Question

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

Answer 1

分析 所求方程組變形后，仿照已知方程組求出解即可．

解答 解：所求方程組整理得：$\left\{\begin{array}{l}{{a}_{1}（2x-1）+_{1}（2-y）={c}_{1}}\\{{a}_{2}（2x-1）+_{2}（2-y）={c}_{2}}\end{array}\right.$，
由方程組$\left\{\begin{array}{l}{{a}_{1}x+_{1}y={c}_{1}}\\{{a}_{2}x+_{2}y={c}_{2}}\end{array}\right.$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：$\left\{\begin{array}{l}{x=-\frac{1}{2}}\\{y=-3}\end{array}\right.$．

點(diǎn)評(píng) 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都成立的未知數(shù)的值．