Important Formulas & Rules for Linear Inequalities

• Equalities vs. Inequalities: Two real numbers or two algebraic expressions related by the symbols '<', '>', '≤' or '≥' form an inequality.
• Rule 1: Equal numbers may be added to (or subtracted from) both sides of an inequality without changing the sign of inequality.
• Rule 2: Both sides of an inequality can be multiplied (or divided) by the same positive number without changing the sign of inequality.
• Rule 3: When both sides are multiplied or divided by a negative number, the sign of the inequality is reversed (e.g., < becomes >, and ≤ becomes ≥).
• Interval Notation:
  • Open Interval (a, b) = {x : a < x < b}
  • Closed Interval [a, b] = {x : a ≤ x ≤ b}
  • Semi-open/Semi-closed Interval [a, b) = {x : a ≤ x < b} or (a, b] = {x : a < x ≤ b}

Question 1:

Solve 24x < 100, when:
(i) x is a natural number (N)
(ii) x is an integer (Z)

Solution:

Given inequality: 24x < 100

Dividing both sides by 24:
x < 100 / 24
x < 25 / 6
x < 4.166...

(i) When x is a natural number (N = {1, 2, 3, 4, ...}):
The natural numbers less than 4.166 are 1, 2, 3, and 4.
Solution Set = {1, 2, 3, 4}

(ii) When x is an integer (Z = {..., -3, -2, -1, 0, 1, 2, ...}):
The integers less than 4.166 are ..., -3, -2, -1, 0, 1, 2, 3, 4.
Solution Set = {..., -3, -2, -1, 0, 1, 2, 3, 4}

Question 2:

Solve -12x > 30, when:
(i) x is a natural number
(ii) x is an integer

Solution:

Given inequality: -12x > 30

Dividing both sides by -12 (reverse the inequality sign):
x < 30 / (-12)
x < -5 / 2
x < -2.5

(i) When x is a natural number:
There is no natural number less than -2.5.
Solution Set = Ø (No solution)

(ii) When x is an integer:
Integers less than -2.5 are ..., -5, -4, -3.
Solution Set = {..., -5, -4, -3}

Question 3:

Solve 5x - 3 < 7, when:
(i) x is an integer
(ii) x is a real number

Solution:

Given: 5x - 3 < 7
Adding 3 to both sides:
5x < 10
Dividing by 5:
x < 2

(i) When x is an integer:
Solution Set = {..., -3, -2, -1, 0, 1}

(ii) When x is a real number:
All real numbers strictly less than 2.
Solution Set = (-∞, 2)

Question 4:

Solve 4x + 3 < 5x + 7 for real x.

Solution:

Given: 4x + 3 < 5x + 7
Subtracting 5x from both sides:
4x - 5x + 3 < 7
-x + 3 < 7
Subtracting 3 from both sides:
-x < 4
Multiplying both sides by -1 (reversing inequality):
x > -4

Solution Set = (-4, ∞)

Question 5:

Solve 3(x - 1) ≤ 2(x - 3) for real x.

Solution:

Expanding the brackets:
3x - 3 ≤ 2x - 6
Subtracting 2x from both sides:
3x - 2x - 3 ≤ -6
x - 3 ≤ -6
Adding 3 to both sides:
x ≤ -3

Solution Set = (-∞, -3]

Question 6:

Solve x + x/2 + x/3 < 11 for real x.

Solution:

Taking LCM of denominators (1, 2, 3) which is 6:
(6x + 3x + 2x) / 6 < 11
11x / 6 < 11
Multiplying both sides by 6:
11x < 66
Dividing by 11:
x < 6

Solution Set = (-∞, 6)

Question 7:

Solve (2x - 1)/3 ≥ (3x - 2)/4 - (2 - x)/5 for real x.

Solution:

RHS LCM of 4 and 5 is 20:
(2x - 1)/3 ≥ [5(3x - 2) - 4(2 - x)] / 20
(2x - 1)/3 ≥ [15x - 10 - 8 + 4x] / 20
(2x - 1)/3 ≥ (19x - 18) / 20
Cross-multiplying positive numbers 3 and 20:
20(2x - 1) ≥ 3(19x - 18)
40x - 20 ≥ 57x - 54
Rearranging terms:
40x - 57x ≥ -54 + 20
-17x ≥ -34
Dividing by -17 (reverse inequality):
x ≤ 2

Solution Set = (-∞, 2]

Question 8 (Word Problem):

Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Solution:

Let the marks obtained by Ravi in the third unit test be x.
Total marks in 3 tests = 70 + 75 + x = 145 + x
Average marks = (145 + x) / 3
According to the problem, average ≥ 60:
(145 + x) / 3 ≥ 60
Multiplying both sides by 3:
145 + x ≥ 180
x ≥ 180 - 145
x ≥ 35

Conclusion: Ravi must obtain a minimum of 35 marks in the third test.

Question 9 (Word Problem):

A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

Solution:

Let x litres of 2% boric acid solution be added.
Total volume of mixture = (640 + x) litres.
Total boric acid in mixture = 8% of 640 + 2% of x = (8/100)*640 + (2/100)*x
We are given:
4% of (640 + x) < [(8/100)*640 + (2/100)*x] < 6% of (640 + x)
Multiplying throughout by 100:
4(640 + x) < 8(640) + 2x < 6(640 + x)
2560 + 4x < 5120 + 2x < 3840 + 6x

Part 1: 2560 + 4x < 5120 + 2x
4x - 2x < 5120 - 2560
2x < 2560 ⇒ x < 1280

Part 2: 5120 + 2x < 3840 + 6x
5120 - 3840 < 6x - 2x
1280 < 4x ⇒ x > 320

Conclusion: More than 320 litres but less than 1280 litres of 2% boric acid solution must be added.

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