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Problems on Linear Inequation

Here we will solve various types of problems on linear inequation.

By applying the law of inequality, we can easily solve simple inequations. This can be seen in the following examples.

1. Solve 4x – 8 ≤ 12

Solution:

4x – 8 ≤ 12

⟹ 4x - 8 + 8 ≤ 12 + 8, [Adding 8 on both sides of the inequation]

⟹ 4x ≤ 20

4x4  ≤ 204, [Dividing both sides by 4]

⟹ x ≤ 5

Therefore, required solution: x ≤ 5

Note: The solution = x ≤ 5. That means, the given inequation is satisfied by 5 and any number less than 5. Here the maximum value of x is 5.


2. Solve the inequation 2(x – 4) ≥ 3x – 5

Solution:

2(x – 4) ≥ 3x – 5

⟹ 2x – 8 ≥ 3x – 5

⟹ 2x – 8 + 8 ≥ 3x – 5 + 8, [Adding 8 on both sides of the inequation]

⟹ 2x ≥ 3x + 3

⟹ 2x – 3x ≥ 3x + 3 – 3x, [Subtracting 3x from both sides of the inequation]

⟹ -x ≥ 3

⟹ x ≤ - 3, [Dividing both sides by -1]

Therefore, the required solution:  x ≤ - 3

Note: As a result of dividing both sides of  - x  ≥  3 by -1, ‘≥’ sign is converted to ‘≤’ sign. Here, find the maximum value of x.


3. Solve the inequation: - 5 ≤ 2x – 7 ≤ 1

Solution:

Here two inequations are given. They are

- 5 ≤ 2x – 7 .............................. (i)

and

2x - 7 ≤ 1 .............................. (ii)

From the inequation (i), we get

- 5 ≤ 2x -7

⟹ -5 + 7 ≤ 2x - 7 + 7, [Adding 7 on both sides of the inequation]

⟹ 2 ≤ 2x

222x2, [Dividing both sides by 2]

⟹ 1 ≤ x

⟹ x ≥ 1

Now from the equation (ii), we get

2x - 7 ≤ 1

⟹ 2x - 7 + 7 ≤ 1 + 7, [Adding 7 on both sides of the inequation]

⟹ 2x ≤ 8

2x282, [Dividing both sides by 2]

⟹ x ≤ 4

Therefore, the required solutions are x ≥ 1, x ≤ 4 i.e., 1 ≤ x ≤ 4.

Note: Here least value of x is 1, and greatest value of x is 4.

We could solve without splitting two inequations.

- 5 ≤ 2x - 7 ≤ 1

⟹ - 5 + 7 ≤ 2x - 7 + 7 ≤ 1 + 7, [Adding 7 on each term of the inequation]

⟹ 2 ≤ 2x ≤ 8

222x282, [Dividing each term by 2]

⟹ 1 ≤ x ≤ 4






10th Grade Math

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