Question
If x + y + xy = 118, such that x < y and both 'x' and
'y' are positive integers, then find minimum value of (x + y) .Solution
x + y + xy = 118
Add 1 on both sides of the equation, we get,
1 + x + y + xy = 118 + 1
Or, (1 + x) + y X (1 + x) = 119
So, (1 + x) X (1 + y) = (7 X 17) or (1 X 119)
When, (1 + x) X (1 + y) = 7 X 17
We can say that, 1 + x = 17 or 1 + y = 17
So, 1 + x = 17 or, 1 + y = 17
So, x = 16 or y = 16
When, x = 16, y = 6 and when y = 16, x = 6
So, for both cases, x + y = 16 + 6 = 22
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