题目内容

19.二元一次方程组$\left\{\begin{array}{l}{x-y=-3}\\{2x+y=0}\end{array}\right.$的解是(  )
A.$\left\{\begin{array}{l}{x=1}\\{y=4}\end{array}\right.$B.$\left\{\begin{array}{l}{x=-1}\\{y=2}\end{array}\right.$C.$\left\{\begin{array}{l}{x=-1}\\{y=4}\end{array}\right.$D.$\left\{\begin{array}{l}{x=3}\\{y=6}\end{array}\right.$

分析 运用加减消元法,两式相加消去y,求出x的值,把x的值代入①求出y的值,得到方程组的解.

解答 解:$\left\{\begin{array}{l}{x-y=-3①}\\{2x+=0②}\end{array}\right.$,
①+②得:3x=-3,即x=-1,
把x=-1代入①得:y=2,
则方程组的解为$\left\{\begin{array}{l}{x=-1}\\{y=2}\end{array}\right.$,
故选:B.

点评 此题考查了解二元一次方程组,利用了消元的思想,掌握加减消元法的步骤是解题的关键.

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