In the xy-plane, point P has coordinates \((-6, 2)\). For any point \((\mathrm{x}, \mathrm{y})\), define u = x + 8...
GMAT Algebra : (Alg) Questions
In the xy-plane, point P has coordinates \((-6, 2)\). For any point \((\mathrm{x}, \mathrm{y})\), define \(\mathrm{u} = \mathrm{x} + 8\) and \(\mathrm{v} = \mathrm{y} + 1\). Which of the following systems of inequalities is satisfied by the values \((\mathrm{u}, \mathrm{v})\) corresponding to point P?
\(\mathrm{u} \gt 0\) and \(\mathrm{v} \gt 0\)
\(\mathrm{u} \gt 0\) and \(\mathrm{v} \lt 0\)
\(\mathrm{u} \lt 0\) and \(\mathrm{v} \gt 0\)
\(\mathrm{u} \lt 0\) and \(\mathrm{v} \lt 0\)
\(\mathrm{u} = 0\) and \(\mathrm{v} \gt 0\)
1. TRANSLATE the problem information
- Given information:
- Point P has coordinates \((-6, 2)\)
- For any point \((\mathrm{x},\mathrm{y})\): \(\mathrm{u} = \mathrm{x} + 8\) and \(\mathrm{v} = \mathrm{y} + 1\)
- What this tells us: We need to substitute P's coordinates into the u and v definitions
2. SIMPLIFY by calculating u and v for point P
- Substitute \(\mathrm{x} = -6\) and \(\mathrm{y} = 2\):
- \(\mathrm{u} = \mathrm{x} + 8 = -6 + 8 = 2\)
- \(\mathrm{v} = \mathrm{y} + 1 = 2 + 1 = 3\)
3. SIMPLIFY by evaluating the inequalities
- Check each inequality:
- Is \(\mathrm{u} \gt 0\)? Yes, because \(\mathrm{u} = 2 \gt 0\)
- Is \(\mathrm{v} \gt 0\)? Yes, because \(\mathrm{v} = 3 \gt 0\)
- Therefore: \(\mathrm{u} \gt 0\) and \(\mathrm{v} \gt 0\)
Answer: A
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students may not understand that the definitions \(\mathrm{u} = \mathrm{x} + 8\) and \(\mathrm{v} = \mathrm{y} + 1\) mean they should substitute P's specific coordinates into these expressions. Instead, they might try to work with the general forms or get confused about what x and y represent.
This leads to confusion and guessing rather than systematic calculation.
Second Most Common Error:
Poor SIMPLIFY execution: Students make arithmetic errors, particularly with negative numbers. For example, they might calculate \(\mathrm{u} = -6 + 8\) as -14 (treating it like \(-6 - 8\)) or make sign errors that result in \(\mathrm{u} \lt 0\).
This may lead them to select Choice C (\(\mathrm{u} \lt 0\) and \(\mathrm{v} \gt 0\)) or Choice D (\(\mathrm{u} \lt 0\) and \(\mathrm{v} \lt 0\)) depending on which calculations they get wrong.
The Bottom Line:
This problem tests whether students can correctly apply function-like definitions to specific coordinate points. The arithmetic is straightforward, but students need to clearly understand what the transformation definitions mean and execute the substitutions accurately.
\(\mathrm{u} \gt 0\) and \(\mathrm{v} \gt 0\)
\(\mathrm{u} \gt 0\) and \(\mathrm{v} \lt 0\)
\(\mathrm{u} \lt 0\) and \(\mathrm{v} \gt 0\)
\(\mathrm{u} \lt 0\) and \(\mathrm{v} \lt 0\)
\(\mathrm{u} = 0\) and \(\mathrm{v} \gt 0\)