NCERT Solutions for Class 11 Maths Chapter 5 Linear-Inequalities

Given that, 3(x – 1) ≤ 2 (x – 3)

By multiplying, the above inequality can be written as

3x – 3 ≤ 2x – 6

Now, by adding 3 to both the sides, we get

3x – 3+ 3 ≤ 2x – 6+ 3

3x ≤ 2x – 3

Again, by subtracting 2x from both the sides,

3x – 2x ≤ 2x – 3 – 2x

x ≤ -3

Therefore, the solutions of the given inequality are defined by all the real numbers less than or equal to -3.

Hence, the required solution set is (-∞, -3]

 

Let us assume that x is the marks obtained by Ravi in his third unit test.

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

(70 + 75 + x)/3 ≥ 60

= 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

 

(i) Given that 24x < 100

Now we have to divide the inequality by 24 then we get x < 25/6

Now when x is a natural integer then

It is clear that the only natural number less than 25/6 are 1, 2, 3, 4.

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

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

(ii) Given that 24x < 100

Now we have to divide the inequality by 24 then we get x < 25/6

now when x is an integer then

It is clear that the integer number less than 25/6 are…-1, 0, 1, 2, 3, 4.

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

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

 

 

(i) Given that, – 12x > 30

Now, by dividing the inequality by -12 on both sides we get, x < -5/2

When x is a natural integer then

It is clear that there is no natural number less than -2/5 because -5/2 is a negative number and natural numbers are positive numbers.

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

(ii) Given that, – 12x > 30

Now by dividing the inequality by -12 on both sides we get, x < -5/2

When x is an integer then

It is clear that the integer number less than -5/2 are…, -5, -4, – 3

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

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

 

 

(i) Given that, 5x – 3 < 7

Now by adding 3 on both sides, we get,

5x – 3 + 3 < 7 + 3

The above inequality becomes

5x < 10

Again, by dividing both sides by 5 we get,

5x/5 < 10/5

x < 2

When x is an integer, then

It is clear that that the integer number less than 2 are…, -2, -1, 0, 1.

Thus, solution of 5x – 3 < 7 is …,-2, -1, 0, 1, when x is an integer.

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

(ii) Given that, 5x – 3 < 7

Now by adding 3 on both sides, we get,

5x – 3 + 3 < 7 + 3

Above inequality becomes

5x < 10

Again, by dividing both sides by 5, we get,

5x/5 < 10/5

x < 2

When x is a real number, then

It is clear 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.

Hence the solution set is x ∈ (-∞, 2)

 

 

(i) Given that, 3x + 8 > 2

Now by subtracting 8 from both sides, we get,

3x + 8 – 8 > 2 – 8

The above inequality becomes,

3x > – 6

Again by dividing both sides by 3, we get,

3x/3 > -6/3

Hence x > -2

When x is an integer, then

It is clear that the integer numbers greater than -2 are -1, 0, 1, 2,…

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

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

(ii) Given that, 3x + 8 > 2

Now by subtracting 8 from both sides we get,

3x + 8 – 8 > 2 – 8

The above inequality becomes,

3x > – 6

Again, by dividing both sides by 3, we get,

3x/3 > -6/3

Hence x > -2

When x is a real number.

It is clear that the solutions of 3x + 8 >2 will be given by x > -2 which means all the real numbers that are greater than -2.

Therefore the solution set is x ∈ (-2, ∞)

 

Given that, 4x + 3 < 5x + 7

Now by subtracting 7 from both the sides, we get

4x + 3 – 7 < 5x + 7 – 7

The above inequality becomes,

4x – 4 < 5x

Again, by 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.

The required solution set is (-4, ∞)

 

Given that,

3x – 7 > 5x – 1

Now, by adding 7 to both the sides, we get

3x – 7 +7 > 5x – 1 + 7

3x > 5x + 6

Again, by subtracting 5x from both the sides,

3x – 5x > 5x + 6 – 5x

-2x > 6

Dividing both sides by -2 to simplify, we get

-2x/-2 < 6/-2

x < -3

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

Hence the required solution set is (-∞, -3)

 

On computing we get

= 15x ≥ 2 (4x – 1)

= 15x ≥ 8x -2

= 15x -8x ≥ 8x -2 -8x

= 7x ≥ -2

= x ≥ -2/7

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

 

 

Given that, 3 (2 – x) ≥ 2 (1 – x)

By multiplying, we get

6 – 3x ≥ 2 – 2x

Now, by adding 2x to both the sides,

6 – 3x + 2x ≥ 2 – 2x + 2x

6 – x ≥ 2

Again, by subtracting 6 from both the sides, we get

6 – x – 6 ≥ 2 – 6

– x ≥ – 4

Multiplying throughout inequality by negative sign, we get

x ≤ 4

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

Hence the required solution set is (- ∞, 4]

 

 

x < 6

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

Hence the solution set is (-∞, 6)

 

 

 

– x/6 > 1

– x > 6

x < – 6

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

Hence, the required solution set is (-∞, -6)

 

Given that,

Now by cross-multiplying the denominators, we get

9(x- 2) ≤ 25 (2 – x)

9x – 18 ≤ 50 – 25x

Now 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,

34x/34 ≤ 68/34

x ≤ 2

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

Required solution set is (-∞, 2]

 

 

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.

 

Given that,

2 (2x + 3) – 10 < 6 (x – 2)

By multiplying, we get

4x + 6 – 10 < 6x – 12

On simplifying, we get

4x – 4 < 6x – 12

4x – 6x < -12 + 4

-2x < -8

Dividing by 2, we get;

-x < -4

Multiply by “-1” and change the sign.

x > 4

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

Hence, the required solution set is (4, ∞).

 

Given that, 37 – (3x + 5) ≥ 9x – 8 (x – 3)

On simplifying, we get

= 37 – 3x – 5 ≥ 9x – 8x + 24

= 32 – 3x ≥ x + 24

On rearranging,

= 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, (-∞, 2] will be the solution for the given inequality

 

 

= 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, (4, ∞) will be the solution for the given inequality

 

= 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, (-∞, 2] will be the solution for the given inequality

 

Given,

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,

5x – 3 ≥ 3x – 5

Solving the given inequality, we get

5x – 3 ≥ 3x – 5

On rearranging, we get

= 5x – 3x ≥ -5 + 3

On simplifying,

= 2x ≥ -2

Now, dividing by 2 on both sides, we get

= x ≥ -1

The graphical representation of the solution is as follows:

 

Given,

3 (1 – x) < 2 (x + 4)

Solving the given inequality, we get

3 (1 – x) < 2 (x + 4)

Multiplying, we get

= 3 – 3x < 2x + 8

On rearranging, we get

= 3 – 8 < 2x + 3x

= – 5 < 5x

Now by dividing 5 on both sides, we get

-5/5 < 5x/5

= – 1 < x

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

 

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 obtain an average of 90 marks or more in her five examinations

(87 + 92 + 94 + 95 + x)/5 ≥ 90

= (368 + x)/5 ≥ 90

= 368 + x ≥ 450

= x ≥ 450 – 368

= x ≥ 82

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

 

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

∴ The other integer is = 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 the sum of two integers is more than 11.

∴ x + (x + 2) > 11

2x + 2 > 11

x > 9/2

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 is the smaller of the two consecutive even positive integers.

∴ The 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 < 21/2

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 to be x cm.

∴ According to the question, the 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, x + 3x + (3x – 2) cm ≥ 61 cm

= 7x – 2 ≥ 61

= 7x ≥ 63

Now dividing by 7, we get

= 7x/7 ≥ 63/7

= x ≥ 9

Hence, the minimum length of the shortest side will be 9 cm.

 

Let us assume the length of the shortest piece to 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

= 4x/4 ≤ 88/4

= 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

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. (i.e., origin is included in the region)

The graph is as follows:

 

Given 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

⇒ 0 ≥ 6 (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

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

⇒ 0 ≤ 12 (this is true)

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

The graph is as follows:

 

Given 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)

⇒ 0≤ 8 (this is true)

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

The graph is as follows:

 

Given 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 ≤ 2 (this is true)

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

The graph is as follows:

 

Given 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 > 6 (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

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. (i.e., origin is included in the region)

The graph is as follows:

 

Given y – 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. (i.e., origin is included in the region)

The graph is as follows:

 

Given 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 (this is false)

∴ Solution region of the given inequality is below the line y < – 2. (i.e., away from the origin)

The graph is as follows:

 

Given 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. (i.e., origin is included in the region)

The graph is as follows:

 

Given x ≥ 3……… (i)

y ≥ 2…………… (ii)

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 (i)

0 ≥ 3 which is not true

Putting y = 0 in (ii)

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…………… (i)

3x + 4y ≤ 12 ……………. (ii)

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 lie 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.

 

Given 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……………… (i)

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………… (ii)

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

y = ½ = 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 lie 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 required area would be the area to the right of the 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.

We have x ≥ 0

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

Therefore, the shaded area 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.

We have 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.

Consider 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 not 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.

 

Given 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.

We have 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.

Consider 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 lie 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 the 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 include the origin and the required solution area would be on the right side of the line’s graph.

We have 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.

Consider, 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 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 be on the left of the line’s graph.

We have 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.

Consider, x ≥ 3,

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

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

So, the origin doesn’t lie in the solution area. Hence, the required solution area would be on the right of the line’s graph.

We have x, y ≥ 0

Since it is given both x and y are greater than 0

∴ the solution area would be in the first Ist 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 the 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.

We have 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.

Given 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.

Consider y ≥ 0, x ≥ 0

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

We have 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 in the solution area. The required solution area would be toward the right of the line’s graph.

Consider 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.

Again, we have x ≥ 0, y ≥ 0

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

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

 

According to the question,

The inequality given is,

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

x ∈ (-80/3, -10/3]

 

According to the question,

The inequality given is,

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

Hence, all real numbers x greater than or equal to 2, but less than or equal to 3 are solutions of given equality.

x ∈ [2, 3]

 

According to the question,

The inequality given is,

6 ≤ –3 (2x – 4) < 12

⇒ 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

Hence, all real numbers x greater than 0, but less than or equal to 1 are solutions of given equality.

x ∈ (0, 1]

 

According to the question,

The inequality given is,

– 3 ≤ 4 – 7x/2 ≤ 18

⇒ – 3 – 4 ≤ 4 – 7x/2 – 4 ≤ 18 – 4

⇒ – 7 ≤ – 7x/2 ≤ 18 – 14

Multiplying the inequality by -2,

⇒ 14 ≥ 7x ≥ -28

⇒ -28 ≤ 7x ≤ 14

Dividing the inequality by 7,

⇒ -4 ≤ x ≤ 2

Hence, all real numbers x greater than or equal to -4, but less than or equal to 2 are solutions of given equality.

x ∈ [-4, 2]

 

According to the question,

The inequality given is,

– 15 ≤ 3(x – 2)/5 ≤ 0

⇒ – 15 < 3(x – 2)/5 ≤ 0

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

Hence, all real numbers x greater than -23, but less than or equal to 2 are solutions of given equality.

x ∈ (-23, 2]

 

According to the question,

The inequality given is,

⇒ 14 ≤ 3x + 11 ≤ 22

⇒ 14 – 11 ≤ 3x + 11 – 11 ≤ 22 – 11

⇒ 3 ≤ 3x ≤ 11

⇒ 1 ≤ x ≤ 11/3

Hence, all real numbers x greater than or equal to -4, but less than or equal to 2 are solutions of given equality.

x ∈ [1, 11/3]

Solve the inequalities in Exercises 7 to 11 and represent the solution graphically on number line.

 

According to the question,

The inequalities given are,

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 equations (i) and (ii),

We can infer that the solution of given inequalities is (-5, 5).

 

According to the question,

The inequalities given are,

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 equations (i) and (ii),

We can infer that the solution of given inequalities is (-1, 7).

 

According to the question,

The inequalities given are,

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 equations (i) and (ii),

We can infer that the solution of given inequalities is (5, ∞).

 

According to the question,

The inequalities given are,

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 equations (i) and (ii),

We can infer that the solution of given inequalities is (-7, 11).

 

According to the question,

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

So, we get, 68° < F < 77°

Substituting,

⇒ 20 < C < 25

Hence, we get,

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

 

According to the question,

8% of solution of boric acid = 640 litres

Let the amount of 2% boric acid solution added = x litres

Then we have,

Total mixture = x + 640 litres

We know that,

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)

⇒ (2/100) × x + (8/100) × 640 > (4/100) × (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)

⇒ (2/100) × x + (8/100) × 640 < (6/100) × (x + 640)

⇒ 2x + 5120 < 6x + 3840

⇒ 6x – 2x > 5120 – 3840

⇒ 4x > 1280

⇒ x > 320 ……….(i)

From (i) and (ii)

320 < x < 1280

Therefore, 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.

 

According to the question,

45% of solution of acid = 1125 litres

Let the amount of water added = x litres

Resulting mixture = x + 1125 litres

We know that,

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

Therefore, the number of litres of water that has to be added will have to be more than 562.5 litres but less than 900 litres.

 

According to the question,

Chronological age = CA = 12 years

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

We get that,

80 ≤ IQ ≤ 140

Substituting,

⇒ 9.6 ≤ MA ≤ 16.8

∴ Range of mental age of the group of 12 year-old-children is 9.6 ≤ MA ≤ 16.8