Exercise 12.1
Question 1
(i) $5 x-3=3 x-5$
=$5 x=3 x-5+3$
=$5 x=3 x-2$
=$5 x-3 x=-2$
=$2 x=-2$
$x=-1$
(ii) $3 x-7=3(5-x)$
=$3 x-7=15-3 x$
=$3 x=15+7-3 x$
=$3 x+3 x=22$
=$6 x=22$
=$x=\frac{22}{6}$
Question 2
(i) $4(2 x+1)=3(x-1)+7$
=$8 x+4=3 x-3+7$
=$8 x=3 x-3+7-4$
=$8 x-3 x=-3+3=0$
$5 x=0 \Rightarrow x=0$
(ii)
$\begin{aligned} 3(2 p-1)=5-(3 p-2) & \\ 6 p-3=5 &-3 p+2 \\ 6 p=& 5-3 j+2+3 \\ 6 p+3 p &=10 \\ 9 p &=10 \\ p &=\frac{10}{9}=1 \frac{1}{9} \end{aligned}$
Question 3
(i) $5 y-2[y-3(y-5)]=6$
=$5 y-2 y+6(y-5)=6$
=$3 y+6 y-30=6$
=$9 y=6+30=36$
=$y=\frac{36}{9}=4$
(ii) $0.3(6-x)=0.4(x+8)$
=$1.8-0.3 x=0.4 x+3.2$
=$0.4 x=1.8-0.3 x-3.2$
=$0.4 x+0.3 x=-1.4$
=$0.7 x=-1.4$
=$x=-\frac{1.4}{0.7}=2$
Question 4
(i) $\frac{x-1}{3}=\frac{x+2}{6}+3$
Multiply and divide 2 on L.H.S
=$\frac{2(x-1)}{2 \times 3}=\frac{x+2}{6}+3$
=$\frac{2 x-1}{6}-\frac{x+2}{6}=3$
=$\frac{2 x-1-x-2}{6}=3$
=$\frac{x-3}{6}=3$
=$x-3=3 \times 6$
=$x-3=18$
$x=18+3$
$x=21$
(ii) $\frac{x+7}{3}=1+\frac{3 x-2}{5}$
=$\frac{x+7}{3}=\frac{5}{5}+\frac{3 x-2}{5}$
=$\frac{x+7}{3}=\frac{5+3 x-2}{5}$
=$\frac{x+7}{3}=\frac{3 x+3}{5}$
Cross Multiplying
=$5(x+7)=3(3 x+3)$
=$5 x+35=9 x+9$
=$9 x=5 x+35-9$
=$9 x-5 x=26$
=$4 x=26$
$x=\frac{26}{4}=\frac{13}{2}=6 \frac{1}{2}$
Question 5
(i) $\frac{y+1}{3}-\frac{y-1}{2}=\frac{1+2 y}{3}$
Multiplying both sides by 6 i.e L.C.M of $3,2,3$ we get
=$2(y+1)-3(y-1)=2(1+2 y)$
=$2 y+2-3 y+3=2+4 y$
=$5-y=4 y+2$
=$4 y+y=5-2$
$5 y=3$
$y=3 / 5$
(ii) $\frac{1}{3}+\frac{p}{4}=55-\frac{p+40}{5}$
=$\frac{4 p+3 p}{12}=\frac{55 \times 5}{5}-\frac{p+40}{5}$
=$\frac{7 p}{12}=\frac{275-p-40}{5}$
$\frac{7 p}{12}=\frac{235-P}{5}$
Cross Multiplication
$7 p \times 5=12(235-p)$
$35 p=12 \times 235-12 p$
$35 p+12 p=12 \times 235$
$47 P=12 \times 235$
$P=\frac{12 \times 2355}{47}$
$p=60$
Question 6
(i) $n-\frac{n-1}{2}=1-\frac{\eta-2}{3}$
$\frac{2 n-(n-1)}{2}=\frac{3-(n-2)}{3}$
$\frac{2 n-n+1}{2}=\frac{3-n+2}{3}$
$\frac{n+1}{2}=\frac{5-n}{3}$
Cross multplication
$3 x(n+1)=2(5-n)$
$3 n+3=10-2 n$
$3 n+2 n=10-3$
$5 n=7$
$n=7 / 5$
(ii) $\frac{3 t-2}{3}+\frac{2 t+3}{2}=t+\frac{7}{6}$
$\frac{3 t-2}{3}+\frac{2 t+3}{2}=\frac{6 t+7}{6}$
Multiplying 6 on both sides
2(3 t-2)+3(2 t+3)=1(6 t+3)
6 t-4+6 t+9=6 t+7
6 t+5=7
6 t=7-5=2
t=2 / 6=1 / 3
Question 7
(i)
$\begin{aligned} 4(3 x+2)-5(6 x-1) &=2(x-8)-6(7 x-4) \\ 12 x+8-30 x+5 &=2 x-16-42 x+24 \\ 13-18 x=& 8-40 x \\ 40 x-18 x &+13=8 \\ 22 x &=8-13=-5 \\ x &=\frac{-5}{22} . \end{aligned}$
(ii) $3(5 x+7)+5(2 x-11)=3(8 x-5)-15$
=$15 x+21+10 x-55=24 x-15-15$
=$25 x-34=24 x-30$
=$25 x-24 x=34-30$
$x=4$
Question 8
(i) $\frac{3-2 x}{2 x+5}=-\frac{3}{11}$
Cross Multiplying both sides
$-11(3-2 x)=3(2 x+5)$
$22 x-3]=6 x+15$
$22 x-6 x=33+15$
$16 x=48$
x=3
(ii) $\frac{5 p+2}{8-2 p}=\frac{7}{6}$
cross multiplying on both sides
$\begin{aligned} 6(5 p+2) &=7(8-2 p) \\ 30 p+12 &=56-14 p \\ 30 p+14 p &=56-12 \\ 44 p &=44 \Rightarrow p=1 \end{aligned}$
Question 9
(i) $\frac{5}{x}=\frac{7}{x-4}$
Cross multiplying on both sides
$\begin{array}{c}5(x-4)=7 x \\ 7 x=5 x-20 \\ 2 x=-20 \\ x=-10\end{array}$
(ii) $\frac{4}{2 x+3}=\frac{5}{x+4}$
Cross multiplication on both sides
$\begin{aligned} 5(2 x+3) &=4(x+4) \\ 10 x+15 &=4 x+16 \\ 10 x-4 x &=16-15 \\ 6 x &=1 \Rightarrow x=\frac{1}{6} \end{aligned}$
(ii) $\frac{4}{2 x+3}=\frac{5}{x+4}$
Cross multhpicalion on bolh sides
$\begin{aligned} 5(2 x+3)=4(x+4) \\ 10 x+15=4 x+16 \\ 10 x-4 x &=16-15 \\ 6 x=1 & \Rightarrow x=\frac{1}{6} \end{aligned}$
Question 10
(i) $\frac{2 x+5}{2}-\frac{5 x}{x-1}=x$
=$\frac{(2 x+5)(x-1)-(51 \times 2)}{2(x-1)}=x$
=$2 x^{2}-2 x+5 x-5-10 x=2 x(x-1)$
=$2 x^{2}+3 x-5-101=2 x^{2}-2 x$
$-7 x-5=-2 x$
$-5=7 x-2 x$
$5 x=-5$
$x=-1$
(ii) $\frac{1}{5}\left(\frac{1}{3 x}-5\right)=\frac{1}{3}\left(3-\frac{1}{x}\right)$
$\frac{1}{5}\left(\frac{1-15 x}{3 x}\right)=\left(\frac{3 x-1}{3 x}\right)$
$1-15 x=5(3 x-1)$
$1-15 x=15 x-5$
$15 x+15 x=1+5$
$30 x=6$
$x=\frac{6}{30}=\frac{1}{5}$
Question 11
(i) $\frac{2 x-3}{2 x-1}=\frac{3 x-1}{3 x+1}$
Substralting ' 1' on both Sides
=$\frac{2 x-3}{2 x-1}-1=\frac{3 x-1}{3 x+1}-1$
=$\frac{2 x-3-2 x+1}{2 x-1}=\frac{3 x-1-3 x-1}{3 x+1}$
=$\frac{-2}{2 x-1}=\frac{-2}{3 x+1}$
=$3 x+1=2 x-1$
$3 x-2 x=-1-1$
$x=-2$
(ii) $\frac{2 y+3}{3 y+2}=\frac{4 y+5}{6 y+7}$
Cross multiplication
$(2 y+3)(6 y+7)=(3 y+2)(4 y+5)$
$12 y^{2}+14 y+18 y+21=12y^{2}+15 y+8 y+10$
$32 y+21=23 y+10$
32 y-23 y=10-21
9 y=-11
$y=\frac{-11}{9}$
Question 12
$x=p+1$
$\frac{5 x-30}{2}-\frac{7 p+1}{3}=\frac{1}{4}$
multiplying 12 on both sides
$6(5 x-30)-4(7 p+1)=3$
$30 x-180-28 p-4=3$
$30(p+1)-180-28 p-4-3=0$
$30 p+30-180-28 p-7=0$
$2 p-157=0$
$p=\frac{157}{2}$
Question 13
$\frac{x+3}{3}-\frac{x-2}{2}=1 \quad$ if $\frac{1}{x}+p=1$
Mulfiplying 6 on both sides
$2(x+3)-3(x-2)=6$
$2 x+6-3 x+6=8$
x=6
$\frac{1}{x}+p=1 \Rightarrow p=1-\frac{1}{x}=1-\frac{1}{6}$
$p=\frac{5}{6}$
Exercise 12.2
Question 1
Let the number be 'x'
Three more than twice a number is
= 2x + 3 -------1
Four less than the number = x -4 ----------2
① = ②
$2 x+3=x-4$
$2 x-x=-4-3$
$x=-7$
The number is -7
Question 2
Let the four consecutive integers are $x+1, x+2, x+3$,
x + 4
Given sum of them =46
$\begin{array}{c}x+1+x+2+x+3+x+y=46 \\4 x+10=46 \\4 x=46-10=36\end{array}$
$x=9$
The integers are 9+1,9+2,9+3,9+4
10,11,12,13
Question 3
Let the number be 'x '
Manjula substracts $\frac{7}{3}$ from it $=x-\frac{7}{3}$
The above result is multiplied by 6 i.e $6\left(x-\frac{7}{3}\right)$
Now it is equal 2 less than twice the number 'x'
$6\left(x-\frac{7}{3}\right)=2 x-2$
6 x-14=2 x-2
6 x-2 x=14-2
4 x=12
x=3
Question 4
Let the number be x , 7x
15 is added to both number then
it becomes x + 15 , 7x + 15
Then one new numbers becomes $\frac{5}{2}$ times the other
new number.
$7 x+15=\frac{5}{2}(x+15)$
$2(7 x+15)=5(x+15)$
$14 x+30=5 x+75$
$14 x-5 x=75-30$
9 x=45
x=5
Therefore the number are 5 , 35
Question 5
Let the three conseutive even integers are x, x+2 ,x+4
Given Sum = 0
x+x+2+x+4=0
3 x+6=0
3 x=-6
x=-2
∴ The integers are -2 , -2+2 ,-2+4
= -2, 0 , 2
Question 6
Let the two consecutive odd integers are x+ 1 , x + 3
Given two fifth of smaller exceeds two ninth of greater by 4
$\frac{2}{5}(x+1)+4=\frac{2}{9}(x+3)$
$\frac{2 x+2+20}{5}=\frac{2 x+6}{9}$
Cross Multiplication.
(2 x+22) 9=5(2 x+6)
18 x+198=10 x+30
18 x-10 x=30-198
8 x=-168
x=-21
x+1=-21+1=-20
x+3=-21+3=-18
The consecutive odd integers are -20,-18
Question 7
Given the denominator of a fraction is 1 more than
twice its numerator
It is written as $\frac{x}{2 x+1}$
Given numerator and denominator are both increased by 5
it becomes $\frac{3}{5}$
$\frac{x+5}{2 x+1+5}=\frac{3}{5}$
$\frac{x+5}{2 x+6}=\frac{3}{5}$
Cross mulhpication.
$5(x+5)=3(2 x+6)$
$5 x+25=6 x+18$
$6 x-5 x=25-18$
$x=7$
$\begin{aligned} \therefore \frac{x}{2 x+1}=\frac{7}{2(7)+1} &=\frac{7}{14+1} \\ &=\frac{7}{15} . \end{aligned}$
Therefore the original fraction is $\frac{7}{15}$
Question 8
Let the two number are 2x , 5x
Given their difference is 15
5 x-2 x=15
3 x=15
$x=15 / 3=5$
x=5
Therefore the two positive numbers are 10 , 25
Question 9
Let the number added to each be x
on adding 'x '' to each of the numbers it becomes
12 + x , 22 + x , 42 + x , 72 + x
for the number to be in proportion
$\frac{12+x}{22+x}=\frac{42+x}{72+x}$
Cross Multiplication
$(x+12)(x+72)=(x+22)(x+42)$
$x^{2}+72 x+12 x+864=x^{2}+22 x+42 x+924$
$84 x+864=64 x+924$
$84 x-64 x=924-864$
$20 x=60$
x=3
So the number that must be added is x = 3
Question 10
Let the unit's digit be x
As the difference of both digits is 3 , the ten's digit is x + 3
∴ The number is 10x (x + 3)
By adding both number , We get 143
10(x+3)+x+10 x+(x+3)=143
10 x+30+x+10 x+x+3=143
22 x+33=143
22 x=143-33
22x=110
x=5
∴ $10(x+3)+x=10(5+3)+5$
=80+5 = 85
∴ The two digit number is 85
Question 11
Let the unit's digits be ' x '
then ,the ten's digits number be 11 - x
The two digit number = 10 (11- x ) + x
When we intercharge the digits , the resulting new number
is greater than original number by 63
$\begin{aligned} 10(x)+(11-x) &=10(11-x)+x+63 \\ 10 x+11-x &=110-10 x+x+63 \\ 9 x+11 &=173-9 x \\ 9 x+9 x &=173-11 \\ 18 x &=162 \\ x &=9 \end{aligned}$
$\begin{aligned} \therefore 10(11-x)+x &=10(11-9)+9 \\ &=10(2)+9=20+9=29 . \end{aligned}$
∴ The two digit number is 29
Question 12
Let the Raju'S present age be ' x' year
then , Ritu's present age be 4x years
In four times , Ritu's age will be twice of raju's age
4 x+4=2(x+4)
4 x+4=2 x+8
4 x-2 x=8-4
2 x=4
x=2
∴ The present ages of Raju's , Ritu are 2 , 8 year
Question 13
Let the son's age be ' x ' years
then the father age be 7x years'
Two years ago , father was 13 times as old as his Son'
7 x-2=13(x-2)
7 x-2=13 x-26
13 x-7 x=26-2
6x =24
x=4
Their present age of Son, Father are 4 , 28
Question 14
Let the ages of sona and sonali are 5x , 3x
five years hence , the ratio of their ages were 10 : 7
$\frac{5 x+5}{3 x+5}=\frac{10}{7}$
Cross Multiplicatim
$\begin{array}{r}7(5 x+5)=10(3 x+5) \\ 35 x+35=30 x+50 \\ 35 x-30 x=50-35 \\ 5 x=15 \\ x-3\end{array}$
There fire their present ages are 15 ,9
Question 15
An employee works on a contract of 30 days
for that he will receive Rs 200 for each days and
he will be find Rs 20 fo each day he is absent
Let the number remain absent be 'x ' days
then the no of days he worked are 30 -x days
∴ $200(30-x)-20 x=3,800$
$6,000-200 x-20 x=3,800$
$220 x=6,000-3,80^{\circ}$
$220 x=2,200$
$x=10$
∴The no. of days he remained absent are 10 days
Question 16
Let the no of Rs 5 coins be x
no of Rs 2 coins be 3x
no of Rs 1 coins be 160 -4x
Total Rs 300 in coins of denominaton
∴ $5 x(x)+2(3 x)+1(160-4 x)=300$
$5 x+6 x+160-4 x=300$
$7 x=300-160$
$7 x=140$
$x=20$
Coins of each denominaton are
Rs Coins = 20
Rs 2 Coins = 60
Rs 1 coins = 80
Question 17
Let the no of passenger with Rs 5 tickets be 'X'
The no of passenger with Rs 75 tickets be 40 - x
Total receipts from passenger is Rs 230
$\begin{aligned} 5 x+2.5(40-x) &=230 . \\ 5 x+300-7 \cdot 5 x &=230 \\ 75 x-5 x &=300-230 \\ 2.5 x &=70 \\ x &=\frac{70}{2.5} \\ x &=28 \end{aligned}$
∴ Number of passenger with Rs 5 tickets are 28
Question 18
Let the no of students in the group be 'x '
They paid equally for use a of a ful boat and
pay Rs 10 each i.e 10$\times$x---------1
If there are 3 more students in group , each would have
paid Rs 2 less i.e Rs 8 i.e = 8 $\times$(x +3)
1 = 2
10 x=8(x+3)
10 x=8 x+24
2 x=24
x=12
Question 19
Let the number of deer in the herd be ' x '
half of a herd of deer are grazing in field i,e = $\frac{1}{2} x$
Three fourth of remaining are playing i,e = $\frac{3}{4}\left(\frac{1}{2} x\right)$
$=\frac{3}{8} x$
The rest 9 are drinking water
$x-\left\{\frac{x}{2}+\frac{3 x}{8}\right\}=9$
$x-\frac{4 x+3 x}{8}=9$
$\frac{8 x-7 x}{8}=9$
$\frac{x}{8}=9$
$x=72$
Number of deer in the herd are 72
Question 20
Let the no. of flower in the beginning be 'x'
At 1st temple she offer = $\frac{1}{2} \times x=\frac{x}{2}$
2nd temple she offer =$\frac{1}{2} \times \frac{x}{2}=\frac{x}{4} .$
3rd temple she offer = $\frac{1}{2} \times \frac{1}{4}=\frac{x}{8}$
Now she is left with 6 flower at end
$x-\left\{\frac{x}{2}+\frac{x}{4}+\frac{x}{8}\right\}=6 .$
$x-\frac{4 x+2 x+x}{8}=6$
$\frac{8 x-7 x}{8}=6$
$\frac{x}{8}=6$
$x=48$
ஃNo of flower in the beginning are 48
Question 21
Let the two supplementary angles be $x, 90-x$
These angles differ by $50^{\circ}$.
i.e $90-x-x=50$
$90-2 x=50$
$2 x=90-50$
$2 x=40$
$x=20^{\circ}$
∴ The two supplementary angles are 20 ,70
Question 22
Let the angles of triangle are 5x , 6x , 7x
Sum of angles of triangle= 180
i .e 5x + 6x + 7 x =180
18x = 180
x = 10
∴ The angles of triangles are $50^{\circ}, 60^{\circ}, 70^{\circ}$
Question 23
Two equal sides of an isosceles triangles are $3 x-1,2 x+2$
$3 x-1=2 x+2$
$3 x-2 x=2+1$
$x=3$
The third side is 2x = $2 \times 3=6$ Units
The two sides are $3 x-1=3 \times 3-1=9-1=8$ unith
$2 x+2=2 \times 3+2=6+2=8$ unih
∴ Perimeter of triangle = 6+8+8=22 Units
Question 24
Let the perimeter of given triangle be x cm
As each Side is increased by 4cm , So the
perimeter is increased by $3 \times 4=12 \mathrm{~cm}$
According to given information
$\frac{x+12}{x}=\frac{7}{5}$
Cross Multiplication
$7 x=5(x+12)$
$7 x=5 x+60$
$7 x-5 x=60$
$2 x=60$
x=30
∴ Perimeter of given triangle = 30cm
Question 25
Length of a rectangle is 5cm less than twice its breadth
i .e $l=2 b-5 \Rightarrow$ 2 b=l+5-------(1)
length is decreased by 3cm i .e (l-3)cm
breadth is increased by 2cm i .e (b+2)cm
Resulting perimeter of rectangle is 72cm
$2(1-3+b+2)=72$
$2 l+2 b-2=72$
Frameq (1) we get
$2 l+1+5-2=72$
$3 x+3=72$
$3 x=72-3$
$3 x=69$
$1=23 \mathrm{~cm}$
Length of rectangle $=23 \mathrm{~cm}$
From eq (1) $\quad 2 b=l+5=23+5$
$2 b=28 \Rightarrow b=14 \mathrm{~cm}$
∴Breadth of rectangle = 14 cm
∴Area of rectangle = l $\times$ b = $23 \times 14=322\mathrm{cm}^{2}$
Question 26
Length of rectangle l = 10cm
breadth of rectangle b = 8cm
Each side of rectangle is increased by x cm , its
perimeter is doubled
perimeter of rectangle = $\begin{aligned} 2(10+8) &=2 \times 18 \\ &=36 \mathrm{~cm} . \end{aligned}$
i.e $2[(10+x)+(8+x)]=2 \times 36,[\because$ perimeter is doubled]
$18+2 x=36-0$
$2 x=36-18=18$
$x=18 / 2=9$
=9 cm
Area of new rectangle i .e $\begin{aligned} &(10+9) \times(8+9) \\=& 19 \times 17=323 \mathrm{~cm}^{2} \end{aligned}$
Question 27
Let the speed of streamer in still water be x km\h
Given the speed of stream= 5km\h
The speed of streamer down steam = (x+5) km/h
Speed of streamer upwardstream= ( x-5)km/h
Both upstream and down stream takes same time
time taken by streams for down stream is $\frac{90}{x+5}$ hr
Time taken by streamer for upstream is $\frac{60}{x-5} h r$
i.e $\frac{90}{x+5}=\frac{60}{x-5}$
Cross Multiplication
$3(x-5)=2(x+5)$
$3 x-15=2 x+10$
$3 x-2 x=10+15$
$x=25$
∴ Speed of streamer in still water = 25km/h
Question 28
Let the speed of streamer in still water be 'x'
Given the speed of stream = 1km/h
Speed of streamer down stream= (x+1)km/h
Speed of streamer upstream = (x-1)km/h
Distance Covered by streamer , downstream= $5(x+1) \mathrm{km}$
Distance Covered by streamer upstream = $6(x-1) \mathrm{km}$
According to given information
$5(x+1)=6(x-1)$
$5 x+5=6 x-6$
$6 x-5 x=5+6$
$x=11$
∴ Speed of streamer in Still water is 11 km/h
Distance between two parts =$\begin{aligned} 5(x+1) &=5(11+1) \\ &=60 \mathrm{~km} . \end{aligned}$
Question 29
Let the speed of faster car be x km/hr
then the speed of other car be (x - 8)km/hr
Let the faster car starts from place A and the slower
car starts from place B
Let P and Q be their position after 4hours
( diagram should be added)
$A p=4 x \mathrm{~km}, \quad B Q=4(x-8) \mathrm{km}, p Q=62 \mathrm{~km}$
According to the given p b
$\begin{aligned} A P &+P Q+B Q=350 \\ 4 x+62+4(x-8)=350 \end{aligned}$
$4 x+62+4 x-32=350$
$8 x+30=350$
$8 x=350-30=320$
$x=320 / 8$
x=40 km/hr
∴ Speed of faster car is 40 km/hr and the speed of slower
car is (40-8) i.e 32km/h
Exercise 12.3
Question 1
(i) $x>-2$
Solution set $=\{-1,0,1,3\}$
(ii) $x<-2$
Solation et $=\{-7,-5,-3\}$
(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
It shown by thick dots on number line
(i) $x \leq 4, x \in N$
(diagram should be added)
(ii) $x<5, x \in W$
(diagram should be added)
(iii) $-3 \leq x<3, x \in \mathbb{L}$
(diagram should be added)
Question 3
Replacement Set =$\{-6,-4,-2,0,2,4,6\}$
$-4 \leq x<4$
(diagram should be added)
Question 4
(i) $\{1,2,3 \ldots \ldots .10\}$
(diagram should be added)
(ii) $\{-1,0,1,2,5,8\}$
(diagram should be added)
(iii) $\{-5,10\}$
(diagram should be added)
(iv) $\{5,6,7,8,9,10\}$
(diagram should be added) Solution Set=$\phi$
Question 5
(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
(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 $\{-1,0,1,2,3 \ldots .\}$
Question 7
(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) $-2(p+3)>5 \quad p \in I$
Mulhpliying '- ' 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 $\{\ldots-9,-8,-7,-6\}$
Question 8
(i) $2 x-3<x+2, x \in N$
$2 x<x+2+3$
$2 x-x<5$
$x<5$
As $x \in N$, the Solufion 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) $\frac{3}{2}-\frac{x}{2}>-1, \quad x \in N$
$\frac{3-x}{2}>-1$
$3-x>-2$
Mulhplying 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
$\{-3,-2,-1,0,1,2,3\}$
$\begin{aligned} \frac{3 x-1}{2} &<2 \\ 3 x-1 &<4 \\ 3 x-4 &<4+1 \\ 3 x &<5 \end{aligned}$
$x<5 / 3$
As $x$ should be in replacement set, the solution
set is $\{-3,-2,-1,0,1\}$
(diagram should be added)
Question 10
$\frac{x}{3}+\frac{1}{4}<\frac{x}{6}+\frac{1}{2}, x \in w$
$\frac{x}{3}-\frac{x}{6}+\frac{1}{4}<\frac{1}{2}$
$\frac{x}{3}-\frac{x}{6}<\frac{1}{2}-\frac{1}{4}$
$\frac{2 x-x}{6}<\frac{2-1}{4}$
$\frac{x}{6}<\frac{1}{4}$
$x<\frac{6}{4}$
$x<\frac{3}{2}$
As $x \in W$, the solution set is $\{0,1\}$
Question 11
(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 $\{-1,0,1,2,3\}$
(diagram should be added)
(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 $\{-3,-2,-1,0,1,2,3,4\}$
(diagram should be added)
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