题目内容

17.(1)$\left\{\begin{array}{l}{3x+2y=14}\\{x=y+3}\end{array}\right.$                       
(2)$\left\{\begin{array}{l}{2x+y=-5}\\{4x-5y=11}\end{array}\right.$.

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

解答 解:(1)$\left\{\begin{array}{l}{3x+2y=14①}\\{x=y+3②}\end{array}\right.$,
把②代入①得:3y+9+2y=14,
解得:y=1,
把y=1代入②得:x=4,
则方程组的解为$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$;
(2)$\left\{\begin{array}{l}{2x+y=-5①}\\{4x-5y=11②}\end{array}\right.$,
①×5+②得:14x=-14,即x=-1,
把x=-1代入①得:y=-3,
则方程组的解为$\left\{\begin{array}{l}{x=-1}\\{y=-3}\end{array}\right.$.

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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