题目内容

2.若关于x、y的二元一次方程组$\left\{\begin{array}{l}{3x+my=16}\\{2x+ny=15}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=7}\\{y=3}\end{array}\right.$,则关于x、y的二元一次方程组$\left\{\begin{array}{l}{3(x+y)-m(x-y)=16}\\{2(x+y)+n(x-y)=15}\end{array}\right.$ 的解是$\left\{\begin{array}{l}{x=5}\\{y=2}\end{array}\right.$.

分析 本题先代入解求出得$\left\{\begin{array}{l}{m=-\frac{5}{3}}\\{n=\frac{1}{3}}\end{array}\right.$,再将其代入二元一次方程组$\left\{\begin{array}{l}{3(x+y)-m(x-y)=16}\\{2(x+y)+n(x-y)=15}\end{array}\right.$,解出即可.

解答 解:把$\left\{\begin{array}{l}{x=7}\\{y=3}\end{array}\right.$代入二元一次方程组$\left\{\begin{array}{l}{3x+my=16}\\{2x+ny=15}\end{array}\right.$,
解得:$\left\{\begin{array}{l}{m=-\frac{5}{3}}\\{n=\frac{1}{3}}\end{array}\right.$,
把$\left\{\begin{array}{l}{m=-\frac{5}{3}}\\{n=\frac{1}{3}}\end{array}\right.$代入二元一次方程组$\left\{\begin{array}{l}{3(x+y)-m(x-y)=16}\\{2(x+y)+n(x-y)=15}\end{array}\right.$,
解得:$\left\{\begin{array}{l}{x=5}\\{y=2}\end{array}\right.$,
故答案为:$\left\{\begin{array}{l}{x=5}\\{y=2}\end{array}\right.$.

点评 本题主要考查二元一次方程组的解法,关键是熟练掌握二元一次方程组的解法即代入消元法和加减消元法.

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