Question

Giải hệ phương trình sau:
\(a,\left\{{}\begin{matrix}x+y=0\\4x+3y=2\end{matrix}\right.\)

\(b,\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\)

Answer 1

Lời giải:

a. HPT:

\(\left\{\begin{matrix} x+y=0\\ 4x+3y=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+y=0\\ x+3(x+y)=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+y=0\\ x+0=2\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x+y=0\\ x=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2\\ y=-2\end{matrix}\right.\)

Vậy.......

b.

HPT \(\Leftrightarrow \left\{\begin{matrix} 5x-y=4\\ 3x-y=5\end{matrix}\right.\Rightarrow (5x-y)-(3x-y)=4-5\)

$\Leftrightarrow 2x=-1$

$\Leftrightarrow x=\frac{-1}{2}$

$y=5x-4=5.\frac{-1}{2}-4=\frac{-13}{2}$

Vậy..........