题目内容
9.下列各组数既是方程3x-2y=4的解,又是2x+3y=7的解是( )| A. | $\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{x=1}\\{y=-\frac{1}{2}}\end{array}\right.$ | C. | $\left\{\begin{array}{l}{x=0}\\{y=-2}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{x=-1}\\{y=\frac{3}{2}}\end{array}\right.$ |
分析 联立已知两方程,求出方程组的解即可.
解答 解:联立得:$\left\{\begin{array}{l}{3x-2y=4①}\\{2x+3y=7②}\end{array}\right.$,
①×3+②×2得:13x=26,即x=2,
把x=2代入①得:y=1,
则方程组的解为$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$,
故选A
点评 此题考查了二元一次方程的解,熟练掌握运算法则是解本题的关键.
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