{"id":199,"date":"2020-04-20T17:25:48","date_gmt":"2020-04-20T21:25:48","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/?post_type=chapter&#038;p=199"},"modified":"2024-10-29T14:10:37","modified_gmt":"2024-10-29T18:10:37","slug":"systems-of-equations","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/chapter\/systems-of-equations\/","title":{"raw":"2.5 Systems of Equations\u00a0\u00a0\u00a0\u00a0\u00a0","rendered":"2.5 Systems of Equations\u00a0\u00a0\u00a0\u00a0\u00a0"},"content":{"raw":"If a number of equations involve the same variables, they are called a [pb_glossary id=\"3173\"]system of equations[\/pb_glossary]<em>.<\/em>\r\n<h2>Example 2.5.1<\/h2>\r\nAn agent is to purchase two products, G and H, and send them to the company's warehouse. He has a budget of $70,000 and truck space of 9,000 cubic meters. The product costs and sizes are as follows:\r\n<div align=\"center\">\r\n<table class=\"aligncenter\">\r\n<thead>\r\n<tr>\r\n<td><strong>\u00a0<\/strong><\/td>\r\n<td><strong>Cost per Case<\/strong><\/td>\r\n<td><strong>Volume per Case<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td><strong>G<\/strong><\/td>\r\n<td>$10<\/td>\r\n<td>3 m<sup>3<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>H<\/strong><\/td>\r\n<td>$20<\/td>\r\n<td>2 m<sup>3<\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nThe question to be answered here is: how much of each product should be purchased if both the budget and space are to be used up?\r\n<div>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Takeaways<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">Limitations are called constraints. Each constraint gives an equation.<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\nLimitations such as those for budget and space are called constraints. Each constraint gives an equation. To find the equation from the data given, you should consider the totals for each constraint\u00a0 separately and find the contribution of each\u00a0 variable. In this case; the variables are the amounts of each product to be bought. Let,\r\n<ul>\r\n \t<li>\u00a0<em>g<\/em> = the amount (number of cases) of G<\/li>\r\n \t<li>\u00a0<em>h<\/em> = the amount (number of cases) of H<\/li>\r\n<\/ul>\r\n&nbsp;\r\n\r\nThen the budget constraint equation can be obtained by considering:\r\n\r\n[latex]\\begin{align*}\r\n\\text{Total Spent} &amp;= \\text{total spent on G} + \\text{total spent on H}\\\\\r\n&amp;= \\text{cost per case of G}\\times\\text{(amount of G) + cost per case of H}\\times \\text{(amount of H)}\\\\\r\n&amp;=$10\\times g+$2\\times h\\\\\r\n$70,000 &amp;=$10g +$20h\\\\\r\n\\end{align*}[\/latex]\r\n\r\nSimilarly,\r\n\r\n[latex]\\begin{align*}\r\n\\text{Total Space} &amp;= \\text{space used by G + total space used by H}\\\\\r\n&amp;= \\text{space used by G} \\times \\text{(amount of G)+ space used by} \\times \\text{(amount of H)}\r\n\\end{align*}[\/latex]\r\n\r\nSo,\r\n<p style=\"text-align: center\">[latex]9,000 m^3=3g m^3+2h m^3[\/latex]<\/p>\r\nThus you have the <em>system of equations<\/em>\r\n<p style=\"text-align: center\">[latex]$70,000 = $10g + $20 h\\text{ and } 9,000 = 3g + 2h[\/latex]<\/p>\r\nfor which the graphs are given below.\r\n\r\n[caption id=\"attachment_200\" align=\"aligncenter\" width=\"300\"]<img class=\"wp-image-200 size-medium\" src=\"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-content\/uploads\/sites\/971\/2020\/04\/chap2-sysofeqns-300x196.png\" alt=\"Graph of both lines, showing the intercept\" width=\"300\" height=\"196\" \/> Graph of G and H, Example 6[\/caption]\r\n\r\nNotice that the data for the problem was chiefly in terms of <strong>rates, <\/strong>the costs <strong>per case <\/strong>and the volume <strong>per case. <\/strong>These were used to get equations for totals, using the idea of,\r\n<p style=\"text-align: center\">[latex]Amount = rate \\times base[\/latex]<\/p>\r\nfor each equation. Notice, also, the use of the units of measurement to help keep track of the parts of the equations.\r\n\r\n&nbsp;\r\n\r\nThe values required by the problem\u00a0 are those of the point at which both equations are satisfied - the point on the graph at which the\u00a0 lines cross. This point is called the <em>solution <\/em>of the equations. It can be estimated from the graph and also calculated from the equations.\r\n<h1>Method of Elimination<\/h1>\r\nTo solve the equations, using the <em>method of elimination <\/em>write each one in the form for which <em>h<\/em> is given as a function of g.\r\n\r\nFor the budget: <em>h<\/em> =\u00a0 3,500 \u2212 0.5<em>g<\/em>\r\nFor the space: <em>h<\/em> = 4,500 \u2212 1.5<em>g<\/em>\r\n\r\nNext, note that at the solution, the values of <em>h<\/em> must be the same, so:\r\n<p style=\"text-align: center\">[latex]3,500 - 0.5g = 4,500 - 1.5g[\/latex]<\/p>\r\nBy adding [latex]1.5g[\/latex] to each side, you get:\r\n<p style=\"text-align: center\">[latex]3,500 - 0.5g + 1.5g = 4,500 - 1.5g + 1.5g[\/latex]\r\n[latex]3,500 + 1g= 4,500[\/latex]<\/p>\r\nSubtracting $3,500 from each side gives:\r\n<p style=\"text-align: center\">[latex]g = 4,500 - 3,500[\/latex]\r\n[latex]g = 1,000 \\text{ cases}[\/latex]<\/p>\r\nSubstituting the result in one of the original equations (space):\r\n<p style=\"text-align: center\">[latex]9,000 = 3\\times 1,000 + 2h[\/latex]<\/p>\r\nWhence\r\n<p style=\"text-align: center\">[latex]h = 6,000 + 2 = 3,000 \\text{ cases}[\/latex]<\/p>\r\nThus, both the budget and space allowances will be used up if 1,000 cases of G and 3,000 cases of H are bought.\r\n\r\n&nbsp;\r\n<h1>Method of Substitution<\/h1>\r\nThe [pb_glossary id=\"205\"]<em>method of substitution<\/em>[\/pb_glossary] gives you another way of solving such equations. It consists of the following:\r\n\r\nOnce the first equation has been solved for h, the result is\r\n<p style=\"text-align: center\">[latex]h = 3,500 - 0.5g[\/latex]<\/p>\r\nThis result would be substituted in the second equation, which would produce an equation containing only the variable <em>g<\/em>.\r\n\r\nThus,\r\n\r\n[latex]\\begin{align*}\r\n\r\n9,000 &amp;= 3g + 2\\times (3,500 - 0.5g)\\\\\r\n9,000 &amp;= 3g + 7,000 - g\r\n\r\n\\end{align*}[\/latex]\r\n\r\nand\r\n\r\n[latex]\\begin{align*}\r\n9,000 - 7,000 &amp;= 2g\\\\\r\n2,000 &amp;= 2g\\\\\r\n\r\ng &amp;= 1,000\r\n\r\n\\end{align*}[\/latex]\r\n\r\nSubstituting in the other equation:\r\n\r\n[latex]\\begin{align*}\r\n\r\n70,000 &amp;= 10 \\times 1,000 + 20h\\\\\r\n70,000 -10,000 &amp;= 20h\\\\\r\n60,000 &amp;= 20h\\\\\r\nh &amp;= 3,000 \\text{ cases}\r\n\r\n\\end{align*}[\/latex]\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Knowledge Check 2.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nOne of the production facilities of Jones Furniture Company produces chairs from kits, each of which contains all the parts for the chair. The facility makes two types of chairs: a regular chair for normal inside use and a special chair for outside use. Each chair is assembled and then painted.\r\n\r\nThere is only a limited amount of labor time for each operation: 120 hours per week for assembly, 80 hours per week for painting. Each regular chair requires 0.5 hours for assembly and 0.3 hours for painting. Each special chair requires 0.4 hours for assembly and 0.4 hours for painting.\r\n\r\nFind the number of regular and special chairs that would have to be produced each week in order to use up both the assembly and painting time available for that week.\r\n\r\nLet,\r\n\r\n<em>r<\/em> = number of regular chairs produced per week\r\n<em>s<\/em> = number of special chairs produced per week\r\n\r\n&nbsp;\r\n<ol>\r\n \t<li>Find the equation that shows the constraint for the available assembly time: _______ (total hours) = _______ _(hours\/chair) \u00d7 <em>r<\/em> (chairs) + _______ (hours\/chair) \u00d7 <em>s<\/em> (chairs)<\/li>\r\n \t<li>Find the equation that shows the constraint for the available painting time:\u00a0 ? = ? \u00d7 <em>r <\/em>+ ? \u00d7 <em>s<\/em><\/li>\r\n \t<li>Solve the equation in Question 1 for <em>r<\/em> and substitute the result in the second equation, in Question 2, to get the value for I.<\/li>\r\n \t<li>Substitute the result from Question 3 in one of the equations to get the value of <em>r<\/em>.<\/li>\r\n \t<li>Check your results by placing them in the other equation (but not the one in Question 4) and see that they give the correct value.<\/li>\r\n \t<li>Graph the two equations to make a visual check of your answer.<\/li>\r\n<\/ol>\r\n<a href=\"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/chapter\/learning-activities-answer-key-2\/\">Solutions at the end of the chapter<\/a>\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n<h2>Your Own Notes<\/h2>\r\n<ul>\r\n \t<li>Are there any notes you want to take from this section? Is there anything you'd like to copy and paste below?<\/li>\r\n \t<li>These notes are for you only (they will not be stored anywhere)<\/li>\r\n \t<li>Make sure to download them at the end to use as a reference<\/li>\r\n<\/ul>\r\n[h5p id=\"1\"]","rendered":"<p>If a number of equations involve the same variables, they are called a <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_199_3173\">system of equations<\/a><em>.<\/em><\/p>\n<h2>Example 2.5.1<\/h2>\n<p>An agent is to purchase two products, G and H, and send them to the company&#8217;s warehouse. He has a budget of $70,000 and truck space of 9,000 cubic meters. The product costs and sizes are as follows:<\/p>\n<div style=\"margin: auto;\">\n<table class=\"aligncenter\">\n<thead>\n<tr>\n<td><strong>\u00a0<\/strong><\/td>\n<td><strong>Cost per Case<\/strong><\/td>\n<td><strong>Volume per Case<\/strong><\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><strong>G<\/strong><\/td>\n<td>$10<\/td>\n<td>3 m<sup>3<\/sup><\/td>\n<\/tr>\n<tr>\n<td><strong>H<\/strong><\/td>\n<td>$20<\/td>\n<td>2 m<sup>3<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>The question to be answered here is: how much of each product should be purchased if both the budget and space are to be used up?<\/p>\n<div>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Takeaways<\/p>\n<\/header>\n<div class=\"textbox__content\">Limitations are called constraints. Each constraint gives an equation.<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p>Limitations such as those for budget and space are called constraints. Each constraint gives an equation. To find the equation from the data given, you should consider the totals for each constraint\u00a0 separately and find the contribution of each\u00a0 variable. In this case; the variables are the amounts of each product to be bought. Let,<\/p>\n<ul>\n<li>\u00a0<em>g<\/em> = the amount (number of cases) of G<\/li>\n<li>\u00a0<em>h<\/em> = the amount (number of cases) of H<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<p>Then the budget constraint equation can be obtained by considering:<\/p>\n<p>[latex]\\begin{align*}  \\text{Total Spent} &= \\text{total spent on G} + \\text{total spent on H}\\\\  &= \\text{cost per case of G}\\times\\text{(amount of G) + cost per case of H}\\times \\text{(amount of H)}\\\\  &=$10\\times g+$2\\times h\\\\  $70,000 &=$10g +$20h\\\\  \\end{align*}[\/latex]<\/p>\n<p>Similarly,<\/p>\n<p>[latex]\\begin{align*}  \\text{Total Space} &= \\text{space used by G + total space used by H}\\\\  &= \\text{space used by G} \\times \\text{(amount of G)+ space used by} \\times \\text{(amount of H)}  \\end{align*}[\/latex]<\/p>\n<p>So,<\/p>\n<p style=\"text-align: center\">[latex]9,000 m^3=3g m^3+2h m^3[\/latex]<\/p>\n<p>Thus you have the <em>system of equations<\/em><\/p>\n<p style=\"text-align: center\">[latex]$70,000 = $10g + $20 h\\text{ and } 9,000 = 3g + 2h[\/latex]<\/p>\n<p>for which the graphs are given below.<\/p>\n<figure id=\"attachment_200\" aria-describedby=\"caption-attachment-200\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-200 size-medium\" src=\"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-content\/uploads\/sites\/971\/2020\/04\/chap2-sysofeqns-300x196.png\" alt=\"Graph of both lines, showing the intercept\" width=\"300\" height=\"196\" srcset=\"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-content\/uploads\/sites\/971\/2020\/04\/chap2-sysofeqns-300x196.png 300w, https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-content\/uploads\/sites\/971\/2020\/04\/chap2-sysofeqns-65x43.png 65w, https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-content\/uploads\/sites\/971\/2020\/04\/chap2-sysofeqns-225x147.png 225w, https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-content\/uploads\/sites\/971\/2020\/04\/chap2-sysofeqns-350x229.png 350w, https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-content\/uploads\/sites\/971\/2020\/04\/chap2-sysofeqns.png 594w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-200\" class=\"wp-caption-text\">Graph of G and H, Example 6<\/figcaption><\/figure>\n<p>Notice that the data for the problem was chiefly in terms of <strong>rates, <\/strong>the costs <strong>per case <\/strong>and the volume <strong>per case. <\/strong>These were used to get equations for totals, using the idea of,<\/p>\n<p style=\"text-align: center\">[latex]Amount = rate \\times base[\/latex]<\/p>\n<p>for each equation. Notice, also, the use of the units of measurement to help keep track of the parts of the equations.<\/p>\n<p>&nbsp;<\/p>\n<p>The values required by the problem\u00a0 are those of the point at which both equations are satisfied &#8211; the point on the graph at which the\u00a0 lines cross. This point is called the <em>solution <\/em>of the equations. It can be estimated from the graph and also calculated from the equations.<\/p>\n<h1>Method of Elimination<\/h1>\n<p>To solve the equations, using the <em>method of elimination <\/em>write each one in the form for which <em>h<\/em> is given as a function of g.<\/p>\n<p>For the budget: <em>h<\/em> =\u00a0 3,500 \u2212 0.5<em>g<\/em><br \/>\nFor the space: <em>h<\/em> = 4,500 \u2212 1.5<em>g<\/em><\/p>\n<p>Next, note that at the solution, the values of <em>h<\/em> must be the same, so:<\/p>\n<p style=\"text-align: center\">[latex]3,500 - 0.5g = 4,500 - 1.5g[\/latex]<\/p>\n<p>By adding [latex]1.5g[\/latex] to each side, you get:<\/p>\n<p style=\"text-align: center\">[latex]3,500 - 0.5g + 1.5g = 4,500 - 1.5g + 1.5g[\/latex]<br \/>\n[latex]3,500 + 1g= 4,500[\/latex]<\/p>\n<p>Subtracting $3,500 from each side gives:<\/p>\n<p style=\"text-align: center\">[latex]g = 4,500 - 3,500[\/latex]<br \/>\n[latex]g = 1,000 \\text{ cases}[\/latex]<\/p>\n<p>Substituting the result in one of the original equations (space):<\/p>\n<p style=\"text-align: center\">[latex]9,000 = 3\\times 1,000 + 2h[\/latex]<\/p>\n<p>Whence<\/p>\n<p style=\"text-align: center\">[latex]h = 6,000 + 2 = 3,000 \\text{ cases}[\/latex]<\/p>\n<p>Thus, both the budget and space allowances will be used up if 1,000 cases of G and 3,000 cases of H are bought.<\/p>\n<p>&nbsp;<\/p>\n<h1>Method of Substitution<\/h1>\n<p>The <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_199_205\"><em>method of substitution<\/em><\/a> gives you another way of solving such equations. It consists of the following:<\/p>\n<p>Once the first equation has been solved for h, the result is<\/p>\n<p style=\"text-align: center\">[latex]h = 3,500 - 0.5g[\/latex]<\/p>\n<p>This result would be substituted in the second equation, which would produce an equation containing only the variable <em>g<\/em>.<\/p>\n<p>Thus,<\/p>\n<p>[latex]\\begin{align*}    9,000 &= 3g + 2\\times (3,500 - 0.5g)\\\\  9,000 &= 3g + 7,000 - g    \\end{align*}[\/latex]<\/p>\n<p>and<\/p>\n<p>[latex]\\begin{align*}  9,000 - 7,000 &= 2g\\\\  2,000 &= 2g\\\\    g &= 1,000    \\end{align*}[\/latex]<\/p>\n<p>Substituting in the other equation:<\/p>\n<p>[latex]\\begin{align*}    70,000 &= 10 \\times 1,000 + 20h\\\\  70,000 -10,000 &= 20h\\\\  60,000 &= 20h\\\\  h &= 3,000 \\text{ cases}    \\end{align*}[\/latex]<\/p>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Knowledge Check 2.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>One of the production facilities of Jones Furniture Company produces chairs from kits, each of which contains all the parts for the chair. The facility makes two types of chairs: a regular chair for normal inside use and a special chair for outside use. Each chair is assembled and then painted.<\/p>\n<p>There is only a limited amount of labor time for each operation: 120 hours per week for assembly, 80 hours per week for painting. Each regular chair requires 0.5 hours for assembly and 0.3 hours for painting. Each special chair requires 0.4 hours for assembly and 0.4 hours for painting.<\/p>\n<p>Find the number of regular and special chairs that would have to be produced each week in order to use up both the assembly and painting time available for that week.<\/p>\n<p>Let,<\/p>\n<p><em>r<\/em> = number of regular chairs produced per week<br \/>\n<em>s<\/em> = number of special chairs produced per week<\/p>\n<p>&nbsp;<\/p>\n<ol>\n<li>Find the equation that shows the constraint for the available assembly time: _______ (total hours) = _______ _(hours\/chair) \u00d7 <em>r<\/em> (chairs) + _______ (hours\/chair) \u00d7 <em>s<\/em> (chairs)<\/li>\n<li>Find the equation that shows the constraint for the available painting time:\u00a0 ? = ? \u00d7 <em>r <\/em>+ ? \u00d7 <em>s<\/em><\/li>\n<li>Solve the equation in Question 1 for <em>r<\/em> and substitute the result in the second equation, in Question 2, to get the value for I.<\/li>\n<li>Substitute the result from Question 3 in one of the equations to get the value of <em>r<\/em>.<\/li>\n<li>Check your results by placing them in the other equation (but not the one in Question 4) and see that they give the correct value.<\/li>\n<li>Graph the two equations to make a visual check of your answer.<\/li>\n<\/ol>\n<p><a href=\"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/chapter\/learning-activities-answer-key-2\/\">Solutions at the end of the chapter<\/a><\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Your Own Notes<\/h2>\n<ul>\n<li>Are there any notes you want to take from this section? Is there anything you&#8217;d like to copy and paste below?<\/li>\n<li>These notes are for you only (they will not be stored anywhere)<\/li>\n<li>Make sure to download them at the end to use as a reference<\/li>\n<\/ul>\n<div id=\"h5p-1\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-1\" class=\"h5p-iframe\" data-content-id=\"1\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Key takeaways, notes and comments from this section document tool.\"><\/iframe><\/div>\n<\/div>\n<div class=\"glossary\"><span class=\"screen-reader-text\" id=\"definition\">definition<\/span><template id=\"term_199_3173\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_199_3173\"><div tabindex=\"-1\"><p>A number of equations, involving the same variables<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><template id=\"term_199_205\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_199_205\"><div tabindex=\"-1\"><p>A method of solving a system of equations.<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><\/div>","protected":false},"author":883,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-199","chapter","type-chapter","status-publish","hentry"],"part":40,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapters\/199","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/wp\/v2\/users\/883"}],"version-history":[{"count":25,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapters\/199\/revisions"}],"predecessor-version":[{"id":3953,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapters\/199\/revisions\/3953"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/parts\/40"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapters\/199\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/wp\/v2\/media?parent=199"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/pressbooks\/v2\/chapter-type?post=199"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/wp\/v2\/contributor?post=199"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/businessmathematics\/wp-json\/wp\/v2\/license?post=199"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}