1.

Solve the system of equations in R.|4 – x| + 1 < 3

Answer»

Given as |4 – x| + 1 < 3

Now, let us subtract 1 from both the sides, we get

|4 – x| + 1 – 1 < 3 – 1

|4 – x| < 2

Suppose ‘r’ be a positive real number and ‘a’ be a fixed real number. Now,

|a – x| < r ⟺ a – r < x < a + r

Here, a = 4 and r = 2

4 – 2 < x < 4 + 2

2 < x < 6

∴ x ∈ (2, 6)


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