Giải bài tập 1 trang 21 SGK Toán 9 tập 1 – Chân trời sáng tạo

Đề bài

Giải các hệ phương trình

a) \(\left\{ {\begin{array}{*{20}{c}}{3x + y = 3}\\{2x – y = 7}\end{array}} \right.\)

b) \(\left\{ {\begin{array}{*{20}{c}}{x – y = 3}\\{3x – 4y = 2}\end{array}} \right.\)

c) \(\left\{ {\begin{array}{*{20}{c}}{4x + 5y = – 2}\\{2x – y = – 8}\end{array}} \right.\)

d) \(\left\{ {\begin{array}{*{20}{c}}{3x + y = 3}\\{ – 3y = 5}\end{array}} \right.\)

Lời giải chi tiết

a) \(\left\{ {\begin{array}{*{20}{c}}{3x + y = 3}\\{2x – y = 7}\end{array}} \right.\)

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

Vậy hệ phương trình có nghiệm duy nhất là (2; – 3).

b) \(\left\{ {\begin{array}{*{20}{c}}{x – y = 3}\\{3x – 4y = 2}\end{array}} \right.\)

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

Vậy hệ phương trình có nghiệm duy nhất là (10; 7).

c) \(\left\{ {\begin{array}{*{20}{c}}{4x + 5y = – 2}\\{2x – y = – 8}\end{array}} \right.\)

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{4x + 5.(2x + 8) = – 2}\\{y = 2x + 8}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{14x = – 42}\\{y = 2x + 8}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{x = – 3}\\{y = 2}\end{array}} \right.\end{array}\)

Vậy hệ phương trình có nghiệm duy nhất là (-3; 2).

d) \(\left\{ {\begin{array}{*{20}{c}}{3x + y = 3}\\{ – 3y = 5}\end{array}} \right.\)

\(\begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{3x + \frac{{ – 5}}{3} = 3}\\{y = \frac{{ – 5}}{3}}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{3x = \frac{{14}}{3}}\\{y = \frac{{ – 5}}{3}}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{x = \frac{{14}}{9}}\\{y = \frac{{ – 5}}{3}}\end{array}} \right.\end{array}\)

Vậy hệ phương trình có nghiệm duy nhất là \(\left( {\frac{{14}}{9};\frac{{ – 5}}{3}} \right)\).