Question

16．方程組$\left\{\begin{array}{l}{3x-4y=1}\\{x-4y=7}\end{array}\right.$的解是$\left\{\begin{array}{l}{x=-3}\\{y=-2.5}\end{array}\right.$．

Answer 1

分析 方程組利用加減消元法求出解即可．

解答 解：$\left\{\begin{array}{l}{3x-4y=1①}\\{x-4y=7②}\end{array}\right.$，
①-②得：2x=-6，即x=-3，
把x=-3代入②得：y=-2.5，
則方程組的解為$\left\{\begin{array}{l}{x=-3}\\{y=-2.5}\end{array}\right.$．
故答案為：$\left\{\begin{array}{l}{x=-3}\\{y=-2.5}\end{array}\right.$．

點評 此題考查了解二元一次方程組，利用了消元的思想，消元的方法有：代入消元法與加減消元法．