Question

5．若$\left\{\begin{array}{l}{{a}_{1}x+_{1}y={c}_{1}}\\{{a}_{2}x+_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$，則$\left\{\begin{array}{l}{3{a}_{1}（x-1）+_{1}（y+3）={c}_{1}}\\{3{a}_{2}（x-1）+_{2}（y+3）={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$．

Answer 1

分析 根據(jù)已知得出方程組$\left\{\begin{array}{l}{x-1=3}\\{y+3=4}\end{array}\right.$，求出方程組的解即可．

解答 解：∵$\left\{\begin{array}{l}{{a}_{1}x+_{1}y={c}_{1}}\\{{a}_{2}x+_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x-1=3}\\{y+3=4}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$，
即$\left\{\begin{array}{l}{3{a}_{1}（x-1）+_{1}（y+3）={c}_{1}}\\{3{a}_{2}（x-1）+_{2}（y+3）={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$，
故答案為：$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$．

點評 本題考查了二元一次方程組的解和解二元一次方程組，能根據(jù)題意得出關(guān)于x、y的方程組是解此題的關(guān)鍵．