# Linear Inequalities

## Class 11 NCERT

### NCERT

1   Solve $24x < 100,$ when$\\$ (i) x is a natural number $\\$ (ii) x is an integer.

The given inequality is $24x < 100$$\\ 24x < 100 \\ \implies \dfrac{24x}{24} < \dfrac{100}{24} \\ [\text{Dividing both sides by same positive number}] \\ \implies x < \dfrac{25}{6}$$\\$ (i) It is evident that $1,2,3,$ and $4$ are the only natural numbers less than$\dfrac{25}{6}$$\\ Thus, when x is a natural number, the solutions of the given inequality are 1, 2, 3 and 4 \\ Hence, in this case, the solution set is \{1, 2,3, 4 \} .\\ (ii)The integers less than \dfrac{25}{6} are.....-3,-2,-1,0,1,2,3,4.$$\\$ Thus, when x is an integer , the solutions of the given inequality are ... $- 3, - 2, - 1,0,1, 2,3, 4$$\\ Hence, in this case, the solution set is \{- 3, - 2, - 1,0,1,2,3,4 \} 2 Solve - 12 x > 30, when \\ (i) x is a natural number \\ (ii) x is an integer ##### Solution : The given inequality is -12 x > 30.$$\\$ $-12x >30.\\ \implies \dfrac{-12x}{-12} < \dfrac{30}{-12}\\ [\text{Dividing both sides by same negative number}]\\ \implies x< -\dfrac{5}{2}$$\\ (i) There is no natural number less than (-\dfrac{5}{2}). Thus, when x is a natural number, there is no solution of the given inequality.\\ (ii)The integers less than (-\dfrac{5}{2}) are......-5,-4,-3.$$\\$ Thus, when x is an integer, the solutions of the given inequality are $...., - 5, - 4, - 3$$\\ Hence, in this case, the solution set is \{ ...., - 5, -4, -3 \}. 3 Solve 5 x - 3 < 7 , when \\(i) x is an integer\\ (ii) x is a real number ##### Solution : The given inequality is 5 x - 3 < 7 .$$\\$ $5 x - 3 < 7 \\ \implies 5x-3+3 < 7+3\\ \implies 5x < 10\\ \implies \dfrac{5x}{5} < \dfrac{10}{5}\\ \implies x < 2$$\\ (i) The integers less than 2 are ..., - 4, - 3, - 2, - 1,0,1 .$$\\$ Thus, when $x$ is an integer, the solutions of the given inequality are $..., -4, - 3, - 2, - 1,0,1 .$$\\ Hence, in this case, the solution set is \{..., - 4, - 3, - 2, - 1,0,1 \} .\\ (ii) When x is a real number, the solutions of the given inequality are given by x < 2, that is, all real numbers x which are less than 2.$$\\$ Thus, the solution set of the given inequality is $x \in (-\infty , 2 ).$

4   Solve $3 x + 8 > 2 ,$ when$\\$ (i) x is an integer $\\$ (ii) x is a real number

##### Solution :

The given inequality is $3 x + 8 > 2$$\\ 3 x + 8 > 2\\ \implies 3 x + 8 - 8 > 2 -8\\ \implies 3 x >- 6\\ \implies \dfrac{3x}{3}> \dfrac{-6}{3}\\ \implies x > -2$$\\$ (i) The integers greater than $- 2$ are $- 1,0,1, 2,...$$\\ Thus, when x is an integer, the solutions of the given inequality are - 1,0,1, 2,...$$\\$ Hence, in this case, the solution set is $\{ -1,0,1,2,... \}$ .$\\$ (ii) When $x$ is a real number, the solutions of the given inequality are all the real numbers, which are greater than $-2$. Thus, in this case, the solution set is $(-2, \infty)$ .

5   Solve the given inequality for real $x : 4 x + 3 < 5 x + 7$

$4 x + 3 < 5 x + 7\\ \implies 4 x + 3 - 7 < 5 x + 7 - 7 \\ \implies 4 x - 4 < 5 x\\ \implies 4 x - 4 - 4 x < 5 x - 4 x\\ \implies -4 < x $$\\Thus, all real numbers x, which are greater than - 4 , are the solutions of the given inequality. Hence, the solution set of the given inequality is (-4, \infty). 6 Solve 24 x < 100 , when (i) x is a natural number (ii) x is an integer. ##### Solution : The given inequality is 24 x < 100 \\ 24 x < 100 \\ \Rightarrow \dfrac{24 x}{24} < \dfrac{100}{24 } \quad [\text{Dividing both sides by same positive number}]\\ \Rightarrow x < \dfrac{25}{6}$$\\$ (i) It is evident that $1,2,3$ and $4$ are the only natural numbers less than $\dfrac{25}{6}$$\\ Thus, when x is a natural number, the solutions of the given inequality are 1, 2, 3 and 4$$\\$ Hence, in this case, the solution set is $(1, 2,3, 4 )$ .$\\$ (ii) The integers less than $\dfrac{25}{6}$ are......$-3,-2,-1,0,1,2,3,4$$\\ Thus, when x is an integer , the solutions of the given inequality are ... - 3, - 2, -1,0,1, 2,3, 4$$\\$ Hence, in this case, the solution set is $\{-3, - 2, -1,0,1,2,3,4 \}$

7   Solve $-12 x > 30$, when$\\$ (i) x is a natural number $\\$ (ii) x is an integer

The given inequality is $-12 x > 30.$$\\ -12x > 30\\ \Rightarrow \dfrac{-12x}{-12} < \dfrac{30}{-12} [\text{Dividing both sides by same negative number}]\\ \Rightarrow x < -\dfrac{5}{2}$$\\$ (i) There is no natural number less than $(-\dfrac{5}{2})$$\\ Thus, when x is a natural number, there is no solution of the given inequality.\\ (ii) The integers less than (-\dfrac{5}{2}) are .... -5,-4,-3.$$\\$ Thus, when $x$ is an integer, the solutions of the given inequality are ....,$- 5, - 4, - 3$$\\ Hence, in this case, the solution set is \{...., - 5, - 4, - 3 \} . 8 Solve 5 x -3 < 7 , when\\ (i) x is an integer\\ (ii) x is a real number ##### Solution : The given inequality is 5 x - 3 < 7$$\\$ $5x-3 < 7\\ \Rightarrow 5x -3+3 < 7+3\\ \Rightarrow 5x < 10\\ \Rightarrow \dfrac{5x}{5} < \dfrac{10}{5}\\ \Rightarrow x< 2$$\\ (i) The integers less than 2 are ..., -4, - 3, -2, - 1,0,1 .$$\\$ Thus, when x is an integer, the solutions of the given inequality are ..., $- 4, - 3, - 2, - 1,0,1 .$$\\ Hence, in this case, the solution set is \{ ..., - 4, - 3, - 2, - 1,0,1 \}$$\\$ . (ii) When $x$ is a real number, the solutions of the given inequality are given by $x < 2,$ that is, all real numbers x which are less than $2$.$\\$ Thus, the solution set of the given inequality is $x \in (- \infty, 2 )$ .

9   Solve $3 x + 8 > 2$ , when$\\$ (i) x is an integer$\\$ (ii) x is a real number

##### Solution :

The given inequality is $3 x + 8 > 2$$\\ 3x+8 > 2\\ \Rightarrow 3x+8-8 > 2-8\\ \Rightarrow 3x > -6\\ \Rightarrow \dfrac{3x}{3} > \dfrac{-6}{3}\\ \Rightarrow x > -2$$\\$ (i) The integers greater than $-2$ are $-1,0,1, 2,...$$\\ Thus, when x is an integer, the solutions of the given inequality are - 1,0,1, 2,...$$\\$ Hence, in this case, the solution set is $\{-1,0,1,2,... \}$ .$\\$ (ii) When $x$ is a real number, the solutions of the given inequality are all the real numbers, which are greater than $- 2$.$\\$ Thus, in this case, the solution set is $(-2, \infty)$ .

10   Solve the given inequality for real $x : 4 x + 3 < 5 x + 7$

$4x+3 < 5x +7\\ \Rightarrow 4x+3-7 < 5x +7 -7\\ \Rightarrow 4x -4 < 5x\\ \Rightarrow 4x - 4 - 4x < 5x -4x \\ \Rightarrow -4 < x $$\\ Thus, all real numbers x, which are greater than -4 , are the solutions of the given inequality.\\ Hence, the solution set of the given inequality is (-4, \infty ) . 11 Solve the given inequality for real x : 3 x - 7 > 5 x - 1 ##### Solution : 3 x - 7 > 5 x - 1\\ 3 x - 7 + 7 > 5 x - 1 + 7\\ 3 x >5 x + 6\\ 3 x - 5 x > 5 x +6 - 5 x\\ - 2 x > 6\\ \dfrac{-2x}{-2}<\dfrac{6}{-2}$$\\$ x<-3$\\$ Thus, all real numbers x, which are less than - 3 , are the solutions of the given inequality.$\\$ Hence, the solution set of the given inequality is (-$\infty$ , - 3 )

12   Solve the given inequality for real x : 3 (x -1 )$\leq$ 2 ( x - 3 )

##### Solution :

3 ( x - 1 )$\leq$ 2 ( x - 3 )$\\$ 3 x - 3 $\leq$ 2 x - 6$\\$ 3 x - 3 + 3 $\leq$ 2 x - 6 + 3$\\$ 3 x $\leq$ 2 x - 3$\\$ 3 x - 2 x $\leq$ 2 x - 3 - 2 x$\\$ x $\leq$ - 3$\\$ Thus, all real numbers x, which are less than or equal to - 3 , are the solutions of the given inequality.$\\$ Hence, the solution set of the given inequality is ($-\infty,- 3 ]$ .