典例1 (2023·南京秦淮模拟)若关于x,y的方程组$\left\{\begin{array}{l}2x+y=4,\\ x+2y=m\end{array}\right.$的解满足x+y=1,则m的值为______.
答案
典例1 -1.
典例2 小明和小玲比赛解方程组$\left\{\begin{array}{l}ax+by=2,\\ cx-3y=-2,\end{array}\right.$小玲很细心地算得此方程组的解为$\left\{\begin{array}{l}x=1,\\ y=-1,\end{array}\right.$小明抄错了c解得$\left\{\begin{array}{l}x=2,\\ y=-6.\end{array}\right.$求a,b,c的值.
答案
典例2 把$\begin{cases}x = 1,\\y = -1\end{cases}$代入原方程组,得$\begin{cases}a - b = 2 ①,\\c = -5.\end{cases}$ $\because$小明抄错了$c$解得$\begin{cases}x = 2,\\y = -6\end{cases}$,$\therefore$把$\begin{cases}x = 2,\\y = -6\end{cases}$代入$ax + by = 2$,得$2a - 6b = 2$②. 联立①②,得$\begin{cases}a - b = 2,\\2a - 6b = 2,\end{cases}$解得$\begin{cases}a = \frac{5}{2},\\b = \frac{1}{2}.\end{cases}$ $\therefore a = \frac{5}{2},b = \frac{1}{2},c = -5$.
典例3 已知关于x,y的方程组$\left\{\begin{array}{l}2x-3y=8,\\ (1-2m)x+2y=1-n\end{array}\right.$与关于x,y的方程组$\left\{\begin{array}{l}x-2y=5,\\ nx+y=m+1\end{array}\right.$有相同的解,求m,n的值.
答案
典例3 $\because$关于$x,y$的方程组$\begin{cases}2x - 3y = 8,\\(1 - 2m)x + 2y = 1 - n\end{cases}$与关于$x,y$的方程组$\begin{cases}x - 2y = 5,\\nx + y = m + 1\end{cases}$有相同的解,$\therefore$这两个方程组的解也是方程组$\begin{cases}2x - 3y = 8,\\x - 2y = 5\end{cases}$的解. 解方程组$\begin{cases}2x - 3y = 8,\\x - 2y = 5,\end{cases}$得$\begin{cases}x = 1,\\y = -2.\end{cases}$把$\begin{cases}x = 1,\\y = -2\end{cases}$代入$\begin{cases}(1 - 2m)x + 2y = 1 - n,\\nx + y = m + 1,\end{cases}$整理,得$\begin{cases}2m - n = -4,\\m - n = -3,\end{cases}$解得$\begin{cases}m = -1,\\n = 2.\end{cases}$
典例4 若关于x的不等式3x - m≤0恰有3个正整数解,则m的取值不可以是 ( )
A. 9
B. 10
C. 11
D. 12
A. 9
B. 10
C. 11
D. 12
答案
典例4 D.
典例5 已知关于x的不等式$\frac{x}{a}$<7的解也是关于x的不等式$\frac{2x - 7a}{5}$>$\frac{a}{2}$-1的解,则常数a的取值范围是__________.
答案
典例5 $-\frac{10}{9}\leqslant a < 0$.
典例6 已知关于x的不等式$\frac{2m - mx}{2}$>$\frac{1}{2}x - 1$.
(1) 当m=1时,求该不等式的非负整数解.
(2) 当m取何值时,该不等式有解? 请求出其解集.
(1) 当m=1时,求该不等式的非负整数解.
(2) 当m取何值时,该不等式有解? 请求出其解集.
答案
典例6 (1)当$m = 1$时,原不等式可化为$\frac{2 - x}{2}>\frac{1}{2}x - 1$,解得$x < 2$. $\therefore$该不等式的非负整数解为$0,1$. (2) 对于不等式$\frac{2m - mx}{2}>\frac{1}{2}x - 1$,去分母,得$2m - mx > x - 2$. 整理,得$(m + 1)x < 2(m + 1)$. $\therefore$当$m\neq -1$时,该不等式有解,且当$m > -1$时,该不等式的解集为$x < 2$;当$m < -1$时,该不等式的解集为$x > 2$.
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