8. (1)若二元一次方程组$\begin{cases}4x + 3y = 14, \\ax + (a + 2)y = \frac{4}{7}\end{cases}$的解$x$和$y$相等，则$a =$ ______ .
(2)已知关于$x$，$y$的方程组$\begin{cases}3x + my = 0, \\x - y = 4\end{cases}$的解是$\begin{cases}x = 2, \\y = ■\end{cases}$，其中$y$的值被盖住了，但仍能求出$m$的值，则$m =$ ______ .
(2)已知关于$x$，$y$的方程组$\begin{cases}3x + my = 0, \\x - y = 4\end{cases}$的解是$\begin{cases}x = 2, \\y = ■\end{cases}$，其中$y$的值被盖住了，但仍能求出$m$的值，则$m =$ ______ .
答案：8. (1)$-\frac {6}{7}$解析：因为$x,y$的值相等，所以$4x+3x=14$，解得$x=2$，则$y=2$.所以$2a+2a+4=\frac {4}{7}$，解得$a=-\frac {6}{7}$.
(2)3 解析：把$x=2$代入$x-y=4$中，得$y=-2$，把$\{\begin{array}{l} x=2,\\ y=-2\end{array} $代入$3x+my=0$中，得$6-2m=0$，解得$m=3$.
(2)3 解析：把$x=2$代入$x-y=4$中，得$y=-2$，把$\{\begin{array}{l} x=2,\\ y=-2\end{array} $代入$3x+my=0$中，得$6-2m=0$，解得$m=3$.
9. 甲、乙两人解关于$x$，$y$的方程组$\begin{cases}3x - by = -1, &① \\ax + by = -5 &②\end{cases}$时，甲因看错$a$得到方程组的解为$\begin{cases}x = 1, \\y = 2,\end{cases}$乙将方程②中的$b$写成了它的相反数得到方程组的解为$\begin{cases}x = -1, \\y = -1,\end{cases}$求$a$，$b$的值.
答案：9. 因为甲看错方程组$\{\begin{array}{l} 3x-by=-1,①\\ ax+by=-5②\end{array} $中的$a$，得到方程组的解为$\{\begin{array}{l} x=1,\\ y=2,\end{array} $所以将$\{\begin{array}{l} x=1,\\ y=2\end{array} $代入①，得$3-2b=-1$，所以$b=2$.因为乙把方程②中的$b$看成了它的相反数，得到方程组的解为$\{\begin{array}{l} x=-1,\\ y=-1,\end{array} $所以将$\{\begin{array}{l} x=-1,\\ y=-1\end{array} $代入$ax-2y=-5$中，得$a=7$.
解析：
解：因为甲看错$a$，得到解$\begin{cases}x = 1 \\ y = 2\end{cases}$，将其代入①得：$3×1 - b×2=-1$，即$3 - 2b=-1$，解得$b = 2$。
乙将②中$b$写成相反数，即$-b=-2$，此时方程②为$ax - 2y=-5$，将$\begin{cases}x = -1 \\ y = -1\end{cases}$代入得：$a×(-1)-2×(-1)=-5$，即$-a + 2=-5$，解得$a = 7$。
综上，$a = 7$，$b = 2$。
乙将②中$b$写成相反数，即$-b=-2$，此时方程②为$ax - 2y=-5$，将$\begin{cases}x = -1 \\ y = -1\end{cases}$代入得：$a×(-1)-2×(-1)=-5$，即$-a + 2=-5$，解得$a = 7$。
综上，$a = 7$，$b = 2$。
10. 阅读下面的内容，利用整体思想通过换元的方法来解决问题.
题目：已知方程组$\begin{cases}a_{1}x + b_{1}y = c_{1}, &① \\a_{2}x + b_{2}y = c_{2} &②\end{cases}$的解是$\begin{cases}x = 4, \\y = 6,\end{cases}$求方程组$\begin{cases}2a_{1}x + 3b_{1}y = c_{1}, &② \\2a_{2}x + 3b_{2}y = c_{2} &②\end{cases}$的解.
解：方程组$\begin{cases}2a_{1}x + 3b_{1}y = c_{1}, &② \\2a_{2}x + 3b_{2}y = c_{2} &②\end{cases}$可以变形为方程组$\begin{cases}a_{1} · 2x + b_{1} · 3y = c_{1}, &③ \\a_{2} · 2x + b_{2} · 3y = c_{2} &③\end{cases}$
设$2x = m$，$3y = n$，则方程组③可化为$\begin{cases}a_{1}m + b_{1}n = c_{1}, &④ \\a_{2}m + b_{2}n = c_{2} &④\end{cases}$
比较方程组④与方程组①可得$\begin{cases}m = 4, \= 6,\end{cases}$即$\begin{cases}2x = 4, \\3y = 6,\end{cases}$所以方程组②的解为$\begin{cases}x = 2, \\y = 2.\end{cases}$
参考上述方法，解决下列问题：
(1)若方程组$\begin{cases}5x - 2y = 4, \\2x - 3y = -5\end{cases}$的解是$\begin{cases}x = 2, \\y = 3,\end{cases}$则方程组$\begin{cases}5(x + 1) - 2(y - 2) = 4, \\2(x + 1) - 3(y - 2) = -5\end{cases}$的解为 ______ ；
(2)若方程组$\begin{cases}a_{1}x + b_{1}y = c_{1}, \\a_{2}x + b_{2}y = c_{2}\end{cases}$的解是$\begin{cases}x = -1, \\y = 3,\end{cases}$则方程组$\begin{cases}a_{1}(x - 2) + 2b_{1}y = c_{1}, \\a_{2}(x - 2) + 2b_{2}y = c_{2}\end{cases}$的解为 ______ ；
(3)若方程组$\begin{cases}a_{1}x + b_{1}y = c_{1}, \\a_{2}x + b_{2}y = c_{2}\end{cases}$的解为$\begin{cases}x = 4, \\y = 6,\end{cases}$求方程组$\begin{cases}4a_{1}x + 3b_{1}y = 5c_{1}, \\4a_{2}x + 3b_{2}y = 5c_{2}\end{cases}$的解.
题目：已知方程组$\begin{cases}a_{1}x + b_{1}y = c_{1}, &① \\a_{2}x + b_{2}y = c_{2} &②\end{cases}$的解是$\begin{cases}x = 4, \\y = 6,\end{cases}$求方程组$\begin{cases}2a_{1}x + 3b_{1}y = c_{1}, &② \\2a_{2}x + 3b_{2}y = c_{2} &②\end{cases}$的解.
解：方程组$\begin{cases}2a_{1}x + 3b_{1}y = c_{1}, &② \\2a_{2}x + 3b_{2}y = c_{2} &②\end{cases}$可以变形为方程组$\begin{cases}a_{1} · 2x + b_{1} · 3y = c_{1}, &③ \\a_{2} · 2x + b_{2} · 3y = c_{2} &③\end{cases}$
设$2x = m$，$3y = n$，则方程组③可化为$\begin{cases}a_{1}m + b_{1}n = c_{1}, &④ \\a_{2}m + b_{2}n = c_{2} &④\end{cases}$
比较方程组④与方程组①可得$\begin{cases}m = 4, \= 6,\end{cases}$即$\begin{cases}2x = 4, \\3y = 6,\end{cases}$所以方程组②的解为$\begin{cases}x = 2, \\y = 2.\end{cases}$
参考上述方法，解决下列问题：
(1)若方程组$\begin{cases}5x - 2y = 4, \\2x - 3y = -5\end{cases}$的解是$\begin{cases}x = 2, \\y = 3,\end{cases}$则方程组$\begin{cases}5(x + 1) - 2(y - 2) = 4, \\2(x + 1) - 3(y - 2) = -5\end{cases}$的解为 ______ ；
(2)若方程组$\begin{cases}a_{1}x + b_{1}y = c_{1}, \\a_{2}x + b_{2}y = c_{2}\end{cases}$的解是$\begin{cases}x = -1, \\y = 3,\end{cases}$则方程组$\begin{cases}a_{1}(x - 2) + 2b_{1}y = c_{1}, \\a_{2}(x - 2) + 2b_{2}y = c_{2}\end{cases}$的解为 ______ ；
(3)若方程组$\begin{cases}a_{1}x + b_{1}y = c_{1}, \\a_{2}x + b_{2}y = c_{2}\end{cases}$的解为$\begin{cases}x = 4, \\y = 6,\end{cases}$求方程组$\begin{cases}4a_{1}x + 3b_{1}y = 5c_{1}, \\4a_{2}x + 3b_{2}y = 5c_{2}\end{cases}$的解.
答案：10. (1)$\{\begin{array}{l} x=1,\\ y=5\end{array} $解析：由题意可得$\{\begin{array}{l} x+1=2,\\ y-2=3,\end{array} $得方程组的解为$\{\begin{array}{l} x=1,\\ y=5.\end{array} $
(2)$\{\begin{array}{l} x=1,\\ y=\frac {3}{2}\end{array} $解析：由题意可得$\{\begin{array}{l} x-2=-1,\\ 2y=3,\end{array} $得方程组的解为$\{\begin{array}{l} x=1,\\ y=\frac {3}{2}.\end{array} $
(3)因为方程组$\{\begin{array}{l} a_{1}x+b_{1}y=c_{1},\\ a_{2}x+b_{2}y=c_{2}\end{array} $的解为$\{\begin{array}{l} x=4,\\ y=6,\end{array} $又因为$\{\begin{array}{l} 4a_{1}x+3b_{1}y=5c_{1},\\ 4a_{2}x+3b_{2}y=5c_{2},\end{array} $所以$\{\begin{array}{l} \frac {4}{5}a_{1}x+\frac {3}{5}b_{1}y=c_{1},\\ \frac {4}{5}a_{2}x+\frac {3}{5}b_{2}y=c_{2},\end{array} $所以$\{\begin{array}{l} \frac {4}{5}x=4,\\ \frac {3}{5}y=6,\end{array} $解得$\{\begin{array}{l} x=5,\\ y=10.\end{array} $
(2)$\{\begin{array}{l} x=1,\\ y=\frac {3}{2}\end{array} $解析：由题意可得$\{\begin{array}{l} x-2=-1,\\ 2y=3,\end{array} $得方程组的解为$\{\begin{array}{l} x=1,\\ y=\frac {3}{2}.\end{array} $
(3)因为方程组$\{\begin{array}{l} a_{1}x+b_{1}y=c_{1},\\ a_{2}x+b_{2}y=c_{2}\end{array} $的解为$\{\begin{array}{l} x=4,\\ y=6,\end{array} $又因为$\{\begin{array}{l} 4a_{1}x+3b_{1}y=5c_{1},\\ 4a_{2}x+3b_{2}y=5c_{2},\end{array} $所以$\{\begin{array}{l} \frac {4}{5}a_{1}x+\frac {3}{5}b_{1}y=c_{1},\\ \frac {4}{5}a_{2}x+\frac {3}{5}b_{2}y=c_{2},\end{array} $所以$\{\begin{array}{l} \frac {4}{5}x=4,\\ \frac {3}{5}y=6,\end{array} $解得$\{\begin{array}{l} x=5,\\ y=10.\end{array} $