{"id":6780,"date":"2011-04-14T13:31:13","date_gmt":"2011-04-14T12:31:13","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6780"},"modified":"2022-01-05T16:37:09","modified_gmt":"2022-01-05T16:37:09","slug":"liga-cada-equacao-a-sua-solucao","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6780","title":{"rendered":"Liga cada equa\u00e7\u00e3o \u00e0 sua solu\u00e7\u00e3o"},"content":{"rendered":"<p><ul id='GTTabs_ul_6780' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6780' class='GTTabs_curr'><a  id=\"6780_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6780' ><a  id=\"6780_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_6780'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Liga cada equa\u00e7\u00e3o \u00e0 sua solu\u00e7\u00e3o:<\/p>\n<table class=\"aligncenter\" style=\"width: 70%;\" border=\"2\" cellspacing=\"4\" cellpadding=\"4\" align=\"center\">\n<tbody>\n<tr>\n<td>1<\/td>\n<td>\\[(x-7)-(3x+2)=9\\]<\/td>\n<td>A<\/td>\n<td>\u00a0\\[2,7\\]<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>\u00a0\\[\\frac{x+3}{2}=\\frac{x-5}{3}\\]<\/td>\n<td>B<\/td>\n<td>\u00a0\\[-19\\]<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>\u00a0\\[\\frac{2}{3}(a+1)=\\frac{a}{6}\\]<\/td>\n<td>C<\/td>\n<td>\u00a0\\[-9\\]<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>\u00a0\\[6x-\\frac{3}{2}=5x+\\frac{6}{5}\\]<\/td>\n<td>D<\/td>\n<td>\u00a0\\[-\\frac{4}{5}\\]<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>\u00a0\\[b-\\frac{1}{3}(b-1)=\\frac{b}{4}\\]<\/td>\n<td>E<\/td>\n<td>\u00a0\\[-\\frac{4}{3}\\]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6780' onClick='GTTabs_show(1,6780)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6780'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p>\\[\\begin{array}{*{35}{l}}<br \/>\n(x-7)-(3x+2)=9 &amp; \\Leftrightarrow\u00a0 &amp; x-7-3x-2=9\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; -2x=18\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; x=-9\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\n\u00ad<\/p>\n<p>\\[\\begin{array}{*{35}{l}}<br \/>\n\\frac{x+3}{\\underset{(3)}{\\mathop{2}}\\,}=\\frac{x-5}{\\underset{(2)}{\\mathop{3}}\\,} &amp; \\Leftrightarrow\u00a0 &amp; 3x+9=2x-10\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; x=-19\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\n\u00ad<\/p>\n<p>\\[\\begin{array}{*{35}{l}}<br \/>\n\\frac{2}{3}(a+1)=\\frac{a}{6} &amp; \\Leftrightarrow\u00a0 &amp; \\frac{2a}{\\underset{(2)}{\\mathop{3}}\\,}+\\frac{2}{\\underset{(2)}{\\mathop{3}}\\,}=\\frac{a}{\\underset{(1)}{\\mathop{6}}\\,}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 4a+4=a\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 3a=-4\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; a=-\\frac{4}{3}\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\n\u00ad<\/p>\n<p>\\[\\begin{array}{*{35}{l}}<br \/>\n\\underset{(10)}{\\mathop{6x}}\\,-\\frac{3}{\\underset{(5)}{\\mathop{2}}\\,}=\\underset{(10)}{\\mathop{5x}}\\,+\\frac{6}{\\underset{(2)}{\\mathop{5}}\\,} &amp; \\Leftrightarrow\u00a0 &amp; 60x-15=50x+12\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 10x=27\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; x=2,7\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\n\u00ad<\/p>\n<p>\\[\\begin{array}{*{35}{l}}<br \/>\nb-\\frac{1}{3}(b-1)=\\frac{b}{4} &amp; \\Leftrightarrow\u00a0 &amp; \\underset{(12)}{\\mathop{b}}\\,-\\frac{b}{\\underset{(4)}{\\mathop{3}}\\,}+\\frac{1}{\\underset{(4)}{\\mathop{3}}\\,}=\\frac{b}{\\underset{(3)}{\\mathop{4}}\\,}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 12b-4b+4=3b\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; 5b=-4\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; b=-\\frac{4}{5}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<table style=\"width: 20%;\" border=\"2\" cellspacing=\"4\" cellpadding=\"4\" align=\"center\">\n<caption>CHAVE<\/caption>\n<tbody>\n<tr>\n<td style=\"text-align: center;\">1<\/td>\n<td style=\"text-align: center;\">2<\/td>\n<td style=\"text-align: center;\">3<\/td>\n<td style=\"text-align: center;\">4<\/td>\n<td style=\"text-align: center;\">5<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\">C<\/td>\n<td style=\"text-align: center;\">B<\/td>\n<td style=\"text-align: center;\">E<\/td>\n<td style=\"text-align: center;\">A<\/td>\n<td style=\"text-align: center;\">D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6780' onClick='GTTabs_show(0,6780)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Liga cada equa\u00e7\u00e3o \u00e0 sua solu\u00e7\u00e3o: 1 \\[(x-7)-(3x+2)=9\\] A \u00a0\\[2,7\\] 2 \u00a0\\[\\frac{x+3}{2}=\\frac{x-5}{3}\\] B \u00a0\\[-19\\] 3 \u00a0\\[\\frac{2}{3}(a+1)=\\frac{a}{6}\\] C \u00a0\\[-9\\] 4 \u00a0\\[6x-\\frac{3}{2}=5x+\\frac{6}{5}\\] D \u00a0\\[-\\frac{4}{5}\\] 5 \u00a0\\[b-\\frac{1}{3}(b-1)=\\frac{b}{4}\\] E \u00a0\\[-\\frac{4}{3}\\] Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":19189,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,159],"tags":[160],"series":[],"class_list":["post-6780","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-equacoes-do-1--grau","tag-equacao"],"views":3324,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat75.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6780","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=6780"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6780\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19189"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6780"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6780"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6780"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6780"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}