题目内容

8.解方程组
(1)$\left\{\begin{array}{l}3x+4y=16\\ 5x-6y=33.\end{array}\right.$
(2)$\left\{\begin{array}{l}4(x-y-1)=3(1-y)-2\\ \frac{x}{2}+\frac{y}{3}=2.\end{array}\right.$.

分析 (1)方程组利用加减消元法求出解即可;
(2)方程组整理后,利用代入消元法求出解即可.

解答 解:(1)$\left\{\begin{array}{l}{3x+4y=16①}\\{5x-6y=33②}\end{array}\right.$,
①×3+②×2得:19x=114,即x=6,
把x=6代入①得:y=-$\frac{1}{2}$,
则方程组的解为$\left\{\begin{array}{l}x=6\\ y=-\frac{1}{2}\end{array}\right.$;
(2)原方程组可化为$\left\{\begin{array}{l}{y=4x-5①}\\{3x+2y=12②}\end{array}\right.$,
把①代入②得:3x+8x-10=12,
解得:x=2,
把x=2代入①得y=3,
则方程组的解为$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$.

点评 此题考查了解二元一次方程组,利用了消元的思想,消元的方法有:代入消元法与加减消元法.

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