Question

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

Answer 1

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

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

點評 本題主要考查二元一次方程組的解法，關(guān)鍵是熟練掌握二元一次方程組的解法即代入消元法和加減消元法．