Question

15．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$\left\{\begin{array}{l}x=5\\ y=10\end{array}\right.$；则关于x，y的方程组$\left\{\begin{array}{l}{a_1}x-{b_1}y={a_1}+{c_1}\\{a_2}x-{b_2}y={a_2}+{c_2}\end{array}\right.$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 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$\left\{\begin{array}{l}x=5\\ y=10\end{array}\right.$，得到$\left\{\begin{array}{l}{5{a}_{1}+10{b}_{1}={c}_{1}}\\{5{a}_{2}+10{b}_{2}={c}_{2}}\end{array}\right.$，把方程组$\left\{\begin{array}{l}{a_1}x-{b_1}y={a_1}+{c_1}\\{a_2}x-{b_2}y={a_2}+{c_2}\end{array}\right.$变形为$\left\{\begin{array}{l}{{a}_{1}x-{b}_{1}y=6{a}_{1}+10{b}_{1}}\\{{a}_{2}x-{b}_{2}y=6{a}_{2}+10{b}_{2}}\end{array}\right.$，即可解答．

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$\left\{\begin{array}{l}x=5\\ y=10\end{array}\right.$，
∴$\left\{\begin{array}{l}{5{a}_{1}+10{b}_{1}={c}_{1}}\\{5{a}_{2}+10{b}_{2}={c}_{2}}\end{array}\right.$，
∴方程组$\left\{\begin{array}{l}{a_1}x-{b_1}y={a_1}+{c_1}\\{a_2}x-{b_2}y={a_2}+{c_2}\end{array}\right.$变形为$\left\{\begin{array}{l}{{a}_{1}x-{b}_{1}y=6{a}_{1}+10{b}_{1}}\\{{a}_{2}x-{b}_{2}y=6{a}_{2}+10{b}_{2}}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x=6}\\{y=-10}\end{array}\right.$对符合条件a₁，b₁，a₂，b₂都成立，
故选：B．

点评 本题考查了二元一次方程组的解，解决本题的关键是熟记二元一次方程组的解的定义．