{"id":4547,"date":"2017-06-07T18:50:51","date_gmt":"2017-06-07T18:50:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-solve-single-step-inequalities\/"},"modified":"2017-08-15T15:23:23","modified_gmt":"2017-08-15T15:23:23","slug":"read-solve-single-step-inequalities","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-solve-single-step-inequalities\/","title":{"raw":"Solve Single-Step Inequalities","rendered":"Solve Single-Step Inequalities"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Solve single-step inequalities\r\n<ul>\r\n \t<li>Use the addition and multiplication properties to solve algebraic inequalities and express their solutions graphically and with interval notation<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Solve inequalities with addition and subtraction<\/h2>\r\nYou can solve most inequalities using inverse operations \u00a0as you did for solving equations. \u00a0This is because when you add or subtract the same value from both sides of an inequality, you have maintained the inequality. These properties are outlined in the box below.\r\n<div class=\"textbox shaded\">\r\n<h3>Addition and Subtraction Properties of Inequality<\/h3>\r\nIf [latex]a&gt;b[\/latex],<i>\u00a0<\/i>then [latex]a+c&gt;b+c[\/latex].\r\n\r\nIf\u00a0[latex]a&gt;b[\/latex]<i>, <\/i>then [latex]a\u2212c&gt;b\u2212c[\/latex].\r\n\r\n<\/div>\r\nBecause inequalities have multiple possible solutions, representing the solutions graphically provides a helpful visual of the situation, as we saw in the last section. The example below shows the steps to solve and graph an inequality and express the solution using interval notation.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>x.<\/i>\r\n<p style=\"text-align: center\">[latex] {x}+3\\lt{5}[\/latex]<\/p>\r\n[reveal-answer q=\"952771\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"952771\"]\r\n\r\nIt is helpful to think of this inequality as asking you to find all the values for <em>x<\/em>, including negative numbers, such that when you add three you will get a number less than 5.\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{l}x+3&lt;\\,\\,\\,\\,5\\\\\\underline{\\,\\,\\,\\,\\,-3\\,\\,\\,\\,-3}\\\\x\\,\\,\\,\\,\\,\\,\\,\\,&lt;\\,\\,\\,\\,2\\,\\,\\end{array}[\/latex]<\/p>\r\nIsolate the variable by subtracting 3 from both sides of the inequality.\r\n<h4>Answer<\/h4>\r\nInequality: \u00a0[latex]x&lt;2[\/latex]\r\n\r\nInterval: \u00a0[latex]\\left(-\\infty, 2\\right)[\/latex]\r\n\r\nGraph: <img class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/MITEdevmath\/NROCUnit10_files\/image036.jpg#fixme#fixme\" alt=\"Number line. Open circle around 2. Shaded line through all numbers less than 2.\" width=\"575\" height=\"31\" \/>[\/hidden-answer]\r\n\r\n<\/div>\r\n<p style=\"text-align: left\">The line represents <em>all<\/em> the numbers to which\u00a0you can add 3 and get a number that is less than 5. There's a lot of numbers that solve this inequality!<\/p>\r\nJust as you can check the solution to an equation, you can check a solution to an inequality. First, you check the end point by substituting it in the related equation. Then you check to see if the inequality is correct by substituting any other solution to see if it is one of the solutions. Because there are multiple solutions, it is a good practice to check more than one of the possible solutions. This can also help you check that your graph is correct.\r\n\r\nThe example below shows how you could check that [latex]x&lt;2[\/latex]<i>\u00a0<\/i>is the solution to [latex]x+3&lt;5[\/latex]<i>.<\/i>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nCheck that [latex]x&lt;2[\/latex]<i>\u00a0<\/i>is the solution to [latex]x+3&lt;5[\/latex].\r\n\r\n[reveal-answer q=\"811564\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"811564\"]\r\n\r\nSubstitute the end point 2 into the related equation, [latex]x+3=5[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}x+3=5 \\\\ 2+3=5 \\\\ 5=5\\end{array}[\/latex]<\/p>\r\nPick a value less than 2, such as 0, to check into the inequality. (This value will be on the shaded part of the graph.)\r\n<p style=\"text-align: center\"><img class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/MITEdevmath\/NROCUnit10_files\/image036.jpg#fixme#fixme\" alt=\"Number line. Open circle around 2. Shaded line through all numbers less than 2.\" width=\"575\" height=\"31\" \/><\/p>\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}x+3&lt;5 \\\\ 0+3&lt;5 \\\\ 3&lt;5\\end{array}[\/latex]<\/p>\r\nIt checks!\r\n\r\n[latex]x&lt;2[\/latex] is the solution to [latex]x+3&lt;5[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following examples show inequality problems that include operations with negative numbers. The graph of the solution to the inequality is also shown. Remember to check the solution. This is a good habit to build!\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <em>x<\/em>:\u00a0[latex]x-10\\leq-12[\/latex]\r\n[reveal-answer q=\"815894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"815894\"]\r\n\r\nIsolate the variable by adding 10 to both sides of the inequality.\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}x-10\\le -12\\\\\\underline{\\,\\,\\,+10\\,\\,\\,\\,\\,+10}\\\\x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\le \\,\\,\\,-2\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nInequality: [latex]x\\leq-2[\/latex]\r\nInterval: [latex]\\left(-\\infty,-2\\right][\/latex]\r\nGraph: Notice that a closed circle is used because the inequality is \u201cless than or equal to\u201d [latex]\\left(\\leq\\right)[\/latex]. The blue arrow is drawn to the left of the point [latex]\u22122[\/latex] because these are the values that are less than [latex]\u22122[\/latex].\r\n<img class=\"aligncenter wp-image-3619\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185044\/image038-300x17.jpg\" alt=\"Number line, closed circle on negative 2 and line drawn through all numbers less than negative 2\" width=\"529\" height=\"30\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\nCheck the\u00a0solution to [latex]x-10\\leq -12[\/latex]\r\n[reveal-answer q=\"268062\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"268062\"]\r\n\r\nSubstitute the end point [latex]\u22122[\/latex] into the related equation \u00a0[latex]x-10=\u221212[\/latex]\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}x-10=-12\\,\\,\\,\\\\\\text{Does}\\,\\,\\,-2-10=-12?\\\\-12=-12\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nPick a value less than [latex]\u22122[\/latex], such as [latex]\u22125[\/latex], to check in the inequality. (This value will be on the shaded part of the graph.)\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}x-10\\le -12\\,\\,\\,\\\\\\text{ }\\,\\text{ Is}\\,\\,-5-10\\le -12?\\\\-15\\le -12\\,\\,\\,\\\\\\text{It}\\,\\text{checks!}\\end{array}[\/latex]<\/p>\r\n[latex]x\\leq -2[\/latex]\u00a0is the solution to [latex]x-10\\leq -12[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve for <em>a<\/em>. [latex]a-17&gt;-17[\/latex]\r\n[reveal-answer q=\"343031\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"343031\"]\r\n\r\nIsolate the variable by adding 17 to both sides of the inequality.\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}a-17&gt;-17\\\\\\underline{\\,\\,\\,+17\\,\\,\\,\\,\\,+17}\\\\a\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,&gt;\\,\\,\\,\\,\\,\\,0\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nInequality:\u00a0[latex] \\displaystyle a\\,\\,&gt;\\,0[\/latex]\r\n\r\nInterval: [latex]\\left(0,\\infty\\right)[\/latex] \u00a0Note how we use parentheses on the left to show that the solution does not include 0.\r\n\r\nGraph: Note the open circle to show that the solution does not include 0.\r\n<div class=\"bcc-box bcc-info\">\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/MITEdevmath\/NROCUnit10_files\/image044.jpg#fixme#fixme\" alt=\"Number line. Open circle on zero. Highlight through all numbers above zero.\" width=\"575\" height=\"32\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\nCheck the solution to\u00a0[latex]a-17&gt;-17[\/latex]\r\n[reveal-answer q=\"653357\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"653357\"]\r\n\r\nIs\u00a0[latex] \\displaystyle a\\,\\,&gt;\\,0[\/latex] the correct solution to\u00a0\u00a0[latex]a-17&gt;-17[\/latex]?\r\n\r\nSubstitute the end point 0 into the related equation.\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}a-17=-17\\,\\,\\,\\\\\\text{Does}\\,\\,\\,0-17=-17?\\\\-17=-17\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nPick a value greater than 0, such as 20, to check in the inequality. (This value will be on the shaded part of the graph.)\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}a-17&gt;-17\\,\\,\\,\\\\\\text{Is }\\,\\,20-17&gt;-17?\\\\3&gt;-17\\,\\,\\,\\\\\\\\\\text{It checks!}\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[latex] \\displaystyle a\\,&gt;\\,0[\/latex] is the solution to\u00a0[latex]a-17&gt;-17[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nThe previous examples showed you how to solve a one-step inequality with the variable on the left hand side. \u00a0The following video provides examples of how to solve the same type of inequality.\r\n\r\nhttps:\/\/youtu.be\/1Z22Xh66VFM\r\n\r\nWhat would you do if the variable were on the right side of the inequality? \u00a0In the following example, you will see how to handle this scenario.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve for <em>x<\/em>:\u00a0[latex]4\\geq{x}+5[\/latex]\r\n[reveal-answer q=\"815893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"815893\"]\r\n\r\nIsolate the variable by adding 10 to both sides of the inequality.\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}4\\geq{x}+5 \\\\\\underline{\\,\\,\\,-5\\,\\,\\,\\,\\,-5}\\\\-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\ge \\,\\,\\,x\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Rewrite the inequality with the variable on the left - this makes writing the interval and drawing the graph easier.<\/p>\r\n<p style=\"text-align: center\">[latex]x\\le{-1}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Note how the the pointy part of the inequality is still directed at the variable, so instead of reading as negative one is greater or equal to x, it now reads as x is less than or equal to negative one.<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nInequality: [latex]x\\le{-1}[\/latex] This can also be written as\r\nInterval: [latex]\\left(-\\infty,-1\\right][\/latex]\r\nGraph: Notice that a closed circle is used because the inequality is \u201cless than or equal to\u201d . The blue arrow is drawn to the left of the point [latex]\u22121[\/latex] because these are the values that are less than [latex]\u22121[\/latex].\r\n<img class=\"wp-image-4022 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185047\/Screen-Shot-2016-05-11-at-6.23.24-PM-300x57.png\" alt=\"(-oo,-1]\" width=\"400\" height=\"76\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\nCheck the\u00a0solution to [latex]4\\geq{x}+5[\/latex]\r\n[reveal-answer q=\"568062\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"568062\"]\r\n\r\nSubstitute the end point [latex]\u22121[\/latex] into the related equation \u00a0[latex]4=x+5[\/latex]\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}4=x+5\\,\\,\\,\\\\\\text{Does}\\,\\,\\,4=-1+5?\\\\-1=-1\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nPick a value less than [latex]\u22121[\/latex], such as [latex]\u22125[\/latex], to check in the inequality. (This value will be on the shaded part of the graph.)\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}4\\geq{-5}+5\\,\\,\\,\\\\\\text{ }\\,\\text{ Is}\\,\\,4\\ge 0?\\\\\\text{It}\\,\\text{checks!}\\end{array}[\/latex]<\/p>\r\n[latex]x\\le{-1}[\/latex] is the solution to [latex]4\\geq{x}+5[\/latex]<span style=\"line-height: 1.5\">[\/hidden-answer] <\/span>\r\n\r\n<\/div>\r\nThe following video show examples of solving inequalities with the variable on the right side.\r\nhttps:\/\/youtu.be\/RBonYKvTCLU\r\n<h2 id=\"title4\">Solve inequalities with multiplication and division<\/h2>\r\nSolving an inequality with a variable that has a coefficient other than 1 usually involves multiplication or division. The steps are like solving one-step equations involving multiplication or division EXCEPT for the inequality sign. Let\u2019s look at what happens to the inequality when you multiply or divide each side by the same number.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Let's start with the true statement:\r\n\r\n[latex]10&gt;5[\/latex]<\/td>\r\n<td>Let's try again by starting with the same true statement:\r\n\r\n[latex]10&gt;5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Next, multiply both sides by the same positive number:\r\n\r\n[latex]10\\cdot 2&gt;5\\cdot 2[\/latex]<\/td>\r\n<td>This time, multiply both sides by the same negative number:\r\n\r\n[latex]10\\cdot-2&gt;5 \\\\ \\,\\,\\,\\,\\,\\cdot -2\\,\\cdot-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>20 is greater than 10, so you still have a true inequality:\r\n\r\n[latex]20&gt;10[\/latex]<\/td>\r\n<td>Wait a minute! [latex]\u221220[\/latex] is <i>not <\/i>greater than [latex]\u221210[\/latex], so you have an untrue statement.\r\n\r\n[latex]\u221220&gt;\u221210[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>When you multiply by a positive number, leave the inequality sign as it is!<\/td>\r\n<td>You must \u201creverse\u201d the inequality sign to make the statement true:\r\n\r\n[latex]\u221220&lt;\u221210[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox shaded\">\r\n\r\n<img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07184952\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"80\" height=\"70\" \/>Caution! \u00a0When you multiply or divide by a negative number, \u201creverse\u201d the inequality sign. \u00a0 Whenever you multiply or divide both sides of an inequality by a negative\u00a0number, the inequality sign must be reversed in order to keep a true statement. These rules are summarized in the box below.\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3>Multiplication and Division Properties of Inequality<\/h3>\r\n<table style=\"height: 162px;width: 419px\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Start With<\/strong><\/td>\r\n<td><strong>Multiply By<\/strong><\/td>\r\n<td><strong>Final Inequality<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]a&gt;b[\/latex]<\/td>\r\n<td>[latex]c[\/latex]<\/td>\r\n<td>[latex]ac&gt;bc[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]a&gt;b[\/latex]<\/td>\r\n<td>[latex]-c[\/latex]<\/td>\r\n<td>[latex]ac&lt;bc[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table style=\"height: 77px;width: 418px\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Start With<\/strong><\/td>\r\n<td><strong>Divide By<\/strong><\/td>\r\n<td><strong>Final Inequality<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]a&gt;b[\/latex]<\/td>\r\n<td>[latex]c[\/latex]<\/td>\r\n<td>[latex] \\displaystyle \\frac{a}{c}&gt;\\frac{b}{c}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]a&gt;b[\/latex]<\/td>\r\n<td>[latex]-c[\/latex]<\/td>\r\n<td>[latex] \\displaystyle \\frac{a}{c}&lt;\\frac{b}{c}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nKeep in mind that you only change the sign when you are multiplying and dividing by a <i>negative<\/i> number. If you <em>add or subtract<\/em> by a negative\u00a0number, the inequality stays the same.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>x.\u00a0<\/i>[latex]3x&gt;12[\/latex]\r\n\r\n[reveal-answer q=\"691711\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"691711\"]Divide both sides by 3 to isolate the variable.\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}\\underline{3x}&gt;\\underline{12}\\\\3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\\\x&gt;4\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nCheck your solution by first checking the end point 4, and then checking another solution for the inequality.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}3\\cdot4=12\\\\12=12\\\\3\\cdot10&gt;12\\\\30&gt;12\\\\\\text{It checks!}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n<p style=\"text-align: left\">Inequality: [latex] \\displaystyle x&gt;4[\/latex]<\/p>\r\n<p style=\"text-align: left\">Interval: [latex]\\left(4,\\infty\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left\">Graph: <img class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/MITEdevmath\/NROCUnit10_files\/image050.jpg#fixme#fixme\" alt=\"Number line. Open circle on 4. Highlight through all numbers greater than 4.\" width=\"575\" height=\"31\" \/><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThere was no need to make any changes to the inequality sign because both sides of the inequality were divided by <i>positive<\/i> 3. In the next example, there is division by a negative number, so there is an additional step in the solution!\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <em>x<\/em>. [latex]\u22122x&gt;6[\/latex]\r\n\r\n[reveal-answer q=\"604033\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"604033\"]Divide each side of the inequality by [latex]\u22122[\/latex] to isolate the variable, and change the direction of the inequality sign because of the division by a negative number.\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{r}\\underline{-2x}&lt;\\underline{\\,6\\,}\\\\-2\\,\\,\\,\\,-2\\,\\\\x&lt;-3\\end{array}[\/latex]<\/p>\r\nCheck your solution by first checking the end point [latex]\u22123[\/latex], and then checking another solution for the inequality.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{r}-2\\left(-3\\right)=6 \\\\6=6\\\\ -2\\left(-6\\right)&gt;6 \\\\ 12&gt;6\\end{array}[\/latex]<\/p>\r\nIt checks!\r\n<h4>Answer<\/h4>\r\nInequality: [latex] \\displaystyle x&lt;-3[\/latex]\r\n\r\nInterval: [latex]\\left(-\\infty, -3\\right)[\/latex]\r\n\r\nGraph: <img class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/MITEdevmath\/NROCUnit10_files\/image051.jpg#fixme#fixme\" alt=\"Number line. Open circle on negative 3. Highlight on all numbers less than negative 3.\" width=\"575\" height=\"31\" \/>\r\nBecause both sides of the inequality were divided by a negative number, [latex]\u22122[\/latex], the inequality symbol was switched from &gt; to &lt;.\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows examples of solving one step inequalities using the multiplication property of equality where the variable is on the left hand side.\r\n\r\nhttps:\/\/youtu.be\/IajiD3R7U-0\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nBefore you read the solution to the next example, think about what properties of inequalities you may need to use to solve the inequality. What is different about this example from the previous one? Write your ideas in the box below.\r\n\r\nSolve for <em>x<\/em>. [latex]-\\frac{1}{2}&gt;-12x[\/latex]\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"811465\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"811465\"]\r\n\r\nThis inequality has the variable on the right hand side, which is different from the previous examples. Start the solution process as before, and at the end, you can move the variable to the left to write the final solution.\r\n\r\nDivide both sides by [latex]-12[\/latex] to isolate the variable. Since you are dividing by a negative number, you need to change the direction of the inequality sign.\r\n<p style=\"text-align: center\">[latex]\\displaystyle\\begin{array}{l}-\\frac{1}{2}\\gt{-12x}\\\\\\\\\\frac{-\\frac{1}{2}}{-12}\\gt\\frac{-12x}{-12}\\\\\\end{array}[\/latex]<\/p>\r\nDividing a fraction by an integer requires you to multiply by the reciprocal, and the reciprocal of [latex]-12[\/latex] is [latex]\\frac{1}{-12}[\/latex]\r\n<p style=\"text-align: center\">[latex]\\displaystyle\\begin{array}{r}\\left(-\\frac{1}{12}\\right)\\left(-\\frac{1}{2}\\right)\\lt\\frac{-12x}{-12}\\,\\,\\\\\\\\ \\frac{1}{24}\\lt\\frac{\\cancel{-12}x}{\\cancel{-12}}\\\\\\\\ \\frac{1}{24}\\lt{x}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer<\/h4>\r\n<p style=\"text-align: left\">Inequality: [latex]\\frac{1}{24}\\lt{x}[\/latex] \u00a0This can also be written with the variable on the left as [latex]x\\gt\\frac{1}{24}[\/latex]. \u00a0Writing the inequality with the variable on the left requires a little thinking, but helps you write the interval and draw the graph correctly.<\/p>\r\n<p style=\"text-align: left\">Interval: [latex]\\left(\\frac{1}{24},\\infty\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left\">Graph:\u00a0<img class=\"aligncenter wp-image-3950\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185050\/Screen-Shot-2016-05-10-at-2.09.52-PM-300x57.png\" alt=\"Open dot on zero with a line through all numbers greater than zero.\" width=\"389\" height=\"74\" \/><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video gives examples of how to solve an inequality with the multiplication property of equality where the variable is on the right hand side.\r\n\r\nhttps:\/\/youtu.be\/s9fJOnVTHhs\r\n<h2>Summary<\/h2>\r\nSolving inequalities is very similar to solving equations, except you have to reverse the inequality symbols when you multiply or divide both sides of an inequality by a negative number. There are three ways to represent solutions to inequalities: an interval, a graph, and an inequality.\u00a0Because there is usually more than one solution to an inequality, when you check your answer you should check the end point and one other value to check the direction of the inequality.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Solve single-step inequalities\n<ul>\n<li>Use the addition and multiplication properties to solve algebraic inequalities and express their solutions graphically and with interval notation<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h2>Solve inequalities with addition and subtraction<\/h2>\n<p>You can solve most inequalities using inverse operations \u00a0as you did for solving equations. \u00a0This is because when you add or subtract the same value from both sides of an inequality, you have maintained the inequality. These properties are outlined in the box below.<\/p>\n<div class=\"textbox shaded\">\n<h3>Addition and Subtraction Properties of Inequality<\/h3>\n<p>If [latex]a>b[\/latex],<i>\u00a0<\/i>then [latex]a+c>b+c[\/latex].<\/p>\n<p>If\u00a0[latex]a>b[\/latex]<i>, <\/i>then [latex]a\u2212c>b\u2212c[\/latex].<\/p>\n<\/div>\n<p>Because inequalities have multiple possible solutions, representing the solutions graphically provides a helpful visual of the situation, as we saw in the last section. The example below shows the steps to solve and graph an inequality and express the solution using interval notation.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>x.<\/i><\/p>\n<p style=\"text-align: center\">[latex]{x}+3\\lt{5}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q952771\">Show Solution<\/span><\/p>\n<div id=\"q952771\" class=\"hidden-answer\" style=\"display: none\">\n<p>It is helpful to think of this inequality as asking you to find all the values for <em>x<\/em>, including negative numbers, such that when you add three you will get a number less than 5.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{l}x+3<\\,\\,\\,\\,5\\\\\\underline{\\,\\,\\,\\,\\,-3\\,\\,\\,\\,-3}\\\\x\\,\\,\\,\\,\\,\\,\\,\\,<\\,\\,\\,\\,2\\,\\,\\end{array}[\/latex]<\/p>\n<p>Isolate the variable by subtracting 3 from both sides of the inequality.<\/p>\n<h4>Answer<\/h4>\n<p>Inequality: \u00a0[latex]x<2[\/latex]\n\nInterval: \u00a0[latex]\\left(-\\infty, 2\\right)[\/latex]\n\nGraph: <img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/MITEdevmath\/NROCUnit10_files\/image036.jpg#fixme#fixme\" alt=\"Number line. Open circle around 2. Shaded line through all numbers less than 2.\" width=\"575\" height=\"31\" \/><\/div>\n<\/div>\n<\/div>\n<p style=\"text-align: left\">The line represents <em>all<\/em> the numbers to which\u00a0you can add 3 and get a number that is less than 5. There&#8217;s a lot of numbers that solve this inequality!<\/p>\n<p>Just as you can check the solution to an equation, you can check a solution to an inequality. First, you check the end point by substituting it in the related equation. Then you check to see if the inequality is correct by substituting any other solution to see if it is one of the solutions. Because there are multiple solutions, it is a good practice to check more than one of the possible solutions. This can also help you check that your graph is correct.<\/p>\n<p>The example below shows how you could check that [latex]x<2[\/latex]<i>\u00a0<\/i>is the solution to [latex]x+3<5[\/latex]<i>.<\/i><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Check that [latex]x<2[\/latex]<i>\u00a0<\/i>is the solution to [latex]x+3<5[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q811564\">Show Solution<\/span><\/p>\n<div id=\"q811564\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the end point 2 into the related equation, [latex]x+3=5[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}x+3=5 \\\\ 2+3=5 \\\\ 5=5\\end{array}[\/latex]<\/p>\n<p>Pick a value less than 2, such as 0, to check into the inequality. (This value will be on the shaded part of the graph.)<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/MITEdevmath\/NROCUnit10_files\/image036.jpg#fixme#fixme\" alt=\"Number line. Open circle around 2. Shaded line through all numbers less than 2.\" width=\"575\" height=\"31\" \/><\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}x+3<5 \\\\ 0+3<5 \\\\ 3<5\\end{array}[\/latex]<\/p>\n<p>It checks!<\/p>\n<p>[latex]x<2[\/latex] is the solution to [latex]x+3<5[\/latex].<\/div>\n<\/div>\n<\/div>\n<p>The following examples show inequality problems that include operations with negative numbers. The graph of the solution to the inequality is also shown. Remember to check the solution. This is a good habit to build!<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <em>x<\/em>:\u00a0[latex]x-10\\leq-12[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q815894\">Show Solution<\/span><\/p>\n<div id=\"q815894\" class=\"hidden-answer\" style=\"display: none\">\n<p>Isolate the variable by adding 10 to both sides of the inequality.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}x-10\\le -12\\\\\\underline{\\,\\,\\,+10\\,\\,\\,\\,\\,+10}\\\\x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\le \\,\\,\\,-2\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>Inequality: [latex]x\\leq-2[\/latex]<br \/>\nInterval: [latex]\\left(-\\infty,-2\\right][\/latex]<br \/>\nGraph: Notice that a closed circle is used because the inequality is \u201cless than or equal to\u201d [latex]\\left(\\leq\\right)[\/latex]. The blue arrow is drawn to the left of the point [latex]\u22122[\/latex] because these are the values that are less than [latex]\u22122[\/latex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3619\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185044\/image038-300x17.jpg\" alt=\"Number line, closed circle on negative 2 and line drawn through all numbers less than negative 2\" width=\"529\" height=\"30\" \/><\/p>\n<\/div>\n<\/div>\n<p>Check the\u00a0solution to [latex]x-10\\leq -12[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q268062\">Show Solution<\/span><\/p>\n<div id=\"q268062\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the end point [latex]\u22122[\/latex] into the related equation \u00a0[latex]x-10=\u221212[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}x-10=-12\\,\\,\\,\\\\\\text{Does}\\,\\,\\,-2-10=-12?\\\\-12=-12\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Pick a value less than [latex]\u22122[\/latex], such as [latex]\u22125[\/latex], to check in the inequality. (This value will be on the shaded part of the graph.)<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}x-10\\le -12\\,\\,\\,\\\\\\text{ }\\,\\text{ Is}\\,\\,-5-10\\le -12?\\\\-15\\le -12\\,\\,\\,\\\\\\text{It}\\,\\text{checks!}\\end{array}[\/latex]<\/p>\n<p>[latex]x\\leq -2[\/latex]\u00a0is the solution to [latex]x-10\\leq -12[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve for <em>a<\/em>. [latex]a-17>-17[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q343031\">Show Solution<\/span><\/p>\n<div id=\"q343031\" class=\"hidden-answer\" style=\"display: none\">\n<p>Isolate the variable by adding 17 to both sides of the inequality.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}a-17>-17\\\\\\underline{\\,\\,\\,+17\\,\\,\\,\\,\\,+17}\\\\a\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,>\\,\\,\\,\\,\\,\\,0\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>Inequality:\u00a0[latex]\\displaystyle a\\,\\,>\\,0[\/latex]<\/p>\n<p>Interval: [latex]\\left(0,\\infty\\right)[\/latex] \u00a0Note how we use parentheses on the left to show that the solution does not include 0.<\/p>\n<p>Graph: Note the open circle to show that the solution does not include 0.<\/p>\n<div class=\"bcc-box bcc-info\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/MITEdevmath\/NROCUnit10_files\/image044.jpg#fixme#fixme\" alt=\"Number line. Open circle on zero. Highlight through all numbers above zero.\" width=\"575\" height=\"32\" \/><\/p>\n<\/div>\n<\/div>\n<p>Check the solution to\u00a0[latex]a-17>-17[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q653357\">Show Solution<\/span><\/p>\n<div id=\"q653357\" class=\"hidden-answer\" style=\"display: none\">\n<p>Is\u00a0[latex]\\displaystyle a\\,\\,>\\,0[\/latex] the correct solution to\u00a0\u00a0[latex]a-17>-17[\/latex]?<\/p>\n<p>Substitute the end point 0 into the related equation.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}a-17=-17\\,\\,\\,\\\\\\text{Does}\\,\\,\\,0-17=-17?\\\\-17=-17\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Pick a value greater than 0, such as 20, to check in the inequality. (This value will be on the shaded part of the graph.)<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}a-17>-17\\,\\,\\,\\\\\\text{Is }\\,\\,20-17>-17?\\\\3>-17\\,\\,\\,\\\\\\\\\\text{It checks!}\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>[latex]\\displaystyle a\\,>\\,0[\/latex] is the solution to\u00a0[latex]a-17>-17[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>The previous examples showed you how to solve a one-step inequality with the variable on the left hand side. \u00a0The following video provides examples of how to solve the same type of inequality.<\/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>What would you do if the variable were on the right side of the inequality? \u00a0In the following example, you will see how to handle this scenario.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve for <em>x<\/em>:\u00a0[latex]4\\geq{x}+5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q815893\">Show Solution<\/span><\/p>\n<div id=\"q815893\" class=\"hidden-answer\" style=\"display: none\">\n<p>Isolate the variable by adding 10 to both sides of the inequality.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}4\\geq{x}+5 \\\\\\underline{\\,\\,\\,-5\\,\\,\\,\\,\\,-5}\\\\-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\ge \\,\\,\\,x\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Rewrite the inequality with the variable on the left &#8211; this makes writing the interval and drawing the graph easier.<\/p>\n<p style=\"text-align: center\">[latex]x\\le{-1}[\/latex]<\/p>\n<p style=\"text-align: left\">Note how the the pointy part of the inequality is still directed at the variable, so instead of reading as negative one is greater or equal to x, it now reads as x is less than or equal to negative one.<\/p>\n<h4>Answer<\/h4>\n<p>Inequality: [latex]x\\le{-1}[\/latex] This can also be written as<br \/>\nInterval: [latex]\\left(-\\infty,-1\\right][\/latex]<br \/>\nGraph: Notice that a closed circle is used because the inequality is \u201cless than or equal to\u201d . The blue arrow is drawn to the left of the point [latex]\u22121[\/latex] because these are the values that are less than [latex]\u22121[\/latex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4022 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185047\/Screen-Shot-2016-05-11-at-6.23.24-PM-300x57.png\" alt=\"(-oo,-1&#093;\" width=\"400\" height=\"76\" \/><\/p>\n<\/div>\n<\/div>\n<p>Check the\u00a0solution to [latex]4\\geq{x}+5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q568062\">Show Solution<\/span><\/p>\n<div id=\"q568062\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the end point [latex]\u22121[\/latex] into the related equation \u00a0[latex]4=x+5[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}4=x+5\\,\\,\\,\\\\\\text{Does}\\,\\,\\,4=-1+5?\\\\-1=-1\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Pick a value less than [latex]\u22121[\/latex], such as [latex]\u22125[\/latex], to check in the inequality. (This value will be on the shaded part of the graph.)<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}4\\geq{-5}+5\\,\\,\\,\\\\\\text{ }\\,\\text{ Is}\\,\\,4\\ge 0?\\\\\\text{It}\\,\\text{checks!}\\end{array}[\/latex]<\/p>\n<p>[latex]x\\le{-1}[\/latex] is the solution to [latex]4\\geq{x}+5[\/latex]<span style=\"line-height: 1.5\"><\/div>\n<\/div>\n<p> <\/span><\/p>\n<\/div>\n<p>The following video show examples of solving inequalities with the variable on the right side.<br \/>\n<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 id=\"title4\">Solve inequalities with multiplication and division<\/h2>\n<p>Solving an inequality with a variable that has a coefficient other than 1 usually involves multiplication or division. The steps are like solving one-step equations involving multiplication or division EXCEPT for the inequality sign. Let\u2019s look at what happens to the inequality when you multiply or divide each side by the same number.<\/p>\n<table>\n<tbody>\n<tr>\n<td>Let&#8217;s start with the true statement:<\/p>\n<p>[latex]10>5[\/latex]<\/td>\n<td>Let&#8217;s try again by starting with the same true statement:<\/p>\n<p>[latex]10>5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Next, multiply both sides by the same positive number:<\/p>\n<p>[latex]10\\cdot 2>5\\cdot 2[\/latex]<\/td>\n<td>This time, multiply both sides by the same negative number:<\/p>\n<p>[latex]10\\cdot-2>5 \\\\ \\,\\,\\,\\,\\,\\cdot -2\\,\\cdot-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>20 is greater than 10, so you still have a true inequality:<\/p>\n<p>[latex]20>10[\/latex]<\/td>\n<td>Wait a minute! [latex]\u221220[\/latex] is <i>not <\/i>greater than [latex]\u221210[\/latex], so you have an untrue statement.<\/p>\n<p>[latex]\u221220>\u221210[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>When you multiply by a positive number, leave the inequality sign as it is!<\/td>\n<td>You must \u201creverse\u201d the inequality sign to make the statement true:<\/p>\n<p>[latex]\u221220<\u221210[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07184952\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"80\" height=\"70\" \/>Caution! \u00a0When you multiply or divide by a negative number, \u201creverse\u201d the inequality sign. \u00a0 Whenever you multiply or divide both sides of an inequality by a negative\u00a0number, the inequality sign must be reversed in order to keep a true statement. These rules are summarized in the box below.<\/p>\n<\/div>\n<hr \/>\n<div class=\"textbox shaded\">\n<h3>Multiplication and Division Properties of Inequality<\/h3>\n<table style=\"height: 162px;width: 419px\">\n<tbody>\n<tr>\n<td><strong>Start With<\/strong><\/td>\n<td><strong>Multiply By<\/strong><\/td>\n<td><strong>Final Inequality<\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]a>b[\/latex]<\/td>\n<td>[latex]c[\/latex]<\/td>\n<td>[latex]ac>bc[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]a>b[\/latex]<\/td>\n<td>[latex]-c[\/latex]<\/td>\n<td>[latex]ac<bc[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table style=\"height: 77px;width: 418px\">\n<tbody>\n<tr>\n<td><strong>Start With<\/strong><\/td>\n<td><strong>Divide By<\/strong><\/td>\n<td><strong>Final Inequality<\/strong><\/td>\n<\/tr>\n<tr>\n<td>[latex]a>b[\/latex]<\/td>\n<td>[latex]c[\/latex]<\/td>\n<td>[latex]\\displaystyle \\frac{a}{c}>\\frac{b}{c}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]a>b[\/latex]<\/td>\n<td>[latex]-c[\/latex]<\/td>\n<td>[latex]\\displaystyle \\frac{a}{c}<\\frac{b}{c}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>Keep in mind that you only change the sign when you are multiplying and dividing by a <i>negative<\/i> number. If you <em>add or subtract<\/em> by a negative\u00a0number, the inequality stays the same.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>x.\u00a0<\/i>[latex]3x>12[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q691711\">Show Solution<\/span><\/p>\n<div id=\"q691711\" class=\"hidden-answer\" style=\"display: none\">Divide both sides by 3 to isolate the variable.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}\\underline{3x}>\\underline{12}\\\\3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3\\\\x>4\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Check your solution by first checking the end point 4, and then checking another solution for the inequality.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}3\\cdot4=12\\\\12=12\\\\3\\cdot10>12\\\\30>12\\\\\\text{It checks!}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p style=\"text-align: left\">Inequality: [latex]\\displaystyle x>4[\/latex]<\/p>\n<p style=\"text-align: left\">Interval: [latex]\\left(4,\\infty\\right)[\/latex]<\/p>\n<p style=\"text-align: left\">Graph: <img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/MITEdevmath\/NROCUnit10_files\/image050.jpg#fixme#fixme\" alt=\"Number line. Open circle on 4. Highlight through all numbers greater than 4.\" width=\"575\" height=\"31\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>There was no need to make any changes to the inequality sign because both sides of the inequality were divided by <i>positive<\/i> 3. In the next example, there is division by a negative number, so there is an additional step in the solution!<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <em>x<\/em>. [latex]\u22122x>6[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q604033\">Show Solution<\/span><\/p>\n<div id=\"q604033\" class=\"hidden-answer\" style=\"display: none\">Divide each side of the inequality by [latex]\u22122[\/latex] to isolate the variable, and change the direction of the inequality sign because of the division by a negative number.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{r}\\underline{-2x}<\\underline{\\,6\\,}\\\\-2\\,\\,\\,\\,-2\\,\\\\x<-3\\end{array}[\/latex]<\/p>\n<p>Check your solution by first checking the end point [latex]\u22123[\/latex], and then checking another solution for the inequality.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{r}-2\\left(-3\\right)=6 \\\\6=6\\\\ -2\\left(-6\\right)>6 \\\\ 12>6\\end{array}[\/latex]<\/p>\n<p>It checks!<\/p>\n<h4>Answer<\/h4>\n<p>Inequality: [latex]\\displaystyle x<-3[\/latex]\n\nInterval: [latex]\\left(-\\infty, -3\\right)[\/latex]\n\nGraph: <img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/MITEdevmath\/NROCUnit10_files\/image051.jpg#fixme#fixme\" alt=\"Number line. Open circle on negative 3. Highlight on all numbers less than negative 3.\" width=\"575\" height=\"31\" \/><br \/>\nBecause both sides of the inequality were divided by a negative number, [latex]\u22122[\/latex], the inequality symbol was switched from &gt; to &lt;.\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows examples of solving one step inequalities using the multiplication property of equality where the variable is on the left hand side.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  Solve One Step Linear Inequality by Dividing (Variable Left)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/IajiD3R7U-0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Before you read the solution to the next example, think about what properties of inequalities you may need to use to solve the inequality. What is different about this example from the previous one? Write your ideas in the box below.<\/p>\n<p>Solve for <em>x<\/em>. [latex]-\\frac{1}{2}>-12x[\/latex]<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q811465\">Show Solution<\/span><\/p>\n<div id=\"q811465\" class=\"hidden-answer\" style=\"display: none\">\n<p>This inequality has the variable on the right hand side, which is different from the previous examples. Start the solution process as before, and at the end, you can move the variable to the left to write the final solution.<\/p>\n<p>Divide both sides by [latex]-12[\/latex] to isolate the variable. Since you are dividing by a negative number, you need to change the direction of the inequality sign.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle\\begin{array}{l}-\\frac{1}{2}\\gt{-12x}\\\\\\\\\\frac{-\\frac{1}{2}}{-12}\\gt\\frac{-12x}{-12}\\\\\\end{array}[\/latex]<\/p>\n<p>Dividing a fraction by an integer requires you to multiply by the reciprocal, and the reciprocal of [latex]-12[\/latex] is [latex]\\frac{1}{-12}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle\\begin{array}{r}\\left(-\\frac{1}{12}\\right)\\left(-\\frac{1}{2}\\right)\\lt\\frac{-12x}{-12}\\,\\,\\\\\\\\ \\frac{1}{24}\\lt\\frac{\\cancel{-12}x}{\\cancel{-12}}\\\\\\\\ \\frac{1}{24}\\lt{x}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left\">Answer<\/h4>\n<p style=\"text-align: left\">Inequality: [latex]\\frac{1}{24}\\lt{x}[\/latex] \u00a0This can also be written with the variable on the left as [latex]x\\gt\\frac{1}{24}[\/latex]. \u00a0Writing the inequality with the variable on the left requires a little thinking, but helps you write the interval and draw the graph correctly.<\/p>\n<p style=\"text-align: left\">Interval: [latex]\\left(\\frac{1}{24},\\infty\\right)[\/latex]<\/p>\n<p style=\"text-align: left\">Graph:\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3950\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185050\/Screen-Shot-2016-05-10-at-2.09.52-PM-300x57.png\" alt=\"Open dot on zero with a line through all numbers greater than zero.\" width=\"389\" height=\"74\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video gives examples of how to solve an inequality with the multiplication property of equality where the variable is on the right hand side.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex:  Solve One Step Linear Inequality by Dividing (Variable Right)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/s9fJOnVTHhs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>Solving inequalities is very similar to solving equations, except you have to reverse the inequality symbols when you multiply or divide both sides of an inequality by a negative number. There are three ways to represent solutions to inequalities: an interval, a graph, and an inequality.\u00a0Because there is usually more than one solution to an inequality, when you check your answer you should check the end point and one other value to check the direction of the inequality.<\/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-4547\">\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. <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><\/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 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