Question

Giải hệ phương trình bằng phương pháp thế

\(\begin{cases} x-y=m\\ 2x+y=4 \end{cases}\)

Answer 1

\(\left\{{}\begin{matrix}x-y=m\left(1\right)\\2x+y=4\left(2\right)\end{matrix}\right.\\ \left(1\right)\Leftrightarrow x=y+m\)

Thay \(x=y+m\) vào (2) ta được:

\(2\left(y+m\right)+y=4\\ \Leftrightarrow2y+2m+y=4\\ \Leftrightarrow3y=4-2m\\ \Leftrightarrow y=\dfrac{4-2m}{3}\)

Thay \(y=\dfrac{4-2m}{3}\) vào (1) ta được:

\(x-\dfrac{4-2m}{3}=m\\ \Leftrightarrow\dfrac{3x}{3}-\dfrac{4-2m}{3}=\dfrac{3m}{3}\\ \Leftrightarrow3x-4+2m=3m\\ \Leftrightarrow3x=m+4\\ \Leftrightarrow x=\dfrac{m+4}{3}\)

Vậy hpt có nghiệm là \(\left(x;y\right)=\left(\dfrac{m+4}{3};\dfrac{4-2m}{3}\right)\)

Answer 2

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-m\\2x+y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-m\\2x+x-m=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-m\\3x=m+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=x-m\\x=\dfrac{m+4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m+4}{3}\\y=\dfrac{-2m+4}{3}\end{matrix}\right.\)