{l}{axgeq2}{sqrt{x}-1>a}{3xleq2a+11} सोडवा

$$\left\{ \begin{array} { l } { a x \geq 2 } \\ { \sqrt { x } - 1 > a } \\ { 3 x \leq 2 a + 11 } \end{array} \right.$$

x साठी सोडवा

$$\left\{\begin{matrix}x\in [\frac{2}{a},\frac{2a+11}{3}]\text{, }&\frac{2}{a}>(a+1)^{2}\text{ and }a\geq \frac{1}{2}\\x\in ((a+1)^{2},\frac{2a+11}{3}]\text{, }&a<\frac{2\sqrt{7}-2}{3}\text{ and }\frac{2}{a}<(a+1)^{2}\text{ and }a\geq \frac{1}{2}\\x\in ((a+1)^{2},\frac{2}{a}]\text{, }&(a+1)^{2}<\frac{2}{a}\text{ and }a>-6\text{ and }a<0\\x\in [0,(a+1)^{2})\text{, }&a\geq \frac{-2\sqrt{7}-2}{3}\text{ and }\frac{2}{a}\geq (a+1)^{2}\text{ and }a<-1\\x\in [0,\frac{2a+11}{3}]\text{, }&a\geq -\frac{11}{2}\text{ and }\frac{2}{a}\geq (a+1)^{2}\text{ and }a<\frac{-2\sqrt{7}-2}{3}\end{matrix}\right.$$

a साठी सोडवा

$$\left\{\begin{matrix}a\in [\frac{2}{x},\sqrt{x}-1)\text{, }&x\leq 4\text{ and }\frac{2}{x}<\sqrt{x}-1\text{ and }x>0\\a\in [\frac{3x-11}{2},\sqrt{x}-1)\text{, }&x<\frac{4\sqrt{7}+29}{9}\text{ and }\frac{2}{x}<\sqrt{x}-1\text{ and }x>4\end{matrix}\right.$$