Solve the system of equations in R.2x + 5 ≤ 0, x – 3 ≤ 0

1.	Solve the system of equations in R.2x + 5 ≤ 0, x – 3 ≤ 0
Answer» Given as 2x + 5 ≤ 0 and x – 3 ≤ 0 Now, let us consider the first inequality. 2x + 5 ≤ 0 2x + 5 – 5 ≤ 0 – 5 2x ≤ –5 Dividing both the sides by 2 we get, 2x/2 ≤ –5/2 x ≤ – 5/2 ∴ x ∈ (–∞, -5/2]… (1) Then, let us consider the second inequality. x – 3 ≤ 0 x – 3 + 3 ≤ 0 + 3 x ≤ 3 ∴ x ∈ (–∞, 3]… (2) From the equation (1) and (2), we get x ∈ (–∞, -5/2) ∩ (–∞, 3) x ∈ (–∞, -5/2) Hence, the solution of the given system of inequations is (–∞, -5/2).

Answer»

Given as

2x + 5 ≤ 0 and x – 3 ≤ 0

Now, let us consider the first inequality.

2x + 5 ≤ 0

2x + 5 – 5 ≤ 0 – 5

2x ≤ –5

Dividing both the sides by 2 we get,

2x/2 ≤ –5/2

x ≤ – 5/2

∴ x ∈ (–∞, -5/2]… (1)

Then, let us consider the second inequality.

x – 3 ≤ 0

x – 3 + 3 ≤ 0 + 3

x ≤ 3

∴ x ∈ (–∞, 3]… (2)

From the equation (1) and (2), we get

x ∈ (–∞, -5/2) ∩ (–∞, 3)

x ∈ (–∞, -5/2)

Hence, the solution of the given system of inequations is (–∞, -5/2).

Discussion