Question

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

Answer 1

分析 把原方程组的解代入方程组，求出a，b的值，再代入所求代数式即可．

解答 解：把$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{ax-by=4}\\{ax+by=2}\end{array}\right.$，得$\left\{\begin{array}{l}{2a-b=4}\\{2a+b=2}\end{array}\right.$，
解得$\left\{\begin{array}{l}{a=\frac{3}{2}}\\{b=-1}\end{array}\right.$．
2a-3b=2×$\frac{3}{2}$-3×（-1）=6．
故2a-3b的值是6．

点评 考查了二元一次方程组的解的定义，所谓“方程组”的解，指的是该数值满足方程组中的每一方程．