Question

5．已知方程組$\left\{\begin{array}{l}{{a}_{1}x+y={c}_{1}}\\{{a}_{2}x+y={c}_{2}}\end{array}\right.$解為$\left\{\begin{array}{l}{x=5}\\{y=10}\end{array}\right.$，則關(guān)于x，y的方程組$\left\{\begin{array}{l}{3{a}_{1}x+2y={a}_{1}+{c}_{1}}\\{3{a}_{2}x+2y={a}_{2}+{c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=5}\end{array}\right.$．

Answer 1

分析 根據(jù)方程組解的定義，把x=5，y=10代入即可得出a₁，a₂，c₁，c₂的關(guān)系，再代入計算即可．

解答 解：∵方程組$\left\{\begin{array}{l}{{a}_{1}x+y={c}_{1}}\\{{a}_{2}x+y={c}_{2}}\end{array}\right.$解為$\left\{\begin{array}{l}{x=5}\\{y=10}\end{array}\right.$，
∴$\left\{\begin{array}{l}{5{a}_{1}+10={c}_{1}}\\{5{a}_{2}+10={c}_{2}}\end{array}\right.$，
∵$\left\{\begin{array}{l}{3{a}_{1}x+2y={a}_{1}+{c}_{1}}\\{3{a}_{2}x+2y={a}_{2}+{c}_{2}}\end{array}\right.$，
∴$\left\{\begin{array}{l}{3{a}_{1}x+2y=6{a}_{1}+10①}\\{3{a}_{2}x+2y=6{a}_{2}+10②}\end{array}\right.$，
①-②，得3a₁x-3a₂x=6a₁-6a₂，
∴x=2，
把x=2代入①得，y=5，
∴方程組$\left\{\begin{array}{l}{3{a}_{1}x+2y={a}_{1}+{c}_{1}}\\{3{a}_{2}x+2y={a}_{2}+{c}_{2}}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=2}\\{y=5}\end{array}\right.$，
故答案為$\left\{\begin{array}{l}{x=2}\\{y=5}\end{array}\right.$．

點評 本題考查了解二元一次方程組，掌握方程組的解法是解題的關(guān)鍵．