We examine four cases based on the signs of \(x\) and \(y\):
Case 1: \(x \ge 0\) and \(y \ge 0\). The equations simplify to \(2x + y = 15\) and \(x = 20\). Solving these yields \(x = 20\) and \(y = -35\), which contradicts \(y \ge 0\).
Case 2: \(x<0\) and \(y<0\). The equations simplify to \(y = 15\) and \(x = 20\). This contradicts both \(x<0\) and \(y<0\).
Case 3: \(x \ge 0\) and \(y<0\). The equations are \(2x + y = 15\) and \(x - 2y = 20\). Solving these gives \(x = 10\) and \(y = -5\). This solution satisfies the conditions for this case.
Case 4: \(x<0\) and \(y \ge 0\). The equations simplify to \(y = 15\) and \(x + 2y = 20\). Solving these yields \(x = -10\) and \(y = 15\). This solution satisfies the conditions for this case.
Comparing the valid solutions from Case 3 and Case 4, only Case 3 satisfies both original equations. Therefore, \(x - y = 10 - (-5) = 15\).
The value of \((x - y)\) is 15.