题目内容

20.解下列方程组.
(1)$\left\{\begin{array}{l}{3(y-2)=x+1}\\{2(x-1)=5y-8}\end{array}\right.$                
(2)$\left\{\begin{array}{l}{4x-15y-17=0}\\{6x-25y-23=0}\end{array}\right.$.

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

解答 解:(1)方程组整理得:$\left\{\begin{array}{l}{x-3y=-7①}\\{2x-5y=-6②}\end{array}\right.$,
②-①×2得:y=8,
把y=8代入①得:x=17,
则方程组的解为$\left\{\begin{array}{l}{x=17}\\{y=8}\end{array}\right.$;
(2)方程组整理得:$\left\{\begin{array}{l}{4x-15y=17①}\\{6x-25y=23②}\end{array}\right.$,
①×3-②×2得:5y=5,即y=1,
把y=1代入①得:x=8,
则方程组的解为$\left\{\begin{array}{l}{x=8}\\{y=1}\end{array}\right.$.

点评 此题考查了解二元一次方程组,熟练掌握运算法则是解本题的关键.

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