{"id":3191,"date":"2025-08-15T22:33:36","date_gmt":"2025-08-15T22:33:36","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3191"},"modified":"2026-03-24T17:03:13","modified_gmt":"2026-03-24T17:03:13","slug":"solving-systems-with-gaussian-elimination-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-systems-with-gaussian-elimination-apply-it\/","title":{"raw":"Solving Systems with Gaussian Elimination: Apply It","rendered":"Solving Systems with Gaussian Elimination: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Convert between augmented matrices and systems of equations<\/li>\r\n \t<li>Perform row operations on a matrix.<\/li>\r\n \t<li>Solve a system of linear equations using row operations.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Apply Matrices to Finance<\/h2>\r\n<section class=\"textbox example\" aria-label=\"Example\">Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?[reveal-answer q=\"750431\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"750431\"]We have a system of two equations in two variables. Let [latex]x=[\/latex] the amount invested at 10.5% interest, and [latex]y=[\/latex] the amount invested at 12% interest.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+y=12,000 \\\\ 0.105x+0.12y=1,335 \\end{gathered}[\/latex]<\/p>\r\nAs a matrix, we have\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1&amp; \\hfill 1\\\\ \\hfill 0.105&amp; \\hfill 0.12\\end{array}\\right\\rvert\\begin{array}{r}\\hfill 12,000\\\\ \\hfill 1,335\\end{array}\\right][\/latex]<\/p>\r\nMultiply row 1 by [latex]-0.105[\/latex] and add the result to row 2.\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1&amp; \\hfill 1\\\\ \\hfill 0&amp; \\hfill 0.015\\end{array}\\right\\rvert\\begin{array}{r}\\hfill 12,000\\\\ \\hfill 75\\end{array}\\right][\/latex]<\/p>\r\nThen,\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}0.015y=75 \\\\ y=5,000 \\end{gathered}[\/latex]<\/p>\r\nSo [latex]12,000 - 5,000=7,000[\/latex].\r\n\r\nThus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?[reveal-answer q=\"462603\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"462603\"]We have a system of three equations in three variables. Let [latex]x[\/latex] be the amount invested at 5% interest, let [latex]y[\/latex] be the amount invested at 8% interest, and let [latex]z[\/latex] be the amount invested at 9% interest. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+y+z=10,000 \\\\ 0.05x+0.08y+0.09z=770 \\\\ 2x-z=0 \\end{gathered}[\/latex]<\/p>\r\nAs a matrix, we have\r\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rrr}\\hfill 1&amp; \\hfill 1&amp; \\hfill 1\\\\ \\hfill 0.05&amp; \\hfill 0.08&amp; \\hfill 0.09\\\\ \\hfill 2&amp; \\hfill 0&amp; \\hfill -1\\end{array}\\right\\rvert\\begin{array}{r}\\hfill 10,000\\\\ \\hfill 770\\\\ \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\nNow, we perform Gaussian elimination to achieve row-echelon form.\r\n<p style=\"text-align: center;\">[latex]-0.05{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\left.\\begin{array}{rrrrrr}\\hfill 1&amp; \\hfill &amp; \\hfill 1&amp; \\hfill &amp; \\hfill 1&amp; \\hfill \\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 0.03&amp; \\hfill &amp; \\hfill 0.04&amp; \\hfill \\\\ \\hfill 2&amp; \\hfill &amp; \\hfill 0&amp; \\hfill &amp; \\hfill -1&amp; \\hfill \\end{array}\\right\\rvert\\begin{array}{rr}\\hfill &amp; \\hfill 10,000\\\\ \\hfill &amp; \\hfill 270\\\\ \\hfill &amp; \\hfill 0\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\left.\\begin{array}{rrrrrr}\\hfill 1&amp; \\hfill &amp; \\hfill 1&amp; \\hfill &amp; \\hfill 1&amp; \\hfill \\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 0.03&amp; \\hfill &amp; \\hfill 0.04&amp; \\hfill \\\\ \\hfill 0&amp; \\hfill &amp; \\hfill -2&amp; \\hfill &amp; \\hfill -3&amp; \\hfill \\end{array}\\right\\rvert\\begin{array}{rr}\\hfill &amp; \\hfill 10,000\\\\ \\hfill &amp; \\hfill 270\\\\ \\hfill &amp; \\hfill -20,000\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{0.03}{R}_{2}={R}_{2}\\to \\left[\\left.\\begin{array}{rrrrrr}\\hfill 0&amp; \\hfill &amp; \\hfill 1&amp; \\hfill &amp; \\hfill 1&amp; \\hfill \\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1&amp; \\hfill &amp; \\hfill \\frac{4}{3}&amp; \\hfill \\\\ \\hfill 0&amp; \\hfill &amp; \\hfill -2&amp; \\hfill &amp; \\hfill -3&amp; \\hfill \\end{array}\\right\\rvert\\begin{array}{rr}\\hfill &amp; \\hfill 10,000\\\\ \\hfill &amp; \\hfill 9,000\\\\ \\hfill &amp; \\hfill -20,000\\end{array}\\right][\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]2{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\left.\\begin{array}{rrrrrr}\\hfill 1&amp; \\hfill &amp; \\hfill 1&amp; \\hfill &amp; \\hfill 1&amp; \\hfill \\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 1&amp; \\hfill &amp; \\hfill \\frac{4}{3}&amp; \\hfill \\\\ \\hfill 0&amp; \\hfill &amp; \\hfill 0&amp; \\hfill &amp; \\hfill -\\frac{1}{3}&amp; \\hfill \\end{array}\\right\\rvert\\begin{array}{rr}\\hfill &amp; \\hfill 10,000\\\\ \\hfill &amp; \\hfill 9,000\\\\ \\hfill &amp; \\hfill -2,000\\end{array}\\right][\/latex]<\/p>\r\nThe third row tells us [latex]-\\frac{1}{3}z=-2,000[\/latex]; thus [latex]z=6,000[\/latex].\r\n\r\nThe second row tells us [latex]y+\\frac{4}{3}z=9,000[\/latex]. Substituting [latex]z=6,000[\/latex], we get\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill y+\\frac{4}{3}\\left(6,000\\right)=9,000\\\\ \\hfill y+8,000=9,000\\\\ \\hfill y=1,000\\end{array}[\/latex]<\/p>\r\nThe first row tells us [latex]x+y+z=10,000[\/latex]. Substituting [latex]y=1,000[\/latex] and [latex]z=6,000[\/latex], we get\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+1,000+6,000=10,000 \\\\ x=3,000 \\end{gathered}[\/latex]<\/p>\r\nThe answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">\r\n<div class=\"bcc-box bcc-success\">\r\n\r\n[ohm_question hide_question_numbers=1]312556[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Convert between augmented matrices and systems of equations<\/li>\n<li>Perform row operations on a matrix.<\/li>\n<li>Solve a system of linear equations using row operations.<\/li>\n<\/ul>\n<\/section>\n<h2>Apply Matrices to Finance<\/h2>\n<section class=\"textbox example\" aria-label=\"Example\">Carolyn invests a total of $12,000 in two municipal bonds, one paying 10.5% interest and the other paying 12% interest. The annual interest earned on the two investments last year was $1,335. How much was invested at each rate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q750431\">Show Solution<\/button><\/p>\n<div id=\"q750431\" class=\"hidden-answer\" style=\"display: none\">We have a system of two equations in two variables. Let [latex]x=[\/latex] the amount invested at 10.5% interest, and [latex]y=[\/latex] the amount invested at 12% interest.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+y=12,000 \\\\ 0.105x+0.12y=1,335 \\end{gathered}[\/latex]<\/p>\n<p>As a matrix, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 0.105& \\hfill 0.12\\end{array}\\right\\rvert\\begin{array}{r}\\hfill 12,000\\\\ \\hfill 1,335\\end{array}\\right][\/latex]<\/p>\n<p>Multiply row 1 by [latex]-0.105[\/latex] and add the result to row 2.<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rr}\\hfill 1& \\hfill 1\\\\ \\hfill 0& \\hfill 0.015\\end{array}\\right\\rvert\\begin{array}{r}\\hfill 12,000\\\\ \\hfill 75\\end{array}\\right][\/latex]<\/p>\n<p>Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}0.015y=75 \\\\ y=5,000 \\end{gathered}[\/latex]<\/p>\n<p>So [latex]12,000 - 5,000=7,000[\/latex].<\/p>\n<p>Thus, $5,000 was invested at 12% interest and $7,000 at 10.5% interest.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Ava invests a total of $10,000 in three accounts, one paying 5% interest, another paying 8% interest, and the third paying 9% interest. The annual interest earned on the three investments last year was $770. The amount invested at 9% was twice the amount invested at 5%. How much was invested at each rate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q462603\">Show Solution<\/button><\/p>\n<div id=\"q462603\" class=\"hidden-answer\" style=\"display: none\">We have a system of three equations in three variables. Let [latex]x[\/latex] be the amount invested at 5% interest, let [latex]y[\/latex] be the amount invested at 8% interest, and let [latex]z[\/latex] be the amount invested at 9% interest. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+y+z=10,000 \\\\ 0.05x+0.08y+0.09z=770 \\\\ 2x-z=0 \\end{gathered}[\/latex]<\/p>\n<p>As a matrix, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\left[\\left.\\begin{array}{rrr}\\hfill 1& \\hfill 1& \\hfill 1\\\\ \\hfill 0.05& \\hfill 0.08& \\hfill 0.09\\\\ \\hfill 2& \\hfill 0& \\hfill -1\\end{array}\\right\\rvert\\begin{array}{r}\\hfill 10,000\\\\ \\hfill 770\\\\ \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<p>Now, we perform Gaussian elimination to achieve row-echelon form.<\/p>\n<p style=\"text-align: center;\">[latex]-0.05{R}_{1}+{R}_{2}={R}_{2}\\to \\left[\\left.\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0.03& \\hfill & \\hfill 0.04& \\hfill \\\\ \\hfill 2& \\hfill & \\hfill 0& \\hfill & \\hfill -1& \\hfill \\end{array}\\right\\rvert\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 270\\\\ \\hfill & \\hfill 0\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]-2{R}_{1}+{R}_{3}={R}_{3}\\to \\left[\\left.\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0.03& \\hfill & \\hfill 0.04& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill -2& \\hfill & \\hfill -3& \\hfill \\end{array}\\right\\rvert\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 270\\\\ \\hfill & \\hfill -20,000\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{0.03}{R}_{2}={R}_{2}\\to \\left[\\left.\\begin{array}{rrrrrr}\\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill \\frac{4}{3}& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill -2& \\hfill & \\hfill -3& \\hfill \\end{array}\\right\\rvert\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 9,000\\\\ \\hfill & \\hfill -20,000\\end{array}\\right][\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]2{R}_{2}+{R}_{3}={R}_{3}\\to \\left[\\left.\\begin{array}{rrrrrr}\\hfill 1& \\hfill & \\hfill 1& \\hfill & \\hfill 1& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 1& \\hfill & \\hfill \\frac{4}{3}& \\hfill \\\\ \\hfill 0& \\hfill & \\hfill 0& \\hfill & \\hfill -\\frac{1}{3}& \\hfill \\end{array}\\right\\rvert\\begin{array}{rr}\\hfill & \\hfill 10,000\\\\ \\hfill & \\hfill 9,000\\\\ \\hfill & \\hfill -2,000\\end{array}\\right][\/latex]<\/p>\n<p>The third row tells us [latex]-\\frac{1}{3}z=-2,000[\/latex]; thus [latex]z=6,000[\/latex].<\/p>\n<p>The second row tells us [latex]y+\\frac{4}{3}z=9,000[\/latex]. Substituting [latex]z=6,000[\/latex], we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\hfill y+\\frac{4}{3}\\left(6,000\\right)=9,000\\\\ \\hfill y+8,000=9,000\\\\ \\hfill y=1,000\\end{array}[\/latex]<\/p>\n<p>The first row tells us [latex]x+y+z=10,000[\/latex]. Substituting [latex]y=1,000[\/latex] and [latex]z=6,000[\/latex], we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x+1,000+6,000=10,000 \\\\ x=3,000 \\end{gathered}[\/latex]<\/p>\n<p>The answer is $3,000 invested at 5% interest, $1,000 invested at 8%, and $6,000 invested at 9% interest.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\">\n<div class=\"bcc-box bcc-success\">\n<p><iframe loading=\"lazy\" id=\"ohm312556\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=312556&theme=lumen&iframe_resize_id=ohm312556&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n","protected":false},"author":67,"menu_order":13,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":514,"module-header":"apply_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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3191"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/67"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3191\/revisions"}],"predecessor-version":[{"id":5991,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3191\/revisions\/5991"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/514"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3191\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3191"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3191"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3191"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3191"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}