{"id":1451,"date":"2016-10-24T22:54:42","date_gmt":"2016-10-24T22:54:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1451"},"modified":"2017-04-04T16:23:31","modified_gmt":"2017-04-04T16:23:31","slug":"putting-it-together-complex-numbers-and-equation-solving-techniques","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/putting-it-together-complex-numbers-and-equation-solving-techniques\/","title":{"raw":"Putting It Together: Equations and Inequalities","rendered":"Putting It Together: Equations and Inequalities"},"content":{"raw":"At the beginning of this module, we explored a few problems associated with the construction of a school. First we had to set up a mathematical model that would accurately reflect the following requirements for the foundation of the building.\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">The shape must be rectangular.<\/li>\r\n \t<li style=\"font-weight: 400;\">The length must be 100 ft. longer than the width.<\/li>\r\n \t<li style=\"font-weight: 400;\">There must be a total of 12,000 sq. ft. in the foundation.<\/li>\r\n<\/ul>\r\n<img class=\"size-medium wp-image-3659 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/03\/16220820\/rectangle-area-300x173.jpg\" alt=\"Blue rectangle with width (x) and length (x + 100) labeled.\" width=\"300\" height=\"173\" \/>\r\n\r\n&nbsp;\r\n\r\nLetting [latex]x[\/latex] stand for the width of the rectangle, the length must then be [latex]x+100[\/latex]. The area of any rectangle is equal to the product of its length and width, and since the area must be 12,000 sq. ft., this leads to the equation,\r\n<p style=\"text-align: center;\">[latex]\\left(x+100\\right)\\left(x\\right)=12,000[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x^2+100x=12,000[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x^2+100x-12,000=0[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThe equation is <strong>quadratic<\/strong>, that is, degree 2. \u00a0In this case, the trinomial can be factored.\r\n<p style=\"text-align: center;\">[latex]\\left(x-300\\right)\\left(x+400\\right)=0[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThis implies either [latex]x=300[\/latex], or [latex]x=-400[\/latex]. However, since negative width does not make much sense (have you ever tried to order a <em>negative<\/em> 8 foot fence post?), we only use the positive answer, [latex]x=300[\/latex]. This means the width is 300 ft., and the length is 300 + 100 = 400 ft. \u00a0A quick check verifies that the dimensions yield the correct square footage:\r\n<p style=\"text-align: center;\">[latex]300\\times400=12,000[\/latex]<\/p>\r\n&nbsp;\r\n\r\nOk, now that we have our foundation planned, let\u2019s move on to the walls. Recall, this project requires at least 1800 concrete blocks. \u00a0The budget for this material is $2,400, and each block costs $1.20. \u00a0Furthermore, there is a fee of $100 for hauling the blocks to the construction site, regardless of how many blocks are ordered.\r\n\r\n&nbsp;\r\n\r\nLet [latex]n[\/latex] stand for the number of blocks that would be purchased. Then two inequalities can be set up based on the constraints in the problem.\r\n<p style=\"text-align: center;\">[latex]n\\ge1800[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex](1.20)n+100\\le2400[\/latex]<\/p>\r\nWhat is the maximum number of blocks that you can purchase? Let\u2019s solve the second inequality to find out.\r\n<p style=\"text-align: center;\">[latex](1.20)n\\le2400-100=2300[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]n\\le{\\Large\\frac{2300}{1.20}}\\approx1916.7[\/latex]<\/p>\r\nThere is no use for a fraction of a block, so we might as well say that [latex]n\\le1916[\/latex]. A maximum of 1916 whole blocks can be purchased. Notice that [latex]1916\\ge1800[\/latex], so the first inequality is also satisfied, meaning that our budget allows for more than enough blocks to complete this phase of the project.\r\n\r\n&nbsp;\r\n\r\nWith the help of equations and inequalities, as well as specialized techniques for solving them, the plans for the new school are coming together nicely!\r\n\r\n&nbsp;","rendered":"<p>At the beginning of this module, we explored a few problems associated with the construction of a school. First we had to set up a mathematical model that would accurately reflect the following requirements for the foundation of the building.<\/p>\n<ul>\n<li style=\"font-weight: 400;\">The shape must be rectangular.<\/li>\n<li style=\"font-weight: 400;\">The length must be 100 ft. longer than the width.<\/li>\n<li style=\"font-weight: 400;\">There must be a total of 12,000 sq. ft. in the foundation.<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-3659 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/03\/16220820\/rectangle-area-300x173.jpg\" alt=\"Blue rectangle with width (x) and length (x + 100) labeled.\" width=\"300\" height=\"173\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Letting [latex]x[\/latex] stand for the width of the rectangle, the length must then be [latex]x+100[\/latex]. The area of any rectangle is equal to the product of its length and width, and since the area must be 12,000 sq. ft., this leads to the equation,<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+100\\right)\\left(x\\right)=12,000[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x^2+100x=12,000[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x^2+100x-12,000=0[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The equation is <strong>quadratic<\/strong>, that is, degree 2. \u00a0In this case, the trinomial can be factored.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x-300\\right)\\left(x+400\\right)=0[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>This implies either [latex]x=300[\/latex], or [latex]x=-400[\/latex]. However, since negative width does not make much sense (have you ever tried to order a <em>negative<\/em> 8 foot fence post?), we only use the positive answer, [latex]x=300[\/latex]. This means the width is 300 ft., and the length is 300 + 100 = 400 ft. \u00a0A quick check verifies that the dimensions yield the correct square footage:<\/p>\n<p style=\"text-align: center;\">[latex]300\\times400=12,000[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Ok, now that we have our foundation planned, let\u2019s move on to the walls. Recall, this project requires at least 1800 concrete blocks. \u00a0The budget for this material is $2,400, and each block costs $1.20. \u00a0Furthermore, there is a fee of $100 for hauling the blocks to the construction site, regardless of how many blocks are ordered.<\/p>\n<p>&nbsp;<\/p>\n<p>Let [latex]n[\/latex] stand for the number of blocks that would be purchased. Then two inequalities can be set up based on the constraints in the problem.<\/p>\n<p style=\"text-align: center;\">[latex]n\\ge1800[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex](1.20)n+100\\le2400[\/latex]<\/p>\n<p>What is the maximum number of blocks that you can purchase? Let\u2019s solve the second inequality to find out.<\/p>\n<p style=\"text-align: center;\">[latex](1.20)n\\le2400-100=2300[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]n\\le{\\Large\\frac{2300}{1.20}}\\approx1916.7[\/latex]<\/p>\n<p>There is no use for a fraction of a block, so we might as well say that [latex]n\\le1916[\/latex]. A maximum of 1916 whole blocks can be purchased. Notice that [latex]1916\\ge1800[\/latex], so the first inequality is also satisfied, meaning that our budget allows for more than enough blocks to complete this phase of the project.<\/p>\n<p>&nbsp;<\/p>\n<p>With the help of equations and inequalities, as well as specialized techniques for solving them, the plans for the new school are coming together nicely!<\/p>\n<p>&nbsp;<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1451\">\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>Putting It Together: Equations and Inequalities. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Shaded Rectangle Depicting Area. <strong>Authored by<\/strong>: S.V. Ault for Lumen. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":18,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Putting It Together: Equations and Inequalities\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Shaded Rectangle Depicting Area\",\"author\":\"S.V. Ault for Lumen\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"8801f453-1950-454c-bb93-bff45607b0ad","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1451","chapter","type-chapter","status-publish","hentry"],"part":1382,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1451","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1451\/revisions"}],"predecessor-version":[{"id":3785,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1451\/revisions\/3785"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1382"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1451\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/media?parent=1451"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1451"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1451"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/license?post=1451"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}