Giải các hệ phương trình:

a, $\left\{\begin{array}{l}2(x+y)+3(x-y)=4\\ (x+y)+2(x-y)=5\end{array}\right.$;          b, $\left\{\begin{array}{l}2(x-2)+3(1+y)=-2\\ 3(x-2)-2(1+y)=-3\end{array}\right.$

Giải:

a, Cách 1: Đặt x+y=u, x-y=v, ta có hệ phương trình (ẩn u,v):

$$\begin{gathered} \left\{\begin{array}{l}2u+3v=4\\ u+2v=5\end{array}\right. \\ nên \left\{\begin{array}{l}2u+3v=4\\ u+2v=5\end{array}\right.\iff \left\{\begin{array}{l}2u+3v=4\\ 2u+4v=10\end{array}\right.\iff \left\{\begin{array}{l}2u+3v=4\\ -v=-6\end{array}\right.\iff \left\{\begin{array}{l}2u+3v=4\\ v=6\end{array}\right.\iff \left\{\begin{array}{l}2u=4-3.6\\ v=6\end{array}\right.\iff \left\{\begin{array}{l}u=-7\\ v=6\end{array}\right. \\ Suy ra hệ đã cho tương đương với: \\ \left\{\begin{array}{l}x+y=-7\\ x-y=6\end{array}\right.\iff \left\{\begin{array}{l}2x=-1\\ x-y=6\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{-1}{2}\\ y=\frac{-13}{2}\end{array}\right. \end{gathered}$$

Cách 2: Thu gọn vế trái của hai phương trình ta được hệ tương đương:

$\left\{\begin{array}{l}\left\{2x+2y+3x-3y\right.=4\\ x+y+2x-2y=5\end{array}\right.\iff \left\{\begin{array}{l}5x-y=4\\ 3x-y=5\end{array}\right.\iff \left\{\begin{array}{l}2x=-1\\ 3x-y=5\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{-1}{2}\\ y=3x-5\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{-1}{2}\\ y=3(-\frac{1}{2})-5\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{-1}{2}\\ y=\frac{-13}{2}\end{array}\right.$

b,

Cách 1: Đặt x-2=u, 1+y=v ta có hệ phương trình ẩn u,v:

$$\begin{gathered} \left\{\begin{array}{l}2u+3v=-2\\ 3u-2v=-3\end{array}\right. \\ nên \left\{\begin{array}{l}2u+3v=-2\\ 3u-2v=-3\end{array}\right.\iff \left\{\begin{array}{l}6u+9v=-6\\ 6u-4v=-6\end{array}\right.\iff \left\{\begin{array}{l}6u+9v=-6\\ 13v=0\end{array}\right.\iff \left\{\begin{array}{l}6u=-6+9v\\ v=0\end{array}\right.\iff \left\{\begin{array}{l}6u=-6\\ v=0\end{array}\iff \left\{\begin{array}{l}u=-1\\ v=0\end{array}\right.\right. \\ Suy ra hệ phương trình tương đương với: \\ \left\{\begin{array}{l}x-2=-1\\ 1+y=0\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ y=-1\end{array}\right. \end{gathered}$$

Cách 2: Thu gọn vế trái của hai phương trình:

$\left\{\begin{array}{l}2(x-2)+3(1+y)=-2\\ 3(x-2)-2(1+y)=-3\end{array}\right.\iff \left\{\begin{array}{l}2x-4+3+3y=-2\\ 3x-6-2-2y=-3\end{array}\right.\iff \left\{\begin{array}{l}2x+3y=-1\\ 3x-2y=5\end{array}\iff \left\{\begin{array}{l}6x+9y=-3\\ 6x-4y=10\end{array}\right.\right.\iff \left\{\begin{array}{l}6x+9y=-3\\ 13y=-13\end{array}\right.\iff \left\{\begin{array}{l}6x=-3-9y\\ y=-1\end{array}\right.\iff \left\{\begin{array}{l}6x=6\\ y=-1\end{array}\iff \left\{\begin{array}{l}x=1\\ y=-1\end{array}\right.\right.$