18.阅读探索
(1)知识累计
解方程组$\left\{\begin{array}{l}{(a-1)+2(b+2)=6}\\{2(a-1)+(b+2)=6}\end{array}\right.$
解:设a-1=x,b+2=y,原方程组可变为$\left\{\begin{array}{l}{x+2y=6}\\{2x+y=6}\end{array}\right.$
解方程组得:$\left\{\begin{array}{l}{x=2}\\{y=2}\end{array}\right.$ 即$\left\{\begin{array}{l}{a-1=2}\\{b+2=2}\end{array}\right.$
所以$\left\{\begin{array}{l}{a=3}\\{b=0}\end{array}\right.$
此种解方程组的方法叫换元法.
(2)拓展提高
运用上述方法解下列方程组:$\left\{\begin{array}{l}{(\frac{a}{3}-1)+2(\frac{b}{5}+2)=4}\\{2(\frac{a}{3}-1)+(\frac{b}{5}+2)=5}\end{array}\right.$
(3)能力运用
已知关于x,y的方程组$\left\{\begin{array}{l}{{a}_{1}x+{b}_{1}y={c}_{1,}}\\{{a}_{2}x+{b}_{2}y={c}_{2}}\end{array}\right.$的解为$\left\{\begin{array}{l}{x=5}\\{y=3}\end{array}\right.$,直接写出关于m、n的方程组$\left\{\begin{array}{l}{5{a}_{1}(m+3)+3{b}_{1}(n-2)={c}_{1}}\\{5{a}_{2}(m+3)+3{b}_{2}(n-2)={c}_{2}}\end{array}\right.$的解为$\left\{\begin{array}{l}{m=}\\{n=}\end{array}\right.$${\;}_{3}^{-2}$.