Question-1
Solution:
3x â€“ 7 > x + 3
3x â€“ 7 + 7 > x + 3 + 7
3x > x + 10
3x â€“ x > x + 10 - x
2x > 10
x > 5
Question-2
Solution:
x + 12 < 4x â€“ 2
x + 12 - 12< 4x â€“ 2 â€“ 12
x < 4x â€“ 14
x â€“ 4x < 4x â€“ 14 â€“ 4x
- 3x < -14
- 3x/-3 > -14/-3
x > 14/3
Question-3
Solution:
4x â€“ 7 < 3 â€“ x
4x â€“ 7 + 7 < 3 â€“ x + 7
4x < 10 â€“ x
4x + x < 10 â€“ x + x
5x < 10
x < 2
Question-4
Solution:
3x + 17 â‰¤ 2(1 - x)
3x â‰¤ - 2x - 15
3x + 2x â‰¤ - 2x â€“ 15 + 2x
5x â‰¤ â€“ 15
5x/5 â‰¤ â€“ 15/5
x â‰¤ â€“ 3
Question-5
Solution:
â€“2x + 6 â‰¥ 5x â€“ 4
â€“2xâ‰¥ 5x â€“ 10
â€“2x â€“ 5x â‰¥ 5x â€“ 10 â€“ 5x
â€“7x â‰¥ â€“ 10
â€“7x/-7 â‰¤ â€“ 10/-7
x â‰¤ 10/7
Question-6
Solution:
â€“(x - 3) + 4 > - 2x + 5
â€“x + 7 > - 2x + 5
â€“x + 7 â€“ 7 > - 2x + 5 â€“ 7
â€“x > - 2x - 2
x < 2x + 2
x â€“ 2x < 2x + 2 â€“ 2x
- x < 2
x > -2
Question-7
Solution:
2(2x + 3) â€“ 10 < 6(x - 2)
4x + 6 â€“ 10 < 6x - 12
4x â€“ 4 < 6x - 12
4x â€“ 4 + 4 < 6x â€“ 12 + 4
4x < 6x - 8
4x â€“ 6x < 6x â€“ 8 â€“ 6x
- 2x < â€“ 8
- 2x/-2 > -8 /-2
x > 4
Question-8
Solution:
2 - 3x â‰¥ 2x + 12
2 - 3x â€“ 2 â‰¥ 2x + 12 â€“ 2
- 3x â‰¥ 2x + 10
- 3x â€“ 2x â‰¥ 2x + 10 â€“ 2x
- 5x â‰¥ 10
- 5x/-5 â‰¤ 10/-5
x â‰¤ -2
Question-9
Solution:
37 â€“ 3x - 5 â‰¥ 9x â€“ 8x + 24
32 â€“ 3x â‰¥ x + 24
32 â€“ 3x - 32 â‰¥ x + 24 â€“ 32
â€“ 3x â‰¥ x - 8
â€“ 3x â€“ x â‰¥ x â€“ 8 â€“ x
- 4x â‰¥ â€“ 8
- 4x/-4 â‰¥ â€“ 8/-4
x â‰¥ 2
Question-10
Solution:
+â‰¥
10x + 3x â‰¥ 39
13x â‰¥ 39
13x/13 â‰¥ 39/13
x â‰¥ 3
Question-11
Solution:
â‰¥ - 3
2(4 + 2x) â‰¥ 3x - 18 (Multiplying by 6 both sides)
8 + 4x â‰¥ 3x â€“ 18
8 + 4x â€“ 8 â‰¥ 3x â€“ 18 â€“ 8
4x â‰¥ 3x â€“ 26
4x â€“ 3x â‰¥ 3x â€“ 26 â€“ 3x
x â‰¥ â€“ 26
Question-12
Solution:
â‰¥
9(x â€“ 2) â‰¥ 25(2 - x) (Multiplying by 15 both sides)
9x â€“ 18 â‰¥ 50 - 25x
9x â€“ 18 + 18 â‰¥ 50 - 25x + 18
9x â‰¥ 68 - 25x
9x + 25x â‰¥ 68 - 25x + 25x
34x â‰¥ 68
34x/34 â‰¥ 68/34
x â‰¥ 2
Question-13
Solution:
<
15x < 20(5x - 2) â€“ 12(7x - 3) (Multiplying by 60 both sides)
15x < 100x - 40 â€“ 84x + 36
15x < 16x - 4
15x â€“ 16x < 16x â€“ 4 â€“ 16x
â€“ x < â€“ 4
x > 4
Question-14
Solution:
2(5 â€“ 2x) â‰¤ x â€“ 30 (Multiplying by 6 both sides)
10 â€“ 4x â‰¤ x â€“ 30
10 â€“ 4x â€“ 10 â‰¤ x â€“ 30 â€“ 10
â€“ 4x â‰¤ x â€“ 40
â€“ 4x â€“x â‰¤ x â€“ 40 â€“x
â€“ 5x â‰¤ â€“ 40
â€“ 5x/ -5 â‰¤ â€“ 40/ -5
x â‰¥ 8
Question-15
Solution:
(Multiplying by 6 both sides)
3(3x + 20) â‰¥ 10(x - 6) (Multiplying by 5 both sides)
9x + 60 â‰¥ 10x - 60
9x + 60 â€“ 60 â‰¥ 10x - 60 â€“ 60
9x â‰¥ 10x â€“ 120
9x â€“ 10x â‰¥ 10x â€“ 120 - 10x
-x â‰¥ â€“ 120
x â‰¤ 120
Question-16
Solution:
x â€“2 > 0 .
3x < 18
x â€“2 > 0
x > 2...................(1)
3x < 18
x < 6 ..................(2)
From (1) and (2), solutions of the given system are, therefore, given by 2 < x < 6
Hence the solution of the system is 2 < x < 6.
Question-17
Solution:
5x + 1 > -24
5x â€“ 1 < 24
5x + 1 > -24
5x > - 25
x > - 5 .....................(1)
5x â€“ 1 < 24
5x < 25
x < 5.......................(2)
From (1) and (2), solutions of the given system are, therefore, given by -5 < x < 5
Hence the solution of the system of is â€“5 < x < 5.
Question-18
Solution:
x + 2 â‰¤ 5
3x â€“ 4 > - 2 + x
x + 2 â‰¤ 5
x â‰¤ 3 .......................(1)
3x â€“ 4 > - 2 + x
3x > 2 + x
2x > 2
x > 1 .....................(2)
From (1) and (2), solutions of the given system are, therefore, given by 1 < x â‰¤ 3.
Hence the solution of the system of is 1 < x â‰¤ 3.
Question-19
Solution:
4x + 5 > 3x
-(x + 3) + 4 â‰¤ â€“2x + 5
4x + 5 > 3x
4x > 3x - 5
x > - 5 ....................(1)
-(x + 3) + 4 â‰¤ â€“2x + 5
-x - 3 + 4 â‰¤ â€“2x + 5
-x + 1 â‰¤ â€“2x + 5
-x â‰¤ â€“2x + 4
x â‰¤ 4 ......................(2)
From (1) and (2), solutions of the given system are, therefore, given by -5 < x â‰¤ 4
Hence the solution of the system of is -5 < x â‰¤ 4.
Question-20
Solution:
16x â€“ 27 < 12x + 9 (Multiplying by 12 both sides)
16x < 12x + 36
4x< 36
x< 9 .........................(1)
2(7x - 1) â€“ (7x + 2) > 6x (Multiplying by 6 both sides)
14x â€“ 2 â€“ 7x â€“ 2 > 6x
7x â€“ 4 > 6x
7x > 6x + 4
x > 4 .......................(2)
From (3) and (4), solutions of the given system are, therefore, given by 4 < x < 9
Hence the solution of the system of is 4 < x < 9.
Question-21
Solution:
2(x + 1) < x + 5
3(x + 2) > 2 â€“ x
2(x + 1) < x + 5
2x + 2 < x + 5
2x < x + 3
x < 3........................ (1)
3(x + 2) > 2 â€“ x
3x + 6 > 2 â€“ x
3x > -4 â€“ x
4x > -4
x > -1 ......................(2)
From (1) and (2), solutions of the given system are, therefore, given by -1 < x < 3
Hence the solution of the system of is -1 < x < 3.
Question-22
Solution:
3x â€“ 1 â‰¥ 5
x + 2 > -1
3x â€“ 1 â‰¥ 5
3x â‰¥ 5 + 1
3x â‰¥ 6
x â‰¥ 2 ...................(1)
x + 2 > -1
x > -1 â€“ 2
x > -3 .................(2)
From (1) and (2), solutions of the given system are, therefore, given by x â‰¥ 2
Hence the solution of the system of is x â‰¥ 2.
Question-23
Solution:
3x â€“ 7 > 2(x - 6)
6 â€“ x > 11 â€“ 2x
3x â€“ 7 > 2x - 12
3x â€“ 7 + 7 > 2x - 12 + 7
3x > 2x - 5
3x â€“ 2x > 2x â€“ 5 â€“ 2x
x > â€“ 5 ............................(1)
6 â€“ x > 11 â€“ 2x
6 â€“ x â€“ 6 > 11 â€“ 2x â€“ 6
- x > 5 â€“ 2x
â€“ x + 2x > 5 â€“ 2x + 2x
x > 5 ...............................(2)
From (1) and (2), solutions of the given system are, therefore, given by x > 5
Hence the solution of the system is x > 5.
Question-24
Solution:
â€“2 - â‰¤
3 â€“ x < 4(x - 3)
â€“2 - â‰¤
â€“24 â€“ 3x â‰¤ 4(1 + x) (Multiplying by 12 both sides)
â€“24 â€“ 3x â‰¤ 4 + 4x
â€“ 3x â‰¤ 4x + 28
â€“ 7x â‰¤ 28
x â‰¥ -4 .......................... (1)
3 â€“ x < 4x - 12
- x < 4x â€“ 15
- 5x < - 15
x > 3 ........................(2)
From (1) and (2), solutions of the given system are, therefore, given by x > 3
Hence the solution of the system is x > 3.
Question-25
5(2x - 7) â€“ 3(2x + 3) â‰¤ 0, 2x + 19 â‰¤ 6x + 47
Solution:
5(2x - 7) â€“ 3(2x + 3) â‰¤ 0
2x + 19 â‰¤ 6x + 47
5(2x - 7) â€“ 3(2x + 3) â‰¤ 0
10x - 35 â€“ 6x - 9 â‰¤ 0
4x - 44 â‰¤ 0
4x â‰¤ 44
x â‰¤ 11 ........................... (3)
2x + 19 â‰¤ 6x + 47
2x â‰¤ 6x + 28
-4x â‰¤ 28
x â‰¥ -7 ............................ (4)
From (1) and (2), solutions of the given system are, therefore, given by -7 â‰¤ x â‰¤ 11
Hence the solution of the system is -7 â‰¤ x â‰¤ 11.
Question-26
Solution:
2x â€“ 7 <11
3x + 4 < - 5
2x â€“ 7 <11
2x <18
x < 9 ...........................(1)
3x + 4 < - 5
3x < - 9
x < - 3 .......................(2)
From (1) and (2), solutions of the given system are, therefore, given by x < - 3
Hence the solution of the system is x < - 3.
Question-27
Solution:
4 â€“ 5x > -11
4x + 11 â‰¤ -13
4 â€“ 5x > -11
â€“ 5x > -15
x < 3 ............................ (1)
4x + 11 â‰¤ -13
4x â‰¤ - 24
x â‰¤ - 6 .......................... (2)
From (1) and (2), solutions of the given system are, therefore, given by x â‰¤ - 6.
Hence the solution of the system is x â‰¤ - 6
Question-28
Solution:
4x â€“ 5 < 11
-3x â€“ 4 â‰¥ 8
4x â€“ 5 < 11
4x < 16
x < 4 ...............................(1)
-3x â€“ 4 â‰¥ 8
-3x â‰¥ 12
x â‰¤ -4................................(2)
From (1) and (2), solutions of the given system are, therefore, given by x â‰¤ -4.
Hence the solution of the system is x â‰¤ -4.
Question-29
Solution:
5x â€“ 7 < 3 (x + 3)
1 â€“ â‰¥ x â€“ 4
5x â€“ 7 < 3 (x + 3)
5x â€“ 7 < 3x + 9
5x < 3x + 16
2x < 16
x < 8.............................(1)
1 â€“ â‰¥ x â€“ 4 (Multiplying both sides by 2)
2 â€“ 3x â‰¥ 2x â€“ 8
- 3x â‰¥ 2x â€“ 10
- 5x â‰¥ â€“ 10
x â‰¤ 2 ............................(2)
From (1) and (2), solutions of the given system are, therefore, given by x â‰¤ 2.
Hence the solution of the system is x â‰¤ 2.
Question-30
Solution:
2(2x + 3) â€“ 10 < 6(x - 2)
2(2x + 3) â€“ 10 < 6(x - 2)
4x + 6 â€“ 10 < 6x - 12
4x â€“ 4 < 6x â€“ 12
4x < 6x â€“ 8
- 2x < â€“ 8
x > 4 .................................. (1)
3(2x - 3) + 72 â‰¥ 24 + 16x (Multiplying both sides by 12)
6x - 9 + 72 â‰¥ 24 + 16x
6x + 63 â‰¥ 24 + 16x
6x â‰¥ 16x â€“ 39
-10x â‰¥ - 39
x â‰¤ 39/10............................(2)
From (1) and (2), the system has no solution.
Question-31
Solution:
We draw the graph of the equation x â€“ 2y + 4 = 0
x |
0 |
-4 |
y |
2 |
0 |
Put x = 0
Then 0 â€“ 2y + 4 â‰¤ 0
or -2y â‰¤ - 4
or y â‰¥ 2
Put x = 0, y = 0
Then 0 â€“ 2(0) + 4 â‰¤ 0
or 4 â‰¤ 0, which is false.
Hence, half plane II is not the solution of the given inequation.
Therefore, the shaded half plane I is the solution region of the inequation including points on the line
x â€“ 2y + 4 = 0.
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
Question-32
Solution:
We draw the graph of the equation 2x + y = 3
X |
0 |
3/2 |
Y |
3 |
0 |
Put x = 0
Then 2(0) + y > 3
or y > 3
Put x = 0, y = 0
Then 2(0) + 0 > 3
or 0 > 3, which is false.
Hence, half plane I is not the solution of the given inequation.
Therefore, the shaded half plane II is the solution region of the inequation excluding points on the line
2x + y = 3.
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
Question-33
Solution:
We draw the graph of the equation 3x â€“ 4y = 12
x |
0 |
4 |
y |
-3 |
0 |
Put x = 0
Then 3(0) â€“ 4y < 12
or - y < 3
or y > -3
Put x = 0, y = 0
Then 3(0) â€“ 4(0) < 12
or 0 < 12, which is true.
Hence, half plane II is not the solution of the given inequation.
Therefore, the shaded half plane I is the solution region of the inequation excluding points on the line
3x â€“ 4y = 12.
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
Question-34
Solution:
We draw the graph of the equation y + 8 = 2x
x |
2 |
4 |
y |
-4 |
0 |
Put x = 0
Then y + 8 â‰¥ 2x
or y + 8 â‰¥ 0
Put x = 0, y = 0
Then 0 + 8 â‰¥ 2(0)
or 8 â‰¥ 0, which is true.
Hence, half plane I is not the solution of the given inequation.
Therefore, the shaded half plane II is the solution region of
the inequation including points on the line
y + 8 = 2x.
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
Question-35
Solution:
We draw the graph of the equation 2x = 6 â€“ 3y
x |
0 |
3 |
y |
2 |
0 |
Put x = 0
Then 2(0) â‰¤ 6 â€“ 3y
or y â‰¤ 2
Put x = 0, y = 0
Then 2(0) â‰¤ 6 â€“ 3(0)
or 0 â‰¤ 6, which is true.
Hence, half plane I is not the solution of the given inequation.
Therefore, the shaded half plane II is the solution region of the inequation including points on the line 2x = 6 â€“ 3y
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
Question-36
Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation 2x â€“ 5y + 10 = 0
x |
0 |
-5 |
y |
2 |
0 |
Put x = 0
Then 0 â‰¤ 2(0) â€“ 5y + 10
or -10 â‰¤ â€“ 5y
Put x = 0, y = 0
Then 0 â‰¤ 2(0) â€“ 5(0) + 10
or 0 â‰¤ 10, which is false.
Hence, half plane I is not the solution of the given inequation.
Therefore, the shaded half plane II is the solution region of the inequation including points on the line
2x â€“ 5y + 10 = 0.
Question-37
Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation 2x â€“ 3y = 6
x |
0 |
3 |
y |
-2 |
0 |
Put x = 0
Then 2(0) â€“ 3y < 6
or y < -2
Put x = 0, y = 0
Then 2(0) â€“ 3(0) < 6
or 0 < 6, which is true.
Hence, half plane II is not the solution of the given inequation.
Therefore, the shaded half plane I is the solution region of the inequation excluding points on the line
2x â€“ 3y = 6
Question-38
Solution:
1cm = 1unit along x-axis
1cm = 1unit along y-axis
The graph of the equation x = - 2 is vertical line parallel to y â€“ axis.
Put x = 0
Then 0 > -2, which is true.
Hence, the solution region is the shaded region on the right hand side of the line x = -2 containing the origin.
Hence every point on the shaded region is the solution of the given inequation.
Question-39
Solution:
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation y = -2
Put y = 0
Then 0 < -2, which is false.
Hence, the solution region is the shaded region below line y = -2 containing the origin.
Hence every point below the shaded region is the solution of the given inequation.
Question-40
Solution:
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation x = 8 â€“ 4y
x |
0 |
8 |
y |
2 |
0 |
Put x = 0
Then 0 â‰¤ 8 â€“ 4y
or y â‰¤ 2
Put x = 0, y = 0
Then (0) â‰¤ 8 â€“ 4(0)
or 0 â‰¤ 8, which is true.
Hence, half plane I is not the solution of the given inequation.
Therefore, the shaded half plane II is the solution region of the inequation including points on the line x = 8 â€“ 4y.
Question-41
x + 2y â‰¥ 20, 3x + y â‰¤ 15.
Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation x + 2y = 20
x |
0 |
20 |
y |
10 |
0 |
Put x = 0
Then 0 + 2y â‰¥ 20
or y â‰¥ 10
Put x = 0, y = 0
Then 0 + 2(0) â‰¥ 20
or 0 â‰¥ 20, which is false.
3x + y â‰¤ 15
We draw the graph of the equation 3x + y = 15
x |
0 |
5 |
y |
15 |
0 |
Put x = 0
Then 3x + y â‰¤ 15
or y â‰¤ 15
Put x = 0, y = 0
Then 3(0) + (0) â‰¤ 15
or 0 â‰¤ 15, which is true.
Every point in the common shaded region represents a solution of the given system of inequations.
Question-42
4x + 3y â‰¥ 12, 4x â€“ 5y â‰¥ -20.
Solution:
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation 4x + 3y = 12
x |
0 |
3 |
y |
4 |
0 |
Put x = 0
Then 4(0) + 3y â‰¥ 12
or y â‰¥ 4
Put x = 0, y = 0
Then 4(0) + 3(0) â‰¥ 12
or 0 â‰¥ 12, which is false.
3x + y â‰¤ 15
We draw the graph of the equation 4x â€“ 5y = -20
x |
0 |
-5 |
y |
4 |
0 |
Put x = 0
Then 4(0) â€“ 5y â‰¥ -20
or y â‰¤ 4
Put x = 0, y = 0
Then 4(0) â€“ 5(0) â‰¥ -20
or 0 â‰¥ -20, which is true.
Every point in the common shaded region represents a solution of the given system of inequations.
Question-43
2x + y â‰¥ 8, x + 2y â‰¥ 10.
Solution:
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation 2x + y = 8
x |
0 |
4 |
y |
8 |
0 |
Put x = 0
Then 2(0) + y â‰¥ 8
or y â‰¥ 8
Put x = 0, y = 0
Then 2(0) + (0) â‰¥ 8
or 0 â‰¥ 8, which is false.
We draw the graph of the equation x + 2y =10
x |
0 |
10 |
y |
5 |
0 |
Put x = 0
Then 0 + 2y â‰¥ 10
or y â‰¥ 5
Put x = 0, y = 0
Then 0 + 2(0) â‰¥ 10
or 0 â‰¥ 10, which is false.
Every point in the common shaded region represents a solution of the given system of inequations.
Question-44
y â‰¤ 4, x â‰¥ 1
Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation y = 4
Put y = 0
Then 0 â‰¤ 4, which is true.
We draw the graph of the equation x = 1
Put x = 0
Then 0 â‰¥ 1, which is false.
Question-45
5x + 6y â‰¥ 30, x â‰¥ 0, y â‰¥ 0
Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation 5x + 6y = 30.
x |
0 |
6 |
y |
5 |
0 |
Put x = 0
Then 5(0) + 6y â‰¥ 30
or y â‰¥ 5
Put x = 0, y = 0
Then 5(0) + 6(0) â‰¥ 30
or 0 â‰¥ 30, which is false.
We draw the graph of the equation x â‰¥ 0.
Put x = 0
Then 0 â‰¥ 0, which is false.
We draw the graph of the equation y = 0.
Put y = 0
Then 0 â‰¥ 0, which is false.
Every point in the common shaded region represents a solution of the given system of inequations.
Question-46
x + y â‰¤ 9, y > x, x â‰¥ 1.
Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation x + y = 9
x |
0 |
9 |
y |
9 |
0 |
Put x = 0
Then 0 + y â‰¤ 9
or y â‰¤ 9
Put x = 0, y = 0
Then 0 + 0 â‰¤ 9
or 0 â‰¤ 9, which is true.
We draw the graph of the equation y = x.
Put x = 0
Then y > 0
Put x = 0, y = 0
Then 0 > 0, which is false.
We draw the graph of the equation x = 1.
Put x = 0
Then 0 â‰¥ 1, which is false.
Every point in the common shaded region represents a solution of the given system of inequations.
Question-47
x + 3y â‰¤ 12, 3x + y â‰¤ 12, x â‰¥ 0, y â‰¥ 0.
Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation x + 3y = 12
x |
0 |
12 |
y |
4 |
0 |
Put x = 0
Then 0 + 3y â‰¤ 12
or y â‰¤ 4
Put x = 0, y = 0
Then 0 + 3(0) â‰¤ 12
or 0 â‰¤ 12, which is true.
We draw the graph of the
equation 3x + y = 12
X |
0 |
4 |
Y |
12 |
0 |
Put x = 0
Then 0 + y â‰¤ 12
or y â‰¤ 12
Put x = 0, y = 0
Then 3(0) + 0 â‰¤ 12
or 0 â‰¤ 12, which is true.
We draw the graph of the equation x = 0.
Put x = 0
Then 0 â‰¥ 0, which is false.
We draw the graph of the equation y = 0.
Put y = 0
Then 0 â‰¥ 0, which is false.
Every point in the common shaded region represents a solution of the given system of inequations.
Question-48
2x + y â‰¥ 4, x + y â‰¤ 3, 2x â€“ 3y â‰¤ 6.
Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation 2x + y = 4
x |
0 |
2 |
y |
4 |
0 |
Put x = 0
Then 2(0) + y â‰¥ 4
or y â‰¥ 4
Put x = 0, y = 0
Then 2(0) + (0) â‰¥ 4
or 0 â‰¥ 4, which is false.
We draw the graph of the equation x + y = 3
x |
0 |
3 |
y |
3 |
0 |
Put x = 0
Then 0 + y â‰¤ 3
or y â‰¤ 3
Put x = 0, y = 0
Then 0 + 0 â‰¤ 3
or 0 â‰¤ 3, which is true.
We draw the graph of the equation 2x â€“ 3y = 6
x |
0 |
3 |
y |
-2 |
0 |
Put x = 0
Then 2(0) â€“ 3y â‰¤ 6
or y â‰¥ -2
Put x = 0, y = 0
Then 2(0) â€“ 3(0) â‰¤ 6
or 0 â‰¤ 6, which is true.
Every point in the common shaded region represents a solution of the given system of inequations.
Question-49
x + y < 6, 7x + 4y â‰¤ 28, x â‰¥ 0, y â‰¥ 0.
Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation x + y = 6
x |
0 |
6 |
y |
6 |
0 |
Put x = 0
Then 0 + y < 6
or y < 6
Put x = 0, y = 0
Then 0 + 0 < 6
or 0 < 6, which is true.
We draw the graph of the equation 7x + 4y = 28
x |
0 |
7 |
y |
4 |
0 |
Put x = 0
Then 7(0) + 4y â‰¤ 28
or y â‰¥ 4
Put x = 0, y = 0
Then 7(0) + 4(0) â‰¤ 28
or 0 â‰¤ 28, which is true.
We draw the graph of the equation x = 0.
Put x = 0
Then 0 â‰¥ 0, which is false.
We draw the graph of the equation y = 0.
Put y = 0
Then 0 â‰¥ 0, which is false.
Every point in the common shaded region represents a solution of the given system of inequations.
Question-50
3x + 2y â‰¤ 24, x + 2y â‰¤ 16, x + y â‰¤ 10, x â‰¥ 0, y â‰¥ 0.
Solution:
Scale
1cm = 1unit along x-axis
1cm = 1unit along y-axis
We draw the graph of the equation 3x + 2y = 24
x |
0 |
8 |
y |
12 |
0 |
Put x = 0
Then 3(0) + 2y â‰¤ 24
or y â‰¤ 12
Put x = 0, y = 0
Then 3(0) + 2(0) â‰¤ 24
or 0 â‰¤ 24, which is true.
We draw the graph of the equation x + 2y = 16
x |
0 |
16 |
y |
8 |
0 |
Put x = 0
Then 0 + 2y â‰¤ 16
or y â‰¤ 8
Put x = 0, y = 0
Then 0 + 2(0) â‰¤ 16
or 0 â‰¤ 16, which is false.
We draw the graph of the equation x + y =10
x |
0 |
10 |
y |
10 |
0 |
Put x = 0
Then x + y â‰¤ 10
or y â‰¤ 10
Put x = 0, y = 0
Then 0 + 0 â‰¤ 10
or 0 â‰¤ 10, which is true.
We draw the graph of the equation x = 0.
Put x = 0
Then 0 â‰¥ 0, which is false.
We draw the graph of the equation y = 0.
Put y = 0
Then 0 â‰¥ 0, which is false.
Every point in the common shaded region represents a solution of the given system of inequations.