Question

7．已知方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=6}\\{y=8}\end{array}\right.$，则方程组$\left\{\begin{array}{l}{3{a}_{1}x+4{b}_{1}y=5{c}_{1}}\\{3{a}_{2}x+4{b}_{2}y=5{c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=10}\\{y=10}\end{array}\right.$．

Answer 1

分析 根据方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=6}\\{y=8}\end{array}\right.$，可得方程组$\left\{\begin{array}{l}{3{a}_{1}x+4{b}_{1}y=5{c}_{1}}\\{3{a}_{2}x+4{b}_{2}y=5{c}_{2}}\end{array}\right.$中的解是$\left\{\begin{array}{l}{3x=5×6}\\{4y=5×8}\end{array}\right.$，从而求解．

解答 解：∵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$\left\{\begin{array}{l}{x=6}\\{y=8}\end{array}\right.$，
∴方程组$\left\{\begin{array}{l}{3{a}_{1}x+4{b}_{1}y=5{c}_{1}}\\{3{a}_{2}x+4{b}_{2}y=5{c}_{2}}\end{array}\right.$中的解是$\left\{\begin{array}{l}{3x=5×6}\\{4y=5×8}\end{array}\right.$，
即$\left\{\begin{array}{l}{x=10}\\{y=10}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{x=10}\\{y=10}\end{array}\right.$．

点评 考查了解二元一次方程组，关键是由方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=6}\\{y=8}\end{array}\right.$，得到方程组$\left\{\begin{array}{l}{3{a}_{1}x+4{b}_{1}y=5{c}_{1}}\\{3{a}_{2}x+4{b}_{2}y=5{c}_{2}}\end{array}\right.$中的解是$\left\{\begin{array}{l}{3x=5×6}\\{4y=5×8}\end{array}\right.$．