Question

10．已知方程組$\left\{\begin{array}{l}{a_1}x+y={c_1}\\{a_2}x+y={c_2}\end{array}\right.$的解是$\left\{\begin{array}{l}x=5\\ y=10\end{array}\right.$；則關(guān)于x，y的方程組$\left\{\begin{array}{l}{a_1}x-y={a_1}+{c_1}\\{a_2}x-y={a_2}+{c_2}\end{array}\right.$的解是（　　）

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

Answer 1

分析 把$\left\{\begin{array}{l}x=5\\ y=10\end{array}\right.$代入方程組$\left\{\begin{array}{l}{a_1}x+y={c_1}\\{a_2}x+y={c_2}\end{array}\right.$得$\left\{\begin{array}{l}{5{a}_{1}+10={c}_{1}}\\{5{a}_{2}+10={c}_{2}}\end{array}\right.$，方程組$\left\{\begin{array}{l}{a_1}x-y={a_1}+{c_1}\\{a_2}x-y={a_2}+{c_2}\end{array}\right.$變形為：$\left\{\begin{array}{l}{{a}_{1}x-y=6{a}_{1}+10}\\{{a}_{2}x-y=6{a}_{2}+10}\end{array}\right.$，即可解答．

解答 解：把$\left\{\begin{array}{l}x=5\\ y=10\end{array}\right.$代入方程組$\left\{\begin{array}{l}{a_1}x+y={c_1}\\{a_2}x+y={c_2}\end{array}\right.$得：$\left\{\begin{array}{l}{5{a}_{1}+10={c}_{1}}\\{5{a}_{2}+10={c}_{2}}\end{array}\right.$，
∴方程組$\left\{\begin{array}{l}{a_1}x-y={a_1}+{c_1}\\{a_2}x-y={a_2}+{c_2}\end{array}\right.$變形為：$\left\{\begin{array}{l}{{a}_{1}x-y=6{a}_{1}+10}\\{{a}_{2}x-y=6{a}_{2}+10}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x=6}\\{y=-10}\end{array}\right.$對(duì)符合a₁，a₂，c₁，c₂都成立，
故選：B．

點(diǎn)評(píng) 本題考查了二元一次方程組的解，解決本題的關(guān)鍵是明確二元一次方程組的解的定義．