Linear Inequalities Class 11

Ex5.1, 1 teachoo.com
Solve 24x < 100, when
(i) x is a natural number
24x < 100
< 100
YS 4
<2
X<G
x< 4,16
Since x is a natural number i.e. 1, 2, 3, 4, 5, 6, 7,....
We need to find values of x which is less than or equal to 4
i.e. x can be {1, 2, 3, 4}

Ex 5.1, 2 teachoo.com
Solve -12x > 30, when
(i) xis a natural number
—12x > 30
-x> 22
x?
—-x> 2.5
Since x is negative, we multiply both sides by
-1 & change the signs
(-x) x (-1) < 2.5 x (-1)
x< -2.5
{i) Since x is a natural number,
it can be only positive number starting from 1 (i.e. 1, 2, 3, 4, 5,....)

Ex 5.1, 3 teachoo.com
Solve 5x- 3 <7, when
(i) xis an integer

5x-3<7

5x <7+3

5x < 10

5

x<2
Since x is an integer (......,-3,-2,-1,0,1,2,3....)
We need to find values of x which is less than 2
ie. x can be......-3,-2,-1,0,1

= {.....—3, -2, -1, 0, 1}

Ex 5.1, 4 teachoo.com
Solve 3x + 8 > 2, when
(i) xis an integer
3x+8>2
integers: .....,-2, -1, 0, 1, 2, 3,....
3x>2-8
3x >-6
-6
x >
3
x>-2
Since x is an integer (......,-3, -2, -1, 0, 1, 2, 3...)
We need to find values of x which is greater than -2
i.e. x can be -1, 0, 1, 2, 3, 4)...
= {-1, 0, 1, 2, 3, 4,......}

Ex5.1, 5 teachoo.com
Solve the given inequality for real x: 4x + 3<5x+7
4x+3<5x+7
4x -5x< 7-3
—-x<4
Since x is negative, we multiply both sides by -1 & change the signs
-1x (-x) > (-1) x4
x>-4
Since x is a real number greater than — 4
Thus x & (4, ©}

Ex5.1, 6 teachoo.com
Solve the given inequality for real x: 3x -7 >5x-1
3x-7>5x-1
-—2x>6
_x>$
2
—x>3
Since x is negative, we multiply both sides by -1 & change the signs
-1 x (-x) > (-1) x3
x<-3
Since x is a real number less than —3
Thus x € (-° , -3)

Ex5.1,7 teachoo.com
Solve the given inequality for real x: 3(x — 1) < 2 (x - 3)
3(x — 1) < 2(x- 3)
3x-3S52x-6
3x-2x< -6+3
xs-3
Since x is a real number less than or equal to — 3
Thus x € (-° , -3]

Ex5.1, 8 teachoo.com
Solve the given inequality for real x: 3(2 — x) > 2(1-x)
3(2 —x) = 2(1-x)
6- 3x 22-2x
6-3x+2x22
6-x22
-x 22-6
—x2—-4
Since x is negative, we multiply both sides by -1 & change the signs
(-1)x (-*)s (-1)x (-4)
xs4
Hence x is a real number which is less than or equal to 4,
Thus, x € (-©, 4].

Ex 5.1, 9 teackoo.com
Solve the given inequality for real x: x + 5 + ; <11
X+ 5 + ; <1l1
x(14+5+5)<11
x (SAS?) ca
x(=)<11
x<11x (=)
x<6
Hence x is a real number which is less than 6,
Thus, x € (-2°, 6}

Ex 5.1, 10 teachoo.com
x x
Solve the given inequality for real x: 375t 1
x xX
zesti
3° 2
x x
saz? 1
3 2
2x —3x
— >1
6
—x
—>l
6
—x>6
Since x is negative, we multiply both sides by -1 & change the signs
(- 1) x (-x) <(-1) x (6)
x<-6

Ex 5.1, 11 teachoo.com
3(x-2) _5(2-
Solve the given inequality for real x: << < “eo
3(x —-2) < 5(2 -x)
5 3
3x 3(x-2) <5 x 5(2-x)
9(x—2) < 25 (2-x)
9x - 18 < 50- 25x
9x + 25x = 50+ 18
34x < 68
68
xs
34
xS2
Hence, x is a real number which is less than or equal to 2
Hence, x € (—°°, 2] is the solution

Ex 5.1, 12 teachoo.com
Solve the given inequality for real x: ; (= + 4) 2 ; (x — 6)
1 {3x 1
a(E+4) 250-8)
3x
3(=+4) = 2 (x-6)
3x 43x42 2x-2x6
= +12 2 2x-12
= = 2x 2-12-12
9x -— 5(2x)
~~ = —-24
ox -— 10x > -24
5
=> -24
5

Ex 5.1, 13 teachoo.com
Solve the given inequality for real x: 2(2x + 3}-— 10 < 6 (x — 2}
2(2x + 3) -10 < 6 (x- 2)
4x+6-10<6x-12
4x-6x< -12+4
-2x< -8
-+x< 2
2
—x< —4
Since x is negative, we multiply both sides by-1 & change the signs
(-1) x (-x)> (-1)x (-8)
x>4
Since x is a real number which is greater than 4
Hence, x € (4,°°) is the solution

Ex5.1, 14 teackoo.com
Solve the given inequality for real x: 37 — (3x + 5) = 9x — 8(x - 3)

37 — (3x + 5) > 9x — 8(x - 3)

37 - 3x-5 2 9x — 8x + 24

32-3x2x+24

-3x -—x 224-32

-4x2-8

x> 3

4

—-x 2-2
Since x is negative, we multiply both sides by -1 & change the signs

(-1)x (-x)s (-1)x (-2)

xs 2

Ex 5.1, 15 teachoo.com
. : : 5x —2 -3
Solve the given inequality for real x: *< on?) ~a=3
x. Ge-2)_ (x-3)
4 3 5
x < 5(5x — 2) -3 (7x - 3)
4 3(5)
x 25x -10-21x4+9
TT
4 15
x 25x —-21x+9-10
ST
4 15
x Ax-1
4M
4 15
15x < 4 (4x - 1)
15x < 16x-4
15x -16x<-4
—x<—4
Since x is negative, we multiply both sides by -1 & change the signs

Ex 5.1, 16 teachoo.com
2x-1 3x-2 2-
Solve the given inequality for real x: a = ao? - eo
(2x —1) > (3x — 2) _ (2 — x)
30° 4 5
(2x —1) > 5(3x —2)—-4(2 — x)
307 4(5)
@x-De 15x-10-8+4x
30 7 20
@x-De 15x+4x-10-8
30 7 20
(2x -1) > 19x —18
3 20
20 (2x — 1) > 3 (19x - 18)
40x-20 > 57x-54

Ex 5.1, 17 teachoo.com
Solve the given inequality and show the graph of the solution on
number line: 3x —2 < 2x +1

3x-2<2x+1

3x-—2x<1+2

x<3
Hence, solution is (—e, 3}
The graphical representation is

——_—_—_—AaseoOoSoS oo

4-3-2 -10 1 2 3

Here, 3 is not included in the shaded graph

Ex 5.1, 18 teachoo.com
Solve the given inequality and show the graph of the solution
on number line: 5x-323x-5

5x-323x-5

5x—-3x2 -5 +3

2x 2-2

2x -2

—>—

27 2

x2-1
The graphical representation is

ERE e——_— >
4-3-2101 23 4

Here -1 is included in the shaded graph

Ex5.1, 19 teachoo.com
Solve the given inequality and show the graph of the solution on
number line: 3(1-x)< 2 (x+4)
3(1-x) <2 (x +4)
3-3x<2x+8
-—3x-2x< 8-3
-5x<5
Dividing both sides by 5
—5x 5
3 55
—x<1
Since x is negative, we multiply both sides by -1 & change the signs
(-1)x (-x) > (1) (1)
x>—-1

Ex 5.1, 20 teachoo.com
Solve the given inequality and show the graph of the solution on
number line:
x. Gx 2) (7x — 3)
2 3 5

x x= 2) (7x — 3)

2° 3 5

xe 5(5x — 2) — 3(7x — 3)

20 3x5

% 25x 10-2149

20 15

x 4x-1

>2—_—

2 15

15x > 2(4x — 1)

15x 2 8x -2

Ex5.4, 21 teachoo.com
Ravi obtained 70 and 75 marks in first two unit test. Find the
minimum marks he should get in the third test to have an

average of at least 60 marks.

Marks in 15 test = 70

Marks in 2"9 test = 75

Let marks in 3 test = x

Let x be the marks obtained by Ravi in the third unit test.

Average marks = ae

Ex 5.1, 22 teachoo.com

To receive Grade ‘A’ in a course, one must obtain an average of 90

marks or more in five examinations (each of 100 marks). If Sunita’s

marks in first four examinations are 87, 92, 94 and 95, find

minimum marks that Sunita must obtain in fifth examination to get

grade ‘A’ in the course.

Marks in 1%t examination = 87

Marks in 2"¢ examination = 92

Marks in 3 examination = 94

Marks in 4" examination = 95

Let x be the marks obtained by Sunita in the 5" examination
87+924+94495 +x

Average marks = 7

Ex 5.1, 23 teachoo.com
Find all pairs of consecutive odd positive integers both of
which are smaller than 10 such that their sum is more than 11.
Let the smaller odd positive integer be x
Since the larger integer is consecutive odd,
Given,
Both integers are smaller than 10, | Sum of the two integers is more
ie.x<10 & x+2<10 than 11.
x< 10-2 xt (x42) > 11
x<8 2x+2>11
2x>11-2
Sincex<10&x<8 x>9
x<8
9
X>=
2
x>4.5

Ex 5.1, 24 teachoo.com
Find all pairs of consecutive even positive integers, both of
which are larger than 5 such that their sum is less than 23.
Let the smaller even positive integer be x
Since the larger integer is consecutive even ,
it will be x +2
Given that, Also, given that
both integers are larger than 5, | Sum of the two integers is less
ie.x>5 than 23.
&x4+2>5 “XxX + (x +2) < 23
x>5-2 2x+2<23
x>3 2x < 23-2
2x<21
Sincex>3&x>5 xc
>5
x x< 10.5

Ex 5.1, 25 teachoo.com
The longest side of a triangle is 3 times the shortest side and the
third side is 2 cm shorter than the longest side. If the perimeter of
the triangle is at least 61 cm, find the minimum length of the
shortest side.
Let the length of the shortest side be x cm.
So, length of longest side = 3 times shorter side
= 3x
& Length of the third side = (3x— 2) cm
Now,
Perimeter = Sum of three sides
=x + 3x + (3x-2)

Ex 5.1, 26 teachoo.com
A man wants to cut three lengths from a single piece of board of
length 91 cm. The second length is to be 3 cm longer than the
shortest and the third length is to be twice as long as the
shortest. What are the possible lengths of the shortest board if
the third piece is to be at least 5 cm longer than the second?
[Hint: If x is the length of the shortest board, then x, (x + 3) and
2x are the lengths of the second and third piece, respectively.
Thus, x + (x +3) + 2x $91 and 2x 2 (x +3) +5]
Let the length of the shortest board be x cm.
Length of second board = 3 cm longer than the shortest side

=x+3 cm
& Length of the third board = Twice the shortest board

=2xcm

Example 1 teachoo.com
Solve 30 x < 200 when
(i) x is a natural number
30 x < 200
200
*< 30
<2
FSG
x <6.66
Since x is a natural number i.e. 1, 2, 3, 4, 5, 6, 7,....
We need to find values of x which is less than or equal to 6
i.e. x can be {1, 2, 3, 4, 5, 6}

Example 2 teachoo.com
Solve 5x — 3 < 3x +1 when
(i) x is an integer,

5x-3<3x+1

5x - 3x <1+3

2x<4

x<4

2

x<2
Since x is an integer (......,-3, -2, -1, 0, 1, 2, 3...)
We need to find values of x which is less than 2
ie. x can be ......-3, -2,-1,0, 1

= (140-3, —2, -1, 0, 1}

Example 3 teackoo.com
Solve 4x + 3 < 6x +7.
4x+3<6x+7
4x -6x< 7-3
-2x<4
+x<i
2
—x<2
Since x is negative, we multiply both sides by -1 & change the signs
-1 x (-x) > (-1)x 2
X>-2
Since x is a real number greater than — 2
Thus x € (-2, °)

Example 4 teachoo.com
5-2x x
Solve ——< - —5.
3 6

5-—2x x
>. £2.75
3 6
5-—2x x —30
—_ < —————_
3 6
5-2x
6x —— <x-30
3
2(5 — 2x) < x- 30.
10- 4x < x-30
-4x-x <- 30-10
-5x <=- 40
<t0
XS 5

Example 5 teachoo.com
Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number
line.
7X+3<5x+9
7x-—5x < 9-3
2x <6
2x 6
ee
2°72
x<3
The graphical representation is
——_—_—_—_—_—s———
4 -3 -2 -1 0 1 2 3
Here 3 is not included in the shaded graph

Example 6 teachoo.com
Solve at = = —1. Show the graph of the solutions on
number line.
3x-4 > x+1 -1
2 4
3x — 45 241-@
2 4
3x-4 > x-3
2 4
A(3x — 4) = 2(x - 3)
12x-16 >2x-6
12x-—2x >-6+16
10x > 10
> 10
x2 10
x21

Example 7 teackoo.com
The marks obtained by a student of Class XI in first and second
terminal examination are 62 and 48, respectively. Find the
minimum marks he should get in the annual examination to
have an average of at least 60 marks.
Marks in first terminal examination = 62
Marks in second terminal examination = 48
Let x be the marks obtained in annual examination

62 +48
Average marks = a

Example 8 teachoo.com
Find all pairs of consecutive odd natural numbers, both of
which are larger than 10, such that their sum is less than 40.
Let the smaller odd natural number be x
Since the larger integer is consecutive odd,
it will bex+2
Given that,
Both integers are larger than 10 |Sum of the numbers is less than 40
ie.x>10 &x+2>10 “x + (x+2)< 40
x>10-2 2x+2< 40
x>8 2x < 40-2
Sincex>10&x>8 2x < 38
>
x> 10 x<19

Example 9 teackoo.com
Solve -8 <5x-3<7.

-8<5x-3<7
Adding 3 all sides (Eliminating -3)

-8+3<5x-3+3<7+3

-5 <5x<10
Dividing 5 all sides (Eliminating 5)

—-5 5x 10

eee

5° 5 5

-1<x<2
Thus, xis a real number which is less than 2 and greater than
or equal to -1
Hence, x € [-1, 2) is the solution

Example 10 teackoo.com

Solve -5 <a> <3.
-5<-=<sg

Multiplying 2 all sides (Eliminating 2)
-5x2<5-3x<8x2
-10<5-3x<16

Subtracting 5 all sides (Eliminating 5)
-10-5<5-3x-5<16-5
-10-5<5-3x-5<16-5
-15<-3x<11

Dividing 3 all sides (Eliminating 3}
15 -3x 11
3 Sa Sa
S<-xS >

teachoo.co
Example 11 eae om
Solve the system of inequalities:
3x-7<5+x
11-5x<1
and represent the solutions on the number line.
Solving Solving
3x-7<5+x 11-5xs1
3x-x<5+7 —5xs1-11
2x< 12 —5xs— 10
x< rte
12 ~xS
x<>
-xs-2
x<6 . . . .
Since x is negative, we multiply both
sides by -1 & change the signs
(-1)x (-x) 2 (-1) x (-2)
x22

Example 12 (Method 1) teachoo.com
In an experiment, a solution of hydrochloric acid is to be kept
between 30° and 35° Celsius. What is the range of temperature in
degree Fahrenheit if conversion formula is given by C = ; (F - 32),
where C and F represent temperature in degree
9
F=2C+32
Given that the solution is to be kept between 30°C and 35°C,
Putting C = 30 Putting C = 35
F=2x30+32 F=2x 35 +32
=9x6+32 =9x7+32
= 54+ 32 =63 + 32
= 86 =95

Example 13 teachoo.com
A manufacturer has 600 litres of a 12% solution of acid. How many
litres of a 30% acid solution must be added to it so that acid content
in the resulting mixture will be more than 15% but less than 18%?
Quantity of existing solution = 600 litres
Amount of acid in existing solution = 600 x 12%
12
= 600 x rn}
= 72 litres
Let the quantity of 30% acid solution to be added be x
Amount of acid in the added solution = 30% of x
30
= 700 ** = 0.3x

teachoo.com
Example 9
Solve 3x + 2y > 6 graphically.
3x+2y>6
Lets first draw graph of
x
3x+2y=6
y
Putting x = Oin (1) Putting y = O in (1)
3(0} + 2y=6 3x + 2(0) =6
0+2y=6 3x+0=6
2y=6 3x=6
6 -6
Y=3 x3
y=3 x=2

teachoo.com
Example 10
Solve 3x — 6 2 0 graphically in two dimensional plane.
3x-62 30
Lets first draw graph of
3x-6=0
3x= 6
| x eee
x=2 0 -13
3
x=2
At x = 2, y can have any value

teachoo.com

Example 11
Solve y < 2 graphically.

y<2
Lets first draw graph of

y=2 Bago -14

222

At y = 2, x can have any value

teachoo.com
Example 12
Solve the following system of linear inequalities
graphically. x+y25,x-ys3
First we solve x+y25
Lets first draw graph of nn
x+y=5 Ly | 5/0 |
Putting y = 0 in (1} Putting x = O in (1)
x+0=5 O+y=5
x=5 y=5

Example 13 teackoo.com
Solve the following system of inequalities graphically:
5x + 4y $40,x22,y23
First we solve 5x + 4y s 40
> oe
Lets first draw graph of 10 0
5x + 4y = 40
Putting x = 0 in (1) Putting y = 0 in (1)
5(0) + 4y = 40 5x + 4(0) = 40
0+ 4y= 40 5x +0=40
4y = 40
Y 5x = 40
_ 40
YoU 40
xT
y= 10
x=8

Example 14 feachoo.com
Solve the following system of inequalities : 8x + 3y < 100,x20,y
20
First we solve 8x + 3y 2 100
Lets first draw graph of
8x + 3y = 100 | x |_o has]
Putting x = 0 in (1) Putting y = Oin (1)
8(0) + 3y = 100 8x + 3(0) = 100
0+ 3y=100 8x +0=100
3y = 100 8x = 100
_ 100 _ 100
yu XeG
y= 33.33 x=12.5

Example 15 teackoo.com
Solve the following system of inequalities graphically
x+2ys8,2x+ys8 ,x20,y20
First we solve x+ 2y< 8
Lets first draw graph of
x+2y=8 (1) | | 0 8
| y [eae
Putting x =Oin(1) | Putting y = Oin (1)
0+2y=8 x + 2(0) =8
2y=8 x+0=8
8 x=8
Y=
y=4

Misc 1 teachoo.com
Solve the inequality 2< 3x-4<5
2<3x-4<5
Adding 4 both sides (Eliminating 4)
24+4<3x-44+455+4
6<3x<9
Dividing 3 all sides (Eliminating 3)
6 63% 9
3°33
26x83
Thus, x is a real number which is less than or equal to 2 and
greater than or equal to 3
“x € [2, 3] is the solution

Misc 2 teachoo.com
Solve the inequality 6 < —3(2x — 4) < 12
6 <— 3(2x—4)<12
6<-—6x+12<12
Subtracting 12 all sides (Eliminating 12)
6-12 <—6x+12-12<12-12
-6<- 6x <0
Dividing 6 all sides (Eliminating 6)
=6 .-6% 0
6 6 6
—-1<-x<0
Since x is negative, we multiply both sides by -1 & change the signs
(-1)« (-1) 2 (-1)x (>) >(-1)~ (0)
12> x>0

Misc 3 teachoo.com

Solve the inequality -3 <4 -= <= 18
3<4-2<18

Subtracting 4 all sides (Eliminating 4)
-3-4<4-%-4<18-4
7s<-teu

Multiplying 2 all sides (Eliminating 2)
-7x2<-2x2<14x2
-14< —7x <28

Dividing 7 all sides (Eliminating 7}
a4 Tt 8

7 7 7

Misc 4 teachoo.com
3(x-2
Solve the inequality —15 < <2 <0
3(x-2
15 <2) 6g
5
Multiplying 5 all sides (Eliminating 5)
3(x — 2)
-15 x5<5x ~~ 5x0
-75 < 3(x-2)<0
Dividing 3 all sides (Eliminating 3)
TTS 372) 60
3 3 3
—25<x-2<0
Adding 2 all sides (Eliminating 2)
-254+2<x-2+2s0+2
—23<x<2

Misc 5 teachoo.com
. . 3x
Solve the inequality —12 <4 — = <2
12<4 —-— 3x <2
—5
Subtracting 4 all sides (Eliminating 4)
3x
-12-4<4 — = -4s2-4
3x
—16<— a <-—2
3x
—16< 7 <-2
Multiplying 5 all sides (Eliminating 4)
3x
—16x5<— x5s—2x5
-80< 3x <-10

Misc 6 teachoo.com
Solve the inequality 7 < ot <11
73S eu
Multiplying 2 all sides(Eliminating 2)
7x2<(3x+11)< 112
14< 3x+11<5 22
Subtracting 11 all sides(Eliminating 11)
14-11 <3x—-115 22-11
3<3x<11
Dividing 3 all sides(Eliminating 3}
ist
t<x<>
3

Misc 7 teachoo.com
Solve the inequalities and represent the solution graphically
onnumberline: 5x+1>-24,5x-1<24
Solving Solving
5x+1>—-24 5x-1<24
5x >-24—1 5x <24+1
5x >-25 5x<25
-25
Xo x< 2
5
x>—5
x<5
Thus, x>-5&x<5

Misc 8 teachoo.com
Solve the inequalities and represent the solution graphically on
numberline: = 2(x-—1)< x +5, 3(x+2)>2-x
Solving Solving
2(x—1)<x45 3(x+2)>2-x
Wx-2<xt5 3x+6>2—x
x—-x<542 3x+x>2-6
xe? 4x >-4
-4
x>—
4
x>-1
Thus,x<7 & x>-1

. teachoo.com
Misc 9
Solve the following inequalities and represent the solution
graphically on number line:
3x-7 > 2(x-6),6-—x>11-2x
Solving Solving
3x-7 > 2(x-6) 6-x>11-2x
3x—7>2x—-12 -x+2x>11-6
3x-2x>-12+7 x>5
x>-5
Thus,x>-5&x>5

Misc 10 teachoo.com
Solve the inequalities and represent the solution graphically on
number line: 5(2x — 7) — 3(2x + 3) < 0, 2x +19 < 6x +47
Solving Solving
§(2x — 7} — 3(2x + 3} $0 2x +19 $ 6x +47
10x-— 35-6x-9<0 2x — 6x < 47-19
4x-44<0 Ax < 28
A(x—11) <0 oe
4
x “4 -xs7
x—11<0 Since x is negative, we multiply
xsi both sides by -1 & change the signs
(1) x (-x) 2 (-1) (7)
x2-7

Misc 11 (Method 1) teachoo.com
A solution is to be kept between 68°F and 77°F. What is the range in
temperature in degree Celsius (C) if the Celsius/Fahrenheit (F)
conversion formula is given by F = - C+32?
9
F=-C+32
Given that the solution is to be kept between 68°F and 77°F,
Putting F = 68 Putting F = 77
68 == C +32 77 =2C +32
9 9
68-32=—-C 77 —32=-C
9 9
36==C 45 =—=C
5 5
C=36x 3 C=45x 3
c=4x5 C=5x5
c=20 C=25

Misc 12 teachoo.com
A solution of 8% boric acid is to be diluted by adding a 2% boric
acid solution to it. The resulting mixture is to be more than 4% but
less than 6% boric acid. If we have 640 litres of the 8% solution,
how many litres of the 2% solution will have to be added?

Let x litres of 2% boric acid solution is required to be added.

Then, total mixture = (x + 640) litres

This resulting mixture is to be more than 4% but less than 6%
boric acid.

«= 2% of x + 8% of 640 > 4% of (x + 640)

And, 2% of x + 8% of 640 < 6% of (x + 640)

Misc 13 teachoo.com
How many litres of water will have to be added to 1125 litres of
the 45% solution of acid so that the resulting mixture will
contain more than 25% but less than 30% acid content?
Volume of existing solution = 1125 litres
Amount of acid in it = 45% of 1125
Hence, amount of water in it = 55% of 1125
55

= Too * 1125
Let amount of water added be x litres
So, Volume of new solution = 1125 + x

Misc 14(Method 1) teachoocom
IQ of a person is given by the formula 1Q = = x 100, where MA is
mental age and CA is chronological age. If 830 < 1Q < 140 fora
group of 12 years old children, find the range of their mental age.
We know that

MA
IQ= aX 100,
Current age of children = 12
CA =12
Calculating Mental Age at IQ = 80 & IQ = 140

Ex 6.2,1 teachoo.com
Solve the following inequalities graphically in two-dimensional
plane:xt+y<5
xty<5
Egos
Lets first draw graph of 5 0
x+y=5
Putting x = 0 in (1) Putting y = 0 in (1)
O+y=5 x+0=5
y=5 x=5

teachoo.com
Ex6.2, 2
Solve the given inequality graphically in two-dimensional plane:
2x+y26
2x+y2 6
x
Lets first draw graph of y
2x+y=6
Putting x = 0 in (1) Putting y = 0 in (1)
2(0}+y=6 2x + (0)=6
O+y=6 2x=6
y=6 x=2
2
x=3

teachoo.com
Ex 6.2, 3
Solve the given inequality graphically in two-dimensional
plane: 3x + 4y < 12
3x + 4y < 12 | x | 04
Ra: ©
Lets first draw graph of La 3
3x + 4y = 12
Putting x = 0 in (1) Putting y = 0 in (1)
3(0) + 4y = 12 3x + 4(0) = 12
0+4y=12 3x+0=12
4y =12 3x=12
12 _12
ven x3
y=3 x=4

Ex 6.2, 4 teachoo.com
Solve the given inequality graphically in two-dimensional plane:
y+822x
y+82 2x
Ego.
Lets first draw graph of -8 0
y+8=2x
Putting x = 0 in (1) Putting y = Oin (1)
y+8=2(0) 0+8=2x
y+8=0 8 = 2x
y=0-8 2x=8
y=-8 8
x=>
2
x=4

teachoo.com
Ex6.2, 5
Solve the given inequality graphically in two-dimensional
plane: x-y<2
x-ys2 x
y
Lets first draw graph of
x-y =2
Putting x = 0 in (1) Putting y = 0 in (1}
O-y =2 x-0 =2
-y=2 x-0 =2
=2
y=-2 *

Ex 6.2, 6 teachoo.com
Solve 2x — 3y > 6 graphically.
2x-3y>6
Ego 3
Lets first draw graph of
2x—3y=6
Putting x = 0 in (1} Putting y = O in (1)
2(0)—3y =6 2x — 3(0) =6
0-3y=6 2x-0=6
—3y=6 2x=6
6 6
y= 3 x= z
y=-2 x=3

Ex 6.2,7 teachoo.com
Solve the given inequality graphically in two-dimensional plane: —
3x + 2y 2-6
-3x + 2y 2-6
Ego 2
Lets first draw graph of ra: 0
—3x+ 2y =-6
Putting x = 0 in (1) Putting y = 0 in (1)
-3(0) + 2y =-6 -3x + 2(0) = -6
0+ 2y=-6 -3x+0=-6
2y =-6 -3x = -6
y= x=o8
2 -3
y=-3 x=2

Ex 6.2, 8 teachoo.com
Solve the given inequality graphically in two-dimensional plane:
3y — 5x < 30
3y— 5x < 30 Lx | 0 -6
Lets first draw graph of ly | 10 0
3y— 5x = 30
Putting x = 0 in (1) Putting y = 0 in (1)
3y—5(0) = 30 3(0) — 5x = 30
3y—0= 30 0-5x = 30
3y = 30 —5x = 30
_30 x= 6
Yr
y=10

teachoo.com

Ex6.2, 9
Solve the given inequality graphically in two-dimensional
plane: y < -2

y <-2
Lets first draw graph of

y= -2 y

At y =-2, x can have any value

teachoo.com
Ex6.2, 10
Solve the given inequality graphically in two-dimensional plane:
X>—3
x >-3
Lets first draw graph of
x=-3
x
At x = -3, y can have any value

Ex 6.3, 1 teachoo.com
Solve the following system of inequalities graphically:
x23,y22
First we solve x 2 3
| x | 3.3 3

Lets first draw graph of

x=3
At x = 3, y can have any value

teachoo.com
Ex 6.3, 2
Solve the following system of inequalities graphically: 3x + 2y < 12,
x21y22
First we solve 3x + 2y s 12
Lets first draw graph of lx |o| 4]
3x+2y=12 by | so)
Putting x = 0 in (1} Putting y = 0 in (1)
3(0) + 2y = 12 3x + 2(0} = 12
O+2y= 12 3x+0=12
2y=12
Y 3x =12
_ 12
vez 12
x= -
y= 6
x=4

teachoo.com
Ex6.3, 3
Solve the following system of inequalities graphically: 2x + y> 6,
3x + 4y $12
First we solve 2x +y > 6
Lets first draw graph of |x [0] 3|
Ly | 6} 0]
ax+y=6
Putting y=Oin(1) | Putting x = O in (1)
2x + (O)=6 2(0)+y=6
2x =6 O+y=6
xa8 y=6
2
x=3

teachoo.com
Ex6.3, 4
Solve the following system of inequalities graphically:
x+y24,2x-y<0
First we solve x+y 24
Lets first draw graph of |x | 0] 4|
x+y=3 | y | 4] 0]

Putting x = 0 in (1) Putting y = O in (1)

O+y=4 Xx+0=4

y=4 x=4

Ex6.3, 5 teachoo.com
Solve the following system of inequalities graphically: 2x
-y>1,x-2y<-1
First we solve 2x-y>1
| x | ofos |
Ly |i] 0 |
Lets first draw graph of
ax-y=1
Putting x = 0 in (1) Putting y = Oin (1)
2(0)-y=1 2x-(0)=1
O-y=1 2x=1
-y=il yet
2
yet x=0.5

teachoo.com
Ex 6.3, 6
Solve the following system of inequalities graphically:
x+ys6,x+y24
First we solve x+y $6
Lets first draw graph of | x | 0 6
6 0
xt+y=6
Putting y = 0 in (1) Putting x = 0 in (1)
x+0=6 O+ry=6
x=6 y=6

Ex 6.3,7 teachoo.com
Solve the following system of inequalities graphically:
2x+y28,x+2y210
First we solve 2x+y2 8
Lets first draw graph of | ° ;
2x+y=8
Putting x = 0 in (1) Putting y = 0 in (1)
2(0)+y=6 2x +{0)=6
O+y=6 2x=8
y=8 xa
2
x=4

teachoo.com
Ex 6.3, 8
Solve the following system of inequalities graphically:
Xx+ys9,y>x,x20
First we solve x+ y $9
|<

Lets first draw graph of ly | 9 0

x+y=9 (1)
Putting x = 0 in (1) Putting y = 0 in (1)

O+ry=9 x+0=9

y=9 x=9

Ex6.3, 9 teachoo.com
Solve the following system of inequalities graphically: 5x + 4y < 20,
x21y22
First we solve 5x + 4y s 20 rx a
Lets first draw graph of Ly} 5 0.
5x + 4y = 20 (1)
Putting x = 0 in (1) Putting y = 0 in (1)
5(0) + 4y = 20 5x + 4(0) = 20
O+4y= 20 5x +0=20
Ay = 20
Y 5x = 20
_ 20
yaa 20
x ==
y=5
x=4

Ex 6.3, 10 teachoo.com
Solve the following system of inequalities graphically:
3x + 4y < 60,x + 3y $30,x20,y20
Now we solve 3x + 4y < 60
Lets first draw graph of | x | 0 20
3x + 4y = 60

Putting x = 0 in (1) Putting y = 0 in (1)

3(0) + 4y = 60 3x + 4(0) = 60

0+ 4y=60 3x+0=60

Ay = 60 3x = 60

60 60
Yr XE>
y=15 x=20

Ex 6.3, 11 teachoo.com
Solve the following system of inequalities graphically:
2x+y24,x+ys3,2x—-3y<6
First we solve 2x+y2 4
Lets first draw graph of x | 02
40
ax+y=4
Putting x = 0 in (1) Putting y = 0 in (1)
2(0)+y=4 2x + (0) =4
O+y=4 2x=4
y=4 4
xX=>
2
x=2

Ex 6.3, 12 teachoo.com
Solve the following system of inequalities graphically:
x—2y $3,3x+4y212,x20,y21
First we solve x—2y > 3
Lets first draw graph of | x | 0 3
x-2y=3
Putting x = 0 in (1) Putting y = 0 in (1)
O-2y=3 x—2(0) =3
-2y=3 x-0=3
3 =
yes x=3
y=-1.5

Ex6.3, 13 feachoo.com
Solve the following system of inequalities graphically:
4x + 3y $ 60, y 2 2x, x23,x,y20
Now we solve 4x + 3y < 60 | x | 0 [45|
Lets first draw graph of | y [20] 0 |
4x+3y=60 _ ...(1)
Putting x = Oin (1) Putting y = 0 in (1)
(0) + 3y = 60 4x + 3(0) = 60
0+ 3y=60 4x+0=60
3y = 60 4x = 60
60 - 60
y= 7 x= Z
y =20 x=15

Ex 6.3, 14 teachoo.com
Solve the following system of inequalities graphically:
3x + 2y $150, x + 4y $ 80,x<15,y20,x20
Now we solve 3x + 2y < 150
Lets first draw graph of | x | 0 50
3x+2y=150 __ ...(1) 750
Putting x = 0 in (1) Putting y = 0 in (1)
3(0) + 2y = 150 3x + 2(0) = 150
0+ 2y=150 3x+0=150
2y = 150 3x = 150
_ 150 _ 150
Yr Xe"
y=75 x=50

Ex 6.3, 15 teachoo.com
Solve the following system of inequalities graphically:
X+2y<s10,x+y21,x-ys0,x20,y20
First we solve x+2y< 10
Lets first draw graph of x | 0 10
x+2y=10 (2) 5 0
Putting x = 0 in (1) Putting y = 0 in (1}
0+2y=10 x + 2(0) = 10
2y =10 x+0=10
y= 10 x=10
2
y=5