Linear Inequalities

It is given in the question that,

24x < 100

Dividing the inequality by 24 on both sides we get,

(i) When x is a natural integer.

It can be clearly observed that the only natural number less than are 1, 2, 3, 4

Thus, 1, 2, 3, 4 will be the solution of the given inequality when x is a natural number.

∴ Here, {1, 2, 3, 4} is the solution set.

(ii) When x is an integer

It can be clearly observed that the integer number less thanare…, -1, 0, 1, 2, 3, 4.

Thus, solution of 24x<100 are…,-1, 0, 1, 2, 3, 4, when x is an integer.

∴ {…, -1, 0, 1, 2, 3, 4} is the solution set.

It is given in the question that,

– 12x > 30

Dividing the inequality by -12 on both sides we get,

(i) When x is a natural integer.

It can be clearly observed that there is no natural number less thanbecause- is a negative number and natural numbers are positive numbers.

∴there would be no solution of the given inequality when x is a natural number.

(ii) When x is an integer

It can be clearly observed that the integer number less thanare…, -5, -4, -3

Thus, solution of– 12x > 30is …,-5, -4, -3, when x is an integer.

∴ {…, -5, -4, -3} is the solution set.

It is given in the question that,

5x – 3 < 7

Adding 3 both side we get,

5x – 3 + 3 < 7 + 3

5x < 10

Dividing both sides by 5 we get,

x < 2

When x is an integer

It can be clearly observed that the integer number less than 2 are…, -2, -1, 0, 1.

Thus, solution of5x – 3 < 7is …,-2, -1, 0, 1, when x is an integer.

∴ {…, -2, -1, 0, 1} is the solution set.

(i) When x is a real number.

It can be clearly observed that the solutions of 5x – 3 < 7 will be given by x < 2 which states that all the real numbers that are less than 2.

∴x (-∞, 2) is the solution set.

It is given in the question that,

3x + 8 >2

Subtracting 8 from both sides we get,

3x + 8 – 8 >2 – 8

3x >- 6

Dividing both sides by 3 we get,

x > -2

(i) When x is an integer

It can be clearly observed that the integer number greater than -2 are -1, 0, 1, 2,...

Thus, solution of 3x + 8 >2is -1, 0, 1, 2,… when x is an integer.

∴ {-1, 0, 1, 2,…} is the solution set.

(ii) When x is a real number.

It can be clearly observed that the solutions of 3x + 8 >2 will be given by x >-2 which states that all the real numbers that are greater than -2.

∴ x ∈ (-2, ∞) is the solution set.

It is given in the question that,

4x + 3 < 5x + 7

Subtracting 7 from both the sides,

4x + 3 – 7 < 5x + 7 – 7

4x – 4 < 5x

Subtracting 4x from both the sides,

4x – 4 – 4x < 5x – 4x

x > -4

∴The solutions of the given inequality are defined by all the real numbers greater than -4.

Thus, (-4, ∞) is the required solution set.

It is given in the question that,

3x – 7 > 5x – 1

Adding 7 to both the sides,

3x – 7 + 7 > 5x – 1 + 7

3x> 5x + 6

Subtracting 5x from both the sides,

3x – 5x > 5x + 6 – 5x

-2x > 6

Dividing both sides by -2

x < -3

∴ The solutions of the given inequality are defined by all the real numbers less than -3.

Thus, (-∞, -3) is the required solution set.

It is given in the question that,

3(x – 1) ≤ 2 (x – 3)

3x – 3 ≤ 2x – 6

Adding 3 to both the sides,

3x – 3+ 3 ≤ 2x – 6+ 3

3x ≤ 2x – 3

Subtracting 2x from both the sides,

3x – 2x ≤ 2x – 3 – 2x
x ≤ -3
∴ The solutions of the given inequality are defined by all the real numbers less than or equal to -3.

Thus, (-∞, -3] is the required solution set.

It is given in the question that,

3 (2 – x) ≥ 2 (1 – x)

6 – 3x ≥ 2 – 2x

Adding 2x to both the sides,

6 – 3x+ 2x ≥ 2 – 2x+ 2x

6 – x≥ 2

Subtracting 6 from both the sides,

6 – x – 6 ≥2 – 6

- x ≥ -4

x ≤ 4

∴ The solutions of the given inequality are defined by all the real numbers greater than or equal to 4.

Thus, (- ∞, 4] is the required solution set.

It is given in the question that,

Dividing by 11 on both sides,

x < 6

∴ The solutions of the given inequality are defined by all the real numbers less than 6.

Thus, (-∞, 6) is the required solution set.

It is given in the question that,

-x > 6

x < - 6

∴ The solutions of the given inequality are defined by all the real numbers less than - 6.

Thus, (-∞, -6) is the required solution set.

It is given in the question that,

Cross - multiplying the denominators, we get:

9(x- 2) ≤ 25(2 – x)

9x – 18≤ 50 – 25x

Adding 25x both the sides,

9x – 18 + 25x ≤ 50 – 25x + 25x

34x – 18≤ 50

Adding 25x both the sides,

34x – 18 + 18 ≤ 50 + 18

34x ≤ 68

Dividing both sides by 34,

x ≤ 2

∴ The solutions of the given inequality are defined by all the real numbers less than or equal to 2.

Thus, (-∞, 2] is the required solution set.

It is given in the question that,

Cross - multiplying the denominators, we get:

120 ≥ x

∴ The solutions of the given inequality are defined by all the real numbers less than or equal to 120.

Thus, (-∞, 120] is the required solution set.

It is given in the question that,

2 (2x + 3) – 10 < 6 (x – 2)

4x + 6 – 10 < 6x – 12

4x – 4 < 6x – 12

-4 + 12 < 6x – 4x

8 < 2x

4 < x

∴ The solutions of the given inequality are defined by all the real numbers greater than or equal to 4.

Thus, (4, -∞) is the required solution set.

We have,

37 - (3x + 5) ≥ 9x - 8 (x - 3)

= 37 - 3x – 5 ≥ 9x - 8x + 24

= 32 - 3x ≥ x + 24

= 32 - 24 ≥ x + 3x

= 8 ≥ 4x

= 2 ≥ x

∴ All the real numbers of x which are less than or equal to 2 are the solutions of the given inequality

Hence, will be the solution for the given inequality

We have,

= 15x < 4 (4x – 1)

= 15x < 16x – 4

= 4 < x

∴ All the real numbers of x which are greater than 4 are the solutions of the given inequality

Hence, will be the solution for the given inequality

We have,

=20 (2x-1) ≥ 3 (19x-18)

=40x-20 ≥ 57x-54

= - 20+54 ≥57x-40x

= 34 ≥17x

= 2 ≥x

∴ All the real numbers of x which are less than or equal to 2 are the solutions of the given inequality

Hence, will be the solution for the given inequality

We have,

3x – 2 < 2x + 1

Solving the given inequality, we get:

3x – 2 < 2x + 1

= 3x – 2x < 1 + 2

= x < 3

Now, the graphical representation of the solution is as follows:

We have,

Solving the given inequality, we get:

5x – 3 ≥ 3x – 5

= 5x - 3x ≥ -5 + 3

= 2x ≥ -2

= x ≥ -1

Now, the graphical representation of the solution is as follows:

We have,

3 (1 – x) < 2 (x + 4)

Solving the given inequality, we get:

3 (1 – x) < 2 (x + 4)

= 3 – 3x < 2x + 8

= 3 – 8 < 2x + 3x

= - 5 < 5x

= - 1 < x

Now, the graphical representation of the solution is as follows:

We have,

Solving the given inequality, we get:

= 15x ≥ 2 (4x - 1)

= 15x ≥ 8x -2

= 15x -8x ≥ 8x -2 -8x

= 7x ≥ -2

Now, the graphical representation of the solution is as follows:

Let us assume, x be the marks obtained by Ravi in his third unit test

According to question, the entire students should have an average of at least 60 marks

∴

= 145 + x ≥ 180

= x ≥ 180 - 145

= x ≥ 35

Hence, all the students must obtain 35 marks in order to have an average of at least 60 marks

Let us assume Sunita scored x marks in her fifth examination

Now, according to the question in order to receive A grade in the course she must have to obtain average 90 marks or more in her five examinations

∴

= 368 + x ≥ 450

= x ≥ 450 - 368

= x ≥ 82

Hence, she must have to obtain 82 or more marks in her fifth examination

Let us assume x be the smaller of the two consecutive odd positive integers

∴ Other integer = x + 2

It is also given in the question that, both the integers are smaller than 10

∴ x + 2 < 10

x < 8 (i)

Also, it is given in the question that sum off two integers is more than 11

∴ x + (x + 2) > 11

2x + 2 > 11

x > 4.5 (ii)

Thus, from (i) and (ii) we have x is an odd integer and it can take values 5 and 7

Hence, possible pairs are (5, 7) and (7, 9)

Let us assume x be the smaller of the two consecutive even positive integers

∴ Other integer = x + 2

It is also given in the question that, both the integers are larger than 5

∴ x > 5 ....(i)

Also, it is given in the question that the sum of two integers is less than 23.

∴ x + (x + 2) < 23

2x + 2 < 23

x < 10.5 .... (ii)

Thus, from (i) and (ii) we have x is an even number and it can take values 6, 8 and 10

Hence, possible pairs are (6, 8), (8, 10) and (10, 12).

Let us assume the length of the shortest side of the triangle be x cm

∴ According to the question, length of the longest side = 3x cm

And, length of third side = (3x – 2) cm

As, the least perimeter of the triangle = 61 cm

Thus,

= 7x – 2 ≥ 61

= 7x ≥ 63

= x ≥ 9

Hence, the minimum length o the shortest side will be 9 cm

Let us assume the length of the shortest piece be x cm

∴ According to the question, length of the second piece = (x + 3) cm

And, length of third piece = 2x cm

As all the three lengths are to be cut from a single piece of board having a length of 91 cm

∴ x + (x + 3) + 2x ≤ 91 cm

= 4x + 3 ≤ 91

= 4x ≤ 88

= x ≤ 22 …(i)

Also, it is given in the question that, the third piece is at least 5 cm longer than the second piece

∴ 2x ≥ (x+3) + 5

2x ≥ x + 8

x ≥ 8 …(ii)

Thus, from equation (i) and (ii) we have:

8 ≤ x ≥ 22

Hence, it is clear that the length of the shortest board is greater than or equal to 8 cm and less than or equal to 22 cm

Given: x + y < 5

Consider: x + y = 5

Now draw a dotted line x + y = 5 in the graph (∵ x + y = 5 is excluded in the given question)

Now Consider x + y < 5

Select a point (0,0)

⇒ 0 + 0 < 5

⇒ 0 < 5 (this is true)

∴ Solution region of the given inequality is below the line x + y = 5. (That is origin is included in the region)

The graph is as follows:

Given: 2x + y 6

Consider: 2x + y = 6

Now draw a solid line 2x + y = 6 in the graph (∵2x + y = 6 is included in the given question)

Now Consider 2x + y 6

Select a point (0,0)

⇒ 2 × (0) + 0 6

⇒ 05 (this is false)

∴ Solution region of the given inequality is above the line 2x + y = 6. (Away from the origin)

The graph is as follows:

Given: 3x + 4y 12

Consider: 3x + 4y = 12

Now draw a solid line 3x + 4y = 12 in the graph (∵3x + 4y = 12 is included in the given question)

Now Consider 3x + 4y 12

Select a point (0,0)

⇒ 3 × (0) + 4 × (0) 12

⇒ 012 (this is true)

∴ Solution region of the given inequality is below the line 3x + 4y = 12. (That is origin is included in the region)

The graph is as follows:

Given: y + 8 ≥ 2x

Consider: y + 8 = 2x

Now draw a solid line y + 8 = 2x in the graph (∵y + 8 = 2x is included in the given question)

Now Consider y + 8 2x

Select a point (0,0)

⇒ (0) + 8 2 × (0)

⇒ 08 (this is true)

∴ Solution region of the given inequality is above the line y + 8 = 2x. (That is origin is included in the region)

The graph is as follows:

Given: x – y 2

Consider: x – y = 2

Now draw a solid line x – y = 2 in the graph (∵ x – y = 2 is included in the given question)

Now Consider x – y ≤ 2

Select a point (0,0)

⇒ (0) – (0) ≤ 2

⇒ 0 ≤ 12 (this is true)

∴ Solution region of the given inequality is above the line x – y = 2. (That is origin is included in the region)

The graph is as follows:

Given: 2x – 3y > 6

Consider: 2x – 3y = 6

Now draw a dotted line 2x – 3y = 6 in the graph (∵2x – 3y = 6 is excluded in the given question)

Now Consider 2x – 3y > 6

Select a point (0,0)

⇒ 2 × (0) – 3 × (0) > 6

⇒ 0>5 (this is false)

∴ Solution region of the given inequality is below the line 2x – 3y > 6. (Away from the origin)

The graph is as follows:

Given: – 3x + 2y ≥ – 6

Consider: – 3x + 2y = – 6

Now draw a solid line – 3x + 2y = – 6 in the graph (∵– 3x + 2y = – 6 is included in the given question)

Now Consider – 3x + 2y ≥ – 6

Select a point (0,0)

⇒ - 3 × (0) + 2 × (0) ≥ – 6

⇒ 0 ≥ – 6 (this is true)

∴ Solution region of the given inequality is above the line – 3x + 2y – 6. (That is origin is included in the region)

The graph is as follows:

Given: 3y – 5x < 30

Consider: 3y – 5x = 30

Now draw a dotted line 3y – 5x = 30 in the graph (∵3y – 5x = 30 is excluded in the given question)

Now Consider 3y – 5x < 30

Select a point (0,0)

⇒ 3 × (0) – 5 × (0) < 30

⇒ 0 < 30 (this is true)

∴ Solution region of the given inequality is below the line 3y – 5x < 30. (That is origin is included in the region)

The graph is as follows:

Given: y < – 2

Consider: y = – 2

Now draw a dotted line y = – 2 in the graph (∵ y = – 2 is excluded in the given question)

Now Consider y < – 2

Select a point (0,0)

⇒ 0 < – 2

⇒ 0 < 30 (this is false)

∴ Solution region of the given inequality is below the line y < – 2. (That is Away from the origin)

The graph is as follows:

Given: x > – 3

Consider: x = – 3

Now draw a dotted line x = – 3 in the graph (∵x = – 3 is excluded in the given question)

Now Consider x > – 3

Select a point (0,0)

⇒ 0 > – 3

⇒ 0 > – 3 (this is true)

∴ Solution region of the given inequality is right to the line x > – 3. (That is origin is included in the region)

The graph is as follows:

Given x ≥ 3………1

y ≥ 2……………2

Since x ≥ 3 means for any value of y the equation will be unaffected so similarly for y ≥ 2, for any value of x the equation will be unaffected.

Now putting x = 0 in the 1

0 ≥3 which is not true

Putting y = 0 in 2

0 ≥ 2 which is not true again

This implies the origin doesn’t satisfy in the given inequalities. The region to be included will be on the right side of the two equalities drawn on the graphs.

The shaded region is the desired region.

Given 3x+ 2y ≤ 12

Solving for the value of x and y by putting x = 0 and y = 0 one by one

We get

y = 6 and x = 4

So the points are (0,6) and (4,0)

Now checking for (0,0)

0 ≤ 12 which is also true,

Hence the origin lies in the plane and the required area is toward the left of the equation.

Now checking for x ≥ 1,

The value of x would be unaffected by any value of y

The origin would not lie on the plane

⇒ 0 ≥ 1 which is not true

The required area to be included would be on the left of the graph x ≥1

Similarly, for y ≥ 2

Value of y will be unaffected by any value of x in the given equality. Also, the origin doesn’t satisfy the given inequality.

⇒ 0 ≥ 2 which is not true, hence origin is not included in the solution of the inequality.

The region to be included in the solution would be towards the left of the equality y≥ 2

The shaded region in the graph will give the answer to the required inequalities as it is the region which is covered by all the given three inequalities at the same time satisfying all the given conditions.

Given 2x + y 6……………1

3x + 4y 12 …………….2

2x + y ≥ 6

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 6 and x = 3

So the point for the (0,6) and (3,0)

Now checking for (0,0)

0 ≥ 6 which is not true, hence the origin does not lies in the solution of the equality. The required region is on the right side of the graph.

Checking for 3x + 4y 12

Putting value of x= 0 and y = 0 one by one in equation

We get y = 3, x = 4

The points are (0, 3) , (4, 0)

Now checking for origin (0, 0)

0 ≤ 12 which is true,

so the origin lies in solution of the equation.

The region on the right of the equation is the region required.

The solution is the region which is common to the graphs of both the inequalities.

The shaded region is the required region.

Solving for x + y ≥ 4

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 4 and x = 4

The points for the line are (0, 4) and (4, 0)

Checking for the origin (0, 0)

0 ≥ 4

This is not true,

So the origin would not lie in the solution area. The required region would be on the right of line`s graph.

2x – y < 0

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y= 0 and x = 0

Putting x = 1 we get y = 2

So the points for the given inequality are (0, 0) and (1,2)

Now that the origin lies on the given equation we will check for (4,0) point to check which side of the line`s graph will be included in the solution.

⇒ 8 < 0 which is not true, hence the required region would be on the left side of the

line 2x-y < 0

The shaded region is the required solution of the inequalities.

Given 2x – y >1………………1

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = -1 and x = 1/2 = 0.5

The points are (0,-1) and (0.5 , 0)

Checking for the origin, putting (0,0)

0 >1 , which is false

Hence the origin does not lie in the solution region. The required region would be on the right of the line`s graph.

x – 2y < – 1…………2

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 1/2 = 0.5 and x = -1

The required points are (0, 0.5) and (-1, 0)

Now checking for the origin, (0,0)

0 < -1 which is false

Hence the origin does not lies in the solution area, the required area would be on the left side of the line`s graph.

∴ the shaded area is the required solution of the given inequalities.

Given. x + y ≤ 6,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

Y = 6 and x = 6

The required points are (0,6 ) and (6,0)

Checking further for origin (0,0)

We get 0 ≤ 6, this is true.

Hence the origin would be included in the area of the line`s graph. So the required solution of the equation would be on the left side of the line graph which will be including origin.

x + y 4

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 4 and x = 4

The required points are (0,4 ) and (4,0)

Checking for the origin (0, 0)

0 ≥ 4 which is false

So the origin would not be included in the required area. The solution area will be above the line graph or the area on the right of line graph.

Hence the shaded area in the graph is required graph area.

Given 2x + y 8

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 8 and x = 4

The required points are (0, 8) and (4, 0)

Checking if the origin is included in the line`s graph (0, 0)

0 ≥ 8, which is false

Hence the origin is not included in the solution area and the requires area would be the area to the right of line`s graph.

x + 2y 10

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 5 and x = 10

The required points are (0, 5) and (10 , 0)

Checking for the origin (0, 0)

0 ≥ 10 which is false,

Hence the origin would not lie in the required solution area. The required area would be to the left of the line graph.

The shaded area in the graph is the required solution of the given inequalities.

Given x + y ≤ 9,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 9 and x = 9

The required points are (0,9) and (9,0)

Checking if the origin is included in the line`s graph (0,0)

0 9

Which is true , so the required area would be including the origin and hence will lie on the left side of the line`s graph.

y > x,

Solving for y = x

we get x= 0, y = 0 so the origin lies on the line`s graph.

The other points would be (0, 0) and (2, 2)

Checking for (9, 0) in y > x,

we get 0 > 9 which is false, since the area would not include the area below the line`s graph and hence would be on the left side of the line.

x 0

The area of the required line`s graph would be on the right side of the line`s graph.

∴ the shaded are is the required solution of the given inequalities.

Given 5x + 4y 20,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 5 and x= 4

The required points are (0, 5) and (4, 0)

Checking If the origin lies in the solution area (0, 0)

0 ≤ 20

Which is true, hence the origin would lie in the solution area. The required area of the line`s graph is on the left side of the graph.

x ≥ 1,

for all the values of y , x would be 1,

The required points would be (1, 0) , (1, 2) and so on.

Checking for origin (0, 0)

0 ≥ 1, which is not true

So the origin would not lie in the required area. The required area on the graph will be on the right side of the line`s graph.

y ≥ 2

Similarly for all the values of x, y would be 2 .

The required points would be ( 0,2) , (1,2) and so on.

Checking for origin (0, 0)

0 2, this is no true

Hence the required area would be on the right side of the line`s graph.

The shaded area on the graph shows the required solution of the given inequalities.

3x + 4y ≤ 60,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 15 and x = 20

The required points are (0, 15) and (20, 0)

Checking if the origin lies in the required solution area (0,0)

0 ≤ 60, this is true.

Hence the origin would lie in the solution area of the line`s graph.

The required solution area would be on the left of the line`s graph.

x +3y ≤ 30,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 10 and x = 30

The required points are (0, 10) and (30, 0)

Checking for the origin (0, 0)

0 ≤ 30, this is true.

Hence the origin lies in the solution area which is given by the left side of the line`s graph.

x ≥ 0,

y ≥ 0,

The given inequalities imply the solution lies in the first quadrant only.

Hence the solution of the inequalities is given by the shaded region in the graph.

Given 2x + y ≥ 4,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 4 and x =2

The required points are (0, 4) and (2, 0)

Checking for origin (0, 0)

0 ≥ 4, this is not true

Hence the origin doesn’t lies in the solution area of the line`s graph. The solution area would be given by the right side of the line`s graph.

x + y 3,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 3 and x = 3

The required points are (0, 3) and (3, 0)

Checking for the origin (0, 0)

0 ≤ 3, this is true

Hence the solution area would include the origin and hence would be on the left side of the line`s graph.

2x – 3y 6

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = -2 and x = 3

The required points are (0, -2), (3, 0)

Checking for the origin (0, 0)

0 ≤ 6 this is true

So the origin lies in the solution area and the area would be on the left of the line`s graph.

Hence the shaded area in the graph is the required solution area for the given inequalities.

Given,

x – 2y 3

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = -3/2 = -1.5 and x = 3

The required points are (0, -1.5) and (3, 0)

Checking for the origin (0, 0)

0 ≤ 3, this is true.

Hence the solution area would be on the left of the line`s graph

3x + 4y 12,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 3 and x = 4

The required points are ( 0,3) and (4,0)

Checking for the origin (0, 0)

0 ≥ 12 , this is not true

So the solution area would of include the origin and the required solution area would be on the right side of the line`s graph.

x 0 ,

For all the values of y, the value of x would be same in the given inequality, which would be the region above the x axis on the graph.

y 1

For all the values of x, the value of y would be same in the given inequality.

The solution area of the line would be not include origin as 0 ≥ 1 is not true.

The solution area would be on the left side of the line`s graph.

The shaded area in the graph is the required solution area which satisfies all the given inequalities at the same time.

Given,

4x + 3y ≤ 60,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 20 and x = 15

The required points are (0, 20) and (15, 0)

Checking for the origin (0, 0)

0 ≤ 60, this is true.

Hence the origin would lie in the solution area. The required area would include be on the left of the line`s graph.

y ≥ 2x,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 0 and x = 0

Hence the line would pass through origin.

To check which side would be included in the line`s graph solution area, we would check for point (15, 0)

⇒ 0 ≥ 15, this is not true so the required solution area would be to the left of the line’s graph.

x ≥ 3,

For any value of y, the value of x would be same.

Also the origin (0, 0) doesn’t satisfies the inequality as 0 ≥ 3

So the origin doesn’t lies in the solution area, hence the required solution area would be the right of the line`s graph.

x, y ≥ 0

Since given both x and y are greater than 0

∴ the solution area would be in the first I^st quadrant only.

The shaded area in the graph shows the solution area for the given inequalities

Given,

3x + 2y ≤ 150

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 75 and x = 50

The required points are (0, 75) and (50, 0)

Checking for the origin (0, 0)

0 ≤ 150, this is true

Hence the solution area for the line would be on the left side of the line`s graph which would be including the origin too.

x + 4y 80,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 20 and x = 80

The required points are (0, 20) and (80, 0)

Checking for the origin (0, 0)

0 ≤ 80, this is also true so the origin lies in the solution area.

The required solution area would be toward the left of the line`s graph.

x 15,

For all the values of y, x would be same

Checking for the origin (0, 0)

0 ≤ 15, this is true so the origin would be included in the solution area. The required solution area would be towards the left of the line`s graph.

y 0, x 0

Since x and y are greater than 0, the solution would lie in the 1^st quadrant.

The shaded area in the graph satisfies all the given inequalities and hence is the solution area for given inequalities.

Given,

x + 2y 10,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 5 and x = 10

The required points are (0, 5) and (10, 0)

Checking for the origin (0, 0)

0 ≤ 10, this is true.

Hence the solution area would be toward origin including the same. The solution area would be toward the left of the line`s graph.

x + y 1,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 1 and x = 1

The required points are (0, 1) and (1, 0)

Checking for the origin (0, 0)

0 ≥ 1, this is not true.

Hence the origin would not be included into the solution area. The required solution area would be toward right of the line`s graph.

x – y 0,

Putting value of x = 0 and y = 0 in equation one by one, we get value of

y = 0 and x = 0

Hence the origin would lie on the line.

To check which side of the line graph would be included in the solution area we would check for the (10, 0)

10 ≤ 0 which is not true hence the solution area would be on the left side of the line`s graph.

x 0, y 0

Since both x and y are greater than 0, the solution area would be in the 1^st quadrant.

Hence, the solution area for the given inequalities would be the shaded area of the graph satisfying all the given inequalities.

Given inequality 2 ≤ 3x – 4 ≤ 5

⇒ 2 ≤ 3x – 4 ≤ 5

⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4

⇒ 6 ≤ 3x ≤ 9

⇒ 6/3 ≤ 3x/3 ≤ 9/3

⇒ 2 ≤ x ≤ 3

∴ all real numbers x greater than or equal to 2 but less than or equal to 3 are solution of given equality.

x ∈ [2, 3]

Given inequality

⇒ 6 ≤ -3 (2x – 4) < 12

Dividing the inequality by 3 we get.

⇒ 2 ≤ - (2x – 4) < 4

Multiplying the inequality by -1.

⇒ -2 ≥ 2x – 4 > -4 [multiplying the inequality with -1 changes the inequality sign.]

⇒ -2 + 4 ≥ 2x – 4 + 4 > -4 + 4

⇒ 2 ≥ 2x > 0

Dividing the inequality by 2

⇒ 0 < x ≤ 1

∴ all real numbers x greater than 0 but less than or equal to 1 are solution of given equality.

x ∈ (0, 1]

Given inequality

⇒

Multiplying the inequality by -2.

⇒ 14 ≥ 7x ≥ -28

⇒ -28 ≤ 7x ≤ 14

Dividing the inequality by 7

⇒ -4 ≤ x ≤ 2

∴ all real numbers x greater than or equal to -4 but less than or equal to 2 are solution of given equality.

x ∈ [-4, 2]

Given inequality

⇒

Multiplying the inequality by 5.

⇒ -75 < 3(x – 2) ≤ 0

Dividing the inequality by 3 we get

⇒ -25 < x – 2 ≤ 0

⇒ – 25 + 2 < x – 2 + 2 ≤ 0 + 2

⇒ – 23 < x ≤ 2

∴ all real numbers x greater than -23 but less than or equal to 2 are solution of given equality.

x ∈ (-23, 2]

Given inequality

⇒

Multiplying the inequality by 5.

⇒ -80 < 3x ≤ -10

∴ all real numbers x greater than -80/3 but less than or equal to -10/3 are solution of given equality.

x ∈ (-80/3, -10/3]

Given inequality

⇒

Multiplying the inequality by 2.

⇒ 14 ≤ 3x + 11 ≤ 22

⇒ 14 – 11 ≤ 3x + 11 – 11 ≤ 22 – 11

⇒ 3 ≤ 3x ≤ 11

∴ all real numbers x greater than or equal to -4 but less than or equal to 2 are solution of given equality.

x ∈ [1, 11/3]

Given inequalities 5x + 1 > -24 and 5x – 1 < 24

5x + 1 > -24

⇒ 5x > -24 – 1

⇒ 5x > -25

⇒ x > -5 ………(I)

5x – 1 < 24

⇒ 5x < 24 + 1

⇒ 5x < 25

⇒ x < 5 ……….(II)

From (I) and (II) we conclude that solution of given inequalities is (-5, 5).

Given inequalities 2 (x – 1) < x + 5 and 3 (x + 2) > 2 – x

2 ( x – 1) < x + 5

⇒ 2x – 2 < x + 5

⇒ 2x – x < 5 + 2

⇒ x < 7 ………(I)

3 (x + 2) > 2 – x

⇒ 3x + 6 > 2 – x

⇒ 3x + x > 2 – 6

⇒ 4x > -4

⇒ x > -1 ……….(II)

From (I) and (II) we conclude that solution of given inequalities is (-1, 7).

Given inequalities 3x – 7 > 2(x – 6) and 6 – x > 11 – 2x

3x – 7 > 2(x – 6)

⇒ 3x – 7 > 2x – 12

⇒ 3x – 2x > 7 – 12

⇒ x > -5 …………(I)

6 – x > 11 – 2x

⇒ 2x – x > 11 – 6

⇒ x > 5 ……….(II)

From (I) and (II) we conclude that solution of given inequalities is (5, ∞).

Given inequalities 5(2x – 7) – 3(2x + 3) ≤ 0 and 2x + 19 ≤ 6x + 47

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ 4x ≤ 44

⇒ x ≤ 11 ……(I)

2x + 19 ≤ 6x +47

⇒ 6x – 2x ≥ 19 – 47

⇒ 4x ≥ -28

⇒ x ≥ -7 ……….(II)

From (I) and (II) we conclude that solution of given inequalities is (-7, 11).

The solution has to be kept between 68° F and 77° F

i.e 68° < F < 77°

Putting

⇒ 20 < C < 25

The range of temperature in degree Celsius is between 20° C to 25° C.

8% of solution of boric acid = 640 litres

Let x litres of 2% boric acid solution is added

Total mixture = x + 640 litres

The resulting mixture has 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)

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

⇒ 2x + 5120 > 4x + 2560

⇒ 5120 – 2560 > 4x – 2x

⇒ 2560 > 2x

⇒ x < 1280 ….(I)

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

⇒ 2x + 5120 < 6x + 3840

⇒ 6x – 2x > 5120 – 3840

⇒ 4x > 1280

⇒ x > 320 ……….(II)

From (I) and (II)

320 < x < 1280

⇒ the number of litres of 2% of boric acid solution that has to be added will be more than 320 litres but less than 1280 litres.

45% of solution of acid = 1125 litres

Let x litres of water is added

Resulting mixture = x + 1125 litres

The resulting mixture has to be more than 25% but less than 30% acid content.

Amount of acid in resulting mixture = 45% of 1125 litres.

∴ 45% of 1125 < 30% of (x + 1125) and 45% of 1125 > 25% of (x + 1125)

45% of 1125 < 30% of (x + 1125)

⇒ 45 × 1125 < 30x + 30 × 1125

⇒ (45 – 30) × 1125 < 30x

⇒ 15 × 1125 < 30x

⇒ x > 562.5 ………..(I)

45% of 1125 > 25% of (x + 1125)

⇒ 45 × 1125 > 25x + 25 × 1125

⇒ (45 – 25) × 1125 > 25x

⇒ 25x < 20 × 1125

⇒ x < 900 …..(II)

∴ 562.5 < x < 900

Number of litres of water that has to be added will have to be more than 562.5 litres but less tan 900 litres.

Chronical age = CA = 12 years

IQ for age group of 12 is 80 ≤ IQ ≤ 140.

Let MA denote Mental Age

80 ≤ IQ ≤ 140

Putting

⇒ 9.6 ≤ MA ≤ 16.8

∴ Range of mental age of the group of 12 years old children is

9.6 ≤ MA ≤ 16.8