题目内容

20.用加减法解下列方程组:
(1)$\left\{\begin{array}{l}{3m+b=11}\\{-4m-b=11}\end{array}\right.$;(2)$\left\{\begin{array}{l}{0.6x-0.4y=1.1}\\{0.2x-0.4y=2.3}\end{array}\right.$.

分析 (1)利用加减法,即可解答;
(2)利用加减法,即可解答.

解答 解:(1)$\left\{\begin{array}{l}{3m+b=11①}\\{-4m-b=11②}\end{array}\right.$
①+②得:-m=22,
解得:m=-22.
把m=-22代入①得:-66+b=11,
解得:b=77,
∴方程组的解为:$\left\{\begin{array}{l}{m=-22}\\{b=77}\end{array}\right.$.
(2)$\left\{\begin{array}{l}{0.6x-0.4x=1.1①}\\{0.2x-0.4y=2.3②}\end{array}\right.$
 ①-②得:0.4x=-1.2,
解得:x=-3,
把x=-3代入①得:-1.8-0.4x=1.1,
解得:y=7.25
∴方程组的解为:$\left\{\begin{array}{l}{x=-3}\\{y=7.25}\end{array}\right.$.

点评 本题考查了解二元一次方程组,解决本题的关键是利用加减法解答.

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