{"id":1490,"date":"2021-04-11T15:42:51","date_gmt":"2021-04-11T15:42:51","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=1490"},"modified":"2021-04-11T15:42:52","modified_gmt":"2021-04-11T15:42:52","slug":"cyclic-system-of-equations","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=1490","title":{"rendered":"Cyclic System of Equations"},"content":{"rendered":"<p>Find $abc$.<br \/>\n$$<br \/>\n\\begin{array}{ll}<br \/>\na+\\dfrac{1}{bc} = \\dfrac{7}{6}\\\\<br \/>\nb+\\dfrac{1}{ca} = \\dfrac{7}{3}\\\\<br \/>\nc+\\dfrac{1}{ab} = \\dfrac{7}{2}<br \/>\n\\end{array}<br \/>\n$$<br \/>\n<button onclick=\"document.getElementById(&quot;solution&quot;).style.display=&quot;block&quot;\">Solution<\/button><\/p>\n<div id=\"solution\" style=\"display:none;\">\nMultiply the first equation by $bc$, the second by $ca$, and the third by $ab$:<br \/>\n$$<br \/>\n\\begin{array}{ll}<br \/>\nabc+1 = \\dfrac{7}{6}bc\\\\<br \/>\nabc+1 = \\dfrac{7}{3}ac\\\\<br \/>\nabc+1 = \\dfrac{7}{2}ab<br \/>\n\\end{array}<br \/>\n$$<br \/>\nSince each expression is equal to $abc+1$, we can set up another equation:<br \/>\n$$bc=2ac=3ab$$<br \/>\nThis means that the ratio $a:b:c = 1:2:3$. We can now substitute $b$ and $c$ and write it in terms of a: $b=2a, c=3a$. Plugging it back into the system, we get<br \/>\n$$<br \/>\n\\begin{array}{ll}<br \/>\na+\\dfrac{1}{6a^2} = \\dfrac{7}{6}\\\\<br \/>\n2a+\\dfrac{1}{3a^2} = \\dfrac{7}{3}\\\\<br \/>\n3a+\\dfrac{1}{2a^2} = \\dfrac{7}{2}<br \/>\n\\end{array}<br \/>\n$$<br \/>\nThese are the same equation, so the final equation for $a$ is $$6a+\\dfrac{1}{a^2} = 7$$, or $$6a^3-7a^2+1=0$$<br \/>\nWe can factor the cubic into<br \/>\n$$<br \/>\n(a-1)(2a-1)(3a+1)=0<br \/>\n$$<br \/>\nTherefore, $$a=1, \\dfrac{1}{2}, -\\dfrac{1}{3}$$ As $abc=6a^3$, $$abc=\\boxed{6,\\dfrac{3}{4},-\\dfrac{2}{9}}$$<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Find $abc$. $$ \\begin{array}{ll} a+\\dfrac{1}{bc} = \\dfrac{7}{6}\\\\ b+\\dfrac{1}{ca} = \\dfrac{7}{3}\\\\ c+\\dfrac{1}{ab} = \\dfrac{7}{2} \\end{array} $$ Solution Multiply the first equation by $bc$, the second by $ca$, and the third by $ab$: $$ \\begin{array}{ll} abc+1 = \\dfrac{7}{6}bc\\\\ abc+1 = \\dfrac{7}{3}ac\\\\ abc+1 &hellip; <a href=\"https:\/\/mathfun4kids.com\/mlog\/?p=1490\">Continue reading <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13,5,6],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1490"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1490"}],"version-history":[{"count":19,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1490\/revisions"}],"predecessor-version":[{"id":1509,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1490\/revisions\/1509"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1490"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1490"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1490"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}