Chapter 6: Linear Inequalities

According to the question,

Multiplying each term by (x + 1)

⇒ 4 ≤ 3(x + 1) ≤ 6

⇒ 4 ≤ 3x + 3 ≤ 6

Subtracting each term by 3, we get,

⇒ 1 ≤ 3x ≤ 3

Dividing each term by 3, we get,

⇒ (1/3) ≤ x ≤ 1

According to the question,

Hence,

1 ≤ y < 2

⇒ 1 ≤ |x – 2| < 2

Here, there are two cases

⇒ 1 ≤ x – 2 < 2

⇒ 3 ≤ x < 4

And

⇒ 1 ≤ -(x – 2) < 2

⇒ 1 ≤ – x + 2 < 2

Multiplying each term by -1,

⇒ -2 ≤ x – 2 < -1

Adding 2 to each term,

⇒ 0 ≤ x < 1

∴ Hence, the solution is [0, 1) ∪ [3, 4)

According to the question,

⇒ 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,

1:-

x – 1 ≤ 5

Adding 1 to LHS and RHS

⇒ x ≤ 6

2:-

⇒ -(x – 1) ≤ 5

⇒ -x + 1 ≤ 5

Subtracting 1 from LHS and RHS,

⇒ -x ≤ 4

⇒ x ≥ -4

From cases 1 and 2, we have

⇒ -4 ≤ x ≤ 6 …[i]

Also,

|x| ≥ 2

⇒ x ≥ 2 and

⇒ -x ≥ 2

⇒ x ≤ -2

⇒ x ∈ (∞, -2] ∪ [2, ∞) …[ii]

Combining equations [i] and [ii], we get

x ∈ [-4, -2] ∪ [2, 6]

According to the question,

Multiplying each term by 4, we get

⇒ -20 ≤ 2 – 3x ≤ 36

Adding -2 each term, we get

⇒ -22 ≤ -3x ≤ 34

Dividing each term by 3, we get

⇒ -22/3 ≤ -x ≤ 34/3

We know that,

Multiplication by -1 inverts the inequality.

So, multiplying each term by -1, we get

⇒  -34/3≤ x ≤ 22/3

According to the question,

4x + 3 ≥ 2x + 17

⇒ 4x – 2x ≥ 17 – 3

⇒ 2x ≥ 14

⇒ x ≥ 7 …(i)

Also,

3x – 5 < – 2

⇒ 3x < 3

⇒ x < 1 …(2)

Since, equations [i] and [ii] cannot be possible simultaneously,

We conclude that x has no solution.

We know that,

Profit = Revenue – cost

The requirement is, profit > 0

According to the question,

Revenue, R(x) = 43 x

Cost, 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

Therefore, the company should sell more than 2000 cassettes to realise profit.

According to the question,

First reading = 8.48

Second reading = 8.35

Now, let the third reading be ‘x’

Average pH should be between 8.2 and 8.5

Average pH = (8.48 + 8.35 + x)/3

Multiplying each term by 3, we get

⇒ 24.6 < 16.83 + x < 25.5

Subtracting 16.83 from each term, we get

⇒ 7.77 < x < 8.67

Therefore, from the above equation,

We get that,

The third pH reading should be between 7.77 and 8.67

According to the question,

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

Then, we get,

Total solution = (460 + x) litres

The total acid content in the resulting solution

= (460 × 9/100 + x × 3/100)

= (41.4 + 0.03x)%

According to the question, we have,

The resulting mixture should be more than 5% acidic but less than 7% acidic

So we get,

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

⇒ 5/100 × (460 + x) < 41.4 + 0.03x < 7/100 × (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

i.e., 0.02x < 18.4 and 0.04x > 9.2

⇒ 2x < 1840 and 4x > 920

⇒ x < 920 and x > 230

⇒ 230 < x < 920

Hence, solution between 230 l and 920 l should be added.

Let temperature in Celsius be C

Let temperature in Fahrenheit be F

According to the question,

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

⇒ 40 < C < 45

Multiplying each term by 9/5, we get

⇒ 72 < 9/5 C < 81

Adding 32 to each term, we get

⇒ 104 < 9/5 C + 32 < 113

⇒ 104 < F < 113

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