{"id":4743,"date":"2017-12-26T22:42:21","date_gmt":"2017-12-26T22:42:21","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/?post_type=chapter&#038;p=4743"},"modified":"2023-11-08T13:18:32","modified_gmt":"2023-11-08T13:18:32","slug":"absolute-value","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/chapter\/absolute-value\/","title":{"raw":"1.2 - Absolute Value","rendered":"1.2 &#8211; Absolute Value"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>(1.2.1) - Evaluating expressions with absolute value signs<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1>(1.2.1) - Evaluating expressions with absolute value signs<\/h1>\r\n<ul>\r\n \t<li><strong>Absolute value:<\/strong> a number's distance from zero; it's always positive:\r\n<ul>\r\n \t<li>\u00a0[latex]|3| = 3[\/latex]<\/li>\r\n \t<li>\u00a0[latex]|-5| = 5[\/latex]<\/li>\r\n \t<li>\u00a0[latex]|0| = 0[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\nRecall that the absolute value of a quantity is always positive or 0.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFind\u00a0[latex]|7- 10|[\/latex].\r\n\r\n[reveal-answer q=\"630537\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"630537\"][latex]|7-10|=|\u22123| =3[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]472[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFind\u00a0[latex]-|3-2|[\/latex].\r\n\r\n[reveal-answer q=\"333674\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"333674\"][latex]-|3-2|=-|-1|=-1[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]348[\/ohm_question]\r\n\r\n<\/div>\r\nWhen you see an absolute value expression included within a larger expression, treat the absolute value like a grouping symbol and evaluate the expression within the absolute value sign first. Then take the absolute value of that expression. The example below shows how this is done.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify [latex]\\Large\\frac{3+\\left|2-6\\right|}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}[\/latex].\r\n\r\n[reveal-answer q=\"572632\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"572632\"]This problem has absolute values, decimals, multiplication, subtraction, and addition in it.\r\n\r\nGrouping symbols, including absolute value, are handled first. Simplify the numerator, then the denominator.\r\n\r\nEvaluate [latex]\\left|2\u20136\\right|[\/latex].\r\n<p style=\"text-align: center\">[latex]\\Large\\begin{array}{c}\\frac{3+\\left|2-6\\right|}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}\\\\\\\\\\frac{3+\\left|-4\\right|}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}\\end{array}[\/latex]<\/p>\r\nTake the absolute value of [latex]\\left|\u22124\\right|[\/latex].\r\n<p style=\"text-align: center\">[latex]\\Large\\begin{array}{c}\\frac{3+\\left|-4\\right|}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}\\\\\\\\\\frac{3+4}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}\\end{array}[\/latex]<\/p>\r\nAdd the numbers in the numerator.\r\n<p style=\"text-align: center\">[latex]\\Large\\begin{array}{c}\\frac{3+4}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}\\\\\\\\\\frac{7}{2\\left| 3\\cdot 1.5 \\right|-(-3)}\\end{array}[\/latex]<\/p>\r\nNow that the numerator is simplified, turn to the denominator.\r\n\r\nEvaluate the absolute value expression first. [latex]3 \\cdot 1.5 = 4.5[\/latex], giving\r\n<p style=\"text-align: center\">\u00a0[latex]\\Large\\begin{array}{c}\\frac{7}{2\\left|{3\\cdot{1.5}}\\right|-(-3)}\\\\\\\\\\frac{7}{2\\left|{ 4.5}\\right|-(-3)}\\end{array}[\/latex]<\/p>\r\nThe expression \u201c[latex]2\\left|4.5\\right|[\/latex]\u201d reads \u201c2 times the absolute value of 4.5.\u201d Multiply 2 times 4.5.\r\n<p style=\"text-align: center\">[latex]\\Large\\begin{array}{c}\\frac{7}{2\\left|4.5\\right|-\\left(-3\\right)}\\\\\\\\\\frac{7}{9-\\left(-3\\right)}\\end{array}[\/latex]<\/p>\r\nSubtract.\r\n<p style=\"text-align: center\">[latex]\\Large\\begin{array}{c}\\frac{7}{9-\\left(-3\\right)}\\\\\\\\\\frac{7}{12}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\Large\\frac{3+\\left|2-6\\right|}{2\\left|3\\cdot1.5\\right|-3\\left(-3\\right)}=\\frac{7}{12}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video uses the order of operations to simplify an expression in fraction form that contains absolute value terms. Note how the absolute values are treated like parentheses and brackets when using the order of operations.\r\n\r\nhttps:\/\/youtu.be\/6wmCQprxlnU","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>(1.2.1) &#8211; Evaluating expressions with absolute value signs<\/li>\n<\/ul>\n<\/div>\n<h1>(1.2.1) &#8211; Evaluating expressions with absolute value signs<\/h1>\n<ul>\n<li><strong>Absolute value:<\/strong> a number&#8217;s distance from zero; it&#8217;s always positive:\n<ul>\n<li>\u00a0[latex]|3| = 3[\/latex]<\/li>\n<li>\u00a0[latex]|-5| = 5[\/latex]<\/li>\n<li>\u00a0[latex]|0| = 0[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>Recall that the absolute value of a quantity is always positive or 0.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Find\u00a0[latex]|7- 10|[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q630537\">Show Answer<\/span><\/p>\n<div id=\"q630537\" class=\"hidden-answer\" style=\"display: none\">[latex]|7-10|=|\u22123| =3[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm472\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=472&theme=oea&iframe_resize_id=ohm472&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Find\u00a0[latex]-|3-2|[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q333674\">Show Answer<\/span><\/p>\n<div id=\"q333674\" class=\"hidden-answer\" style=\"display: none\">[latex]-|3-2|=-|-1|=-1[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm348\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=348&theme=oea&iframe_resize_id=ohm348&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>When you see an absolute value expression included within a larger expression, treat the absolute value like a grouping symbol and evaluate the expression within the absolute value sign first. Then take the absolute value of that expression. The example below shows how this is done.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify [latex]\\Large\\frac{3+\\left|2-6\\right|}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q572632\">Show Solution<\/span><\/p>\n<div id=\"q572632\" class=\"hidden-answer\" style=\"display: none\">This problem has absolute values, decimals, multiplication, subtraction, and addition in it.<\/p>\n<p>Grouping symbols, including absolute value, are handled first. Simplify the numerator, then the denominator.<\/p>\n<p>Evaluate [latex]\\left|2\u20136\\right|[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\Large\\begin{array}{c}\\frac{3+\\left|2-6\\right|}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}\\\\\\\\\\frac{3+\\left|-4\\right|}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}\\end{array}[\/latex]<\/p>\n<p>Take the absolute value of [latex]\\left|\u22124\\right|[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\Large\\begin{array}{c}\\frac{3+\\left|-4\\right|}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}\\\\\\\\\\frac{3+4}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}\\end{array}[\/latex]<\/p>\n<p>Add the numbers in the numerator.<\/p>\n<p style=\"text-align: center\">[latex]\\Large\\begin{array}{c}\\frac{3+4}{2\\left|3\\cdot1.5\\right|-\\left(-3\\right)}\\\\\\\\\\frac{7}{2\\left| 3\\cdot 1.5 \\right|-(-3)}\\end{array}[\/latex]<\/p>\n<p>Now that the numerator is simplified, turn to the denominator.<\/p>\n<p>Evaluate the absolute value expression first. [latex]3 \\cdot 1.5 = 4.5[\/latex], giving<\/p>\n<p style=\"text-align: center\">\u00a0[latex]\\Large\\begin{array}{c}\\frac{7}{2\\left|{3\\cdot{1.5}}\\right|-(-3)}\\\\\\\\\\frac{7}{2\\left|{ 4.5}\\right|-(-3)}\\end{array}[\/latex]<\/p>\n<p>The expression \u201c[latex]2\\left|4.5\\right|[\/latex]\u201d reads \u201c2 times the absolute value of 4.5.\u201d Multiply 2 times 4.5.<\/p>\n<p style=\"text-align: center\">[latex]\\Large\\begin{array}{c}\\frac{7}{2\\left|4.5\\right|-\\left(-3\\right)}\\\\\\\\\\frac{7}{9-\\left(-3\\right)}\\end{array}[\/latex]<\/p>\n<p>Subtract.<\/p>\n<p style=\"text-align: center\">[latex]\\Large\\begin{array}{c}\\frac{7}{9-\\left(-3\\right)}\\\\\\\\\\frac{7}{12}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\Large\\frac{3+\\left|2-6\\right|}{2\\left|3\\cdot1.5\\right|-3\\left(-3\\right)}=\\frac{7}{12}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video uses the order of operations to simplify an expression in fraction form that contains absolute value terms. Note how the absolute values are treated like parentheses and brackets when using the order of operations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Simplify an Expression in Fraction Form with Absolute Values\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/6wmCQprxlnU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n","protected":false},"author":60342,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4743","chapter","type-chapter","status-web-only","hentry"],"part":359,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/4743","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/users\/60342"}],"version-history":[{"count":23,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/4743\/revisions"}],"predecessor-version":[{"id":5365,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/4743\/revisions\/5365"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/359"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/4743\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/media?parent=4743"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=4743"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/contributor?post=4743"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/license?post=4743"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}