Question

16．已知二元一次方程组$\left\{\begin{array}{l}{ax+by=7}\\{bx-ay=5}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=1}\\{y=-2}\end{array}\right.$，求4a-3b的值．

Answer 1

分析 首先把$\left\{\begin{array}{l}{x=1}\\{y=-2}\end{array}\right.$代入$\left\{\begin{array}{l}{ax+by=7}\\{bx-ay=5}\end{array}\right.$得关于a、b的方程组，解方程组可得a、b的值，进而可得4a-3b的值．

解答 解：把$\left\{\begin{array}{l}{x=1}\\{y=-2}\end{array}\right.$代入$\left\{\begin{array}{l}{ax+by=7}\\{bx-ay=5}\end{array}\right.$得：$\left\{\begin{array}{l}{a-2b=7}\\{b+2a=5}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=\frac{17}{5}}\\{b=-\frac{9}{5}}\end{array}\right.$，
4a-3b=4×$\frac{17}{5}$+3×$\frac{9}{5}$=19．

点评 此题主要考查了二元一次方程组的解，关键是掌握二元一次方程组的解能同时是两个方程成立．