Escreve na forma canónica
Equações literais e sistemas: Matematicamente Falando 8 - Pág. 202 Ex. 5
Enunciado
Escreve na forma canónica os seguintes sistemas e, em seguida, resolve-os, utilizando o método de substituição.
\(\left\{ {\begin{array}{*{20}{l}}{x – 3y = 4 + x}\\{2\left( {x – 3} \right) = 3\left( {y – 1} \right)}\end{array}} \right.\)
\(\left\{ {\begin{array}{*{20}{l}}{3\left( {x + y} \right) – 2 = x}\\{y = 5 – x}\end{array}} \right.\)
\(\left\{ {\begin{array}{*{20}{l}}{\frac{p}{2} + \frac{q}{3} – 4 = 0}\\{\frac{p}{4} – \frac{q}{2} + 2 = 0}\end{array}} \right.\)
\(\left\{ {\begin{array}{*{20}{l}}{3\left( {a – 2} \right) = \frac{1}{2}a + b}\\{a – 2b = 2a + b}\end{array}} \right.\)
\(\left\{ {\begin{array}{*{20}{l}}{\frac{{3\left( {a – b} \right)}}{2} – \frac{{4\left( {2b – 1} \right)}}{3} = 6}\\{a – \frac{{1 – b}}{6} = 0}\end{array}} \right.\)
\(\left\{ {\begin{array}{*{20}{l}}{\frac{{m – 4}}{3} – \frac{{10n + 4}}{{10}} = m – n}\\{\frac{{2m – 5}}{5} – \frac{{2n – 4}}{4} = m – 12}\end{array}} \right.\)
Resolução
\[\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{l}}{x – 3y = 4 + x}\\{2\left( {x – 3} \right) = 3\left( {y – 1} \right)}\end{array}} \right.}& \Leftrightarrow &{\underbrace {\left\{ {\begin{array}{*{20}{l}}{ – 3y = 4}\\{2x – 3y = 3}\end{array}} \right.}_{{\rm{F}}{\rm{.}}\;{\rm{C}}{\rm{.}}}}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{ – 3y = 4}\\{2x + 4 = 3}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{x = – \frac{1}{2}}\\{y = – \frac{4}{3}}\end{array}} \right.}\end{array}\]
\[\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{l}}{3\left( {x + y} \right) – 2 = x}\\{y = 5 – x}\end{array}} \right.}& \Leftrightarrow &{\underbrace {\left\{ {\begin{array}{*{20}{l}}{2x + 3y = 2}\\{x + y = 5}\end{array}} \right.}_{{\rm{F}}{\rm{.}}\;{\rm{C}}{\rm{.}}}}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{x = 5 – y}\\{10 – 2y + 3y = 2}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{y = – 8}\\{x = 13}\end{array}} \right.}\end{array}\]
\[\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{l}}{\frac{p}{2} + \frac{q}{3} – 4 = 0}\\{\frac{p}{4} – \frac{q}{2} + 2 = 0}\end{array}} \right.}& \Leftrightarrow &{\underbrace {\left\{ {\begin{array}{*{20}{l}}{3p + 2q = 24}\\{p – 2q = – 8}\end{array}} \right.}_{{\rm{F}}{\rm{.}}\;{\rm{C}}{\rm{.}}}}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{2q = 24 – 3p}\\{p – 24 + 3p = – 8}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{p = 4}\\{q = 6}\end{array}} \right.}\end{array}\]
\[\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{l}}{3\left( {a – 2} \right) = \frac{1}{2}a + b}\\{a – 2b = 2a + b}\end{array}} \right.}& \Leftrightarrow &{\underbrace {\left\{ {\begin{array}{*{20}{l}}{5a – 2b = 12}\\{a + 3b = 0}\end{array}} \right.}_{{\rm{F}}{\rm{.}}\;{\rm{C}}{\rm{.}}}}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{a = – 3b}\\{ – 15b – 2b = 12}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{b = – \frac{{12}}{{17}}}\\{a = \frac{{36}}{{17}}}\end{array}} \right.}\end{array}\]
\[\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{l}}{\frac{{3\left( {a – b} \right)}}{2} – \frac{{4\left( {2b – 1} \right)}}{3} = 6}\\{a – \frac{{1 – b}}{6} = 0}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{9a – 9b – 16b + 8 = 36}\\{6a – 1 + b = 0}\end{array}} \right.}& \Leftrightarrow &{\underbrace {\left\{ {\begin{array}{*{20}{l}}{9a – 25b = 28}\\{6a + b = 1}\end{array}} \right.}_{{\rm{F}}{\rm{.}}\;{\rm{C}}{\rm{.}}}}& \Leftrightarrow \\{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{b = 1 – 6a}\\{9a – 25 + 150a = 28}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{a = \frac{1}{3}}\\{b = – 1}\end{array}} \right.}&{}\end{array}\]
\[\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{l}}{\frac{{m – 4}}{3} – \frac{{10n + 4}}{{10}} = m – n}\\{\frac{{2m – 5}}{5} – \frac{{2n – 4}}{4} = m – 12}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{10m – 40 – 30n – 12 = 30m – 30n}\\{8m – 20 – 10n + 20 = 20m – 240}\end{array}} \right.}& \Leftrightarrow &{\underbrace {\left\{ {\begin{array}{*{20}{l}}{5m = – 13}\\{6m + 5n = 120}\end{array}} \right.}_{{\rm{F}}{\rm{.}}\;{\rm{C}}{\rm{.}}}}& \Leftrightarrow \\{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{m = – \frac{{13}}{5}}\\{5n = 120 + \frac{{78}}{5}}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{m = – \frac{{13}}{5}}\\{n = \frac{{678}}{{25}}}\end{array}} \right.}&{}\end{array}\]
