Linear Inequalities Questions

Linear inequalities questions and solutions help the students to solve various problems on inequalities in mathematics. In this article, you will get solved questions on linear inequalities, and additional questions for practice. You will find detailed explanations to the solutions of linear inequalities in one variable and two variables.

What are Linear inequalities?

In maths, linear inequalities are the equations in which both sides of the equation are separated by the inequality symbol, such as <, >, ≤ and ≥. That means LHS and RHS are not equal. For example, 2x – 4 < 12, x – y > 7, 3x – 4 ≤ 0, and so on.

Also, check: Linear inequalities

Linear Inequalities Questions and Answers

1. Solve for x:

3(x – 1) ≤ 2 (x – 3)

Solution:

Given,

3(x – 1) ≤ 2 (x – 3)

The above inequality can be written as,

3x – 3 ≤ 2x – 6

Adding 3 to both the sides, we get;

3x – 3+ 3 ≤ 2x – 6+ 3

3x ≤ 2x – 3

Subtracting 2x from both the sides,

3x – 2x ≤ 2x – 3 – 2x

x ≤ -3

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

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

2. Solve the inequality: (x – 2)/(x + 5) > 2

Solution:

(x – 2)/(x + 5) > 2

Subtracting 2 from both sides, we get;

(x – 2)/(x + 5) – 2 > 0

[(x – 2) – 2(x + 5)]/ (x + 5) > 0

(x – 2 – 2x – 10)/(x + 5) > 0

-(x + 12)/(x + 5) > 0

Multiplying -1 on both sides, we get;

(x + 12)/(x + 5) < 0

⇒ x + 12 < 0 and x + 5 > 0 (or) x + 12 > 0 and x + 5 < 0

⇒ x < -12 and x > -5 (or) x > -12 and x < -5

⇒ -12 < x < -5

Therefore, x ∈ (-12, -5).

3. Solve for x from the following:

1/(|x| – 3) ≤ ½

Solution:

Given,

1/(|x| – 3) ≤ ½

Subtracting ½ from both sides, we get;

[1/ (|x| – 3)] – (½) ≤ 0

(2 – |x| + 3)/ 2(|x| – 3) ≤ 0

(5 – |x|)/(|x| – 3) ≤ 0

⇒ 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, ∞)

4. Solve for x: |x + 1| + |x| > 3

Solution:

Given,

|x + 1| + |x| > 3

In LHS, we have two terms with modulus.

So, let’s equate the given expression within the modulus to 0.

Then, we get critical points x = -1, 0.

Thus, we can write the real line intervals for these critical points as: (-∞, -1), [-1, 0), [0, ∞)

Case 1:

When – ∞ < x < – 1

|x + 1| + |x| > 3

⇒ – x – 1 – x > 3

⇒ x < – 2.

Case 2:

When – 1 ≤ x < 0,

|x + 1| + |x| > 3

⇒ x + 1 – x > 3

⇒ 1 > 3 (not possible)

Case 3:

When 0 ≤ x < ∞,

|x + 1| + |x| > 3

⇒ x + 1 + x > 3

⇒ x > 1.

From the three cases written above, we get x ∈ (– ∞ , – 2) ∪ (1, ∞).

5. The longest side of a triangle is 3 times the shortest side, and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Solution:

Let x cm be the length of the shortest side of the triangle.

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

Length of the third side = (3x – 2) cm

The least perimeter of the triangle = 61 cm (given)

Thus, x + 3x + (3x – 2) cm ≥ 61 cm

= 7x – 2 ≥ 61

= 7x ≥ 63

Dividing by 7 on both sides, we get;

= 7x/7 ≥ 63/7

= x ≥ 9

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

6. Solve (3x – 4)/2 ≥ (x + 1)/4 – 1. Show the graph of the solutions on the number line.

Solution:

Given,

(3x – 4)/2 ≥ (x + 1)/4 – 1

(3x – 4)/2 ≥ [(x + 1) – 4]/4

(3x – 4)/2 ≥ (x – 3)/4

Multiplying by 4 on both sides,

(3x – 4) ≥ (x – 3)/2

2(3x – 4) ≥ (x – 3)

6x – 8 ≥ x – 3

Adding 8 on both sides,

6x ≥ x + 5

Subtracting x from both sides, we get;

5x ≥ 5

i.e., x ≥ 1

Therefore, the graphical representation of this solution (on the number line) is given as:

7. Solve the following system of linear inequalities graphically.

x + y ≥ 5

x – y ≤ 3

Solution:

Given

x + y ≥ 5….(i)

x – y ≤ 3….(ii)

From (i), x + y ≥ 5

Let x + y = 5

y = 5 – x

From (ii), x – y ≤ 3

Let x – y = 3

y = x – 3

Now, let’s draw the graph for the above two linear equations, i.e., x + y = 5 and x – y = 3.

Now, the solution of inequality (i) can be represented by shading the region towards the right of the line x + y = 5, including the points on the line.

On the same set of axes, we draw the graph of the equation x – y = 3. The inequality (ii) represents the shaded region towards the left side of the line x – y = 3, including the points on the line.

The double shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

8. A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution must be added to it so that the acid content in the resulting mixture will be more than 15% but less than 18%?

Solution:

Let x litres of 30% acid solution be required to be added.

Total mixture = (x + 600) litres

Thus, 30% x + 12% of 600 > 15% of (x + 600)

and

30% x + 12% of 600 < 18% of (x + 600)

⇒ (30x/100) + (12/100) × (600) > (15/100) (x + 600)

And

(30x/100) + (12/100) × (600) < (18/100) (x + 600)

⇒ 30x + 7200 > 15x + 9000 and 30x + 7200 < 18x + 10800

⇒ 15x > 1800 and 12x < 3600

⇒ x > 120 and x < 300,

i.e., 120 < x < 300

Hence, the number of litres of the 30% acid solution will have to be more than 120 litres but less than 300 litres.

9. Solve the following system of inequalities graphically.

x + y ≥ 4, 2x – y < 0

Solution:

Given,

x + y ≥ 4….(i)

2x – y < 0….(ii)

From (i), x + y ≥ 4

Let x + y = 4

y = 4 – x

When x = 0, y = 4

When x = 4, y = 0

From (ii), 2x – y < 0

Let 2x – y = 0

y = 2x

When x = 0, y = 0

When x = 1, y = 2

From the above computation we can draw the graph of linear equations x + y = 4 and 2x – y = 0.

Now, shade the region for x + y ≥ 4, on the right of the line of x + y = 4.

For (ii), the shaded region will be on the left side of the line 2x – y = 0.

These can be represented graphically as:

Hence, the combined shaded region will be the solution region for the given system of inequalities.

10. Solve the following inequalities and represent the solution set on the number line.

5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

Solution:

Given,

5(2x – 7) – 3(2x + 3) ≤ 0….(i)

2x + 19 ≤ 6x + 47….(ii)

From (i),

5(2x – 7) – 3(2x + 3) ≤ 0

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ 4x ≤ 44

⇒ x ≤ 11 ……(iii)

From (ii),

2x + 19 ≤ 6x +47

⇒ 6x – 2x ≥ 19 – 47

⇒ 4x ≥ -28

⇒ x ≥ -7 ……….(iv)

From equations (iii) and (iv), the solution of the given inequalities is (-7, 11).

This can be shown on the number line as:

Practice Questions on Linear Inequalities

1. The cost and revenue functions of a product are given by C(x) = 20 x + 4000 and R(x) = 60x + 2000, respectively, where x is the number of items produced and sold. How many items must be sold to realise some profit?
2. Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
3. Solve the following inequalities and represent the solution set graphically on the number line.

5x + 1 > – 24, 5x – 1 < 24

4. Show that the following system of linear inequalities has no solution:
   
   x + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
5. Solve for x: -5 ≤ (5 – 3x)/2 ≤ 8