Question 1: If  $x \in \{-3,-2,-1,0,1,2,3 \}$ , find the solution set of each of the following:

i)  $x+2<1$     ii)  $2x-1 < 4$     iii)  $2/3 x<1$     iv)  $1-x>0$

v)  $3-5x<-1$     vi)  $2-3x>1$     vii)  $-6 \geq 2x-4$     viii)  $3x-5 \geq -12$

ix)  $14-2x<6$

 i) $x+2<1$$x+2<1$ $\Rightarrow x \leq -1$$\Rightarrow x \leq -1$ $\displaystyle \text{Solution Set } = \{-3,-2\}$$\displaystyle \text{Solution Set } = \{-3,-2\}$ ii) $2x-1 < 4$$2x-1 < 4$ $\Rightarrow 2x < 5$$\Rightarrow 2x < 5$ $\displaystyle \Rightarrow x < \frac{5}{2}$$\displaystyle \Rightarrow x < \frac{5}{2}$ $\displaystyle \text{Solution Set } = \{-3,-2,-1,0,1,2 \}$$\displaystyle \text{Solution Set } = \{-3,-2,-1,0,1,2 \}$ iii) $2/3 x<1$$2/3 x<1$ $\Rightarrow x<3/2$$\Rightarrow x<3/2$ $\displaystyle \text{Solution Set } = \{-3,-2,-1,0,1 \}$$\displaystyle \text{Solution Set } = \{-3,-2,-1,0,1 \}$ iv) $1-x>0$$1-x>0$ $\Rightarrow x<1$$\Rightarrow x<1$ $\displaystyle \text{Solution Set } = \{-3,-2,-1,0 \}$$\displaystyle \text{Solution Set } = \{-3,-2,-1,0 \}$ v) $3-5x<-1$$3-5x<-1$ $\Rightarrow 5x>4$$\Rightarrow 5x>4$ $\Rightarrow x>4/5$$\Rightarrow x>4/5$ $\displaystyle \text{Solution Set } = \{ 1,2,3 \}$$\displaystyle \text{Solution Set } = \{ 1,2,3 \}$ vi) $2-3x>1$$2-3x>1$ $\Rightarrow 3x<1$$\Rightarrow 3x<1$ $\Rightarrow x<1/3$$\Rightarrow x<1/3$ $\displaystyle \text{Solution Set } = \{-3,-2,-1,0 \}$$\displaystyle \text{Solution Set } = \{-3,-2,-1,0 \}$ vii) $-6 \geq 2x-4$$-6 \geq 2x-4$ $\Rightarrow 2x \leq 2$$\Rightarrow 2x \leq 2$ $\Rightarrow x \leq 1$$\Rightarrow x \leq 1$ $\displaystyle \text{Solution Set } = \{-3,-2,-1,0,1 \}$$\displaystyle \text{Solution Set } = \{-3,-2,-1,0,1 \}$ viii) $3x-5 \geq -12$$3x-5 \geq -12$ $\Rightarrow 3x \geq -7$$\Rightarrow 3x \geq -7$ $\Rightarrow x \geq (-7)/3$$\Rightarrow x \geq (-7)/3$ $\displaystyle \text{Solution Set } = \{-2,-1, 0, 1, 2, 3 \}$$\displaystyle \text{Solution Set } = \{-2,-1, 0, 1, 2, 3 \}$ ix) $14-2x<6$$14-2x<6$ $\Rightarrow 2x>8$$\Rightarrow 2x>8$ $\Rightarrow x>4$$\Rightarrow x>4$ $\displaystyle \text{Solution Set } = \phi$$\displaystyle \text{Solution Set } = \phi$

$\\$

Question 2: If $x \in N$; find the solution set of each of the following in equation: $N=\{1, 2, 3, ... \}$

i)  $3x-8<0$     ii)  $7x+3 \leq 17$     iii)  $5-x>1$

iv)  $1-3x>-4$      v)  $\displaystyle \frac{3}{2}-\frac{x}{2}>-1$     vi)  $\displaystyle \frac{-1}{4} \leq \frac{1}{2} - \frac{2}{3}$

 i) $3x-8<0$$3x-8<0$ $\Rightarrow 3x<8$$\Rightarrow 3x<8$ $\displaystyle \Rightarrow x<\frac{8}{3}$$\displaystyle \Rightarrow x<\frac{8}{3}$ $\displaystyle \text{Solution Set } = \{1,2 \}$$\displaystyle \text{Solution Set } = \{1,2 \}$ ii) $7x+3 \leq 17$$7x+3 \leq 17$ $\Rightarrow 7x \leq 14$$\Rightarrow 7x \leq 14$ $\Rightarrow x \leq 2$$\Rightarrow x \leq 2$ $\displaystyle \text{Solution Set } = \{1,2 \}$$\displaystyle \text{Solution Set } = \{1,2 \}$ iii) $5-x>1$$5-x>1$ $\Rightarrow x<4$$\Rightarrow x<4$ $\displaystyle \text{Solution Set } = \{1,2,3 \}$$\displaystyle \text{Solution Set } = \{1,2,3 \}$ iv) $1-3x>-4$$1-3x>-4$ $\Rightarrow 3x<5$$\Rightarrow 3x<5$ $\displaystyle \Rightarrow x<\frac{5}{3}$$\displaystyle \Rightarrow x<\frac{5}{3}$ $\displaystyle \text{Solution Set } = \{1 \}$$\displaystyle \text{Solution Set } = \{1 \}$ v) $\frac{3}{2}-\frac{x}{2}>-1$$\frac{3}{2}-\frac{x}{2}>-1$ $\displaystyle \Rightarrow \frac{x}{2}<\frac{5}{2}$$\displaystyle \Rightarrow \frac{x}{2}<\frac{5}{2}$ $\Rightarrow x<5$$\Rightarrow x<5$ $\displaystyle \text{Solution Set } = \{1,2,3,4 \}$$\displaystyle \text{Solution Set } = \{1,2,3,4 \}$ vi) $\displaystyle \frac{-1}{4} \leq \frac{1}{2} - \frac{2}{3}$$\displaystyle \frac{-1}{4} \leq \frac{1}{2} - \frac{2}{3}$ $\displaystyle \Rightarrow \frac{-3}{4} \leq \frac{-x}{3}$$\displaystyle \Rightarrow \frac{-3}{4} \leq \frac{-x}{3}$ $\displaystyle \Rightarrow \frac{x}{3} \leq \frac{3}{4}$$\displaystyle \Rightarrow \frac{x}{3} \leq \frac{3}{4}$ $\displaystyle \Rightarrow x \leq \frac{9}{4}$$\displaystyle \Rightarrow x \leq \frac{9}{4}$ $\displaystyle \text{Solution Set } = \{1,2 \}$$\displaystyle \text{Solution Set } = \{1,2 \}$

$\\$

Question 3: If $x \in Z$ , find the solution set of the following in equations: $Z = \{...-3, -2, -1, 0, 1, 2, 3, ...\}$

i)  $9x-7 \leq 25+3x$     ii)  $-17<9x-8$     iii)  $-4(x+5)>10$

iv)  $4-3x<13+x$     v)  $5-4x<10-x$     i)  $10-2(1+4x)<20$

 i) $9x-7 \leq 25+3x$$9x-7 \leq 25+3x$ $\Rightarrow 6x \leq 32$$\Rightarrow 6x \leq 32$ $\displaystyle \Rightarrow x \leq \frac{32}{6}$$\displaystyle \Rightarrow x \leq \frac{32}{6}$ $\displaystyle \text{Solution Set } = \{...-3, -2, -1, 0, 1, 2, 3, 4, 5...\}$$\displaystyle \text{Solution Set } = \{...-3, -2, -1, 0, 1, 2, 3, 4, 5...\}$ ii) $-17<9x-8$$-17<9x-8$ $\Rightarrow 9x>-9$$\Rightarrow 9x>-9$ $\Rightarrow x>-1$$\Rightarrow x>-1$ $\displaystyle \text{Solution Set } = \{0, 1, 2, 3...\}$$\displaystyle \text{Solution Set } = \{0, 1, 2, 3...\}$ iii) $-4(x+5)>10$$-4(x+5)>10$ $\Rightarrow -4x>30$$\Rightarrow -4x>30$ $\displaystyle \Rightarrow x<\frac{(-15)}{2}$$\displaystyle \Rightarrow x<\frac{(-15)}{2}$ $\displaystyle \text{Solution Set } = \{...-10, -9, -8 \}$$\displaystyle \text{Solution Set } = \{...-10, -9, -8 \}$ iv) $4-3x<13+x$$4-3x<13+x$ $\Rightarrow 4x>-9$$\Rightarrow 4x>-9$ $\displaystyle \Rightarrow x>\frac{(-9)}{4}$$\displaystyle \Rightarrow x>\frac{(-9)}{4}$ $\displaystyle \text{Solution Set } = \{-2, -1, 0, 1, 2, 3, ...\}$$\displaystyle \text{Solution Set } = \{-2, -1, 0, 1, 2, 3, ...\}$ v) $5-4x<10-x$$5-4x<10-x$ $\Rightarrow -5<3x$$\Rightarrow -5<3x$ $\displaystyle \Rightarrow x >\frac{(-5)}{3}$$\displaystyle \Rightarrow x >\frac{(-5)}{3}$ $\displaystyle \text{Solution Set } = \{-1, 0, 1, 2, ...\}$$\displaystyle \text{Solution Set } = \{-1, 0, 1, 2, ...\}$ vi) $10-2(1+4x)<20$$10-2(1+4x)<20$ $\Rightarrow 10-2-20<8x$$\Rightarrow 10-2-20<8x$ $\Rightarrow 8x>-12$$\Rightarrow 8x>-12$ $\Rightarrow x>-3/2$$\Rightarrow x>-3/2$ $\displaystyle \text{Solution Set } = \{-1, 0, 1, 2, ...\}$$\displaystyle \text{Solution Set } = \{-1, 0, 1, 2, ...\}$

$\\$

Question 4: Find the Solution Set of each of the following in equations and represent the solution on a real line.

i)  $1-4x\geq{}-1, x \in N$     ii)  $-3\leq{}4x+1<9, x \in N$

iii)    $0<2x-5<5, x \in W \ \ \ \ \ W= \{0,1,2,3\ldots{}.\}$     iv)  $\displaystyle -3< \frac{x}{2} - 1<1, x \in Z$

v)    $\displaystyle -4< \frac{2x}{5} +1<-3, x\in Z$     vi)    $\displaystyle 3+ \frac{x}{4} < \frac{2x}{3} +5, x\in R$

vii)    $\displaystyle \frac{(3x+1)}{4} \leq \frac{(5x-2)}{3} , x\in R$     viii)  $\displaystyle \frac{1}{3} (4x-1)+3\leq 4+ \frac{2}{5} (6x+2)+ \frac{4}{5}$

i)  $1-4x\geq{}-1, x \in N$

$\Rightarrow{} 4x\leq{}2$

$\displaystyle \Rightarrow{} x\leq{} \frac{1}{2}$

$\displaystyle \text{Solution Set } = \varnothing{}$

ii)  $-3\leq{}4x+1<9, x \in N$

$\Rightarrow{} -4\leq{}4x<8$

$\Rightarrow{} -1\leq{}x<2$

$\displaystyle \text{Solution Set } = \{-1,0,1\}$

iii)    $0<2x-5<5, x \in W \ \ \ \ \ W= \{0,1,2,3\ldots{}.\}$

$\Rightarrow{} 5<2x<10$

$\displaystyle \Rightarrow{} \frac{5}{2}

$\displaystyle \text{Solution Set } = \{3,4\}$

iv)  $\displaystyle -3< \frac{x}{2} - 1<1, x \in Z$

$\displaystyle \Rightarrow{} -2< \frac{x}{2} <2$

$\Rightarrow{} -4

$\displaystyle \text{Solution Set } = \{-3,-2,-1,0,1,2,3\}$

v)    $\displaystyle -4< \frac{2x}{5} +1<-3, x\in Z$

$\displaystyle \Rightarrow{} -5< \frac{2x}{5} <-4$

$\Rightarrow{} 25<2x<-20$

$\displaystyle \Rightarrow{} \frac{-25}{2}

$\displaystyle \text{Solution Set } = \{-12,-11\}$

vi)   $\displaystyle 3+ \frac{x}{4} < \frac{2x}{3} +5, x\in R$

$\displaystyle \Rightarrow{} \frac{x}{4} < \frac{2x}{3} +2$

$\displaystyle \Rightarrow{} \frac{2x}{3} - \frac{x}{4} >-2$

$\Rightarrow{} 5x>-24$

$\displaystyle \Rightarrow{} x> \frac{(-24)}{5}$

$\displaystyle \text{Solution Set } = \{x\in R :x> \frac{(-24)}{5} \}$

vii)   $\displaystyle \frac{(3x+1)}{4} \leq{} \frac{(5x-2)}{3} , x\in R$

$\Rightarrow{} 9x+3\leq{}20x-8$

$\Rightarrow{} 11x\geq{}11$

$\Rightarrow{} x\geq{}1$

$\text{Solution Set } = \{x \in R: x\geq{}1\}$

viii)   $\displaystyle \frac{1}{3} (4x-1)+3\leq{}4+ \frac{2}{5} (6x+2)+ \frac{4}{5}$

$\displaystyle \Rightarrow{} \frac{4}{3} x- \frac{1}{3} +3\leq{}4+ \frac{12}{5} x+ \frac{4}{5}$

$\displaystyle \Rightarrow{} \frac{(-32)}{15} \leq{} \frac{16}{15} x$

$\Rightarrow{} x\geq{}-2$

$\text{Solution Set } = \{x \in R: x\geq{}-2\}$