《2025年天府前沿八年级数学上册北师大版》

解方程组$\left\{\begin{array}{l} x+2y= 5,①\\ 3x-4y= -5.②\end{array} \right.$
解：由①可得$x=$$5 - 2y$.③
将③代入②,得$3(5 - 2y) - 4y = - 5$.
解得$y=$$2$.
把$y = 2$代入③,得$x=$$5 - 2×2$.
$\therefore x=$$1$.
$\therefore$原方程组的解为$\begin{cases}x = 1,\\y = 2\end{cases}$.

答案： $5 - 2y$ $3(5 - 2y) - 4y = - 5$ $2$ $y = 2$ $5 - 2×2$ $1$ $\begin{cases}x = 1,\\y = 2\end{cases}$

1. (1)(2025·编写)将方程$3y-x= 2$变形成用含$y$的代数式表示$x$,则$x=$$3y - 2$.
(2)(2025·编写)已知$x-2y-1= 0$,用含$x$的代数式表示$y$,则$y=$$\frac{x - 1}{2}$.

答案：
(1)$3y - 2$
(2)$\frac{x - 1}{2}$

2. (1)(2025·编写)若$\left\{\begin{array}{l} x= 2,\\ y= 1\end{array} \right.$是关于x,$y的方程x+ay= 3$的解,则$a$的值为____1____.
(2)(2025·编写)方程组$\left\{\begin{array}{l} x-y= 1,\\ 2x+y= 5\end{array} \right.$的解是____$\begin{cases}x = 2,\\y = 1\end{cases}$____.

答案：
(1)$1$
(2)$\begin{cases}x = 2,\\y = 1\end{cases}$

3. (1)(2025·编写)已知$\left\{\begin{array}{l} x= -1,\\ y= 3\end{array} \right.$是方程组$\left\{\begin{array}{l} ax-2y= 1,\\ x+by= 5\end{array} \right.$的解,则$a+b=$$-5$.
(2)(2025·编写)方程组$\left\{\begin{array}{l} x+y= 10,\\ 2x+y= 16\end{array} \right.$的解是$\begin{cases}x = 6,\\y = 4\end{cases}$.

答案：
(1)$- 5$
(2)$\begin{cases}x = 6,\\y = 4\end{cases}$

4. (1)(2025·编写)已知方程组$\left\{\begin{array}{l} 2x+y= 3,\\ x-2y= 5,\end{array} \right.则2x+6y$的值是$-4$.
(2)(2025·编写)同时满足二元一次方程$x-y= 9和4x+3y= 1的x= $$4$,$y= $$-5$.

答案：
(1)$- 4$
(2)$4$ $- 5$