Question

6．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$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$，则方程组$\left\{\begin{array}{l}{2{a}_{1}x+4{b}_{1}y=5{c}_{1}}\\{2{a}_{2}x+4{b}_{2}y=5{c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=5}\\{y=\frac{15}{4}}\end{array}\right.$．

Answer 1

分析 首先分析题干中给的例子得到解题规律：将求解方程组的未知量系数化成与已知方程组未知量的系数相同，除去相同的系数部分即等于已知方程组的解，由此等式求出需求解的方程组的解．

解答 解：把$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$得：$\left\{\begin{array}{l}{2{a}_{1}+3{b}_{1}={c}_{1}}\\{2{a}_{2}+3{b}_{2}={c}_{2}}\end{array}\right.$，
把$\left\{\begin{array}{l}{2{a}_{1}+3{b}_{1}={c}_{1}}\\{2{a}_{2}+3{b}_{2}={c}_{2}}\end{array}\right.$代入方程组$\left\{\begin{array}{l}{2{a}_{1}x+4{b}_{1}y=5{c}_{1}}\\{2{a}_{2}x+4{b}_{2}y=5{c}_{2}}\end{array}\right.$得：
$\left\{\begin{array}{l}{2{a}_{1}x+4{b}_{1}y=10{a}_{1}+1{5b}_{1}}\\{2{a}_{2}x+4{b}_{2}y=10{a}_{2}+15{b}_{2}}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=5}\\{y=\frac{15}{4}}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{x=5}\\{y=\frac{15}{4}}\end{array}\right.$．

点评 本题考查了二元一次方程组的解，难点在于根据题干的例子得出求解规律．对于需求解的方程组，将其未知量的系数化成和已知解的方程组未知量系数相同．然后让需求解方程组未知量除去相同部分等于已知方程组的解进而求出需求解的方程组的解．