Linear Inequalities

Given,

Multiplication by (x + 1)

⇒ 4 ≤ 3(x + 1) ≤ 6

⇒ 4 ≤ 3x + 3 ≤ 6

Subtracting each side 3, we get

⇒ 1 ≤ 3x ≤ 3

Dividing each side by 3,

Let y = |x – 2|, then

Now, if y < 1, then

y – 1 < 0 and y – 2 < 0

and

This is not required

if y > 2, then

y – 1 > 0 and y – 2 > 0

and

This is not required

if 1 ≤ y < 2, then

y – 1 ≥ 0 and y – 2 < 0

and

This is required answer,

Hence

1 ≤ y < 2

⇒ 1 ≤ |x – 2| < 2

Therefore, there are two cases

⇒ 1 ≤ x – 2 < 2

⇒ 3 ≤ x < 4

and

⇒ 1 ≤ -(x – 2) < 2

⇒ 1 ≤ - x + 2 < 2

Multiplying with -1 all sides

⇒ -2 ≤ x – 2 < -1

Adding 2 both side

⇒ 0 ≤ x < 1

∴ Hence, solution is [0, 1) ∪ [3, 4)

Given,

⇒ 5 - |x| ≤ 0 and |x| - 3 > 0 or 5 - |x| ≥ 0 and |x| - 3 < 0

⇒ |x| ≥ 5 and |x| > 3 or |x| ≤ 5 and |x| < 3

⇒ |x| ≥ 5 or |x| < 3

⇒ x ∈ (- ∞ , - 5] or [5, ∞) or x ∈ ( -3 , 3)

⇒ x ∈ (- ∞ , - 5] ∪ ( -3 , 3) ∪ [5, ∞)

|x – 1|≤ 5

There are two cases,

Case 1:

x – 1 ≤ 5

⇒ x ≤ 6 [adding 1 both side]

Case 2:

⇒ -(x – 1) ≤ 5

⇒ -x + 1 ≤ 5

⇒ -x ≤ 4 [Subtracting 1 both side]

⇒ x ≥ -4

From both cases, we have

⇒ -4 ≤ x ≤ 6 [1]

Also,

|x| ≥ 2

⇒ x ≥ 2 and

⇒ -x ≥ 2

⇒ x ≤ -2

⇒ x ∈ (∞, -2] ∪ [2, ∞) [2]

Combining [1] and [2] we get

x ∈ [-4, -2] ∪ [2, 6]

Given,

Multiplication by 4, we get

⇒ -20 ≤ 2 – 3x ≤ 36

Adding -2 each side,

⇒ -22 ≤ -3x ≤ 34

Divide by 3, we get

Multiplication by -1, we get

[Multiplication by -1 inverts the inequality]

Given,

4x + 3 ≥ 2x + 17

⇒ 4x – 2x ≥ 17 – 3

⇒ 2x ≥ 14

⇒ x ≥ 7 [1]

Also,

3x – 5 < - 2

⇒ 3x < 3

⇒ x < 1 [2]

As [1] and [2] are not possible together.

So, x has no solution.

Profit = Revenue – cost

Requirement is, profit > 0

Also, Given

Revenue, R(x) = 43 x

and C(x) = 26,000 + 30 x

where x is number of cassettes

⇒ Profit = 43x – (26,000 + 30x) > 0

⇒ 13x – 26,000 > 0

⇒ 13x > 26000

⇒ x > 2000

Hence, the company should sell more than 2000 cassettes

Given,

First reading = 8.48

Second reading = 8.35

Let the third reading be ‘x’

Now, average pH should be between 8.2 and 8.2

Average pH

Multiplying by 3, we get

⇒ 24.6 < 16.83 + x < 25.5

Subtracting 16.83, we get

⇒ 7.77 < x < 8.67

Hence, third pH reading should be between 7.77 and 8.67

Let the x litres of 3% solution is to be added to 460 liters of the 9% of solution, then

Total solution = (460 + x) litres

Total acid content in resulting solution

= (41.4 + 0.03x)%

It is given that resulting mixture should be more than 5% acidic but less than 7% acidic

⇒ 5 % of (460 + x) < 41.4 + 0.03x < 7% of (460 + x)

⇒ 23 + 0.05 x < 41.4 + 0.03x < 32.2 + 0.07x

Now, we have

⇒ 23 + 0.05x < 41.4 + 0.03x and 41.4 + 0.03x < 32.2 + 0.07x

⇒ 0.02x < 18.4 and 0.04x > 9.2

⇒ 2x < 1840 and 4x > 920

⇒ x < 920 and x > 230

⇒ 230 < x < 920

Hence, solution must be added between 230 l and 920 l

Let temperature in Celsius be C and temperature in Fahrenheit be F

Given,

Solution should be kept between 40° C and 45°C

⇒ 40 < C < 45

Multiplying by , we get

Adding 32, we get

⇒ 104 < F < 113

Hence, temperature in Fahrenheit should be between 104° F and 113° F.

Let the length of shortest side be ‘x’ cm

Given, the longest side of a triangle is twice the shortest side

⇒ length of largest side = 2x

Also, the third side is 2 cm longer than the shortest side

⇒ length of third side = (x + 2) cm

Perimeter of triangle = sum of three sides

= x + 2x + x + 2

= 4x + 2 cm

Now, perimeter is more than 166 cm

⇒ 4x + 2 ≥ 166

⇒ 4x ≥ 164

⇒ x ≥ 41

Hence, minimum length of the shortest side should be 41 cm.

Given,

T = 30 + 25(x – 3), 3 ≤ x ≤ 15

Where, T = temperature and

x = depth inside the earth

Now, Temperature should be between 155°C and 205°C

⇒ 155 < T < 205

⇒ 155 < 30 + 25(x – 3) < 205

⇒ 155 < 30 + 25x – 75 < 205

⇒ 155 < 25x – 45 < 205

Adding 45, we get

⇒ 200 < 25x < 250

Dividing by 25, we get

⇒ 8 < x < 10

Hence, between 8 to 10 cm depth, temperature varies from 155° C to 205° C.

Given,

Subtracting 5 both side, we get

Hence, for above fraction be greater than 0, either both denominator and numerator should be greater than 0 or both should be less than 0

⇒ 6 – 33x > 0 and 7x – 1 > 0

⇒ 33x < 6 and 7x > 1

and

⇒ [1]

Or

⇒ 6 – 33x < 0 and 7x – 1 < 0

⇒ 33x > 6 and 7x < 1

and

⇒ [not possible as ]

Also,

Subtracting 2 both sides, we get

Hence, for above fraction be greater than 0, either both denominator and numerator should be greater than 0 or both should be less than 0

⇒ 23 – x > 0 and x – 8 > 0

⇒ x < 23 and x > 8

⇒ 8 < x < 23 [2]

Or

23 – x < 0 and x – 8 < 0

⇒ x > 23 and x < 8

⇒ 23 < x < 8 [Not possible, as 23 > 8]

From [1] and [2], it is clear that there is no solution satisfying both inequalities, hence the given system has no solution

First consider the linear equation 3x + 2y = 48, Observe that the shaded region and the origin both are on the same side of the graph of the line and (0, 0) satisfy the constraint 3x + 2y ≤ 48.

Now, consider x + y = 20, observe that the shaded region and the origin both are on the same side of the graph of the line and (0, 0) satisfy the constraint x + y ≤ 20.

Also, shaded region is in the first quadrant i.e. x ≥ 0 and y ≥ 0,

Hence, the linear inequalities are

3x + 2y ≤ 48

x + y ≤ 20

x ≥ 0

y ≥ 0

First consider the linear equation x + y = 8, Observe that the shaded region and the origin both are on the same side of the graph of the line and (0, 0) satisfy the constraint x + y ≤ 8.

Now, consider x + y = 4, observe that the origin is on the opposite side of the shaded region and (0, 0) doesn’t satisfy the constraint x + y ≥ 4, therefore required constraint is x + y ≥ 4

Also, shaded region is in the first quadrant i.e. x ≥ 0 and y ≥ 0,

Also, shades region is below the line y = 5 and left to the line x = 5

⇒ y ≤ 5 and x ≥ 5

Hence, the linear inequalities are

x + y ≤ 8

x + y ≥ 4

x ≥ 0

y ≥ 0

x ≤ 5

y ≤ 5

Let’s plot the region of each inequality and then find the common region of all

x + 2y ≤ 3

Line: x + 2y = 3

Also, (0, 0) satisfies the x + 2y ≤ 3, hence region is towards the origin

3x + 4y ≤ 12

Line: 3x + 4y = 12

Also, (0, 0) satisfies the 3x + 4y ≤ 3, hence region is towards the origin

x ≥ 0 implies that region is right to the y-axis and y ≥ 1 implies that region is up above the line x = 1, Therefore graph is

It is clear from the graph the above system has no common region as solution

Let’s plot the region of each inequality and then find the common region of all

3x + 2y ≥ 24

Line: 3x + 2y = 24

Also, (0, 0) doesn’t satisfy the 3x + 2y ≥ 24, hence region is away from the origin

3x + y ≤ 15

Line: 3x + y = 15

Also, (0, 0) satisfies the 3x + y ≤ 15, hence region is towards the origin

x ≥ 4 implies that region is right to the line x = 4, therefore graph is

It is clear from the graph the above system has no common region as solution

Let’s plot the region of each inequality and then find the common region of all

2x + y ≥ 8

Line: 2x + y = 8

Also, (0, 0) doesn’t satisy the 2x + y ≥ 8, hence region is away from the origin

x + 2y ≥ 10

Line: x + 2y = 10

Also, (0, 0) doesn’t satisy the x + 2y ≥ 10, hence region is away from the origin

x ≥ 0, and y ≥ 0 implies that region is in first quadrant, therefore graph is

Clearly shaded region is unbounded.

x < 5

Multiplication or dividing by negative number inverts the inequality sign.

⇒ -x > -5

Given,

x < y

Now, b < 0, and multiplying/dividing by a negative inequality inverts the inequality sign

-3x + 17 < -13

⇒ -3x < -30 [Subtracting 17 both side]

⇒ x > 10 [Division by negative number inverts the inequality sign]

⇒ x ∈ (10, ∞)

|x| < 3

Hence, there are two cases,

x < 3 [1]

and

-x < 3

⇒ x > -3 [2]

From [1] and [2], we get

⇒ -3 < x < 3

|x| > b

Hence, there are two cases,

x > b

⇒ x ∈ (b, ∞) [1]

and

-x > b

⇒ x < -b

⇒ x ∈ (-∞, -b) [2]

From [1] and [2], we get

⇒ x ∈ (-∞, -b) ∪ (b, ∞)

Given,

|x – 1| > 5

There are two cases,

(x – 1) > 5

⇒ x > 6

⇒ x ∈ (6, ∞) [1]

And

-(x – 1) > 5

⇒ -x + 1 > 5

⇒ -x > 4

⇒ x < -4 [Multiplication by negative inverts the inequality sign]

⇒ x ∈ (-∞, -4) [2]

From [1] and [2], we get

⇒ x ∈ (-∞, -4) ∪ (6, ∞)

Given,

|x + 2| ≤ 9

There are two cases,

(x + 2) ≤ 9

⇒ x ≤ 7

⇒ x ∈ (-∞,7] [1]

And

-(x + 2) ≤ 9

⇒ -x – 2 ≤ 9

⇒ -x ≤ 11

⇒ x ≥ -11 [Multiplication by negative inverts the inequality sign]

⇒ x ∈ [-11, ∞) [2]

From [1] and [2], we get

⇒ x ∈ [-11, 7]

|x| < 5

Hence, there are two cases,

x < 5 [1]

and

-x < 5

⇒ x > -5 [2]

From [1] and [2], we get

⇒ -5 < x < 5

And clearly, graph represents the graph of this function

The above graph represents all values of x greater than 5 excluding 5, hence

x > 5 i.e. x ∈ (5, ∞)

The above graph represents all values of x greater than including , hence

i.e.

The above graph represents all values of x less than excluding , hence

i.e.

The above graph represents all values of x less tha -2 including -2, hence

x ≤ 2

⇒ x ∈ (-∞, -2]

(i) False

Explanation:

Given,

x < y

Now, b < 0, and multiplying/dividing by a negative inequality inverts the inequality sign

(ii) False

Explanation:

We know, If xy > 0

Then, either x > 0 and y > 0

Or

x < 0 and y < 0

(iii) True

Explanation:

We know, If xy > 0

Then, either x > 0 and y > 0

Or

x < 0 and y < 0

(iv) False

Explanation:

We know, If xy < 0

Then, either x < 0 and y > 0

Or

x > 0 and y < 0

(v) True

Explanation:

x < -5, i.e.

x ∈ (-∞, -5) [1]

and

x < -2, i.e.

x ∈ (-∞, -2) [2]

Taking intersection from [1] and [2], we get

x ∈ (-∞, -5)

(vi) False

Explanation:

x < -5, i.e.

x ∈ (-∞, -5) [1]

and

x > 2, i.e.

x ∈ (2, ∞) [2]

From [1] and [2], we get

x has no common solution

(vii) True

Explanation:

x > -2, i.e.

x ∈ (-2, ∞) [1]

and

x < 9, i.e.

x ∈ (-∞, 9) [2]

Taking intersection from [1] and [2], we get

x ∈ (-2, 9)

(viii) True

Explanation:

|x| > 5

Hence, there are two cases,

x > 5

⇒ x ∈ (5, ∞) [1]

and

-x > 5

⇒ x < -5

⇒ x ∈ (-∞, -5) [2]

From [1] and [2], we get

⇒ x ∈ (-∞, -5) ∪ (5, ∞)

(ix) True

Explanation:

|x| ≤ 4

Hence, there are two cases,

x ≤ 4

⇒ x ∈ (-∞, 4] [1]

and

-x ≤ 4

⇒ x ≥ -4

⇒ x ∈ [-4, ∞) [2]

From [1] and [2], we get

⇒ x ∈ [-4, 4]

(x) True

Explanation:

Line is x = 3, and

since origin i.e.

(0, 0) satisfies the inequality x < 3, the above graph is correct

(xi) True

Explanation:

Since, x ≥ 0 represents the positive value of x, the region must be on positive side of line x = 0 i.e. y-axis

(xii) false

Explanation:

Since, y ≤ 0 represents the negative value of y, the region must be on negative side of line y = 0 i.e. x-axis

(xiii) False

Explanation:

x ≥ 0 and y ≤ 0 represents the 4^th quadrant, while the region shaded is first quadrant

(xiv) False

Explanation:

x ≥ 0 implies that, region is on the left side of y axis, and

y ≤ 1 implies that, region is below the line y = 1, therefore graph must be

(xv) True
Explanation:

If we take any point above the line x + y = 0, say (3, 2) it satisfy the inequality

x + y ≥ 0 [as, 3 + 2 = 5 > 0]

Hence, region should be above the line x + y = 0

(i) ≤

-4x ≥ 12

Dividing by 4, we get

⇒ -x ≥ 3

Now, taking negative both sides inverts the inequality sign

⇒ x ≤ 3

(ii) ≤

Multiplying by 4, we get

⇒ -3x ≥ -12

Now, taking negative both sides invert the inequality sign

⇒ 3x ≤ 12

⇒ x ≤ 4

(iii) >

Now, for above to be greater than 0,

x + 2 > 0

⇒ x > -2

(iv) >

x > -5

Multiplying by 4 both sides, we get

⇒ 4x > -20

(v) >

x > y

Since, z < 0, i.e. z is negative and multiplying by negative inverts the inequality sign

⇒ xz < yz

Now, taking negative both sides invert the inequality sign

⇒ -xz > -yz

(vi) >

q < 0

Now, taking negative both sides invert the inequality sign

⇒ -q > 0

Adding p both side, we get

⇒ p – q > p [since, p>0 inequality sign retains the same]

(vii) <, >

Given,

|x + 2| > 5

∴ there are two cases

⇒ x + 2 > 5 and –(x + 2) < 5

⇒ x > 3 and –x – 2 < 5

⇒ x > 3 and –x > 7

⇒ x > 3 and x < -7

[Taking negative both sides invert the inequality sign]

(viii) ≤

-2x + 1 ≥ 9

Adding -1 both side, we get

⇒ -2x ≥ 8

⇒ x ≤ -4 [Dividing by negative inverts the inequality sign]