{"id":7101,"date":"2011-10-23T19:06:59","date_gmt":"2011-10-23T18:06:59","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7101"},"modified":"2022-01-11T00:59:38","modified_gmt":"2022-01-11T00:59:38","slug":"resolve-os-seguintes-sistemas","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7101","title":{"rendered":"Resolve os seguintes sistemas"},"content":{"rendered":"<p><ul id='GTTabs_ul_7101' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7101' class='GTTabs_curr'><a  id=\"7101_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7101' ><a  id=\"7101_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_7101'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolve os seguintes sistemas de equa\u00e7\u00f5es:<\/p>\n<ol>\n<li>\u00a0 \\(\\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 2(x-1)-4y=1\u00a0 \\\\ \u00a0\u00a0 3y=2\u00a0 \\\\ \\end{array} \\right.\\)<br \/>\n\u00ad<\/li>\n<li>\u00a0 \\(\\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 2x+3y=10\u00a0 \\\\ \u00a0\u00a0 4x-y=-1\u00a0 \\\\ \\end{array} \\right.\\)<br \/>\n\u00ad<\/li>\n<li>\u00a0 \\(\\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 x+y=7\u00a0 \\\\ \u00a0\u00a0 \\frac{2x}{5}=\\frac{3y}{7}\u00a0 \\\\ \\end{array} \\right.\\)<br \/>\n\u00ad<\/li>\n<li>\u00a0 \\(\\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 5(x+1)+3(y-2)=4\u00a0 \\\\ \u00a0\u00a0 8(x+1)+5(y-2)=9\u00a0 \\\\ \\end{array} \\right.\\)<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7101' onClick='GTTabs_show(1,7101)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7101'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Resolvendo o sistema, temos:<br \/>\n\\[\\begin{array}{*{35}{l}} \u00a0\u00a0 \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 2(x-1)-4y=1\u00a0 \\\\ \u00a0\u00a0 3y=2\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 2x-2-4y=1\u00a0 \\\\ \u00a0\u00a0 y=\\frac{2}{3}\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 2x-2-4\\times \\frac{2}{3}=1\u00a0 \\\\ \u00a0\u00a0 y=\\frac{2}{3}\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0\u00a0 \\\\ \u00a0\u00a0 {} &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 6x-6-8=3\u00a0 \\\\ \u00a0\u00a0 y=\\frac{2}{3}\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 x=\\frac{17}{6}\u00a0 \\\\ \u00a0\u00a0 y=\\frac{2}{3}\u00a0 \\\\ \\end{array} \\right. &amp; {}\u00a0 \\\\ \\end{array}\\]<br \/>\nO conjunto solu\u00e7\u00e3o \u00e9 $S=\\left\\{ \\frac{17}{6},\\frac{2}{3} \\right\\}$.<br \/>\n\u00ad<\/li>\n<li>Resolvendo o sistema, temos:<br \/>\n\\[\\begin{array}{*{35}{l}} \u00a0\u00a0 \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 2x+3y=10\u00a0 \\\\ \u00a0\u00a0 4x-y=-1\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 y=4x+1\u00a0 \\\\ \u00a0\u00a0 2x+3(4x+1)=10\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 y=4x+1\u00a0 \\\\ \u00a0\u00a0 2x+12x+3=10\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0\u00a0 \\\\ \u00a0\u00a0 {} &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 y=4x+1\u00a0 \\\\ \u00a0\u00a0 14x=7\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 x=\\frac{1}{2}\u00a0 \\\\ \u00a0\u00a0 y=3\u00a0 \\\\ \\end{array} \\right. &amp; {}\u00a0 \\\\ \\end{array}\\]<br \/>\nO conjunto solu\u00e7\u00e3o \u00e9 $S=\\left\\{ \\frac{1}{2},3 \\right\\}$.<br \/>\n\u00ad<\/li>\n<li>Resolvendo o sistema, temos:<br \/>\n\\[\\begin{array}{*{35}{l}} \u00a0\u00a0 \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 x+y=7\u00a0 \\\\ \u00a0\u00a0 \\frac{2x}{5}=\\frac{3y}{7}\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 y=-x+7\u00a0 \\\\ \u00a0\u00a0 14x=15y\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 y=-x+7\u00a0 \\\\ \u00a0\u00a0 14x=15(-x+7)\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; {}\u00a0 \\\\ \u00a0\u00a0 {} &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 y=-x+7\u00a0 \\\\ \u00a0\u00a0 14x=-15x+105\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 x=\\frac{105}{29}\u00a0 \\\\ \u00a0\u00a0 y=-\\frac{105}{29}+\\frac{203}{29}\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 x=\\frac{105}{29}\u00a0 \\\\ \u00a0\u00a0 y=\\frac{98}{29}\u00a0 \\\\ \\end{array} \\right.\u00a0 \\\\ \\end{array}\\]<br \/>\nO conjunto solu\u00e7\u00e3o \u00e9 $S=\\left\\{ \\frac{105}{29},\\frac{98}{29} \\right\\}$.<br \/>\n\u00ad<\/li>\n<li>Resolvendo o sistema, temos:<br \/>\n\\[\\begin{array}{*{35}{l}} \u00a0\u00a0 \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 5(x+1)+3(y-2)=4\u00a0 \\\\ \u00a0\u00a0 8(x+1)+5(y-2)=9\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 5x+5+3y-6=4\u00a0 \\\\ \u00a0\u00a0 8x+8+5y-10=9\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 5x+3y=5\u00a0 \\\\ \u00a0\u00a0 8x+5y=11\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0\u00a0 \\\\ \u00a0\u00a0 {} &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 x=1-\\frac{3y}{5}\u00a0 \\\\ \u00a0\u00a0 8(1-\\frac{3y}{5})+5y=11\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 x=1-\\frac{3y}{5}\u00a0 \\\\ \u00a0\u00a0 8-\\frac{24y}{5}+5y=11\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0\u00a0 \\\\ \u00a0\u00a0 {} &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 x=1-\\frac{3y}{5}\u00a0 \\\\ \u00a0\u00a0 40-24y+25y=55\u00a0 \\\\ \\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 y=15\u00a0 \\\\ \u00a0\u00a0 x=-8\u00a0 \\\\ \\end{array} \\right. &amp; {}\u00a0 \\\\ \\end{array}\\]<br \/>\nO conjunto solu\u00e7\u00e3o \u00e9 $S=\\left\\{ -8,15 \\right\\}$.<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_7101' onClick='GTTabs_show(0,7101)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Resolve os seguintes sistemas de equa\u00e7\u00f5es: \u00a0 \\(\\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 2(x-1)-4y=1\u00a0 \\\\ \u00a0\u00a0 3y=2\u00a0 \\\\ \\end{array} \\right.\\) \u00ad \u00a0 \\(\\left\\{ \\begin{array}{*{35}{l}} \u00a0\u00a0 2x+3y=10\u00a0 \\\\ \u00a0\u00a0 4x-y=-1\u00a0 \\\\ \\end{array} \\right.\\) \u00ad \u00a0&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":14061,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,240],"tags":[426,243],"series":[],"class_list":["post-7101","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9--ano","category-aplicando","category-sistemas-de-equacoes","tag-9-o-ano","tag-sistemas-de-equacoes-2"],"views":3058,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat06.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7101","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=7101"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7101\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14061"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7101"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7101"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7101"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7101"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}