Question 14 Marks
Solve and graph the solution set on a number line : $ 8x – 8 \leq – 24 ; x \in Z$
Answer$8x – 8 \leq – 24$
$\Rightarrow 8x - 8 + 8 \leq - 24 + 8 ..($Adding $8$ to both sides$)$
$\Rightarrow 8x \leq -16$
$\Rightarrow \frac{8 x}{8} \leq \frac{-16}{8} \quad..($Dividing both sides by $8 )$
$\Rightarrow x \leq -2$
$\therefore$ The required graph is:
View full question & answer→Question 24 Marks
Solve and graph the solution set on a number line : $ 7 > 3x – 8 ; x \in W$
Answer$7 \geq 3 x-8$
$\Rightarrow 7+8 \geq 3 x-88 ($Adding $8$ to both sides$)$
$\Rightarrow 15 \geq 3 x$
$\Rightarrow \frac{15}{3} \geq \frac{3 \mathrm{x}}{3} ($Dividing both sides by $3 )$
$\Rightarrow 5 \geq x$
$\therefore$ The required graph is:
View full question & answer→Question 34 Marks
Solve and graph the solution set on a number line : -$3x + 12 < -15 ; x \in R.$
Answer$-3 x+12<-15$
$\Rightarrow-3 x+12-12<-15-12 ($Subtracting $12$ from both sides$)$
$\Rightarrow-3 x<-27$
$\Rightarrow \frac{-3 \mathrm{x}}{-3}>\frac{-27}{-3} ($Dividing both sides by $-3 )$
Note: Division by a negative number reverses the inequality.
$\Rightarrow x>9$
$\therefore$ The required graph is:
View full question & answer→Question 44 Marks
Solve and graph the solution set on a number line : $3x – 1 > 5 ; x \in W$
Answer$3 x-1>5$
$ \Rightarrow 3 x-1+1>5+1 ($Adding $1$ to both sides$)$
$ \Rightarrow 3 x>6$
$ \Rightarrow \frac{3 x}{3}>\frac{6}{3} ($Dividing both sides by $3) $
$ \Rightarrow x>2$
$\therefore$ The required graph is:
View full question & answer→Question 54 Marks
For the following inequation, represent the solution on a number line : $-2(x + 8) \leq 8, x \in R$
Answer$ -2(x+8) \leq 8, x \in R$
$ -2 x-16 \leq 8$
$ \Rightarrow-2 x \leq 8+16$
$ \Rightarrow-2 x \leq 24$
$ \Rightarrow x \geq \frac{-24}{2}$
$ x \geq-12$
View full question & answer→Question 64 Marks
For the following inequation, represent the solution on a number line : $\frac{4-x}{2}<3, x \in R$
Answer$ \frac{4-x}{2}<3, x \in R$
$ \Rightarrow 4-x<6$
$ \Rightarrow-x<6-4$
$ \Rightarrow-x<2$
$ \Rightarrow x>-2$
$ \therefore x>-2$
View full question & answer→Question 74 Marks
For the following inequation, represent the solution on a number line : $4(3x + 1) > 2(4x - 1), x$ is a negative integer
Answer$4(3 x+1)>2(4 x-1), x$ is a negative integer
$\Rightarrow 12 x+4>8 x-2$
$ \Rightarrow 12 x-8 x>-2-4$
$ \Rightarrow 4 x>-6$
$ \Rightarrow x>\frac{-6}{4}$
$ \Rightarrow x>-1.5$
$\therefore \mathrm{x}=\{-1\}$
View full question & answer→Question 84 Marks
For the following inequation, represent the solution on a number line : $2(4 - 3x) \leq 4(x - 5), x \in W$
Answer$ 2(4-3 x) \leq 4(x-5), x \in W$
$ 8-6 x \leq-20-8$
$ \Rightarrow-6 x-4 x \leq-20-8$
$ \Rightarrow-10 x \leq-28$
$ \Rightarrow 10 x \geq 28$
$ \Rightarrow x \geq \frac{28}{10}$
$ \Rightarrow x \geq 2.8$
$ \therefore x=\{3,4,5, \ldots .\}$
View full question & answer→Question 94 Marks
For the following inequation, represent the solution on a number line : $3(2x -1) \geq 2(2x + 3), x \in Z$
Answer$ 3(2 x-1) \geq 2(2 x+3), x \in Z$
$ \Rightarrow 6 x-3 \geq 4 x+6, x \in Z$
$ \Rightarrow 6 x-4 x \geq 6+3$
$ \Rightarrow 2 x \geq 9$
$ \Rightarrow x \geq \frac{9}{2}$
$ \Rightarrow x \geq 4 \frac{1}{2}$
$ \therefore x=\{5,6,7, \ldots .\}$
View full question & answer→Question 104 Marks
For the following inequation, represent the solution on a number line : $\frac{5}{2}-2 \mathrm{x} \geq \frac{1}{2}, \mathrm{x} \in \mathrm{W}$
Answer$ \frac{5}{2}-2 \mathrm{x} \geq \frac{1}{2}, \mathrm{x} \in \mathrm{W}$
$ \Rightarrow-2 \mathrm{x} \geq \frac{1}{2}-\frac{5}{2}$
$ \Rightarrow-2 \mathrm{x} \geq \frac{-4}{2}$
$ \Rightarrow \mathrm{x} \leq 1$
$ \therefore \mathrm{x}=\{0,1\}$
View full question & answer→Question 114 Marks
Solve the inequation $5(x – 2) > 4 (x + 3) – 24$ and represent its solution on a number line. Given the replacement set is $ \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}.$
Answer$5(x-2)>4(x+3)-24$
$ \Rightarrow 5 x-10>4 x+12-24$
$ \Rightarrow 5 x-4 x>10-12$
$ \Rightarrow x>-2$
Since replacement set $=\{-4,-3,-2,-1,0,1,2,3,4\}$
$\therefore$ Solution set $=\{-1,0,1,2,3,4\}$
$\therefore$ Solution set on a number line is shown below.
View full question & answer→Question 124 Marks
Solve the inequation $18 – 3 (2x – 5) > 12; x \in W.$
Answer$18-3(2 x-5)>12 ; x \in W$
$\Rightarrow 18-6 x+15>12$
$\Rightarrow 33-12>6 x$
$\Rightarrow 21>6 x$
$\Rightarrow 6 x<21$
$\Rightarrow \mathrm{x}<\frac{21}{6}+\frac{7}{2}=3 \frac{1}{2}$
But $x \in W, x=0,1,2,3$
$\therefore$ Solution set $=\{0,1,2,3\}$
View full question & answer→Question 134 Marks
Solve the inquation $8 – 2x \geq x – 5; x \in N.$
Answer$8-2 x \geq x-5 ; x \in N$
$ \Rightarrow 8+5 \geq 2 x+x$
$ \Rightarrow 13 \geq 3 x $
$\Rightarrow 3 x \leq 13$
$ \Rightarrow x \leq \frac{13}{3}=4 \frac{1}{3}$
$ x=1,2,3,4(x \in N)$
Solution set $=\{1,2,3,4\}$
View full question & answer→Question 144 Marks
Solve : $-5(x+4)>30 ; x \in Z$
Answer$-5(x+4)>30$
$ \Rightarrow \frac{-5(x+4)}{-5}<\frac{30}{-5} \ldots ($Dividing by $-5)$
Note: Division by a negative number reverses the equality.
$\Rightarrow x+4<-6$
$\Rightarrow x+4-4<-6-4 \ldots ($Subtracting $4)$
$\Rightarrow \mathrm{x}<-10$
$\therefore$ Required answer $=\{-11,-12,-13, \ldots\}$
View full question & answer→Question 154 Marks
Solve : $\frac{2}{3} \mathrm{x}+8<12 ; \mathrm{x} \in \mathrm{W}$
Answer$ \frac{2}{3} \mathrm{x}+8<12$
$ \Rightarrow \frac{2}{3} \mathrm{x}+8-8<12-8$
$ \Rightarrow \frac{2}{3} \mathrm{x}<4$
$ \Rightarrow \frac{2}{3} \mathrm{x} \times \frac{3}{2}<4 \times \frac{3}{2} ($Multiplying by $\frac{3}{2} \text { ) }$
$ \Rightarrow \mathrm{x}<6$
$ \therefore$ Required answer $=\{0,1,2,3,4,5\}$
View full question & answer→Question 164 Marks
Solve : $9x - 7 \leq 5: 28 + 4x ; x \in W.$
Answer$9x – 1 \leq 28 + 4x\Rightarrow 9x – 4x – 7 \leq 28 + 4x – 4x ($Subtracting $4x)$
$\Rightarrow 5x – 7 \leq 28$
$\Rightarrow 5x – 7 + 7 \leq 28 + 7 ($Adding $7)$
$\Rightarrow 5x \leq 35$
$\Rightarrow x \leq 7 ($Dividing by $5)$
Required answer $= \{0, 1, 2, 3, 4, 5, 6, 7\}$
View full question & answer→Question 174 Marks
Solve : $\frac{2 \mathrm{x}+3}{5}>\frac{4 \mathrm{x}-1}{2}, \mathrm{x} \in \mathrm{W}$.
Answer$\frac{2 x+3}{5}>\frac{4 x-1}{2}, x \in W$
$ \Rightarrow 2(2 x+3)>5(4 x-1)$
$ \Rightarrow 4 x+6>20 x-5$
$ \Rightarrow 4 x-20 x>-5-6$
$ \Rightarrow-16 x>-11$
$\Rightarrow x<\frac{-11}{-16} ($Dividing by $-16 )$
$\Rightarrow \mathrm{x}<\frac{11}{16}$
$\therefore \mathrm{x}=0$
$\therefore$ Solution set $=\{0\}$
View full question & answer→Question 184 Marks
Solve : $15 – 2(2x – 1) < 15, x \in Z.$
Answer$ 15-2(2 x-1)<15, x \in Z$
$ \Rightarrow 15-4 x+2<15$
$ \Rightarrow 17-4 x<15$
$ \Rightarrow-4 x<15-17$
$ \Rightarrow-4 x<-2$
$ \Rightarrow \frac{-4}{-4} x>\frac{-2}{-4}=\frac{1}{2} ($Dividing by $-4)$
$ \therefore x=1,2,3,4,5, \ldots . .$
$ \therefore$ Solution set $=\{1,2,3,4,5, \ldots\}$
View full question & answer→Question 194 Marks
Solve the inequation $\frac{2 \mathrm{x}+1}{3}+15 \leq 17 ; \mathrm{x} \in \mathrm{W}$.
Answer$\frac{2 x+1}{3}+15 \leq 17 ; x \in W$
$ \Rightarrow \frac{2 x+1}{3} \leq 17-15=2$
$ \Rightarrow 2 x+1 \leq 6$
$ \Rightarrow 2 x \leq 5$
$ \Rightarrow x \leq \frac{5}{2}=2 \frac{1}{2}$
But $x \in W$
$\therefore \mathrm{x}=0,1,2$
$\therefore$ Solution set is $=\{0,1,2\}$
View full question & answer→Question 204 Marks
Solve and graph the solution set on a number line : $\frac{x}{2}>-1+\frac{3 x}{4} ; x \in N$
Answer$\frac{x}{2}>-1+\frac{3 x}{4}$
$\Rightarrow \frac{\mathrm{x}}{2} \times 4>-1 \times 4+\frac{3 \mathrm{x}}{4} \times 4..($Multiplying both sides by $4 )$
$\Rightarrow 2 x>-4+3 x$
$\Rightarrow 2 \mathrm{x}-2 \mathrm{x}>-4+3 \mathrm{x}-2 \mathrm{x}...($Subtracting $2 \mathrm{x}$ from both sides$)$
$\Rightarrow 0>-4+x$
$\Rightarrow 0+4>-4+4+x \ldots ($Adding $4$ to both sides$)$
$\Rightarrow 4>x$
$\therefore$ The required graph is :
View full question & answer→Question 214 Marks
Solve and graph the solution set on a number line : $\frac{2 \mathrm{x}}{5}+1<-3 ; \mathrm{x} \in \mathrm{R}$
Answer$\frac{2 \mathrm{x}}{5}+1<-3$
$ \Rightarrow \frac{2 \mathrm{x}}{5}+1-1<-3-1 \ ($Subtracting from both sides$)$
$ \Rightarrow \frac{2 \mathrm{x}}{5}<-4$
$ \Rightarrow \frac{2 \mathrm{x}}{5} \times 5<-4 \times 5 \ldots ($Multiplying both sides by $5)$
$ \Rightarrow 2 \mathrm{x}<-20$
$ \Rightarrow \frac{2 \mathrm{x}}{2}<\frac{-20}{2} \ldots ($Dividing both sides by $2)$
$ \Rightarrow \mathrm{x}<-10$
$\therefore$ The required graph is :
View full question & answer→Question 224 Marks
Solve and graph the solution set on a number line : $ 5x + 4 > 8x – 11 ; x \in Z$
Answer$5 x+4>8 x-11$
$\Rightarrow 5 x-5 x+4>8 x-5 x-11..($Subtracting $5 x$ from both sides$)$
$\Rightarrow 4>3 x-11$
$\Rightarrow 4+11>3 x-11+11...($Adding $11$ to both sides$)$
$\Rightarrow 15>3 x$
$\Rightarrow \frac{15}{3}>\frac{3 \mathrm{x}}{3} \ldots ($Dividing both sides by $3 )$
$\Rightarrow 5>x$
$\therefore$ The required graph is:
View full question & answer→Question 234 Marks
Solve and graph the solution set on a number line : $8x – 9 > 35 – 3x ; x \in N$
Answer$8 x-9 \geq 35-3 x$
$ \Rightarrow 8 x+3 x-9 \geq 35-3 x+3 x \ldots($Adding $ 3 x$ to both $s)$
$ \Rightarrow 11 x-9 \geq 35$
$ \Rightarrow 11 x-9+9 \geq 35+9 ($Adding $9$ to both sides $)$
$ \Rightarrow 11 x \geq 44$
$ \Rightarrow \frac{11 x}{11} \geq \frac{44}{11} ...($Dividing both side by $11)$
$ \Rightarrow x \geq 4...($Adding $3 x$ to both sides$)$
$\therefore$ The required graph is :
View full question & answer→Question 244 Marks
Solve $\frac{2}{3}(x+1)+4<10$ and represent its solution on a number line. Given replacement set is $\{-8,-6,-4,3,6,8,12\}$.
Answer$\frac{2}{3}(x+1)+4<10$
$\Rightarrow \frac{2}{3}(\mathrm{x}-1)<10-4$
$\Rightarrow \frac{2}{3}(\mathrm{x}-1)<6$
$\Rightarrow 2(x-1)<18$
$\Rightarrow x-1<9$
$\Rightarrow x-1+1<9+1 \ldots ($Adding $1$ to both sides$)$
$\Rightarrow x<10$
Thus $x<10$
Since, replacement set $=\{-8,-6,-4,3,6,8,12\}$
$\Rightarrow$ Solution set $=\{-8,-6,-4,3,6,8\}$ View full question & answer→Question 254 Marks
Solve and graph the solution set on a number line : $\frac{2}{3} \mathrm{x}+5 \leq \frac{1}{2} \mathrm{x}+6 ; \mathrm{x} \in \mathrm{W}$
Answer$\frac{2}{3} x+5 \leq \frac{1}{2} x+6$
$\Rightarrow \frac{2}{3} \mathrm{x} \times 6+5 \times 6 \leq \frac{1}{2} \mathrm{x} \times 6+6 \times 6 \ldots$ $($Multiplying both sides by $6 )$
$\Rightarrow 4 x+30 \leq 3 x+36$
$\Rightarrow 4 x-3 x+30 \leq 3 x-3 x+36 \ldots($ Substracting $3 x$ from both sides$)$
$\Rightarrow x+30 \leq 36$
$\Rightarrow x+30-30 \leq 36-30 \ldots ($Substracting $30$ from both sides$)$
$x \leq 6$
$\therefore$ The required graph is :
View full question & answer→