Sol. The given inequality is

– |y| + x – ≥ 1 …(1)

The domain of admissible values of inequality (1) is defined by the condition x 2 + y 2 –1 ≥ 0, we rewrite

inequality (1) as

x – |y| ≥ 1 + . It follows that x – |y| ≥ 0.

When this condition is fulfilled, both sides of the inequality obtained are non negative on its domain of admissible values. Therefore, when we square both sides of inequality (1), we get

Since x ≥ |y| ≥ 0, the L.H.S. of the first inequality is non positive and the R.H.S. is non negative. Therefore, the system is satisfied only when both sides are zero.

i.e.

The first equation means that either x = 0 or y = 0 If x = 0, then we find from the third inequality that y = 0. But the pair (0, 0) does not belong to the domain of admissible values of the original inequality.

If y = 0, then we get from the second equation (noting that x ≥ 0) that x = 1. The pair (1, 0) satisfies inequality (1) and is its only solution