Question

7．二元一次方程組$\left\{\begin{array}{l}{x-2y=5}\\{5x+4y=-3}\end{array}\right.$的解是（　�。�

• A． $\left\{\begin{array}{l}{x=-1}\\{y=-3}\end{array}\right.$
• B． $\left\{\begin{array}{l}{x=-\frac{7}{5}}\\{y=1}\end{array}\right.$
• C． $\left\{\begin{array}{l}{x=1}\\{y=-2}\end{array}\right.$
• D． $\left\{\begin{array}{l}{x=2}\\{y=-\frac{3}{2}}\end{array}\right.$

Answer 1

分析 利用加減消元法求出方程組的解即可．

解答 解：$\left\{\begin{array}{l}{x-2y=5①}\\{5x+4y=-3②}\end{array}\right.$，
①×2+②得：7x=7，
解得：x=1，
把x=1代入①得：y=-2，
則方程組的解為$\left\{\begin{array}{l}{x=1}\\{y=-2}\end{array}\right.$，
故選C

點(diǎn)評(píng) 此題考查了二元一次方程組的解，方程組的解即為能使方程組中兩方程都成立的未知數(shù)的值．