（2021 秋・安宁市校级期末）解下列方程组:
    (1) $\left\{\begin{array}{l}x-y=3 \\ x+y=1\end{array} ;\right.$
    (2) $\left\{\begin{array}{l}3(x-1)=y+5 \\ 5(y-1)=3(x+5)\end{array}\right.$.

【思路点拨】 (1) 用加减消元法解二元一次方程组即可;

(2) 先化简二元一次方程组, 再用加减消元法解二元一次方程组即可.
【答案】解: (1) $\left\{\begin{array}{l}x-y=3(1) \\ x+y=1(2)\end{array}\right.$,

(1)+(2), 得 $2 x=4$, 解得 $x=2$,

将 $x=2$ 代入(1)得, $y=-1$,

$\therefore$ 方程组的解为 $\left\{\begin{array}{c}x=2 \\ y=-1\end{array} ;\right.$

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

化简方程组可得 $\left\{\begin{array}{l}3 x-y=8(1) \\ 3 x-5 y=-20(2)\end{array}\right.$,

(1) - (2)得, $4 y=28$, 解得 $y=7$,

将 $y=7$ 代入(1)得, $x=5$,

$\therefore$ 方程组的解为 $\left\{\begin{array}{l}x=5 \\ y=7\end{array}\right.$.

【点睛】本题考查二元一次方程组的解, 熟练掌握二元一次方程组的解法是解题的关键
秋・安宁市校级期末）解下列方程组:
    (1) $\left\{\begin{array}{l}x-y=3 \\ x+y=1\end{array} ;\right.$
    (2) $\left\{\begin{array}{l}3(x-1)=y+5 \\ 5(y-1)=3(x+5)\end{array}\right.$.

【思路点拨】 (1) 用加减消元法解二元一次方程组即可;

(2) 先化简二元一次方程组, 再用加减消元法解二元一次方程组即可.
【答案】解: (1) $\left\{\begin{array}{l}x-y=3(1) \\ x+y=1(2)\end{array}\right.$,

(1)+(2), 得 $2 x=4$, 解得 $x=2$,

将 $x=2$ 代入(1)得, $y=-1$,

$\therefore$ 方程组的解为 $\left\{\begin{array}{c}x=2 \\ y=-1\end{array} ;\right.$

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

化简方程组可得 $\left\{\begin{array}{l}3 x-y=8(1) \\ 3 x-5 y=-20(2)\end{array}\right.$,

(1) - (2)得, $4 y=28$, 解得 $y=7$,

将 $y=7$ 代入(1)得, $x=5$,

$\therefore$ 方程组的解为 $\left\{\begin{array}{l}x=5 \\ y=7\end{array}\right.$.

【点睛】本题考查二元一次方程组的解, 熟练掌握二元一次方程组的解法是解题的关键.