Question

13．若方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=-2}\\{y=5}\end{array}\right.$，求方程组$\left\{\begin{array}{l}{2{a}_{1}x-{b}_{1}y={a}_{1}-2{b}_{1}+{c}_{1}}\\{2{a}_{2}x-{b}_{2}y={a}_{2}-2{b}_{2}+{c}_{2}}\end{array}\right.$的解．

Answer 1

分析 所求方程组变形后，仿照已知方程组求出解即可．

解答 解：所求方程组整理得：$\left\{\begin{array}{l}{{a}_{1}（2x-1）+{b}_{1}（2-y）={c}_{1}}\\{{a}_{2}（2x-1）+{b}_{2}（2-y）={c}_{2}}\end{array}\right.$，
由方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=-2}\\{y=5}\end{array}\right.$，得到$\left\{\begin{array}{l}{2x-1=-2}\\{2-y=5}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=-\frac{1}{2}}\\{y=-3}\end{array}\right.$．

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．