Question

18．已知：a、b是常數(shù)，若關(guān)于m、n的二元一次方程組$\left\{\begin{array}{l}{am+bn=9.1}\\{3am+2bn=6.7}\end{array}\right.$的解是$\left\{\begin{array}{l}{m=3.1}\\{n=1.7}\end{array}\right.$，則關(guān)于x、y的二元一次方程組$\left\{\begin{array}{l}{a（x+1）+b（y-2）=9.1}\\{3a（x+1）+2b（y-2）=6.7}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2.1}\\{y=3.7}\end{array}\right.$．

Answer 1

分析 把$\left\{\begin{array}{l}{m=3.1}\\{n=1.7}\end{array}\right.$代入二元一次方程組$\left\{\begin{array}{l}{am+bn=9.1}\\{3am+2bn=6.7}\end{array}\right.$，得出a和b的值后代入二元一次方程組$\left\{\begin{array}{l}{a（x+1）+b（y-2）=9.1}\\{3a（x+1）+2b（y-2）=6.7}\end{array}\right.$解答即可．

解答 解：把$\left\{\begin{array}{l}{m=3.1}\\{n=1.7}\end{array}\right.$代入二元一次方程組$\left\{\begin{array}{l}{am+bn=9.1}\\{3am+2bn=6.7}\end{array}\right.$，
可得：$\left\{\begin{array}{l}{3.1a+1.7b=9.1}\\{9.3a+3.4b=6.7}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=\frac{115}{31}}\\{b=\frac{206}{17}}\end{array}\right.$，
把a(bǔ)和b的值代入二元一次方程組$\left\{\begin{array}{l}{a（x+1）+b（y-2）=9.1}\\{3a（x+1）+2b（y-2）=6.7}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=2.1}\\{y=3.7}\end{array}\right.$．

點評 此題考查了二元一次方程組的解，方程組的解即為能使方程組中兩方程成立的未知數(shù)的值．