{"id":2836,"date":"2024-02-09T19:28:48","date_gmt":"2024-02-09T19:28:48","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/?post_type=chapter&#038;p=2836"},"modified":"2024-02-09T19:28:48","modified_gmt":"2024-02-09T19:28:48","slug":"5-3-evaluating-algebraic-expressions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/5-3-evaluating-algebraic-expressions\/","title":{"raw":"5.3 Evaluating Algebraic Expressions","rendered":"5.3 Evaluating Algebraic Expressions"},"content":{"raw":"<div class=\"wrapper\">\r\n<div id=\"wrap\">\r\n<div id=\"content\" role=\"main\">\r\n<div id=\"post-68\" class=\"standard post-68 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Evaluate algebraic expressions for different values&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:7041,&quot;3&quot;:{&quot;1&quot;:0},&quot;10&quot;:0,&quot;11&quot;:4,&quot;12&quot;:0,&quot;14&quot;:[null,2,0],&quot;15&quot;:&quot;Calibri&quot;}\">Evaluate algebraic expressions using the order of operations<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>\u00a0Evaluating Algebraic Expressions<\/h2>\r\nAny variable in an algebraic expression may take on or be assigned different values. When that happens, the value of the algebraic expression changes. To evaluate an algebraic expression means to determine the value of the expression for a given value of each variable in the expression. Replace each variable in the expression with the given value then simplify the resulting expression using the order of operations. If the algebraic expression contains more than one variable, replace each variable with its assigned value and simplify the expression as before. In the next example we show how to substitute various types of numbers into a mathematical expression.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nEvaluate [latex]x+7[\/latex] when\r\n<ol>\r\n \t<li>[latex]x=3[\/latex]<\/li>\r\n \t<li>[latex]x=-12[\/latex]<\/li>\r\n<\/ol>\r\nSolution:\r\n\r\n1. To evaluate, substitute [latex]3[\/latex] for [latex]x[\/latex] in the expression, and then simplify.\r\n<table id=\"eip-id1166566546426\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression x plus 7. Substitute 3 for x. The expression becomes 3 plus x which is 10.\" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]x+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute.<\/td>\r\n<td>[latex]\\color{red}{3}+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add.<\/td>\r\n<td>[latex]10[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen [latex]x=3[\/latex], the expression [latex]x+7[\/latex] has a value of [latex]10[\/latex].\r\n\r\n2. To evaluate, substitute [latex]-12[\/latex] for [latex]x[\/latex] in the expression, and then simplify.\r\n<table id=\"eip-id1166566410105\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression x plus 7, substitute 12 for x. The expression becomes 12 plus x which is 19.\" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]x+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute.<\/td>\r\n<td>[latex]\\color{red}{(-12)}+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add.<\/td>\r\n<td>[latex]-5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen [latex]x=-12[\/latex], the expression [latex]x+7[\/latex] has a value of [latex]-5[\/latex].\r\n\r\n&nbsp;\r\n\r\nNotice that we got different results for parts 1 and 2 even though we started with the same expression. This is because the values used for [latex]x[\/latex] were different. When we evaluate an expression, the value of the expression varies depending on the value used for the variable.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<iframe id=\"ohm144878\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=144878&amp;theme=oea&amp;iframe_resize_id=ohm144878&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nEvaluate [latex]9x - 2[\/latex] when\r\n<ol>\r\n \t<li>[latex]x=5[\/latex]<\/li>\r\n \t<li>[latex]x=\\frac{1}{9}[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\nRemember [latex]ab[\/latex] means [latex]a[\/latex] times [latex]b[\/latex], so [latex]9x[\/latex] means [latex]9[\/latex] times [latex]x[\/latex].\r\n\r\n1. To evaluate the expression when [latex]x=5[\/latex], we substitute [latex]5[\/latex] for [latex]x[\/latex], and then simplify.\r\n<table id=\"eip-id1168469462966\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression nine x minus 2. Substitute 5 for x. The expression becomes 9 times 5 minus 2. Multiply first. Nine times 5 is 45 and the expression is now 45 minus 2. Subtract to get 43.\" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]9x-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]\\color{red}{5}[\/latex] for x.<\/td>\r\n<td>[latex]9\\cdot\\color{red}{5}-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]45-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract.<\/td>\r\n<td>[latex]43[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n2. To evaluate the expression when [latex]x=\\frac{1}{9}[\/latex], we substitute [latex]\\frac{1}{9}[\/latex] for [latex]x[\/latex], and then simplify.\r\n<table id=\"eip-id1168468440939\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression nine x minus 2. Substitute 1 for x. The expression becomes 9 times 1 minus 2. Multiply first. Nine times 1 is 9 and the expression is now 9 minus 2. Subtract to get 7.\" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]9x-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]\\color{red}{\\frac{1}{9}}[\/latex] for x.<\/td>\r\n<td>[latex]9\\color{red}{\\left (\\frac{1}{9}\\right )}-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]1-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Subtract.<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n&nbsp;\r\n\r\nNotice that in part 1 that we wrote [latex]9\\cdot 5[\/latex] and in part 2 we wrote [latex]9\\left(\\frac{1}{9}\\right)[\/latex]. Both the dot and the parentheses tell us to multiply.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nEvaluate the expression [latex]2x^3 + 7[\/latex] for each value for [latex]x[\/latex].\r\n<em>.<\/em>\r\n<ol>\r\n \t<li>[latex]x=0[\/latex]<\/li>\r\n \t<li>[latex]x=-1[\/latex]<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\n<ul>\r\n \t<li>Substitute [latex]0[\/latex] for [latex]x[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}2x^3+7 \\hfill&amp; = 2\\color{blue}{\\left(0\\right)}^3+7 \\\\ \\hfill&amp; =0+7 \\\\ \\hfill&amp; =7\\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Substitute [latex]-1[\/latex] for [latex]x[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}2x^3+7 \\hfill&amp; = 2\\color{blue}{\\left(-1\\right)}^3+7 \\\\ \\hfill&amp; =-2+7 \\\\ \\hfill&amp; =5\\end{array}[\/latex]<\/div><\/li>\r\n<\/ul>\r\n<\/div>\r\nIf we always put negative numbers inside parentheses when we substitute, we are more likely to follow the order of operations correctly.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<iframe id=\"ohm141843\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141843&amp;theme=oea&amp;iframe_resize_id=ohm141843&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nEvaluate [latex]{x}^{2}[\/latex] when [latex]x=-10[\/latex].\r\n<h4>Solution<\/h4>\r\n[latex]{x}^{2}=\\,\\color{blue}{(-10)}^2=100[\/latex]\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n\r\n&nbsp;\r\n<div id=\"q729694\" class=\"hidden-answer\" style=\"display: none;\">\r\n\r\nSolution\r\n\r\nWe substitute [latex]10[\/latex] for [latex]x[\/latex], and then simplify the expression.\r\n<table id=\"eip-id1168468538199\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression x squared. Substitute 10 for x. The expression becomes 10 squared. By the definition of exponents, 10 squared is 10 times 10. Multiply to get 100.\" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]x^2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]\\color{red}{10}[\/latex] for x.<\/td>\r\n<td>[latex]{\\color{red}{10}}^{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the definition of exponent.<\/td>\r\n<td>[latex]10\\cdot 10[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]100[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen [latex]x=10[\/latex], the expression [latex]{x}^{2}[\/latex] has a value of [latex]100[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<iframe id=\"ohm144879\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=144879&amp;theme=oea&amp;iframe_resize_id=ohm144879&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\n[latex]\\text{Evaluate }{2}^{x}\\text{ when }x=5[\/latex].\r\n<h4>Solution<\/h4>\r\nIn this expression, the variable is an exponent.\r\n<table id=\"eip-id1168469574741\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression 2 to the power of x. Substitute 5 for x. The expression becomes 2 to the fifth power. By the definition of exponents, 2 to the fifth power is 2 times 2 times 2 times 2 times 2, or 5 factors of 2. Multiply from left to right to get 32.\" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]2^x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]\\color{red}{5}[\/latex] for x.<\/td>\r\n<td>[latex]{2}^{\\color{red}{5}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the definition of exponent.<\/td>\r\n<td>[latex]2\\cdot2\\cdot2\\cdot2\\cdot2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]32[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen [latex]x=5[\/latex], the expression [latex]{2}^{x}[\/latex] has a value of [latex]32[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<iframe id=\"ohm144882\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=144882&amp;theme=oea&amp;iframe_resize_id=ohm144882&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\n[latex]\\text{Evaluate }3x+4y - 6\\text{ when }x=10\\text{ and }y=-2[\/latex].\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n<h4>Solution<\/h4>\r\n[latex]\\begin{array}{ll}3x+4y - 6 \\hfill&amp; = 3\\color{blue}{(10)}+4\\color{red}{(-2)}-6 \\\\ \\hfill&amp; = \\color{blue}{30}\\color{red}{-8}-6 \\\\ \\hfill&amp; =16\\end{array}[\/latex]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"qa-wrapper\" style=\"display: block;\">\r\n<div id=\"q769566\" class=\"hidden-answer\" style=\"display: none;\">\r\n\r\nSolution\r\n\r\n&nbsp;\r\n\r\nThis expression contains two variables, so we must make two substitutions.\r\n<table id=\"eip-id1168467158036\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression three x plus four y minus 6. Substitute 10 for x and 2 for y. The expression becomes 3 times 10 plus 4 times 2 minus 6. Perform multiplication from left to right. Three times 10 is 30 and 4 times 2 is 8. The expression becomes 30 plus 8 minus 6. Add and subtract from left to right. Thirty plus 8 is 38. Thirty-eight minus 6 is 32.\" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]3x+4y-6[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]\\color{red}{10}[\/latex] for x and [latex]\\color{blue}{2}[\/latex] for y.<\/td>\r\n<td>[latex]3(\\color{red}{10})+4(\\color{blue}{2})-6[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]30+8-6[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add and subtract left to right.<\/td>\r\n<td>[latex]32[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen [latex]x=10[\/latex] and [latex]y=2[\/latex], the expression [latex]3x+4y - 6[\/latex] has a value of [latex]32[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>TRY\u00a0IT<\/h3>\r\n<iframe id=\"ohm144884\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=144884&amp;theme=oea&amp;iframe_resize_id=ohm144884&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\n[latex]\\text{Evaluate }-2{x}^{2}+3x+8\\text{ when }x=-4[\/latex].\r\n<h4>Solution<\/h4>\r\n<table id=\"eip-id1168466011069\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression two x squared plus three x plus 8. Substitute 4 for each x. The expression becomes 2 times 4 squared plus 3 times 4 plus 8. Simplify exponents first. Four squared is 16 so the expression becomes 2 times 16 plus 3 times 4 plus 8. Next perform multiplication from left to right. Two times 16 is 32 and 3 times 4 is 12. The expression becomes 32 plus 12 plus 8. Add from left to right. Thirty-two plus 12 is 44. Forty-four plus 8 is 52.\" data-label=\"\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]-2x^2+3x+8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute [latex]\\color{red}{4}[\/latex] for each x.<\/td>\r\n<td>[latex]-2{(\\color{red}{(-4)})}^{2}+3(\\color{red}{4})+8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify [latex]{-4}^{2}[\/latex] .<\/td>\r\n<td>[latex]-2(16)+3(-4)+8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply.<\/td>\r\n<td>[latex]-32-12+8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add.<\/td>\r\n<td>[latex]-36[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\n[latex]\\text{Evaluate }\\frac{1}{x-3}\\text{ when }x=5 \\text{ and when }x=3[\/latex].\r\n<h4>Solution<\/h4>\r\n<ul>\r\n \t<li>If we substitute [latex]5[\/latex] for [latex]x[\/latex], the expression becomes [latex]\\frac{1}{\\color{blue}{5}-3}=\\frac{1}{2}[\/latex]. The answer is [latex]\\frac{1}{2}[\/latex]<\/li>\r\n<\/ul>\r\n&nbsp;\r\n<ul>\r\n \t<li>If we substitute [latex]3[\/latex] for [latex]x[\/latex], the expression becomes [latex]\\frac{1}{\\color{blue}{3}-3}=\\frac{1}{0}[\/latex]. A fraction where the denominator is zero is\u00a0 undefined.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nEvaluate: [latex]4\\sqrt{x^2}-\\sqrt{-6x}\\text{ when } x=-3[\/latex]\r\n<h4>Solution<\/h4>\r\n[latex]4\\sqrt{x^2}-\\sqrt{-6x}[\/latex]\r\n[latex]=4\\sqrt{\\color{blue}{(-3)}^2}-\\sqrt{-6\\color{blue}{(-3)}}[\/latex]\r\n\r\n[latex]=4\\sqrt{\\color{blue}{9}}-\\sqrt{\\color{blue}{(18)}}[\/latex]\r\n\r\n[latex]=4\\cdot \\color{blue}{3}-\\sqrt{\\color{blue}{(9\\cdot 2)}}[\/latex]\r\n\r\n[latex]=\\color{blue}{12}-\\sqrt{\\color{blue}{9}}\\cdot\\sqrt{2}[\/latex]\r\n\r\n[latex]=12-\\color{blue}{3}\\cdot\\sqrt{2}[\/latex]\r\n\r\n[latex]=12-3\\sqrt{2}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<iframe id=\"ohm144886\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=144886&amp;theme=oea&amp;iframe_resize_id=ohm144886&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe>\r\n\r\n<\/div>\r\nIn the video below we show more examples of how to substitute a value for variable in an expression, then evaluate the expression.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/dkFIVfJTG9E?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nEvaluate [latex]\\frac{x+y}{x}-\\frac{x-y}{y}[\/latex] when [latex]x=12[\/latex] and [latex]y=-9[\/latex].\r\n\r\nSolution\r\n\r\n[latex]\\frac{x+y}{x}-\\frac{x-y}{y}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Substitute\u00a0[latex]x=\\color{blue}{12}[\/latex] and [latex]y=\\color{red}{-9}[\/latex]\r\n\r\n[latex]=\\frac{\\color{blue}{12}+\\color{red}{(-9)}}{\\color{blue}{12}}-\\frac{\\color{blue}{12}-\\color{red}{(-9)}}{\\color{red}{(-9)}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Simplify the numerators.\r\n\r\n[latex]=\\frac{\\color{purple}{3}}{12}-\\frac{\\color{purple}{21}}{-9}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Look for common factors between the numerator and denominator in each fraction.\r\n\r\n[latex]=\\frac{\\color{green}{3}}{\\color{green}{3}\\cdot 4}-\\frac{\\color{green}{3}\\cdot 7}{-3\\cdot\\color{green}{3}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Cancel common factors.\r\n\r\n[latex]=\\frac{1}{4}\\color{red}{-}\\frac{7}{\\color{red}{-3}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Subtracting a negative is equivalent to adding a positive.\r\n\r\n[latex]=\\frac{1}{4}\\color{red}{+}\\frac{7}{3}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Build equivalent fractions with a common denominator of [latex]12[\/latex].\r\n\r\n[latex]=\\frac{1\\color{blue}{\\cdot 3}}{4\\color{blue}{\\cdot 3}}+\\frac{7\\color{red}{\\cdot 4}}{3\\color{red}{\\cdot 4}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Multiply the numerators and denominators.\r\n\r\n[latex]=\\frac{3}{\\color{purple}{12}}+\\frac{28}{\\color{purple}{12}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Add the fractions by combining the numerators and keeping the common denominator.\r\n\r\n[latex]=\\frac{3+28}{\\color{purple}{12}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Add the numerators.\r\n\r\n[latex]=\\frac{31}{12}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\frac{x^2}{y}+\\frac{y^2}{x}[\/latex] when [latex]x=6[\/latex] and [latex]y=-4[\/latex].\r\n\r\n[reveal-answer q=\"hjm686\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm686\"]\r\n\r\n[latex]-\\frac{19}{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"wrapper\">\n<div id=\"wrap\">\n<div id=\"content\" role=\"main\">\n<div id=\"post-68\" class=\"standard post-68 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Evaluate algebraic expressions for different values&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:7041,&quot;3&quot;:{&quot;1&quot;:0},&quot;10&quot;:0,&quot;11&quot;:4,&quot;12&quot;:0,&quot;14&quot;:[null,2,0],&quot;15&quot;:&quot;Calibri&quot;}\">Evaluate algebraic expressions using the order of operations<\/span><\/li>\n<\/ul>\n<\/div>\n<h2>\u00a0Evaluating Algebraic Expressions<\/h2>\n<p>Any variable in an algebraic expression may take on or be assigned different values. When that happens, the value of the algebraic expression changes. To evaluate an algebraic expression means to determine the value of the expression for a given value of each variable in the expression. Replace each variable in the expression with the given value then simplify the resulting expression using the order of operations. If the algebraic expression contains more than one variable, replace each variable with its assigned value and simplify the expression as before. In the next example we show how to substitute various types of numbers into a mathematical expression.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Evaluate [latex]x+7[\/latex] when<\/p>\n<ol>\n<li>[latex]x=3[\/latex]<\/li>\n<li>[latex]x=-12[\/latex]<\/li>\n<\/ol>\n<p>Solution:<\/p>\n<p>1. To evaluate, substitute [latex]3[\/latex] for [latex]x[\/latex] in the expression, and then simplify.<\/p>\n<table id=\"eip-id1166566546426\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression x plus 7. Substitute 3 for x. The expression becomes 3 plus x which is 10.\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]x+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute.<\/td>\n<td>[latex]\\color{red}{3}+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add.<\/td>\n<td>[latex]10[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When [latex]x=3[\/latex], the expression [latex]x+7[\/latex] has a value of [latex]10[\/latex].<\/p>\n<p>2. To evaluate, substitute [latex]-12[\/latex] for [latex]x[\/latex] in the expression, and then simplify.<\/p>\n<table id=\"eip-id1166566410105\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression x plus 7, substitute 12 for x. The expression becomes 12 plus x which is 19.\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]x+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute.<\/td>\n<td>[latex]\\color{red}{(-12)}+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add.<\/td>\n<td>[latex]-5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When [latex]x=-12[\/latex], the expression [latex]x+7[\/latex] has a value of [latex]-5[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p>Notice that we got different results for parts 1 and 2 even though we started with the same expression. This is because the values used for [latex]x[\/latex] were different. When we evaluate an expression, the value of the expression varies depending on the value used for the variable.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm144878\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=144878&amp;theme=oea&amp;iframe_resize_id=ohm144878&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Evaluate [latex]9x - 2[\/latex] when<\/p>\n<ol>\n<li>[latex]x=5[\/latex]<\/li>\n<li>[latex]x=\\frac{1}{9}[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>Remember [latex]ab[\/latex] means [latex]a[\/latex] times [latex]b[\/latex], so [latex]9x[\/latex] means [latex]9[\/latex] times [latex]x[\/latex].<\/p>\n<p>1. To evaluate the expression when [latex]x=5[\/latex], we substitute [latex]5[\/latex] for [latex]x[\/latex], and then simplify.<\/p>\n<table id=\"eip-id1168469462966\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression nine x minus 2. Substitute 5 for x. The expression becomes 9 times 5 minus 2. Multiply first. Nine times 5 is 45 and the expression is now 45 minus 2. Subtract to get 43.\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]9x-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]\\color{red}{5}[\/latex] for x.<\/td>\n<td>[latex]9\\cdot\\color{red}{5}-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]45-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract.<\/td>\n<td>[latex]43[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>2. To evaluate the expression when [latex]x=\\frac{1}{9}[\/latex], we substitute [latex]\\frac{1}{9}[\/latex] for [latex]x[\/latex], and then simplify.<\/p>\n<table id=\"eip-id1168468440939\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression nine x minus 2. Substitute 1 for x. The expression becomes 9 times 1 minus 2. Multiply first. Nine times 1 is 9 and the expression is now 9 minus 2. Subtract to get 7.\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]9x-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]\\color{red}{\\frac{1}{9}}[\/latex] for x.<\/td>\n<td>[latex]9\\color{red}{\\left (\\frac{1}{9}\\right )}-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]1-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Subtract.<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>Notice that in part 1 that we wrote [latex]9\\cdot 5[\/latex] and in part 2 we wrote [latex]9\\left(\\frac{1}{9}\\right)[\/latex]. Both the dot and the parentheses tell us to multiply.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Evaluate the expression [latex]2x^3 + 7[\/latex] for each value for [latex]x[\/latex].<br \/>\n<em>.<\/em><\/p>\n<ol>\n<li>[latex]x=0[\/latex]<\/li>\n<li>[latex]x=-1[\/latex]<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<ul>\n<li>Substitute [latex]0[\/latex] for [latex]x[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}2x^3+7 \\hfill& = 2\\color{blue}{\\left(0\\right)}^3+7 \\\\ \\hfill& =0+7 \\\\ \\hfill& =7\\end{array}[\/latex]<\/div>\n<\/li>\n<li>Substitute [latex]-1[\/latex] for [latex]x[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}2x^3+7 \\hfill& = 2\\color{blue}{\\left(-1\\right)}^3+7 \\\\ \\hfill& =-2+7 \\\\ \\hfill& =5\\end{array}[\/latex]<\/div>\n<\/li>\n<\/ul>\n<\/div>\n<p>If we always put negative numbers inside parentheses when we substitute, we are more likely to follow the order of operations correctly.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm141843\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=141843&amp;theme=oea&amp;iframe_resize_id=ohm141843&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Evaluate [latex]{x}^{2}[\/latex] when [latex]x=-10[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>[latex]{x}^{2}=\\,\\color{blue}{(-10)}^2=100[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<p>&nbsp;<\/p>\n<div id=\"q729694\" class=\"hidden-answer\" style=\"display: none;\">\n<p>Solution<\/p>\n<p>We substitute [latex]10[\/latex] for [latex]x[\/latex], and then simplify the expression.<\/p>\n<table id=\"eip-id1168468538199\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression x squared. Substitute 10 for x. The expression becomes 10 squared. By the definition of exponents, 10 squared is 10 times 10. Multiply to get 100.\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]x^2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]\\color{red}{10}[\/latex] for x.<\/td>\n<td>[latex]{\\color{red}{10}}^{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the definition of exponent.<\/td>\n<td>[latex]10\\cdot 10[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]100[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When [latex]x=10[\/latex], the expression [latex]{x}^{2}[\/latex] has a value of [latex]100[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm144879\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=144879&amp;theme=oea&amp;iframe_resize_id=ohm144879&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>[latex]\\text{Evaluate }{2}^{x}\\text{ when }x=5[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>In this expression, the variable is an exponent.<\/p>\n<table id=\"eip-id1168469574741\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression 2 to the power of x. Substitute 5 for x. The expression becomes 2 to the fifth power. By the definition of exponents, 2 to the fifth power is 2 times 2 times 2 times 2 times 2, or 5 factors of 2. Multiply from left to right to get 32.\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]2^x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]\\color{red}{5}[\/latex] for x.<\/td>\n<td>[latex]{2}^{\\color{red}{5}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the definition of exponent.<\/td>\n<td>[latex]2\\cdot2\\cdot2\\cdot2\\cdot2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]32[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When [latex]x=5[\/latex], the expression [latex]{2}^{x}[\/latex] has a value of [latex]32[\/latex].<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm144882\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=144882&amp;theme=oea&amp;iframe_resize_id=ohm144882&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>[latex]\\text{Evaluate }3x+4y - 6\\text{ when }x=10\\text{ and }y=-2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<h4>Solution<\/h4>\n<p>[latex]\\begin{array}{ll}3x+4y - 6 \\hfill& = 3\\color{blue}{(10)}+4\\color{red}{(-2)}-6 \\\\ \\hfill& = \\color{blue}{30}\\color{red}{-8}-6 \\\\ \\hfill& =16\\end{array}[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block;\">\n<div id=\"q769566\" class=\"hidden-answer\" style=\"display: none;\">\n<p>Solution<\/p>\n<p>&nbsp;<\/p>\n<p>This expression contains two variables, so we must make two substitutions.<\/p>\n<table id=\"eip-id1168467158036\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression three x plus four y minus 6. Substitute 10 for x and 2 for y. The expression becomes 3 times 10 plus 4 times 2 minus 6. Perform multiplication from left to right. Three times 10 is 30 and 4 times 2 is 8. The expression becomes 30 plus 8 minus 6. Add and subtract from left to right. Thirty plus 8 is 38. Thirty-eight minus 6 is 32.\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]3x+4y-6[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]\\color{red}{10}[\/latex] for x and [latex]\\color{blue}{2}[\/latex] for y.<\/td>\n<td>[latex]3(\\color{red}{10})+4(\\color{blue}{2})-6[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]30+8-6[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add and subtract left to right.<\/td>\n<td>[latex]32[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When [latex]x=10[\/latex] and [latex]y=2[\/latex], the expression [latex]3x+4y - 6[\/latex] has a value of [latex]32[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>TRY\u00a0IT<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm144884\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=144884&amp;theme=oea&amp;iframe_resize_id=ohm144884&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>[latex]\\text{Evaluate }-2{x}^{2}+3x+8\\text{ when }x=-4[\/latex].<\/p>\n<h4>Solution<\/h4>\n<table id=\"eip-id1168466011069\" class=\"unnumbered unstyled\" summary=\"The image shows the given expression two x squared plus three x plus 8. Substitute 4 for each x. The expression becomes 2 times 4 squared plus 3 times 4 plus 8. Simplify exponents first. Four squared is 16 so the expression becomes 2 times 16 plus 3 times 4 plus 8. Next perform multiplication from left to right. Two times 16 is 32 and 3 times 4 is 12. The expression becomes 32 plus 12 plus 8. Add from left to right. Thirty-two plus 12 is 44. Forty-four plus 8 is 52.\" data-label=\"\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]-2x^2+3x+8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute [latex]\\color{red}{4}[\/latex] for each x.<\/td>\n<td>[latex]-2{(\\color{red}{(-4)})}^{2}+3(\\color{red}{4})+8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify [latex]{-4}^{2}[\/latex] .<\/td>\n<td>[latex]-2(16)+3(-4)+8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Multiply.<\/td>\n<td>[latex]-32-12+8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Add.<\/td>\n<td>[latex]-36[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>[latex]\\text{Evaluate }\\frac{1}{x-3}\\text{ when }x=5 \\text{ and when }x=3[\/latex].<\/p>\n<h4>Solution<\/h4>\n<ul>\n<li>If we substitute [latex]5[\/latex] for [latex]x[\/latex], the expression becomes [latex]\\frac{1}{\\color{blue}{5}-3}=\\frac{1}{2}[\/latex]. The answer is [latex]\\frac{1}{2}[\/latex]<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<ul>\n<li>If we substitute [latex]3[\/latex] for [latex]x[\/latex], the expression becomes [latex]\\frac{1}{\\color{blue}{3}-3}=\\frac{1}{0}[\/latex]. A fraction where the denominator is zero is\u00a0 undefined.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Evaluate: [latex]4\\sqrt{x^2}-\\sqrt{-6x}\\text{ when } x=-3[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>[latex]4\\sqrt{x^2}-\\sqrt{-6x}[\/latex]<br \/>\n[latex]=4\\sqrt{\\color{blue}{(-3)}^2}-\\sqrt{-6\\color{blue}{(-3)}}[\/latex]<\/p>\n<p>[latex]=4\\sqrt{\\color{blue}{9}}-\\sqrt{\\color{blue}{(18)}}[\/latex]<\/p>\n<p>[latex]=4\\cdot \\color{blue}{3}-\\sqrt{\\color{blue}{(9\\cdot 2)}}[\/latex]<\/p>\n<p>[latex]=\\color{blue}{12}-\\sqrt{\\color{blue}{9}}\\cdot\\sqrt{2}[\/latex]<\/p>\n<p>[latex]=12-\\color{blue}{3}\\cdot\\sqrt{2}[\/latex]<\/p>\n<p>[latex]=12-3\\sqrt{2}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm144886\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=144886&amp;theme=oea&amp;iframe_resize_id=ohm144886&amp;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the video below we show more examples of how to substitute a value for variable in an expression, then evaluate the expression.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/dkFIVfJTG9E?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Evaluate [latex]\\frac{x+y}{x}-\\frac{x-y}{y}[\/latex] when [latex]x=12[\/latex] and [latex]y=-9[\/latex].<\/p>\n<p>Solution<\/p>\n<p>[latex]\\frac{x+y}{x}-\\frac{x-y}{y}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Substitute\u00a0[latex]x=\\color{blue}{12}[\/latex] and [latex]y=\\color{red}{-9}[\/latex]<\/p>\n<p>[latex]=\\frac{\\color{blue}{12}+\\color{red}{(-9)}}{\\color{blue}{12}}-\\frac{\\color{blue}{12}-\\color{red}{(-9)}}{\\color{red}{(-9)}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Simplify the numerators.<\/p>\n<p>[latex]=\\frac{\\color{purple}{3}}{12}-\\frac{\\color{purple}{21}}{-9}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Look for common factors between the numerator and denominator in each fraction.<\/p>\n<p>[latex]=\\frac{\\color{green}{3}}{\\color{green}{3}\\cdot 4}-\\frac{\\color{green}{3}\\cdot 7}{-3\\cdot\\color{green}{3}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Cancel common factors.<\/p>\n<p>[latex]=\\frac{1}{4}\\color{red}{-}\\frac{7}{\\color{red}{-3}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Subtracting a negative is equivalent to adding a positive.<\/p>\n<p>[latex]=\\frac{1}{4}\\color{red}{+}\\frac{7}{3}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Build equivalent fractions with a common denominator of [latex]12[\/latex].<\/p>\n<p>[latex]=\\frac{1\\color{blue}{\\cdot 3}}{4\\color{blue}{\\cdot 3}}+\\frac{7\\color{red}{\\cdot 4}}{3\\color{red}{\\cdot 4}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Multiply the numerators and denominators.<\/p>\n<p>[latex]=\\frac{3}{\\color{purple}{12}}+\\frac{28}{\\color{purple}{12}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Add the fractions by combining the numerators and keeping the common denominator.<\/p>\n<p>[latex]=\\frac{3+28}{\\color{purple}{12}}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Add the numerators.<\/p>\n<p>[latex]=\\frac{31}{12}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\frac{x^2}{y}+\\frac{y^2}{x}[\/latex] when [latex]x=6[\/latex] and [latex]y=-4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm686\">Show Answer<\/span><\/p>\n<div id=\"qhjm686\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-\\frac{19}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":370291,"menu_order":5,"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-2836","chapter","type-chapter","status-publish","hentry"],"part":2825,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2836","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":1,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2836\/revisions"}],"predecessor-version":[{"id":2837,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2836\/revisions\/2837"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/parts\/2825"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/2836\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/media?parent=2836"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2836"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/contributor?post=2836"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/license?post=2836"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}