{"id":300,"date":"2015-05-16T13:59:30","date_gmt":"2015-05-16T17:59:30","guid":{"rendered":"https:\/\/mathvault.wordpress.com\/?p=300"},"modified":"2023-07-28T21:58:53","modified_gmt":"2023-07-29T01:58:53","slug":"system-linear-equations","status":"publish","type":"post","link":"https:\/\/mathvault.ca\/system-linear-equations\/","title":{"rendered":"How to Solve a System of Linear Equations with an Additional Parameter?"},"content":{"rendered":"<figure><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-6859 size-full aligncenter\" src=\"https:\/\/mathvault.ca\/wp-content\/uploads\/Three-Lines.jpg\" alt=\"Solve the System of Linear Equations with an additional parameter h\" width=\"820\" height=\"820\" title=\"\"><\/figure>\n<p>(For more on matrices, vectors, vector products and other related topics, check out our <a href=\"https:\/\/mathvault.ca\/free-ebooks-series\/\">Linear Algebra Ebook Series<\/a>.)<\/p>\n<p style=\"text-align: justify;\">Today, Connor asked the following question involving a system of linear equation with an additional parameter:<\/p>\n<blockquote>\n<p style=\"text-align: justify;\">Given that $x_1 + hx_2 = 4$\u00a0 and $3x_1 + 6x_2 =8$<span class=\"math_w\">, find all pairs of the form $(x_1,x_2)$ satisfying both equations.<\/span><\/p>\n<\/blockquote>\n<p>OK&#8230;What do we do with that additional $h$? Well, read on!<br \/>\n<!--more--><\/p>\n<div id=\"toc\"><div id=\"ez-toc-container\" class=\"ez-toc-v2_0_82_1 counter-hierarchy ez-toc-counter ez-toc-custom ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title ez-toc-toggle\" style=\"cursor:pointer\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mathvault.ca\/system-linear-equations\/#Ignoring_the_h_First\" >Ignoring the h First<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mathvault.ca\/system-linear-equations\/#Case_1_%E2%80%94_When_h2\" >Case 1 \u2014 When h=2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mathvault.ca\/system-linear-equations\/#Case_2_%E2%80%94_When_h_%E2%89%A0_2\" >Case 2 \u2014 When h \u2260 2<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mathvault.ca\/system-linear-equations\/#Some_Illustrations\" >Some Illustrations<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<\/div>\n<h2 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Ignoring_the_h_First\"><\/span><a href=\"#toc\">Ignoring the h First<\/a><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: justify;\">We are given this system of linear equations from the get-go:<\/p>\n<p>\\begin{cases}x_1+hx_2=4 &amp; (a) \\\\ 3x_1+6x_2=8 &amp; (b) \\end{cases}<\/p>\n<p style=\"text-align: justify;\">What to do? Well, we can try to solve this system of linear equations by regarding $h$ as an ordinary coefficient first, and see what it takes us to as we proceed judiciously. First, dividing the equation $(b)$\u00a0by $3$ on both sides yields the following equivalent system:<\/p>\n<p>\\begin{cases}x_1+hx_2=4 &amp; (1) \\\\ x_1+2x_2=\\frac{8}{3} &amp; (2) \\end{cases}<\/p>\n<p style=\"text-align: justify;\">Turning the 2<sup>nd<\/sup> equation into $(2)-(1)$, we get:<\/p>\n<p>\\begin{align*} \\begin{cases} x_1+hx_2=4 \\\\ (2-h)x_2=\\frac{8}{3}-4 \\end{cases} \\qquad &amp; \\Leftrightarrow \\qquad \\begin{cases} x_1+hx_2=4 &amp; (3) \\\\ (h-2)x_2=4-\\frac{8}{3} = \\frac{4}{3} &amp; (4) \\end{cases} \\end{align*}<\/p>\n<p style=\"text-align: justify;\">At this point, our latest system is still <em>equivalent<\/em> to the original one (i.e., $(a)$ and $(b) \\iff (3)$ and $(4)$) . And with $(4)$, we are definitely one step closer towards solving for $x_2$. But before we do just that, notice that we can&#8217;t solve for $x_2$ by moving $h-2$ to the right, if $h-2=0$ (i.e., $h=2$). This would then open up two cans of worms&#8230;<\/p>\n<h2><span class=\"ez-toc-section\" id=\"Case_1_%E2%80%94_When_h2\"><\/span><a href=\"#toc\">Case 1 \u2014 When h=2<\/a><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: justify;\">Replace $h$ with 2 in the equations\u00a0$(3)$ and $(4)$, the following system pops up:<\/p>\n<p>\\begin{cases} x_1+2x_2=4 \\\\ 0= \\frac{4}{3} \\end{cases}<\/p>\n<p style=\"text-align: justify;\">Oops. Contradiction. That can&#8217;t happen at any rate, so there is no solution in this case.<\/p>\n<p style=\"text-align: justify;\">In fact, we don&#8217;t even have to get to $(3)$ and $(4)$ to see it. This is because if we just plug in 2 into $h$ by the time we got to $(1)$ and $(2)$, we would get the following system:<\/p>\n<p>\\begin{cases} x_1+2x_2=4 \\\\ x_1+2x_2=\\frac{8}{3} \\end{cases}<\/p>\n<p>which, as we can see, is already a system whose equations contradict each other. Here &#8212; a picture is worth a thousand words:<\/p>\n<figure id=\"attachment_314\" aria-describedby=\"caption-attachment-314\" style=\"width: 820px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-314 size-full\" src=\"https:\/\/mathvault.ca\/wp-content\/uploads\/parallel-lines.jpg\" alt=\"System of Linear Equations with two parallel lines\" width=\"820\" height=\"401\" title=\"\"><figcaption id=\"caption-attachment-314\" class=\"wp-caption-text\">You see&#8230;it&#8217;s because they are parallel to each other. Hence no intersection point!<\/figcaption><\/figure>\n<h2><span class=\"ez-toc-section\" id=\"Case_2_%E2%80%94_When_h_%E2%89%A0_2\"><\/span><a href=\"#toc\">Case 2 \u2014 When h \u2260 2<\/a><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Once the above <a href=\"\/math-glossary\/#trivial\">trivial<\/a> hurdle is clear, we can then move on to something more serious. For the record, here is the latest system we have so far:<\/p>\n<p>\\begin{cases} x_1+hx_2=4 &amp; (3)\\\\\u00a0(h-2)x_2=4 &#8211; \\frac{8}{3} = \\frac{4}{3} &amp; (4) \\end{cases}<\/p>\n<p>Since $h \\ne 2$ now, $h-2 \\ne 0$. We can therefore proceed to divide the equation $(4)$ by\u00a0$h-2$ on both sides, in which case we get:<\/p>\n<p>\\begin{cases} x_1+hx_2=4 &amp; (5) \\\\ x_2= \\frac{4}{3(h-2)} &amp; (6) \\end{cases}<\/p>\n<p>Lastly, if we turn $(5)$ into $(5)-h(6)$, we get a very neat system in return:<\/p>\n<p>\\begin{cases} x_1=4 &#8211; \\frac{4h}{3(h-2)}=\\frac{8h-24}{3h-6} &amp; (7) \\\\ x_2= \\frac{4}{3(h-2)}= \\frac{4}{3h-6}&amp; (8) \\end{cases}<\/p>\n<p style=\"text-align: justify;\">So for <em>each<\/em> $h \\ne 2$, we get a unique pair of solution of the form $\\left( \\dfrac{8h-24}{3h-6}, \\dfrac{4}{3h-6} \\right)$.<\/p>\n<h3 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Some_Illustrations\"><\/span><a href=\"#toc\">Some Illustrations<\/a><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\">For example, if $h=0$, then the original system becomes:<\/p>\n<p>\\begin{cases} x_1=4 \\\\ 3x_1+6x_2=8 \\end{cases}<\/p>\n<p>In which case, the unique solution would be:<\/p>\n<p>$$\\left( \\frac{8h-24}{3h-6} , \\frac{4}{3h-6} \\right)_{h=0} = \\left( 4,-\\frac{2}{3} \\right)$$<\/p>\n<p>Here is a picture:<\/p>\n<figure id=\"attachment_338\" aria-describedby=\"caption-attachment-338\" style=\"width: 820px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-338 size-full\" src=\"https:\/\/mathvault.ca\/wp-content\/uploads\/intersection-point-1.jpg\" alt=\"System of linear equations where h=0\" width=\"820\" height=\"464\" title=\"\"><figcaption id=\"caption-attachment-338\" class=\"wp-caption-text\">The intersection point occurs at $(4, -\\frac{2}{3})$.<\/figcaption><\/figure>\n<p>And in the case where $h=10$, the original system becomes:<\/p>\n<p>\\begin{cases} x_1+10x_2=4 \\\\ 3x_1+6x_2=8 \\end{cases}<\/p>\n<p style=\"text-align: justify;\">In which case, the unique solution would be:<\/p>\n<p>$$\\left(\\frac{8h-24}{3h-6}, \\frac{4}{3h-6} \\right)_{h=10} = \\left(\\frac{56}{24}, \\frac{4}{24} \\right) = \\left( \\frac{7}{3}, \\frac{1}{6} \\right)$$<\/p>\n<p>Here is a picture again:<\/p>\n<figure id=\"attachment_340\" aria-describedby=\"caption-attachment-340\" style=\"width: 820px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-340 size-full\" src=\"https:\/\/mathvault.ca\/wp-content\/uploads\/intersection-point-2.jpg\" alt=\"System of linear equations when h=10\" width=\"820\" height=\"524\" title=\"\"><figcaption id=\"caption-attachment-340\" class=\"wp-caption-text\">Notice how the 1<sup>st<\/sup> equation (the <span style=\"color: #a30000;\">red line<\/span>) got rotated counter-clockwise about $(4,0)$ as h increases. The intersection point now occurs at $\\displaystyle \\left( \\frac{7}{3}, \\frac{1}{16} \\right)$.<\/figcaption><\/figure>\n<p>And this, is how we can solving infinitely many system of linear equations all at once! For a last sanity check, here&#8217;s a <a href=\"\/desmos-guide\/\">Desmos<\/a> animation we&#8217;ve put together \u2014 just so that we can put everything we&#8217;ve learned thus far into a single slide.<\/p>\n<p><iframe class=\"iframe\" style=\"margin: 15px auto; border: 1px solid #ccc; min-height: 400px;\" src=\"https:\/\/www.desmos.com\/calculator\/7wmnw640to?embed\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n","protected":false},"excerpt":{"rendered":"<p>How to solve a system of linear equations, when a coefficient of a variable is itself a variable.<\/p>\n","protected":false},"author":1,"featured_media":10788,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[367,6],"tags":[],"class_list":["post-300","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-college-math","category-linear-algebra","post-wrapper","thrv_wrapper"],"_links":{"self":[{"href":"https:\/\/mathvault.ca\/wp-json\/wp\/v2\/posts\/300","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathvault.ca\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathvault.ca\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathvault.ca\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathvault.ca\/wp-json\/wp\/v2\/comments?post=300"}],"version-history":[{"count":0,"href":"https:\/\/mathvault.ca\/wp-json\/wp\/v2\/posts\/300\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/mathvault.ca\/wp-json\/wp\/v2\/media\/10788"}],"wp:attachment":[{"href":"https:\/\/mathvault.ca\/wp-json\/wp\/v2\/media?parent=300"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathvault.ca\/wp-json\/wp\/v2\/categories?post=300"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathvault.ca\/wp-json\/wp\/v2\/tags?post=300"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}