Question

14．已知方程组$\left\{\begin{array}{l}{ax-by=4}\\{ax+by=2}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$，6a+3b的值为6．

Answer 1

分析 把x与y的值代入方程组求出a与b的值，即可确定出6a+3b的值．

解答 解：把$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$代入方程组得：$\left\{\begin{array}{l}{2a-b=4}\\{2a+b=2}\end{array}\right.$，
解得：a=$\frac{3}{2}$，b=-1，
则6a+3b=9-3=6．
故答案为：6

点评 此题考查了二元一次方程组的解，方程组的解即为能使方程组中两方程都成立的未知数的值．