题目内容

19.解下列方程组
(1)$\left\{\begin{array}{l}{x+y=150}\\{0.1x+0.15y=19}\end{array}\right.$
(2)$\left\{\begin{array}{l}\frac{x+2}{3}+\frac{y-1}{2}=2\\ \frac{x+2}{3}+\frac{1-y}{2}=1\end{array}\right.$.

分析 (1)方程组整理后,利用加减消元法求出解即可;
(2)方程组利用换元法求出解即可.

解答 解:(1)方程组整理得:$\left\{\begin{array}{l}{x+y=150①}\\{10x+15y=1900②}\end{array}\right.$,
①×15-②得:5x=350,即x=70,
把x=70代入①得:y=80,
则方程组的解为$\left\{\begin{array}{l}{x=70}\\{y=80}\end{array}\right.$;
(2)设$\frac{x+2}{3}$=a,$\frac{y-1}{2}$=b,方程组变形为$\left\{\begin{array}{l}{a+b=2}\\{a-b=1}\end{array}\right.$,
解得:$\left\{\begin{array}{l}{a=1.5}\\{b=0.5}\end{array}\right.$,即$\left\{\begin{array}{l}{\frac{x+2}{3}=1.5}\\{\frac{y-1}{2}=0.5}\end{array}\right.$,
解得:$\left\{\begin{array}{l}{x=2.5}\\{y=2}\end{array}\right.$.

点评 此题考查了解二元一次方程组,利用了消元的思想,消元的方法有:代入消元法与加减消元法.

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