Question

1．方程組$\left\{{\begin{array}{l}{{a_1}x+{b_1}y={c_1}}\\{{a_2}x+{b_2}y={c_2}}\end{array}}\right.$的解為$\left\{{\begin{array}{l}{x=4}\\{y=6}\end{array}}\right.$，則方程組$\left\{{\begin{array}{l}{4{a_1}x+3{b_1}y=5{c_1}}\\{4{a_2}x+3{b_2}y=5{c_2}}\end{array}}\right.$的解為（　�。�

• 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 先根據(jù)方程組$\left\{{\begin{array}{l}{{a_1}x+{b_1}y={c_1}}\\{{a_2}x+{b_2}y={c_2}}\end{array}}\right.$的解為$\left\{{\begin{array}{l}{x=4}\\{y=6}\end{array}}\right.$，得到$\left\{\begin{array}{l}{20{a}_{1}+30_{1}=5{c}_{1}}\\{20{a}_{2}+30_{2}=5{c}_{2}}\end{array}\right.$，進(jìn)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：∵方程組$\left\{{\begin{array}{l}{{a_1}x+{b_1}y={c_1}}\\{{a_2}x+{b_2}y={c_2}}\end{array}}\right.$的解為$\left\{{\begin{array}{l}{x=4}\\{y=6}\end{array}}\right.$，
∴$\left\{\begin{array}{l}{4{a}_{1}+6_{1}={c}_{1}}\\{4{a}_{2}+6_{2}={c}_{2}}\end{array}\right.$，即$\left\{\begin{array}{l}{20{a}_{1}+30_{1}=5{c}_{1}}\\{20{a}_{2}+30_{2}=5{c}_{2}}\end{array}\right.$，
又∵方程組$\left\{{\begin{array}{l}{4{a_1}x+3{b_1}y=5{c_1}}\\{4{a_2}x+3{b_2}y=5{c_2}}\end{array}}\right.$，
∴$\left\{\begin{array}{l}{20=4x}\\{30=3y}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x=5}\\{y=10}\end{array}\right.$，
故選：C．

點評 本題主要考查了二元一次方程組的解，解題時注意：當(dāng)遇到有關(guān)二元一次方程組的解的問題時，通常采用代入法，即將解代入原方程組，這種方法主要用在求方程中的字母系數(shù)．