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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