Bài 16 trang 16 SGK Toán 9 tập 2

Giải các hệ phương trình sau bằng phương pháp thế.

a) \(\left\{\begin{matrix} 3x - y = 5 & & \\ 5x + 2y = 23 & & \end{matrix}\right.\);        

b) \(\left\{\begin{matrix} 3x +5y = 1 & & \\ 2x -y =-8 & & \end{matrix}\right.\);     

c) \(\left\{\begin{matrix} \dfrac{x}{y} = \dfrac{2}{3}& & \\ x + y - 10 = 0 & & \end{matrix}\right.\)

a) Ta có:

\(\left\{ \matrix{ 3x - y = 5 \hfill \cr 5x + 2y = 23 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ y = 3x - 5 \hfill \cr 5x + 2\left( {3x - 5} \right) = 23 \hfill \cr} \right.\)

\(\Leftrightarrow \left\{ \matrix{ y = 3x - 5 \hfill \cr 5x + 6x - 10 = 23 \hfill \cr} \right.\)

\( \Leftrightarrow \left\{ \matrix{ y = 3x - 5 \hfill \cr 11x = 23 + 10 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ y = 3x - 5 \hfill \cr 11x = 33 \hfill \cr} \right. \)

\(\Leftrightarrow \left\{ \matrix{ y = 3x - 5 \hfill \cr x = 3 \hfill \cr} \right.\)

\( \Leftrightarrow \left\{ \matrix{ y = 3.3 - 5 \hfill \cr x = 3 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ y = 4 \hfill \cr x = 3 \hfill \cr} \right.\)

Vậy hệ có nghiệm duy nhất là \((x; y) = (3; 4)\).

b) Ta có:

\(\left\{ \matrix{ 3x + 5y = 1 \hfill \cr 2x - y = - 8 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 3x + 5y = 1 \hfill \cr y = 2x + 8 \hfill \cr} \right. \)

\(\Leftrightarrow \left\{ \matrix{ 3x + 5\left( {2x + 8} \right) = 1 \hfill \cr y = 2x + 8 \hfill \cr} \right.\)

\( \Leftrightarrow \left\{ \matrix{ 3x + 10x + 40 = 1 \hfill \cr y = 2x + 8 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 13x = 1 - 40 \hfill \cr y = 2x + 8 \hfill \cr} \right.\)

\( \Leftrightarrow \left\{ \matrix{ 13x = - 39 \hfill \cr y = 2x + 8 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = - 3 \hfill \cr y = 2x + 8 \hfill \cr} \right.\)

\(\Leftrightarrow \left\{ \matrix{ x = - 3 \hfill \cr y = 2.\left( { - 3} \right) + 8 \hfill \cr} \right.\)

\( \Leftrightarrow \left\{ \matrix{ x = - 3 \hfill \cr y = 2 \hfill \cr} \right.\)

Vậy hệ có nghiệm \((x; y) = (-3; 2)\).

c) Ta có:

\(\left\{ \matrix{\dfrac{x}{y} = \dfrac{2}{3} \hfill \cr x + y - 10 = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = \dfrac{2y}{3} \hfill \cr \dfrac{2y}{3} + y = 10 \hfill \cr} \right.\)

\(\Leftrightarrow \left\{ \matrix{ x = \dfrac{2y}{3} \hfill \cr {\left( \dfrac{2}{3} + 1 \right)}y = 10 \hfill \cr} \right.\)

\( \Leftrightarrow \left\{ \matrix{ x = \dfrac{2y}{3} \hfill \cr \dfrac{5}{ 3}y = 10 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = \dfrac{2y}{3} \hfill \cr y = 6 \hfill \cr} \right. \)

\(\Leftrightarrow \left\{ \matrix{ x = \dfrac{2.6}{3} \hfill \cr y = 6 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = 4 \hfill \cr y = 6 \hfill \cr} \right.\)

Vậy nghiệm của hệ là \((x; y) = (4; 6)\).