Q.191. Simultaneous equations

Question 191.
If x<0 ,y <0 , x + y + (x/y) = 1/2 and (x + y) * (x/y) = - 1/2, 
then what are the values of x and y?

Answer 191.
Let x + y = m and x/y = n
=> m + n = 1/2 and mn = - 1/2
=> m - n
= ± √[(m + n)^2 - 4mn]
= ± √(1/4 + 2)
= ± 3/2

Case 1: m - n = 3/2

=> (m + n) + (m - n) = 1/2 + 3/2 = 2
=> 2m = 2
=> m = 1
Plugging m = 1 in m + n = 1/2
=> n = 1/2 - 1 = - 1/2

=> x + y = 1 and x/y = - 1/2
Plugging x = (-1/2) y in x + y = 1
=> (-1/2) y + y = 1
=> y = 2 and x = 1 - y = 1 - 2 = - 1

As x < 0 and y < 0, y = 2 is not acceptable. So this solution is discarded.

Case 2: m - n = - 3/2
 => (m + n) + (m - n) = (1/2) - (3/2) = - 1
=> m = - 1/2 and n = m + 3/2 = - 1/2 + 3/2 = 1
=> x + y = - 1/2 and x/y = 1
Plugging x = y in x + y = - 1/2,
2x = - 1/2 => x = - 1/4 and y = - 1/4

Answer: x = y = - 1/4.

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