Question

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

Answer 1

分析 首先分析題干中給的例子得到解題規律：將求解方程組的未知量系數化成與已知方程組未知量的系數相同，除去相同的系數部分即等于已知方程組的解，由此等式求出需求解的方程組的解．

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

點評 本題考查了二元一次方程組的解，難點在于根據題干的例子得出求解規律．對于需求解的方程組，將其未知量的系數化成和已知解的方程組未知量系數相同．然后讓需求解方程組未知量除去相同部分等于已知方程組的解進而求出需求解的方程組的解．