Question

20．下列方程组中，与$\left\{\begin{array}{l}{x+2y=5}\\{2x+5y=7}\end{array}\right.$不同解的是（　　）

• A． $\left\{\begin{array}{l}{x+2y=5}\\{3x+7y=12}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x+3y=2}\\{2x+5y=7}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x+2y=5}\\{x+3y=2}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x+y=3}\\{2x+5y=7}\end{array}\right.$

Answer 1

分析 先解方程组$\left\{\begin{array}{l}{x+2y=5}\\{2x+5y=7}\end{array}\right.$，得到其解，再将该解分别代入各方程组，判断各方程组是否成立即可，若成立，则是同解方程．

解答 解：解方程组$\left\{\begin{array}{l}{x+2y=5}\\{2x+5y=7}\end{array}\right.$可得，$\left\{\begin{array}{l}{x=11}\\{y=-3}\end{array}\right.$，
将$\left\{\begin{array}{l}{x=11}\\{y=-3}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{x+2y=5}\\{3x+7y=12}\end{array}\right.$可得，方程组成立；
将$\left\{\begin{array}{l}{x=11}\\{y=-3}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{x+3y=2}\\{2x+5y=7}\end{array}\right.$可得，方程组成立；
将$\left\{\begin{array}{l}{x=11}\\{y=-3}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{x+2y=5}\\{x+3y=2}\end{array}\right.$可得，方程组成立；
将$\left\{\begin{array}{l}{x=11}\\{y=-3}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{x+y=3}\\{2x+5y=7}\end{array}\right.$可得，方程组不成立；
故选（D）

点评 本题主要考查了二元一次方程组的解，解决问题的关键是判断原方程组的解是否满足各方程组．