{"id":1157,"date":"2024-05-06T21:16:17","date_gmt":"2024-05-06T21:16:17","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1157"},"modified":"2025-08-13T15:32:09","modified_gmt":"2025-08-13T15:32:09","slug":"linear-inequalities-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/linear-inequalities-learn-it-3\/","title":{"raw":"Linear Inequalities: Learn It 3","rendered":"Linear Inequalities: Learn It 3"},"content":{"raw":"<h2>Compound Inequalities<\/h2>\r\nA <strong>compound inequality<\/strong> includes two inequalities in one statement. A statement such as [latex]4 &lt; x\\le 6[\/latex] means [latex]4 &lt; x[\/latex] and [latex]x\\le 6[\/latex]. There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<div>\r\n<h3>compound inequality<\/h3>\r\nA <strong>compound inequality<\/strong> is a mathematical statement that combines two inequalities into one expression using the words \"and\" or \"or\". These are used to express a range of possible solutions or conditions that satisfy more than one requirement simultaneously.\r\n<ul>\r\n \t<li><strong>\"And\" <\/strong>Compound Inequality (Conjunction):\r\n<ul>\r\n \t<li>This type connects two inequalities where both conditions must be true simultaneously.<\/li>\r\n \t<li>For example: [latex]a &lt; x &lt; b[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li><strong>Or\" <\/strong>Compound Inequality (Disjunction)\r\n<ul>\r\n \t<li>This type connects two inequalities where at least one of the conditions must be true. It is used to express that solutions may satisfy any one of multiple criteria.<\/li>\r\n \t<li>For example: [latex]x &lt; a \\text{ or } x &gt; b[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section><section class=\"textbox example\">Solve the compound inequality:\r\n<p style=\"text-align: center;\">[latex]3\\le 2x+2 &lt; 6[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll} \\text{Given inequality:} &amp; 3 \\leq 2x + 2 &lt; 6 \\\\ \\text{Subtract 2 from all parts:} &amp; 3 - 2 \\leq 2x + 2 - 2 &lt; 6 - 2 \\\\ \\text{Simplify:} &amp; 1 \\leq 2x &lt; 4 \\\\ \\text{Divide all parts by 2:} &amp; \\frac{1}{2} \\leq x &lt; 2 \\end{array}[\/latex]<\/p>\r\nIn interval notation, the solution is: [latex][\\frac{1}{2}, 2)[\/latex]\r\n\r\n<\/section><section class=\"textbox example\">Solve the compound inequality:\r\n<p style=\"text-align: center;\">[latex]\\begin{align*} 7x + 1 &amp; &gt; 3x + 5 \\\\ \\text{OR} \\\\ 3(-4x + 5) + 15 &amp;\\geq -6x + 54 \\end{align*}[\/latex]<\/p>\r\n[reveal-answer q=\"91279\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"91279\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll} \\text{First inequality:} &amp; 7x + 1 &gt; 3x + 5 \\\\ \\text{Subtract } 3x \\text{ and } 1: &amp; 4x &gt; 4 \\\\ \\text{Divide by 4:} &amp; x &gt; 1 \\\\\\end{array}[\/latex][latex]\\begin{array}{ll} \\text{Second inequality:} &amp; 3(-4x + 5) + 15 \\geq -6x + 54 \\\\ \\text{Expand and simplify:} &amp; -12x + 15 + 15 \\geq -6x + 54 \\\\ \\text{Combine terms:} &amp; -12x + 30 \\geq -6x + 54 \\\\ \\text{Add } 12x \\text{ and subtract 54:} &amp; 30 - 54 \\geq 6x \\\\ \\text{Simplify:} &amp; -24 \\geq 6x \\\\ \\text{Divide by 6 and reverse inequality:} &amp; x \\leq -4 \\\\ \\end{array}[\/latex]<\/p>\r\nIn interval notation, the solution is: [latex](-\\infty, -4] \\cup (1, \\infty)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">Solve the compound inequality:\r\n<p style=\"text-align: center;\">[latex]3+x &gt; 7x - 2 &gt; 5x - 10[\/latex]<\/p>\r\n[reveal-answer q=\"648248\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"648248\"]\r\n\r\nTo solve the compound inequality with variables on all 3 parts, we'll handle it in parts to isolate [latex]x[\/latex] in both sections of the inequality. We can follow the steps methodically to ensure each part of the compound inequality is addressed correctly.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl} 3 + x &gt; 7x - 2 &amp; \\quad\\text{and}\\quad &amp; 7x - 2 &gt; 5x - 10 \\\\ 3 + x + 2 &gt; 7x -2 +2 &amp; \\quad\\text{Add 2}\\quad &amp; 7x - 2 + 2 &gt; 5x - 10 + 2 \\\\ 5 + x &gt; 7x &amp; \\quad\\text{Simplify}\\quad &amp; 7x &gt; 5x - 8 \\\\ 5 &gt; 6x &amp; \\quad\\text{Isolate } x \\text{ to one side}\\quad &amp; 2x &gt; -8 \\\\ \\frac{5}{6} &gt; x &amp; \\quad\\text{Divide by the coefficient of }x \\quad &amp; x &gt; -4 \\\\ x &lt; \\frac{5}{6} &amp; \\quad\\text{Rewrite}\\quad &amp; -4 &lt; x \\end{array}[\/latex]<\/p>\r\n\r\n<ul>\r\n \t<li>Combining the two parts: [latex]-4 &lt; x &lt; \\frac{5}{6}[\/latex]<\/li>\r\n \t<li>In interval notation, the solution is: [latex](-4, \\frac{5}{6})[\/latex]<\/li>\r\n \t<li>Number Line:\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24225901\/CNX_CAT_Figure_02_07_003.jpg\" alt=\"A number line with the points -4 and 5\/6 labeled. Dots appear at these points and a line connects these two dots.\" width=\"487\" height=\"69\" \/> Number line with open circle on -4 and 5\/6[\/caption]<\/li>\r\n<\/ul>\r\nNotice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right as they appear on a number line.[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18947[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18948[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18949[\/ohm2_question]<\/section>","rendered":"<h2>Compound Inequalities<\/h2>\n<p>A <strong>compound inequality<\/strong> includes two inequalities in one statement. A statement such as [latex]4 < x\\le 6[\/latex] means [latex]4 < x[\/latex] and [latex]x\\le 6[\/latex]. There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods.\n\n\n\n<section class=\"textbox keyTakeaway\">\n<div>\n<h3>compound inequality<\/h3>\n<p>A <strong>compound inequality<\/strong> is a mathematical statement that combines two inequalities into one expression using the words &#8220;and&#8221; or &#8220;or&#8221;. These are used to express a range of possible solutions or conditions that satisfy more than one requirement simultaneously.<\/p>\n<ul>\n<li><strong>&#8220;And&#8221; <\/strong>Compound Inequality (Conjunction):\n<ul>\n<li>This type connects two inequalities where both conditions must be true simultaneously.<\/li>\n<li>For example: [latex]a < x < b[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li><strong>Or&#8221; <\/strong>Compound Inequality (Disjunction)\n<ul>\n<li>This type connects two inequalities where at least one of the conditions must be true. It is used to express that solutions may satisfy any one of multiple criteria.<\/li>\n<li>For example: [latex]x < a \\text{ or } x > b[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Solve the compound inequality:<\/p>\n<p style=\"text-align: center;\">[latex]3\\le 2x+2 < 6[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll} \\text{Given inequality:} & 3 \\leq 2x + 2 < 6 \\\\ \\text{Subtract 2 from all parts:} & 3 - 2 \\leq 2x + 2 - 2 < 6 - 2 \\\\ \\text{Simplify:} & 1 \\leq 2x < 4 \\\\ \\text{Divide all parts by 2:} & \\frac{1}{2} \\leq x < 2 \\end{array}[\/latex]<\/p>\n<p>In interval notation, the solution is: [latex][\\frac{1}{2}, 2)[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">Solve the compound inequality:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align*} 7x + 1 & > 3x + 5 \\\\ \\text{OR} \\\\ 3(-4x + 5) + 15 &\\geq -6x + 54 \\end{align*}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q91279\">Show Answer<\/button><\/p>\n<div id=\"q91279\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll} \\text{First inequality:} & 7x + 1 > 3x + 5 \\\\ \\text{Subtract } 3x \\text{ and } 1: & 4x > 4 \\\\ \\text{Divide by 4:} & x > 1 \\\\\\end{array}[\/latex][latex]\\begin{array}{ll} \\text{Second inequality:} & 3(-4x + 5) + 15 \\geq -6x + 54 \\\\ \\text{Expand and simplify:} & -12x + 15 + 15 \\geq -6x + 54 \\\\ \\text{Combine terms:} & -12x + 30 \\geq -6x + 54 \\\\ \\text{Add } 12x \\text{ and subtract 54:} & 30 - 54 \\geq 6x \\\\ \\text{Simplify:} & -24 \\geq 6x \\\\ \\text{Divide by 6 and reverse inequality:} & x \\leq -4 \\\\ \\end{array}[\/latex]<\/p>\n<p>In interval notation, the solution is: [latex](-\\infty, -4] \\cup (1, \\infty)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Solve the compound inequality:<\/p>\n<p style=\"text-align: center;\">[latex]3+x > 7x - 2 > 5x - 10[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q648248\">Show Answer<\/button><\/p>\n<div id=\"q648248\" class=\"hidden-answer\" style=\"display: none\">\n<p>To solve the compound inequality with variables on all 3 parts, we&#8217;ll handle it in parts to isolate [latex]x[\/latex] in both sections of the inequality. We can follow the steps methodically to ensure each part of the compound inequality is addressed correctly.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl} 3 + x > 7x - 2 & \\quad\\text{and}\\quad & 7x - 2 > 5x - 10 \\\\ 3 + x + 2 > 7x -2 +2 & \\quad\\text{Add 2}\\quad & 7x - 2 + 2 > 5x - 10 + 2 \\\\ 5 + x > 7x & \\quad\\text{Simplify}\\quad & 7x > 5x - 8 \\\\ 5 > 6x & \\quad\\text{Isolate } x \\text{ to one side}\\quad & 2x > -8 \\\\ \\frac{5}{6} > x & \\quad\\text{Divide by the coefficient of }x \\quad & x > -4 \\\\ x < \\frac{5}{6} & \\quad\\text{Rewrite}\\quad & -4 < x \\end{array}[\/latex]<\/p>\n<ul>\n<li>Combining the two parts: [latex]-4 < x < \\frac{5}{6}[\/latex]<\/li>\n<li>In interval notation, the solution is: [latex](-4, \\frac{5}{6})[\/latex]<\/li>\n<li>Number Line:<br \/>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24225901\/CNX_CAT_Figure_02_07_003.jpg\" alt=\"A number line with the points -4 and 5\/6 labeled. Dots appear at these points and a line connects these two dots.\" width=\"487\" height=\"69\" \/><figcaption class=\"wp-caption-text\">Number line with open circle on -4 and 5\/6<\/figcaption><\/figure>\n<\/li>\n<\/ul>\n<p>Notice that when we write the solution in interval notation, the smaller number comes first. We read intervals from left to right as they appear on a number line.<\/p><\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18947\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18947&theme=lumen&iframe_resize_id=ohm18947&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18948\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18948&theme=lumen&iframe_resize_id=ohm18948&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18949\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18949&theme=lumen&iframe_resize_id=ohm18949&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":27,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":75,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1157"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":44,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1157\/revisions"}],"predecessor-version":[{"id":7603,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1157\/revisions\/7603"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/75"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1157\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1157"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1157"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1157"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1157"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}