题目内容

20.(1)$\left\{\begin{array}{l}{y=2x-3}\\{5x+y=11}\end{array}\right.$;   (2)$\left\{\begin{array}{l}{x+3y=-1}\\{3x-y=7}\end{array}\right.$.

分析 (1)利用代入消元法解出方程组;
(2)利用加减消元法解出方程组.

解答 解:(1)$\left\{\begin{array}{l}{y=2x-3①}\\{5x+y=11②}\end{array}\right.$,
把①代入②得,5x+2x-3=11,
解得,x=2,
把x=2代入①得,y=1,
则方程组的解为:$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$;
(2)$\left\{\begin{array}{l}{x+3y=-1①}\\{3x-y=7②}\end{array}\right.$,
①+②×3得,10x=20,
解得,x=2,
把x=2代入①得,y=-1,
则方程组的解为:$\left\{\begin{array}{l}{x=2}\\{y=-1}\end{array}\right.$.

点评 本题考查的是二元一次方程组的解法,掌握代入消元法和加减消元法解二元一次方程组的一般步骤是解题的关键.

练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网