Question

[JaKi Blue]

Giải các hệ phương trình sau bằng phương pháp thế:

a\(\left\{{}\begin{matrix}x+3y=4\\4x=5y=18\end{matrix}\right.\)

b\(\left\{{}\begin{matrix}5x-3y=5\\2x+5y=33\end{matrix}\right.\)

c\(\left\{{}\begin{matrix}\frac{x}{2}-\frac{y}{3}=0\\5x+y=13\end{matrix}\right.\)

d\(\left\{{}\begin{matrix}x+2y=\frac{7}{6}\\4x+6y=4\end{matrix}\right.\)

Answer 1

[Nguyễn Hoàng Minh Vy]

a) Xem lại đề

b) \(\left\{{}\begin{matrix}5x-3y=5\\2x+5y=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5.\frac{33-5y}{2}-3y=5\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}165-25y-6y=10\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}31y=155\\x=\frac{33-5y}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=\frac{33-5.5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=4\end{matrix}\right.\)

Answer 2

[Nguyễn Hoàng Minh Vy]

c)\(\left\{{}\begin{matrix}\frac{x}{2}-\frac{y}{3}=0\\5x+y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\\frac{x}{2}-\frac{13-5x}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\\frac{3x-26+10x}{6}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5x\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=13-5.2\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)