Question

6．方程組$\left\{\begin{array}{l}{{a}_{1}x+_{1}y={c}_{1}}\\{{a}_{2}x+_{2}y={c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$，則方程組$\left\{\begin{array}{l}{2{a}_{1}x+4_{1}y=5{c}_{1}}\\{2{a}_{2}x+4_{2}y=5{c}_{2}}\end{array}\right.$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(guī)律：將求解方程組的未知量系數(shù)化成與已知方程組未知量的系數(shù)相同，除去相同的系數(shù)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$\left\{\begin{array}{l}{{a}_{1}x+_{1}y={c}_{1}}\\{{a}_{2}x+_{2}y={c}_{2}}\end{array}\right.$得：$\left\{\begin{array}{l}{2{a}_{1}+3_{1}={c}_{1}}\\{2{a}_{2}+3_{2}={c}_{2}}\end{array}\right.$，
把$\left\{\begin{array}{l}{2{a}_{1}+3_{1}={c}_{1}}\\{2{a}_{2}+3_{2}={c}_{2}}\end{array}\right.$代入方程組$\left\{\begin{array}{l}{2{a}_{1}x+4_{1}y=5{c}_{1}}\\{2{a}_{2}x+4_{2}y=5{c}_{2}}\end{array}\right.$得：
$\left\{\begin{array}{l}{2{a}_{1}x+4_{1}y=10{a}_{1}+1{5b}_{1}}\\{2{a}_{2}x+4_{2}y=10{a}_{2}+15_{2}}\end{array}\right.$，
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(diǎn)評(píng) 本題考查了二元一次方程組的解，難點(diǎn)在于根據(jù)題干的例子得出求解規(guī)律．對(duì)于需求解的方程組，將其未知量的系數(shù)化成和已知解的方程組未知量系數(shù)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(jìn)而求出需求解的方程組的解．