题目若方程组{a1x+b1y=c1a2x+b2y=c2\begin{cases}a_{1}x+b_{1}y=c_{1}\\a_{2}x+b_{2}y=c_{2}\end{cases}{a1x+b1y=c1a2x+b2y=c2的解是{x=3y=2\begin{cases}x=3\\y=2\end{cases}{x=3y=2求方程组{56a1(x−1)+13b1(y+2)=c1,56a2(x−1)+13b2(y+2)=c2\begin{cases}\dfrac{5}{6}a_{1}(x-1)+\dfrac{1}{3}b_{1}(y+2)=c_{1},\\\dfrac{5}{6}a_{2}(x-1)+\dfrac{1}{3}b_{2}(y+2)=c_{2}\end{cases}⎩⎨⎧65a1(x−1)+31b1(y+2)=c1,65a2(x−1)+31b2(y+2)=c2的解.知识点：二元一次方程组的解、同解方程组章节：第5章 二元一次方程组 / 5.2 求解二元一次方程组
