Question

1．方程组$\left\{{\begin{array}{l}{{a_1}x+{b_1}y={c_1}}\\{{a_2}x+{b_2}y={c_2}}\end{array}}\right.$的解为$\left\{{\begin{array}{l}{x=4}\\{y=6}\end{array}}\right.$，则方程组$\left\{{\begin{array}{l}{4{a_1}x+3{b_1}y=5{c_1}}\\{4{a_2}x+3{b_2}y=5{c_2}}\end{array}}\right.$的解为（　　）

• A． $\left\{{\begin{array}{l}{x=4}\\{y=6}\end{array}}\right.$
• B． $\left\{{\begin{array}{l}{x=5}\\{y=6}\end{array}}\right.$
• C． $\left\{{\begin{array}{l}{x=5}\\{y=10}\end{array}}\right.$
• D． $\left\{{\begin{array}{l}{x=10}\\{y=15}\end{array}}\right.$

Answer 1

分析 先根据方程组$\left\{{\begin{array}{l}{{a_1}x+{b_1}y={c_1}}\\{{a_2}x+{b_2}y={c_2}}\end{array}}\right.$的解为$\left\{{\begin{array}{l}{x=4}\\{y=6}\end{array}}\right.$，得到$\left\{\begin{array}{l}{20{a}_{1}+30{b}_{1}=5{c}_{1}}\\{20{a}_{2}+30{b}_{2}=5{c}_{2}}\end{array}\right.$，进而得到$\left\{\begin{array}{l}{20=4x}\\{30=3y}\end{array}\right.$，求得$\left\{\begin{array}{l}{x=5}\\{y=10}\end{array}\right.$．

解答 解：∵方程组$\left\{{\begin{array}{l}{{a_1}x+{b_1}y={c_1}}\\{{a_2}x+{b_2}y={c_2}}\end{array}}\right.$的解为$\left\{{\begin{array}{l}{x=4}\\{y=6}\end{array}}\right.$，
∴$\left\{\begin{array}{l}{4{a}_{1}+6{b}_{1}={c}_{1}}\\{4{a}_{2}+6{b}_{2}={c}_{2}}\end{array}\right.$，即$\left\{\begin{array}{l}{20{a}_{1}+30{b}_{1}=5{c}_{1}}\\{20{a}_{2}+30{b}_{2}=5{c}_{2}}\end{array}\right.$，
又∵方程组$\left\{{\begin{array}{l}{4{a_1}x+3{b_1}y=5{c_1}}\\{4{a_2}x+3{b_2}y=5{c_2}}\end{array}}\right.$，
∴$\left\{\begin{array}{l}{20=4x}\\{30=3y}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x=5}\\{y=10}\end{array}\right.$，
故选：C．

点评 本题主要考查了二元一次方程组的解，解题时注意：当遇到有关二元一次方程组的解的问题时，通常采用代入法，即将解代入原方程组，这种方法主要用在求方程中的字母系数．