ML AGGARWAL CLASS 8 Chapter 12 Linear Equations and Inqualities In one Variable Exercise 12.3

 Exercise 12.3

Question 1

Sol :

(i) x>-2

Solution set ={-1,0,1,3}

(ii) x<-2

Solution set ={-7,-5,-3}

(iii) x>2

Solution set ={3}

(iv)

$\begin{aligned}-5<x & \leq 5 \\ & \text { Solution set }=\{-3,-1,0,1,3\} \end{aligned}$

(v) 

$\begin{aligned}-8<x &<1 \\ & \text { Solution set }=\{-7,-5,-3,-1,0\} \end{aligned}$

(vi) 

$\begin{aligned} 0 \leq x & \leq 4 \\ & \text { Solution set }=\{0,1,3\} \end{aligned}$

Question 2

Sol :

It shown by thick dots on number line 

(i) $x \leq 4, x \in N$

(ii) $x<5, x \in W$

(iii) $-3 \leq x<3, x \in \mathbb{L}$

Question 3

Sol :

Replacement Set ={-6,-4,-2,0,2,4,6}

$-4 \leq x<4$

Question 4

Sol :

(i) {1,2,3 ...10}

(ii) {-1,0,1,2,5,8}

(iii) {-5,10}

(iv) {5,6,7,8,9,10}

Solution Set=$\phi$

Question 5

Sol :

(i) 2 x-3>7

2 x>7+3

2 x>10

x>5

Solution set ={6,9,12}

(ii) $\begin{aligned} 3 x+8 \leq & 2 \\ 3 x \leq & 2-8 \\ 3 x & \leq-6 \\ x & \leq-2 \end{aligned}$

Solution set ={-6,-3}

(iii) -3<1-2 x

3>2 x-1

2 x-1<3

2 x<3+1

2 x<4

x<2

Solution set ={-6,-3,0}

Question 6

Sol :

(i) $\begin{aligned} 4 x+1 &<17, x \in N 1 \\ 4 x &<17-1 \\ 4 x &<16 \\ x &<4, x \in N \end{aligned}$

As $x \in N$, the Solution set is {1,2,3}

(ii) $\begin{aligned} 4 x+1 & \leq 17, x \in W \\ 4 x & \leq 17-1 \\ 4 x & \leq 16 \\ x & \leq 4 \end{aligned}$

As $x \in W$, the solution set is {0,1,2,3,4}

(iii) $\begin{array}{rl}4>3 x-11 & , x \in N \\ 3 x-11 & <4 \\ 3 x & <4+11 \\ 3 & 3 x<15 \\ & x<5\end{array}$

As $x \in N$, the solution set is $\{1,2,3,4\}$

(iv)$-17 \leq 9 x-8, x \in z$

$9 x-8 \geq-17$

$9 x \geq-17+8$

$9 x \geq-9$

$x \geq-1$

As x+z, the solution set is {-1,0,1,2,3 ...}

Question 7

Sol :

(i) $\frac{2 y-1}{5} \leq 2, y \in N$

$2 y-1 \leq 10$

$2 y \leq 10+1$

$2 y \leq 11$

$y \leq \frac{11}{2}$

As $y \in N$, the solution set is $\{1,2,3,4,5\}$

(ii) $\frac{2 y+1}{3}+1 \leq 3, y \in w$

$\frac{2 y+1}{3} \leq 3-1$

$\frac{2 y+1}{3} \leq 2$

$2 y+1 \leq 6$

$2 y+6-1$

$2 y \leq 5$

$y \leq \frac{5}{2}$

As $y \in W$, the solution set is $\{0,1,2\}$

(iii) $\frac{2}{3} p+s<9, p \in W .$

$\frac{2}{3} p<9-5$

$\frac{21}{3}<4$

2 p<12

p<6

As $p \in W$, the solution set is {0,1,2,3,4,5}

(iv) $-2(p+3)>5 \quad p \in I$

Multiplying '- ' on both sides

2(p+3)<-5

2 p+6<-5

2 p<-5-6

2 p<-11

$p<-\frac{11}{2}$

As $p \in I$, the solution set is {...-9,-8,-7,-6}

Question 8

Sol :

(i) $2 x-3<x+2, x \in N$

2x<x+2+3

2 x-x<5

x<5

As $x \in N$, the Solution set is {1,2,3,4}

(ii) $3-x \leq 5-3 x, \quad x \in W$

$3-x+3 x \leq 5$

$2 x+3 \leq 5$

$2 x \leq 5-3$

$2 x \leq 2$

$x \leq 1$

As $x \in W$, the solution set is {0,1}

(iv) $\frac{3}{2}-\frac{x}{2}>-1, \quad x \in N$

$\frac{3-x}{2}>-1$

3-x>-2

Multiplying with '- ' on son sides

x-3<2

x<2+3

x<5

As $x \in N$, the solution set is {1,2,34}

Question 9

Sol :
{-3,-2,-1,0,1,2,3}

$\begin{aligned} \frac{3 x-1}{2} &<2 \\ 3 x-1 &<4 \\ 3 x-4 &<4+1 \\ 3 x &<5 \end{aligned}$

$x<5 / 3$

As x should be in replacement set, the solution

set is {-3,-2,-1,0,1}

Question 10

Sol :

$\frac{x}{3}+\frac{1}{4}<\frac{x}{6}+\frac{1}{2}, x \in w$

$\frac{x}{3}-\frac{x}{6}+\frac{1}{4}<\frac{1}{2}$

$\frac{x}{3}-\frac{x}{6}<\frac{1}{2}-\frac{1}{4}$

$\frac{2 x-x}{6}<\frac{2-1}{4}$

$\frac{x}{6}<\frac{1}{4}$

$x<\frac{6}{4}$

$x<\frac{3}{2}$

As $x \in W$, the solution set is {0,1}

Question 11

Sol :

(i) 

$\begin{aligned}-4 \leq 4 x<14, & x \in N \\ 4 x \geqslant-4 ; & 4 x<14 \\ x \geqslant-1 ; & x<14 / 4 \\ & x<7 / 2 \end{aligned}$

As $x \in N$, the solution set is {-1,0,1,2,3}

(ii) $-1<\frac{x}{2}+1 \leq 3, \quad x \in I$

$\frac{x}{2}+1>-1 \quad ; \quad \frac{1}{2}+1 \leq 3$

$\frac{x}{2}>-1-1 \quad ; \quad \frac{x}{2} \leq 3-1$

$\frac{x}{2}>-2 ; \quad \frac{x}{2} \leq 2$

$x>-4 ; \quad x \leq 4$

i.e $-4<x \leq 4$

As $x \in I$, the solution set is {-3,-2,-1,0,1,2,3,4}