|2x− 3| ≤ 11, |x− 2| ≥ 3

Question

Solve the following system of in equalities for all x ∈ R.

|2x− 3| ≤ 11, |x− 2| ≥ 3

Answer 1

The given inequalities are

|2x − 3| ≤ 11 ...(i)

And |x − 2| ≥ 3 ..(ii)

Now, |2x − 3| ≤ 11

We know that,

|x − a| ≥ r ⇔ a − r ≤ x ≤ a + r

∴ |2x − 3| ≤ 11 ⇔ 3 − 11 ≤ 2x ≤ 3 + 11

⇔ −8 ≤ 2x ≤ 14

⇔ −4 ≤ x ≤ 7

⇔ x ∈ [−4,7]

And |x − 2| ≥ 3

We know that,

|x − a| ≥ r ⇔ x ≤ a + r

or x ≥ a + r

|x − 2| ≥ 3 ⇔ x ≤ 2 + 3

or x ≥ 2 + 3

⇔ x ≤ −1 or x ≥ 5

⇔ x ∈ (−∞, −1] or x ∈ [5, ∞)

⇔ x ∈ (−∞, −1] ∪ x ∈ [5, ∞)

The solution set of inequalities (i) and (ii) are represented graphically in fig (i) and fig (ii) respectively.

Hence, the solution set of the given system of inequations is [-4,-1] ∪ [5,7]