The Integer Boundary Trap
A set of points \((x, y)\) in the \(xy\)-plane satisfies the following system of inequalities:
\(y > 2x + 10\)
\(y < -3x + 40\)
\(y < -3x + 40\)
If \(x\) and \(y\) are positive integers, what is the maximum possible value of \(x\)?
🚩 The Trap
Finding the intersection point \((x = 6, y = 22)\) and assuming \(x=6\) is a valid solution. But the inequalities are strict (\(<\) and \(>\)), and \(y\) must be an integer!
✅ The Practix Shortcut
- Find the Intersection: Set them equal: \(2x + 10 = -3x + 40 \Rightarrow 5x = 30 \Rightarrow x = 6\).
- Check the Condition: At \(x = 6\), both inequalities become \(y > 22\) and \(y < 22\). There is no \(y\) that satisfies both!
- Test Downward: Try \(x = 5\).
\(y > 2(5) + 10 = 20\)
\(y < -3(5) + 40=25\)
Valid integers for \(y\) are \(\{21, 22, 23, 24\}\). Since valid integers exist, \(x = 5\) is our maximum.
Answer: 5.