Question

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

Answer 1

分析 方程组$\left\{\begin{array}{l}{5{a}_{1}x+3{b}_{1}y=4{c}_{1}}\\{5{a}_{2}x+3{b}_{2}y=4{c}_{2}}\end{array}\right.$变形为$\left\{\begin{array}{l}{\frac{5}{4}{a}_{1}x+\frac{3}{4}{b}_{1}y={c}_{1}}\\{\frac{5}{4}{a}_{2}x+\frac{3}{4}{b}_{2}y={c}_{2}}\end{array}\right.$，然后根据题意得出$\left\{\begin{array}{l}{\frac{5}{4}x=4}\\{\frac{3}{4}y=5}\end{array}\right.$，解方程即可求得原方程的解．

解答 解：方程组$\left\{\begin{array}{l}{5{a}_{1}x+3{b}_{1}y=4{c}_{1}}\\{5{a}_{2}x+3{b}_{2}y=4{c}_{2}}\end{array}\right.$变形为$\left\{\begin{array}{l}{\frac{5}{4}{a}_{1}x+\frac{3}{4}{b}_{1}y={c}_{1}}\\{\frac{5}{4}{a}_{2}x+\frac{3}{4}{b}_{2}y={c}_{2}}\end{array}\right.$，
把$\frac{5}{4}$x和$\frac{3}{4}$y看出整体，根据题意$\left\{\begin{array}{l}{\frac{5}{4}x=4}\\{\frac{3}{4}y=5}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x=\frac{16}{5}}\\{y=\frac{20}{3}}\end{array}\right.$；

点评 本题考查了二元一次方程组的解的应用，方程变形和把$\frac{5}{4}$x和$\frac{3}{4}$y看出整体是本题的关键．