\nEnunciado<\/b><\/span><\/p>\n
Escreve na forma can\u00f3nica os seguintes sistemas e, em seguida, resolve-os, utilizando o m\u00e9todo de substitui\u00e7\u00e3o.<\/p>\n
\n

\\(\\left\\{ {\\begin{array}{*{20}{l}}{x – 3y = 4 + x}\\\\{2\\left( {x – 3} \\right) = 3\\left( {y – 1} \\right)}\\end{array}} \\right.\\)<\/li>\n

\\(\\left\\{ {\\begin{array}{*{20}{l}}{3\\left( {x + y} \\right) – 2 = x}\\\\{y = 5 – x}\\end{array}} \\right.\\)<\/li>\n

\\(\\left\\{ {\\begin{array}{*{20}{l}}{\\frac{p}{2} + \\frac{q}{3} – 4 = 0}\\\\{\\frac{p}{4} – \\frac{q}{2} + 2 = 0}\\end{array}} \\right.\\)<\/li>\n

\\(\\left\\{ {\\begin{array}{*{20}{l}}{3\\left( {a – 2} \\right) = \\frac{1}{2}a + b}\\\\{a – 2b = 2a + b}\\end{array}} \\right.\\)<\/li>\n

\\(\\left\\{ {\\begin{array}{*{20}{l}}{\\frac{{3\\left( {a – b} \\right)}}{2} – \\frac{{4\\left( {2b – 1} \\right)}}{3} = 6}\\\\{a – \\frac{{1 – b}}{6} = 0}\\end{array}} \\right.\\)<\/li>\n

\\(\\left\\{ {\\begin{array}{*{20}{l}}{\\frac{{m – 4}}{3} – \\frac{{10n + 4}}{{10}} = m – n}\\\\{\\frac{{2m – 5}}{5} – \\frac{{2n – 4}}{4} = m – 12}\\end{array}} \\right.\\)<\/li>\n<\/ol>\n
Resolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n
\nResolu\u00e7\u00e3o<\/b><\/span><\/p>\n
\n

\\[\\begin{array}{*{20}{c}}{\\left\\{ {\\begin{array}{*{20}{l}}{x – 3y = 4 + x}\\\\{2\\left( {x – 3} \\right) = 3\\left( {y – 1} \\right)}\\end{array}} \\right.}& \\Leftrightarrow &{\\underbrace {\\left\\{ {\\begin{array}{*{20}{l}}{ – 3y = 4}\\\\{2x – 3y = 3}\\end{array}} \\right.}_{{\\rm{F}}{\\rm{.}}\\;{\\rm{C}}{\\rm{.}}}}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{ – 3y = 4}\\\\{2x + 4 = 3}\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{x = – \\frac{1}{2}}\\\\{y = – \\frac{4}{3}}\\end{array}} \\right.}\\end{array}\\]<\/li>\n

\\[\\begin{array}{*{20}{c}}{\\left\\{ {\\begin{array}{*{20}{l}}{3\\left( {x + y} \\right) – 2 = x}\\\\{y = 5 – x}\\end{array}} \\right.}& \\Leftrightarrow &{\\underbrace {\\left\\{ {\\begin{array}{*{20}{l}}{2x + 3y = 2}\\\\{x + y = 5}\\end{array}} \\right.}_{{\\rm{F}}{\\rm{.}}\\;{\\rm{C}}{\\rm{.}}}}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{x = 5 – y}\\\\{10 – 2y + 3y = 2}\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{y = – 8}\\\\{x = 13}\\end{array}} \\right.}\\end{array}\\]<\/li>\n

\\[\\begin{array}{*{20}{c}}{\\left\\{ {\\begin{array}{*{20}{l}}{\\frac{p}{2} + \\frac{q}{3} – 4 = 0}\\\\{\\frac{p}{4} – \\frac{q}{2} + 2 = 0}\\end{array}} \\right.}& \\Leftrightarrow &{\\underbrace {\\left\\{ {\\begin{array}{*{20}{l}}{3p + 2q = 24}\\\\{p – 2q = – 8}\\end{array}} \\right.}_{{\\rm{F}}{\\rm{.}}\\;{\\rm{C}}{\\rm{.}}}}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{2q = 24 – 3p}\\\\{p – 24 + 3p = – 8}\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{p = 4}\\\\{q = 6}\\end{array}} \\right.}\\end{array}\\]<\/li>\n

\\[\\begin{array}{*{20}{c}}{\\left\\{ {\\begin{array}{*{20}{l}}{3\\left( {a – 2} \\right) = \\frac{1}{2}a + b}\\\\{a – 2b = 2a + b}\\end{array}} \\right.}& \\Leftrightarrow &{\\underbrace {\\left\\{ {\\begin{array}{*{20}{l}}{5a – 2b = 12}\\\\{a + 3b = 0}\\end{array}} \\right.}_{{\\rm{F}}{\\rm{.}}\\;{\\rm{C}}{\\rm{.}}}}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{a = – 3b}\\\\{ – 15b – 2b = 12}\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{b = – \\frac{{12}}{{17}}}\\\\{a = \\frac{{36}}{{17}}}\\end{array}} \\right.}\\end{array}\\]<\/li>\n

\\[\\begin{array}{*{20}{c}}{\\left\\{ {\\begin{array}{*{20}{l}}{\\frac{{3\\left( {a – b} \\right)}}{2} – \\frac{{4\\left( {2b – 1} \\right)}}{3} = 6}\\\\{a – \\frac{{1 – b}}{6} = 0}\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{9a – 9b – 16b + 8 = 36}\\\\{6a – 1 + b = 0}\\end{array}} \\right.}& \\Leftrightarrow &{\\underbrace {\\left\\{ {\\begin{array}{*{20}{l}}{9a – 25b = 28}\\\\{6a + b = 1}\\end{array}} \\right.}_{{\\rm{F}}{\\rm{.}}\\;{\\rm{C}}{\\rm{.}}}}& \\Leftrightarrow \\\\{}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{b = 1 – 6a}\\\\{9a – 25 + 150a = 28}\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{a = \\frac{1}{3}}\\\\{b = – 1}\\end{array}} \\right.}&{}\\end{array}\\]<\/li>\n

\\[\\begin{array}{*{20}{c}}{\\left\\{ {\\begin{array}{*{20}{l}}{\\frac{{m – 4}}{3} – \\frac{{10n + 4}}{{10}} = m – n}\\\\{\\frac{{2m – 5}}{5} – \\frac{{2n – 4}}{4} = m – 12}\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{10m – 40 – 30n – 12 = 30m – 30n}\\\\{8m – 20 – 10n + 20 = 20m – 240}\\end{array}} \\right.}& \\Leftrightarrow &{\\underbrace {\\left\\{ {\\begin{array}{*{20}{l}}{5m = – 13}\\\\{6m + 5n = 120}\\end{array}} \\right.}_{{\\rm{F}}{\\rm{.}}\\;{\\rm{C}}{\\rm{.}}}}& \\Leftrightarrow \\\\{}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{m = – \\frac{{13}}{5}}\\\\{5n = 120 + \\frac{{78}}{5}}\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{m = – \\frac{{13}}{5}}\\\\{n = \\frac{{678}}{{25}}}\\end{array}} \\right.}&{}\\end{array}\\]<\/li>\n<\/ol>\n
<< Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"
Enunciado Resolu\u00e7\u00e3o Enunciado Escreve na forma can\u00f3nica os seguintes sistemas e, em seguida, resolve-os, utilizando o m\u00e9todo de substitui\u00e7\u00e3o. \\(\\left\\{ {\\begin{array}{*{20}{l}}{x – 3y = 4 + x}\\\\{2\\left( {x – 3} \\right) = 3\\left( {y...<\/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":[100,97,714],"tags":[424,345,239],"series":[],"class_list":["post-25933","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-equacoes-literais-e-sistemas","tag-8-o-ano","tag-funcao-afim","tag-sistema-de-equacoes"],"views":95,"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\/25933","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=25933"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/25933\/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=25933"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=25933"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=25933"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=25933"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}