Question

3．若方程组$\left\{{\begin{array}{l}{a_1}x+{b_1}y={c_1}\\{a_2}x+{b_2}y={c_2}\end{array}}\right.$的解是$\left\{\begin{array}{l}x=21\\ y=-10\end{array}\right.$，则方程组$\left\{{\begin{array}{l}7{a_1}x+5{b_1}y=2{c_1}\\ 7{a_2}x+5{b_2}y=2{c_2}\end{array}}\right.$的解为$\left\{\begin{array}{l}{x=6}\\{y=-4}\end{array}\right.$．

Answer 1

分析 把后面的方程组整理为关于$\left\{\begin{array}{l}{{a}_{1}•\frac{7}{2}x+{b}_{1}•\frac{5}{2}y={c}_{1}}\\{{a}_{2}•\frac{7}{2}x+{b}_{2}•\frac{5}{2}y={c}_{2}}\end{array}\right.$，此方程与前面的方程组一样，它是关于$\frac{7}{2}$x和$\frac{5}{2}$y的方程组，所以$\frac{7}{2}$x=21，$\frac{5}{2}$y=-10，然后求出x和y即可．

解答 解：把方程组$\left\{{\begin{array}{l}7{a_1}x+5{b_1}y=2{c_1}\\ 7{a_2}x+5{b_2}y=2{c_2}\end{array}}\right.$变形为$\left\{\begin{array}{l}{{a}_{1}•\frac{7}{2}x+{b}_{1}•\frac{5}{2}y={c}_{1}}\\{{a}_{2}•\frac{7}{2}x+{b}_{2}•\frac{5}{2}y={c}_{2}}\end{array}\right.$，
而方程组$\left\{{\begin{array}{l}{a_1}x+{b_1}y={c_1}\\{a_2}x+{b_2}y={c_2}\end{array}}\right.$的解是$\left\{\begin{array}{l}x=21\\ y=-10\end{array}\right.$，
所以$\frac{7}{2}$x=21，$\frac{5}{2}$y=-10，
解得x=6，y=-4，
所以方程组$\left\{{\begin{array}{l}7{a_1}x+5{b_1}y=2{c_1}\\ 7{a_2}x+5{b_2}y=2{c_2}\end{array}}\right.$的解为$\left\{\begin{array}{l}{x=6}\\{y=-4}\end{array}\right.$．
故答案为$\left\{\begin{array}{l}{x=6}\\{y=-4}\end{array}\right.$．

点评 本题考查了解二元一次方程组：利用加减消元法或代入消元法解二元一次方程组．解决本题的关键把后面的方程组变形为第一个方程组的形式．