Giải bài 2 trang 68 – SGK môn Đại số lớp 10
Giải các hệ phương trình sau
\( a)\,\left\{ \begin{align} & 2x-3y=1 \\ & x+2y=3 \\ \end{align} \right.\)
\(b)\,\left\{ \begin{align} & 3x+4y=5 \\ & 4x-2y=2 \\ \end{align} \right.\)
\(c)\,\left\{ \begin{align} & \frac{2}{3}x+\frac{1}{2}y=\frac{2}{3} \\ & \frac{1}{3}x-\frac{3}{4}y=\frac{1}{2} \\ \end{align} \right.\)
\(d)\,\left\{ \begin{align} & 0,3x-0,2y=0,5 \\ & 0,5x+0,4y=1,2 \\ \end{align} \right.\)
a)
\(\begin{aligned} & \left\{ \begin{aligned} & 2x-3y=1 \\ & x+2y=3 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 2x-3y=1 \\ & 2x+4y=6 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 7y=5 \\ & x+2y=3 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & y=\dfrac{5}{7} \\ & x=\dfrac{11}{7} \\ \end{aligned} \right. \\ \end{aligned}\)
Vậy \(S=\left\{ \left( \dfrac{11}{7};\dfrac{5}{7} \right) \right\}\)
b)
\(\begin{aligned} & \left\{ \begin{aligned} & 3x+4y=5 \\ & 4x-2y=2 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 3x+4y=5 \\ & 8x-4y=4 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 11x=9 \\ & 3x+4y=5 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x=\dfrac{9}{11} \\ & y=\dfrac{7}{11} \\ \end{aligned} \right. \\ \end{aligned}\)
Vậy \(S=\left\{ \left( \dfrac{9}{11};\dfrac{7}{11} \right) \right\}\)
c)
\(\begin{aligned} & \left\{ \begin{aligned} & \dfrac{2}{3}x+\dfrac{1}{2}y=\dfrac{2}{3} \\ & \dfrac{1}{3}x-\dfrac{3}{4}y=\dfrac{1}{2} \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 4x+3y=4 \\ & 4x-9y=6 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 12y=-2 \\ & 4x-9y=6 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & y=-\dfrac{1}{6} \\ & x=\dfrac{9}{8} \\ \end{aligned} \right. \\ \end{aligned}\)
Vậy \(S=\left\{ \left( \dfrac{9}{8};-\dfrac{1}{6} \right) \right\}\)
d
\(\begin{aligned} & \left\{ \begin{aligned} & 0,3x-0,2y=0,5 \\ & 0,5x+0,4y=1,2 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 3x-2y=5 \\ & 5x+4y=12 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 6x-4y=10 \\ & 5x+4y=12 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & 11x=22 \\ & 5x+4y=12 \\ \end{aligned} \right. \\ & \Leftrightarrow \left\{ \begin{aligned} & x=2 \\ & y=\dfrac{1}{2} \\ \end{aligned} \right. \\ \end{aligned}\)
Vậy \(S=\left\{ \left( 2;\dfrac{1}{2} \right) \right\}\)
Ghi nhớ:
Để giải hể phương trình bậc nhất hai ẩn ta thường sử dụng hai phương pháp:
- Phương pháp cộng đại số.
- Phương pháp thế.