一、由解确定解例1 已知关于x、y的方程组(?) 解析:方程组的解只与未知数的系数有关,而与未知数是什么字母无关,对这两个方程组,可知前者的x对应后者的(x-y),前者的y对应后者的(x+y),因此可得(?)即为所求方程组的解．二、两个方程来定解例2 已知关于x、y的方程组(?)有相同的解,求a、b的值．分析:两个方程组有相同的解,故x、y,的值必然同时适合两个方程 First, the solution to determine the solution 1 Known about the x, y equations (?) Analysis: The solution of the equation group is only related to the coefficient of the unknown, and has nothing to do with the unknown number of letters, for the two equations, we can see the former The x corresponds to the latter (xy), and the former y corresponds to the latter (x+y), so the available (?) is the solution of the sought equations. Second, the two equations to determine solution 2 Known about the x, y equations (?) Have the same solution, seeking a, b value. Analysis: The two equations have the same solution, so the values ​​of x and y must fit both equations.