《2025年一遍过七年级数学上册沪科版》

1 [2025 邢台信都区段考]用加减法解方程组$\left\{\begin{array}{l} x+y= -3,①\\ 3x+y= 6,②\end{array} \right. $由② - ①消去未知数 y,所得到的一元一次方程是( )

A.$2x= 9$
B.$2x= 3$
C.$4x= 9$
D.$4x= 3$

答案： A

2 二元一次方程组$\left\{\begin{array}{l} 3x+2y= 9,\\ x-2y= 3\end{array} \right. $的解是( )

A.$\left\{\begin{array}{l} x= 1,\\ y= -1\end{array} \right. $
B.$\left\{\begin{array}{l} x= 3,\\ y= -1\end{array} \right. $
C.$\left\{\begin{array}{l} x= 5,\\ y= -3\end{array} \right. $
D.$\left\{\begin{array}{l} x= 3,\\ y= 0\end{array} \right. $

答案： D $\left\{\begin{array}{l} 3x+2y=9,①\\ x-2y=3,②\end{array}\right. $①+②,得$4x=12$,解得$x=3$.把$x=3$代入②,得$3-2y=3$,解得$y=0$.所以$\left\{\begin{array}{l} x=3,\\ y=0.\end{array}\right. $

3 用加减法解方程组$\left\{\begin{array}{l} 2x+y= 8①,\\ x-y= 1②\end{array} \right. $的步骤如下：
(1) ① + ②,得$3x= 9$,解得$x= 3$.
(2) ① - ② × 2,得$3y= 6$,解得$y= 2$.
所以$\left\{\begin{array}{l} x= 3,\\ y= 2.\end{array} \right. $
则下列说法正确的是( )
A. 步骤(1)(2)都不对
B. 步骤(1)(2)都对
C. 本题不适宜用加减法
D. 加减法不能用两次

答案： B

4 [2024 合肥段考]用加减法解二元一次方程组$\left\{\begin{array}{l} x-y= 5①,\\ 2x-3y= 3②\end{array} \right. $时,下列方法中能消元的是( )

A.① × 2 + ②
B.① × (-2) - ②
C.① × 3 + ②
D.① × (-3) + ②

答案： D ①×2+②,得$4x-5y=13$,不能消元;①×(-2)-②,得$-4x+5y=-13$,不能消元;①×3+②,得$5x-6y=18$,不能消元;①×(-3)+②,得$-x=-12$,能消元.

5 一题多解 若$a+2b= 8,3a+4b= 18$,则$a+b$的值为____.

答案： 5 通解 由题意,得$\left\{\begin{array}{l} a+2b=8,①\\ 3a+4b=18,②\end{array}\right. $①×3-②,得$2b=6$,解得$b=3$.把$b=3$代入①,得$a+6=8$,解得$a=2$.所以$a+b=5.$优解 由题意,得$\left\{\begin{array}{l} a+2b=8,①\\ 3a+4b=18,②\end{array}\right. $②-①,得$2a+2b=10$.所以$a+b=5.$

6 利用加减法解下列方程组：
(1)$\left\{\begin{array}{l} x-y= 2,\\ 2x+y= 7;\end{array} \right. $
(2)$\left\{\begin{array}{l} 4x+8y= 12,\\ 3x-2y= 5;\end{array} \right. $
(3)$\left\{\begin{array}{l} 3(x-1)= y+5,\\ 5(y+1)= 3(x-5);\end{array} \right. $
(4)$\left\{\begin{array}{l} \frac {x}{3}-\frac {y}{4}= 1,\\ 3x-4y= 2.\end{array} \right. $

答案： 6 解:
(1)$\left\{\begin{array}{l} x-y=2,①\\ 2x+y=7.②\end{array}\right. $①+②,得$3x=9$,解得$x=3$.将$x=3$代入②,得$2×3+y=7$,解得$y=1$.所以$\left\{\begin{array}{l} x=3,\\ y=1.\end{array}\right. $
(2)$\left\{\begin{array}{l} 4x+8y=12,①\\ 3x-2y=5.②\end{array}\right. $②×4,得$12x-8y=20$.③①+③,得$16x=32$,解得$x=2$.将$x=2$代入②,得$6-2y=5$,解得$y=\frac {1}{2}$.所以$\left\{\begin{array}{l} x=2,\\ y=\frac {1}{2}.\end{array}\right. $
(3)将原方程组整理,得$\left\{\begin{array}{l} 3x-y=8,①\\ 3x-5y=20.②\end{array}\right. $①-②,得$4y=-12$,解得$y=-3$.把$y=-3$代入①,得$3x+3=8$,解得$x=\frac {5}{3}$.所以$\left\{\begin{array}{l} x=\frac {5}{3},\\ y=-3.\end{array}\right. $
(4)$\left\{\begin{array}{l} \frac {x}{3}-\frac {y}{4}=1,①\\ 3x-4y=2.②\end{array}\right. $由①,得$4x-3y=12$.③②×3,得$9x-12y=6$.④③×4,得$16x-12y=48$.⑤⑤-④,得$7x=42$,解得$x=6$.将$x=6$代入②,得$3×6-4y=2$,解得$y=4$.所以$\left\{\begin{array}{l} x=6,\\ y=4.\end{array}\right. $

7 [2025 西安长安区段考]现定义某种运算“★”,对给定的两个有理数 a,b 有$a★b= 2a-b$.
(1) 求$(-2)★(-4)$的值;
(2) 若$x★3y= -4,2x★y= 2$,求$x★y$的值.

答案： 7 解:
(1)由题意,得$(-2)\ast (-4)=2×(-2)-(-4)=0.$
(2)由$x\ast 3y=-4,2x\ast y=2$,得$2x-3y=-4,4x-y=2$.即解方程组$\left\{\begin{array}{l} 2x-3y=-4,①\\ 4x-y=2,②\end{array}\right. $②×3-①,得$10x=10$,解得$x=1$.把$x=1$代入②,得$4-y=2$,解得$y=2$.所以原方程组的解为$\left\{\begin{array}{l} x=1,\\ y=2.\end{array}\right. $所以$x\ast y=1\ast 2=1×2-2=0.$