Question

9．已知$\left\{\begin{array}{l}{{a}_{1}x+_{1}y={c}_{1}}\\{{a}_{2}x+_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$，則方程組$\left\{\begin{array}{l}{3{a}_{1}（x-1）+4_{1}（y+3）={c}_{1}}\\{3{a}_{2}（x-1）+4_{2}（y+3）={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=-2}\end{array}\right.$．

Answer 1

分析 根據(jù)二元一次方程組的解，即可解答．

解答 解：將$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$代入$\left\{\begin{array}{l}{{a}_{1}x+_{1}y={c}_{1}}\\{{a}_{2}x+_{2}y={c}_{2}}\end{array}\right.$得：$\left\{\begin{array}{l}{3{a}_{1}+4_{1}={c}_{1}}\\{3{a}_{2}+4_{2}={c}_{2}}\end{array}\right.$，
將$\left\{\begin{array}{l}{3{a}_{1}+4_{1}={c}_{1}}\\{3{a}_{2}+4_{2}={c}_{2}}\end{array}\right.$代入方程組$\left\{\begin{array}{l}{3{a}_{1}（x-1）+4_{1}（y+3）={c}_{1}}\\{3{a}_{2}（x-1）+4_{2}（y+3）={c}_{2}}\end{array}\right.$得：
$\left\{\begin{array}{l}{x-1=1}\\{y+3=1}\end{array}\right.$
解得：$\left\{\begin{array}{l}{x=2}\\{y=-2}\end{array}\right.$，
故答案為：$\left\{\begin{array}{l}{x=2}\\{y=-2}\end{array}\right.$．

點評 本題考查了二元一次方程組的解，解決本題的關(guān)鍵是解二元一次方程組．