Resolve os seguintes sistemas
Sistemas de equações: Matematicamente Falando 9 - Parte 1 Pág. 50 Ex. 2
Enunciado
Resolve os seguintes sistemas de equações:
- \(\left\{ \begin{array}{*{35}{l}} 2(x-1)-4y=1 \\ 3y=2 \\ \end{array} \right.\)
- \(\left\{ \begin{array}{*{35}{l}} 2x+3y=10 \\ 4x-y=-1 \\ \end{array} \right.\)
- \(\left\{ \begin{array}{*{35}{l}} x+y=7 \\ \frac{2x}{5}=\frac{3y}{7} \\ \end{array} \right.\)
- \(\left\{ \begin{array}{*{35}{l}} 5(x+1)+3(y-2)=4 \\ 8(x+1)+5(y-2)=9 \\ \end{array} \right.\)
Resolução
- Resolvendo o sistema, temos:
\[\begin{array}{*{35}{l}} \left\{ \begin{array}{*{35}{l}} 2(x-1)-4y=1 \\ 3y=2 \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} 2x-2-4y=1 \\ y=\frac{2}{3} \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} 2x-2-4\times \frac{2}{3}=1 \\ y=\frac{2}{3} \\ \end{array} \right. & \Leftrightarrow \\ {} & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} 6x-6-8=3 \\ y=\frac{2}{3} \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} x=\frac{17}{6} \\ y=\frac{2}{3} \\ \end{array} \right. & {} \\ \end{array}\]
O conjunto solução é $S=\left\{ \frac{17}{6},\frac{2}{3} \right\}$.
- Resolvendo o sistema, temos:
\[\begin{array}{*{35}{l}} \left\{ \begin{array}{*{35}{l}} 2x+3y=10 \\ 4x-y=-1 \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} y=4x+1 \\ 2x+3(4x+1)=10 \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} y=4x+1 \\ 2x+12x+3=10 \\ \end{array} \right. & \Leftrightarrow \\ {} & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} y=4x+1 \\ 14x=7 \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} x=\frac{1}{2} \\ y=3 \\ \end{array} \right. & {} \\ \end{array}\]
O conjunto solução é $S=\left\{ \frac{1}{2},3 \right\}$.
- Resolvendo o sistema, temos:
\[\begin{array}{*{35}{l}} \left\{ \begin{array}{*{35}{l}} x+y=7 \\ \frac{2x}{5}=\frac{3y}{7} \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} y=-x+7 \\ 14x=15y \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} y=-x+7 \\ 14x=15(-x+7) \\ \end{array} \right. & \Leftrightarrow & {} \\ {} & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} y=-x+7 \\ 14x=-15x+105 \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} x=\frac{105}{29} \\ y=-\frac{105}{29}+\frac{203}{29} \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} x=\frac{105}{29} \\ y=\frac{98}{29} \\ \end{array} \right. \\ \end{array}\]
O conjunto solução é $S=\left\{ \frac{105}{29},\frac{98}{29} \right\}$.
- Resolvendo o sistema, temos:
\[\begin{array}{*{35}{l}} \left\{ \begin{array}{*{35}{l}} 5(x+1)+3(y-2)=4 \\ 8(x+1)+5(y-2)=9 \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} 5x+5+3y-6=4 \\ 8x+8+5y-10=9 \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} 5x+3y=5 \\ 8x+5y=11 \\ \end{array} \right. & \Leftrightarrow \\ {} & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} x=1-\frac{3y}{5} \\ 8(1-\frac{3y}{5})+5y=11 \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} x=1-\frac{3y}{5} \\ 8-\frac{24y}{5}+5y=11 \\ \end{array} \right. & \Leftrightarrow \\ {} & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} x=1-\frac{3y}{5} \\ 40-24y+25y=55 \\ \end{array} \right. & \Leftrightarrow & \left\{ \begin{array}{*{35}{l}} y=15 \\ x=-8 \\ \end{array} \right. & {} \\ \end{array}\]
O conjunto solução é $S=\left\{ -8,15 \right\}$.
