theAlbatross wrote:
\(x^{2}-y^{2}< 8\)
x+y>3
If x and y are integers in the above inequalities and 0 < y < x, what is the greatest possible value of x? _____
Since \(x\) and \(y\) are positive integers, \(x^2\) and \(y^2\) are PERFECT SQUARES, implying that \(x^2-y^2\)is equal to the difference of two perfect squares.
Make list of perfect squares:
1, 4,
9, 16, 25...
Since \(x^2 - y^2 < 8\), the value of x will be maximized if \(x^2\) and \(y^2\) are the perfect squares in blue, which have a difference of 7.
Thus:
\(x^2 = 16\), implying that the greatest possible value of \(x=4\).
The inequality in red is not necessary to solve the problem.