5.在学过了二元一次方程组的解法后,课堂上老师又写出了一个题目:$\left\{\begin{array}{l}{\frac{x+y}{6}+\frac{x-y}{10}=3①}\\{\frac{x+y}{6}-\frac{x-y}{10}=-1②}\end{array}\right.$,你会解这个方程组吗?
小明、小刚、小芳争论了一会儿,他们分别写出了一种方法:
小明:把原方程组整理得$\left\{\begin{array}{l}{8x+2y=90③}\\{2x+8y=-30④}\end{array}\right.$
④×4-③得30y=-210,所以y=-7
把y=-7代入③得8x=104,所以x=13,
即$\left\{\begin{array}{l}{x=13}\\{y=-7}\end{array}\right.$
小刚:设$\frac{x+y}{6}$=m,$\frac{x-y}{10}$=n,则$\left\{\begin{array}{l}{m+n=3③}\\{m-n=-1④}\end{array}\right.$
③+④得m=1,
③-④得m=2,
即$\left\{\begin{array}{l}{\frac{x+y}{6}=1}\\{\frac{x-y}{10}=2}\end{array}\right.$,所以$\left\{\begin{array}{l}{x+y=6}\\{x-y=20}\end{array}\right.$,所以$\left\{\begin{array}{l}{x=13}\\{y=-7}\end{array}\right.$.
小芳:①+②得$\frac{2(x+y)}{6}$=2,即x+y=6.③
①-②得$\frac{2(x-y)}{10}$=4,即x-y=20.④
③④组成方程组得x=13
③-④得y=-7,即$\left\{\begin{array}{l}{x=13}\\{y=-7}\end{array}\right.$.
老师看过后,非常高兴,特别是小刚的方法独特,像小刚的这种方法叫做换元法,你能用换元法解下列方程组吗?
$\left\{\begin{array}{l}{\frac{3x-2y}{6}+\frac{2x+3y}{7}=1}\\{\frac{3x-2y}{6}-\frac{2x+3y}{7}=5}\end{array}\right.$.