Resolve os seguintes sistemas de equações

Equações literais e sistemas: Matematicamente Falando 8 - Pág. 204 Ex. 8

by AMMA · Published 24 de Maio de 2023 · Updated 24 de Maio de 2023

Enunciado

Resolve os seguintes sistemas de equações.

\(\left\{ {\begin{array}{*{20}{l}}{2\left( {x – 1} \right) – 4y = 1}\\{3y = 2}\end{array}} \right.\)
\(\left\{ {\begin{array}{*{20}{l}}{2x + 3y = 10}\\{4x – y = – 1}\end{array}} \right.\)
\(\left\{ {\begin{array}{*{20}{l}}{x + y = 7}\\{\frac{{2x}}{5} = \frac{{3y}}{7}}\end{array}} \right.\)
\(\left\{ {\begin{array}{*{20}{l}}{5\left( {x + 1} \right) + 3\left( {y – 2} \right) = 4}\\{8\left( {x + 1} \right) + 5\left( {y – 2} \right) = 9}\end{array}} \right.\)

Resolução

\[\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{2\left( {x – 1} \right) – 4y = 1}\\{3y = 2}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{2x – 2 – 4y = 1}\\{y = \frac{2}{3}}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{2x – 2 – 4 \times \frac{2}{3} = 1}\\{y = \frac{2}{3}}\end{array}} \right.}& \Leftrightarrow \\{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{6x – 6 – 8 = 3}\\{y = \frac{2}{3}}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{x = \frac{{17}}{6}}\\{y = \frac{2}{3}}\end{array}} \right.}&{}\\{}&{}&{}&{}&{}&{}\\{}&{}&{}&{}&{S = \left\{ {\left( {\frac{{17}}{6},\frac{2}{3}} \right)} \right\}}&{}\end{array}\]
\[\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{2x + 3y = 10}\\{4x – y = – 1}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{y = 4x + 1}\\{2x + 3\left( {4x + 1} \right) = 10}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{2x + 12x + 3 = 10}\\{y = 4x + 1}\end{array}} \right.}& \Leftrightarrow \\{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{14x = 7}\\{y = 4x + 1}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{x = \frac{1}{2}}\\{y = 3}\end{array}} \right.}&{}\\{}&{}&{}&{}&{}&{}\\{}&{}&{}&{}&{S = \left\{ {\left( {\frac{1}{2},3} \right)} \right\}}&{}\end{array}\]
\[\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{x + y = 7}\\{\frac{{2x}}{5} = \frac{{3y}}{7}}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{y = 7 – x}\\{14x = 15y}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{14x = 105 – 15x}\\{y = 7 – x}\end{array}} \right.}& \Leftrightarrow \\{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{x = \frac{{105}}{{29}}}\\{y = \frac{{203}}{{29}} – \frac{{105}}{{29}}}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{x = \frac{{105}}{{29}}}\\{y = \frac{{98}}{{29}}}\end{array}} \right.}&{}\\{}&{}&{}&{}&{}&{}\\{}&{}&{}&{}&{S = \left\{ {\left( {\frac{{105}}{{29}},\frac{{98}}{{29}}} \right)} \right\}}&{}\end{array}\]
\[\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{5\left( {x + 1} \right) + 3\left( {y – 2} \right) = 4}\\{8\left( {x + 1} \right) + 5\left( {y – 2} \right) = 9}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{5x + 5 + 3y – 6 = 4}\\{8x + 8 + 5y – 10 = 9}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{5x + 3y = 5}\\{8x + 5y = 11}\end{array}} \right.}& \Leftrightarrow \\{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{y = \frac{{5 – 5x}}{3}}\\{8x + 5 \times \frac{{5 – 5x}}{3} = 11}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{8x + \frac{{25}}{3} – \frac{{25x}}{3} = 11}\\{y = \frac{{5 – 5x}}{3}}\end{array}} \right.}& \Leftrightarrow \\{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{ – \frac{x}{3} = \frac{8}{3}}\\{y = \frac{{5 – 5x}}{3}}\end{array}} \right.}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{l}}{x = – 8}\\{y = 15}\end{array}} \right.}&{}\\{}&{}&{}&{}&{}&{}\\{}&{}&{}&{}&{S = \left\{ {\left( { – 8,15} \right)} \right\}}&{}\end{array}\]