5<x2+y<28...(i)
From the above we can conclude that both |x| and |y| must be less then 6.
Also, |x — y| < 3. ...(ii)
We can see that if (x, y) satisfies (i) and (ii) then (- x. - y) will also definitely satisfy (i) and (ii).
Consider x > 0 and y > 0:
Equation (i) can accept 0 < x < 6.
Simple analysis of (i) and (ii) gives values for x and y:
Same number of cases i.e. 10 will be there when x < 0 and y < 0.
Consider either x or y or both equal to zero:
The absolute value of other variable can be at maximum equal to 2 (from (ii)). This will definitely not satisfy (i). So no such case is possible.
Consider x > 0, y < 0:
The only possible case is (1, - 1) which we can get from equation (ii) but this does not satisfy equation (i). So no such case is possible.
Similarly no case is possible for y > 0, x < 0.
So total cases = 10 + 10 = 20