/p>\n<p>[latex]\\begin{equation}\\begin{aligned}2 & \\lt\\ 5 \\\\ 2 \\color{blue}{-3} & \\lt 5\\color{blue}{-3} \\\\ -1 & \\lt 2 \\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Subtracting the same number to both sides keeps the inequality in tact.<\/p>\n<p>The addition and subtraction property of equality applies to inequalities. But what of the multiplication and division properties of equality? Let&#8217;s consider the inequality [latex]2\\lt 5[\/latex] again, and start by multiplying both sides by a positive number.<\/p>\n<p>[latex]\\begin{equation}\\begin{aligned}2 & \\lt\\ 5 \\\\ 2 \\color{blue}{\\cdot 3} & \\lt 5\\color{blue}{\\cdot 3} \\\\ 6 & \\lt 15 \\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>The inequality still holds. But what about multiplying by a negative number?<\/p>\n<p>[latex]\\begin{equation}\\begin{aligned}2 & \\lt\\ 5 \\\\ 2 \\color{blue}{\\cdot (-4)} & \\lt 5\\color{blue}{\\cdot (-4)} \\\\ -8 & \\lt -20 \\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Thuis results in a false statement! To make the statement true, we must flip the [latex]\\lt[\/latex] sign to a\u00a0[latex]\\gt[\/latex] sign because\u00a0[latex]-8\\gt -20[\/latex]. The same thing happens if we try to divide by a negative number: we must flip the inequality sign.<\/p>\n<div class=\"textbox shaded\">\n<h3>Properties of inequality<\/h3>\n<p style=\"text-align: center;\">For all real numbers [latex]a,\\,b,\\,c[\/latex], if [latex]a<b[\/latex] then [latex]a+c\\lt b+c[\/latex].<\/p>\n<p style=\"text-align: center;\">Adding or subtracting the same term to both sides an inequality keeps the inequality true.<\/p>\n<p style=\"text-align: center;\">For all real numbers [latex]a,\\,b,\\,c[\/latex] with [latex]c\\gt 0[\/latex], if [latex]a<b[\/latex] then [latex]a\\cdot c\\lt b\\cdot c[\/latex].<\/p>\n<p style=\"text-align: center;\">Multiplying or dividing by a <strong>positive term<\/strong> to both\u00a0sides an inequality keeps the inequality true.<\/p>\n<p style=\"text-align: center;\">For all real numbers [latex]a,\\,b,\\,c[\/latex] with [latex]c\\lt 0[\/latex], if [latex]a\\lt b[\/latex] then [latex]a\\cdot c\\gt b\\cdot c[\/latex].<\/p>\n<p style=\"text-align: center;\">To multiply or divide by a <strong>negative term<\/strong>\u00a0on both\u00a0sides of an inequality we must <strong>reverse<\/strong> the inequality sign.<\/p>\n<\/div>\n<p><span style=\"font-size: 1rem; text-align: initial;\">Solving an linear inequality follows the same rules as solving a linear equation. In the first example, we use the multiplication and division property to isolate the variable.<\/span><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve.<\/p>\n<p>1. [latex]3x\\lt 6[\/latex]\u00a0 \u00a0Write your answer in set-builder notation.<\/p>\n<p>2. [latex]-2x - 1\\ge 5[\/latex]\u00a0 \u00a0Write your answer in interval notation.<\/p>\n<p>3. [latex]5-x\\geq 10[\/latex]\u00a0 \u00a0Graph your answer on a number line.<\/p>\n<h4>Solution:<\/h4>\n<p>1.<br \/>\n[latex]\\begin{equation}\\begin{aligned} 3x & < 6 \\\\ \\color{blue}{\\frac{1}{3}}\\left( 3x \\right) & < \\left ( 6\\right )\\color{blue}{\\frac{1}{3}}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Multiply both sides by }\\frac{1}{3} \\text{ (or divide both sides by }3) \\\\ x & \\lt 2 \\end{aligned}\\end{equation}[\/latex]\n\nAnswer: [latex]\\{\\;x\\;\\large |\\normalsize\\;\\;x \\lt 2,\\;x\\in\\mathbb{R}\\;\\}[\/latex]\n\n&nbsp;\n\n2.\n[latex]\\begin{equation}\\begin{aligned}-2x - 1 & \\ge 5\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Add 1 to both sides} \\\\ -2x & \\ge 6\\\\ \\color{blue}{\\left(-\\frac{1}{2}\\right)}\\left(-2x\\right) & \\color{blue}{\\le} \\left(6\\right)\\color{blue}{\\left(-\\frac{1}{2}\\right)}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Multiply both sides by }-\\frac{1}{2} \\text{ and reverse the inequality sign} \\\\ x & \\le -3\\end{aligned}\\end{equation}[\/latex]\n\nAnswer: [latex]x\\in \\left ( -\\infty,\\,-3\\right ][\/latex]\n\n&nbsp;\n\n3.\n[latex]\\begin{equation}\\begin{aligned}5-x & \\geq10 \\\\ -x & \\geq 5\\\\ \\color{blue}{\\left(-1\\right)}\\left(-x\\right) & \\color{blue}{\\leq} \\left(5\\right)\\color{blue}{\\left(-1\\right)}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Multiply both sides by } -1\\text{ and reverse the sign}\\\\ x & \\leq-5\\end{aligned}\\end{equation}[\/latex]\n\nAnswer:\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1884\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5676\/2021\/09\/27235040\/Number-line-x%E2%89%A4-5.png\" alt=\"number line\" width=\"392\" height=\"79\" \/><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve:<\/p>\n<p>1. [latex]4x\\ge -8[\/latex]\u00a0 \u00a0Write your answer in set-builder notation.<\/p>\n<p>2. [latex]-5x\\gt 4[\/latex]\u00a0 \u00a0Write your answer in interval notation.<\/p>\n<p>3. [latex]-3x\\lt -9[\/latex]\u00a0Graph your answer on a number line.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm106\">Show Answer<\/span><\/p>\n<div id=\"qhjm106\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]\\{\\;x\\;\\large |\\normalsize\\;\\;x \\ge 2,\\;x\\in\\mathbb{R}\\;\\}[\/latex]<\/li>\n<li>[latex]x\\in\\left (-\\infty,\\,-\\frac{4}{5}\\right )[\/latex]<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1885 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5676\/2021\/09\/28000018\/x3-number-line.png\" alt=\"number line x&gt;3\" width=\"473\" height=\"96\" \/><\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In our next example, we will use the addition property to solve inequalities.<\/p>\n<div class=\"textbox exercises\" style=\"text-align: left;\">\n<h3>Example<\/h3>\n<p>Solve. Write the answer in interval notation.<\/p>\n<p>1. [latex]x - 15\\lt 4[\/latex]<\/p>\n<p>2. [latex]6\\ge x - 1[\/latex]<\/p>\n<p>3. [latex]x+7\\gt 9[\/latex]<\/p>\n<h4>Solution:<\/h4>\n<p>1.<br \/>\n[latex]\\begin{equation}\\begin{aligned} x - 15 & \\lt 4 \\\\ x - 15 \\color{blue}{+15} & \\lt 4 \\color{blue}{+15}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Add 15 to both sides.}\\\\ x & \\lt 19 \\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Answer: [latex]x\\in\\left ( -\\infty,\\,19\\right )[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>2.<br \/>\n[latex]\\begin{equation}\\begin{aligned}6 & \\geq x - 1 \\\\ 6\\color{blue}{+1} & \\geq x - 1 \\color{blue}{+1} \\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Add 1 to both sides}. \\\\ 7 & \\geq x \\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Answer: [latex]x\\in\\left ( -\\infty,\\,7\\right ][\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>3.<br \/>\n[latex]\\begin{equation}\\begin{aligned}x+7 & \\gt 9\\\\ x+7 \\color{blue}{- 7} & \\gt 9\\color{blue}{- 7}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Subtract 7 from both sides}.\\\\x & \\gt 2\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Answer: [latex]x\\in\\left [ 2,\\,\\infty\\right )[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve. Write the answer in interval notation.<\/p>\n<p>1. [latex]x-8\\lt -6[\/latex]<\/p>\n<p>2. [latex]x+7\\gt 7[\/latex]<\/p>\n<p>3. [latex]x+\\frac{2}{5}\\geq -\\frac{2}{5}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qHJM540\">Show Answer<\/span><\/p>\n<div id=\"qHJM540\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]x\\in\\left (-\\infty,\\,2\\right )[\/latex]<\/li>\n<li>[latex]x\\in\\left (0,\\,\\infty\\right )[\/latex]<\/li>\n<li>[latex]x\\in\\left [-\\frac{4}{5},\\,\\infty\\right )[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The following video shows examples of solving single-step inequalities using the multiplication and addition properties.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Solving One Step Inequalities by Adding and Subtracting (Variable Left Side)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/1Z22Xh66VFM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The following video shows examples of solving inequalities with the variable on the right side.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Solving One Step Inequalities by Adding and Subtracting (Variable Right Side)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/RBonYKvTCLU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 style=\"text-align: left;\">Solving Multi-Step Inequalities<\/h2>\n<p>As the previous examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations. We just have to remember to <em>reverse the sign if we multiply or divide by a negative term<\/em>. To isolate the variable and solve,\u00a0we combine like terms and perform operations with the multiplication and addition properties.<\/p>\n<div class=\"textbox exercises\" style=\"text-align: left;\">\n<h3>Example<\/h3>\n<p>Solve the inequality: [latex]13 - 7x\\ge 10x - 4[\/latex]. Write the solution in interval notation.<\/p>\n<p>&nbsp;<\/p>\n<h4>Solution<\/h4>\n<p>Solving this inequality is similar to solving an equation up until the last step.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}13 - 7x & \\ge 10x - 4 \\\\ 13 - 17x & \\ge -4\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Move variable terms to one side of the inequality by adding }17x\\text{ to both sides}. \\\\-17x & \\ge -17\\;\\;\\;\\;\\;\\;\\;\\;\\text{Isolate the variable term by subtracting }17\\text{ from both sides}. \\\\ x & \\le 1\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Dividing both sides by -17 reverses the inequality}.\\end{aligned}\\end{equation}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>The solution set is given by the interval [latex]\\left(-\\infty ,1\\right][\/latex].<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm143594\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=143594&theme=oea&iframe_resize_id=ohm143594&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the next example, we solve an inequality that contains fractions. Notice how we need to reverse the inequality sign at the end because we multiply by a negative.<\/p>\n<div class=\"textbox exercises\" style=\"text-align: left;\">\n<h3>Example<\/h3>\n<p>Solve the following inequality and write the answer in interval notation: [latex]-\\frac{3}{4} x\\ge -\\frac{5}{8} +\\frac{2}{3} x[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<h4>Solution<\/h4>\n<div style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}-\\frac{3}{4}x & \\ge -\\frac{5}{8}+\\frac{2}{3}x \\\\ -\\frac{3}{4}x-\\frac{2}{3}x & \\ge -\\frac{5}{8}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Make the left side the variable side}. \\\\ -\\frac{9}{12}x-\\frac{8}{12}x & \\ge -\\frac{5}{8}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Build equivalent fractions with a common denominator}. \\\\ -\\frac{17}{12}x & \\ge -\\frac{5}{8} \\\\ x & \\le -\\frac{5}{8}\\left(-\\frac{12}{17}\\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; \\text{Multiplying by a negative number reverses the inequality}.\\\\ x & \\le \\frac{15}{34} \\end{aligned}\\end{equation}[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div>The solution set is the interval [latex]\\left(-\\infty ,\\frac{15}{34}\\right][\/latex]. This can also be written:\u00a0 [latex]x\\in\\left(-\\infty ,\\frac{15}{34}\\right][\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Instead of working with the fractions in this last example, we could clear the fractions by multiplying both sides by a common denominator.<\/p>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Solve the following inequality and write the answer in interval notation: [latex]-\\frac{3}{4} x\\ge -\\frac{5}{8} +\\frac{2}{3} x[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<h4>Solution<\/h4>\n<p>The least common multiple of 4, 8 and 3 is 24. We will multiply both sides of the inequality by 24. This requires applying the distributive property to the two terms on the right side of the inequality.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}\\color{blue}{\\frac{24}{1}}\\left (\\frac{-3}{4}\\right ) x & \\ge\\color{blue}{\\frac{24}{1}}\\left ( \\frac{-5}{8}\\right )+\\color{blue}{\\frac{24}{1}}\\left ( \\frac{2}{3}\\right ) x \\;\\;\\;\\;\\;\\text{Multiply by the least common multiple}\\\\ -18x & \\ge-15+16x \\\\ -18x-16x & \\ge -15\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Subtract }16x\\text{ from both sides.}\\\\ -34x &\\ge -15\\;\\;\\;\\;\\;\\;\\;\\;\\;\\text{Divide both sides by }-34\\text{ and reverse the sign.} \\\\ x & \\le \\frac{15}{34}\\end{aligned}\\end{equation}[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div><\/div>\n<div>The solution set is the interval [latex]\\left(-\\infty ,\\frac{15}{34}\\right][\/latex]. This can also be written:\u00a0 [latex]x\\in\\left(-\\infty ,\\frac{15}{34}\\right][\/latex]<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve the following inequality and write the answer in interval notation: [latex]\\frac{1}{3} x\\le -\\frac{4}{5} +\\frac{3}{4} x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm260\">Show Answer<\/span><\/p>\n<div id=\"qhjm260\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x\\in\\left [\\frac{48}{25},\\,\\infty\\right )[\/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-810\">\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>Revision and Adaptation. Provided by: Lumen Learning. . <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Properties of Inequality; Clearing Fractions Example and Try It. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex. Solving One Step Inequalities by Adding and Subtracting (Variable Left Side). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/1Z22Xh66VFM\">https:\/\/youtu.be\/1Z22Xh66VFM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Solving One Step Inequalities by Adding and Subtracting (Variable Right Side). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/RBonYKvTCLU\">https:\/\/youtu.be\/RBonYKvTCLU<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Solve One Step Linear Inequality by Dividing (Variable Left). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/IajiD3R7U-0\">https:\/\/youtu.be\/IajiD3R7U-0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Solve One Step Linear Inequality by Dividing (Variable Right). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/s9fJOnVTHhs\">https:\/\/youtu.be\/s9fJOnVTHhs<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/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":370291,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation. Provided by: Lumen Learning. \",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex. Solving One Step Inequalities by Adding and Subtracting (Variable Left Side)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/1Z22Xh66VFM\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Solving One Step Inequalities by Adding and Subtracting (Variable Right Side)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/RBonYKvTCLU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Solve One Step Linear Inequality by Dividing (Variable Left)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/IajiD3R7U-0\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Solve One Step Linear Inequality by Dividing (Variable Right)\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/s9fJOnVTHhs\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Properties of Inequality; Clearing Fractions Example and Try It\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"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-810","chapter","type-chapter","status-publish","hentry"],"part":642,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/810","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/users\/370291"}],"version-history":[{"count":15,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/810\/revisions"}],"predecessor-version":[{"id":2180,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/810\/revisions\/2180"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/parts\/642"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/810\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/media?parent=810"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=810"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/contributor?post=810"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/license?post=810"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}