Bài 2:
a. Lấy PT $(1)$ + PT $(2)$ ta có:
$4x=-4$
$\Leftrightarrow x=-1$
$2y=1-3x=1-3(-1)=4\Rightarrow y=2$
Vậy $(x,y)=(-1,2)$
b.
Lấy 3.PT$(1)$ + 2.PT$(2)$ thu được:
\(3(3x-2y)+2(4x+3y)=3.12+2(-1)\)
\(\Leftrightarrow 17x=34\Leftrightarrow x=2\)
\(2y=3x-12=3.2-12=-6\Rightarrow y=-3\)
2.
a. HPT \(\Leftrightarrow \left\{\begin{matrix} xy-2x-y+2=xy-3x+y-3\\ xy+4x-5y-20=xy+x-4y-4\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x-2y=-5\\ 3x-y=16\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-2y=-5\\ 6x-2y=32\end{matrix}\right.\)
Lấy PT $(1)$ - PT $(2)$ tì:
$-5x=-37$
$\Leftrightarrow x=\frac{37}{5}$
$y=3x-16=\frac{31}{5}$
b.
HPT \(\Leftrightarrow \left\{\begin{matrix} (x+2)(y+3)=xy+100\\ (x-2)(y-2)=xy-64\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 3x+2y=94\\ 2x+2y=68\end{matrix}\right.\)
Lấy PT $(1)$ - PT $(2)$ thu được:
$x=26$
$y=\frac{68-2.26}{2}=8$
Bài 1:
a: Ta có: \(\left\{{}\begin{matrix}x-2y=-5\\3x+2y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=-4\\x-2y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2y=x+5=-1+5=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
