5x + 3y = 38x + 3y = 10In the solution \((\mathrm{x}, \mathrm{y})\) to the system of equations above, what...
GMAT Algebra : (Alg) Questions
\(5\mathrm{x} + 3\mathrm{y} = 38\)
\(\mathrm{x} + 3\mathrm{y} = 10\)
In the solution \((\mathrm{x}, \mathrm{y})\) to the system of equations above, what is the value of \(\mathrm{x}\)?
1. TRANSLATE the problem information
- Given information:
- First equation: \(\mathrm{5x + 3y = 38}\)
- Second equation: \(\mathrm{x + 3y = 10}\)
- Need to find: the value of x
2. INFER the best solution approach
- Notice both equations have the same y-coefficient (\(\mathrm{3y}\))
- This makes elimination the most efficient method
- Strategy: Subtract one equation from the other to eliminate y
3. SIMPLIFY using elimination
- Subtract equation (2) from equation (1):
\(\mathrm{(5x + 3y) - (x + 3y) = 38 - 10}\) - Distribute the subtraction:
\(\mathrm{5x + 3y - x - 3y = 28}\) - Combine like terms:
\(\mathrm{4x = 28}\) - Divide both sides by 4:
\(\mathrm{x = 7}\)
4. Verify the solution (optional but recommended)
- Substitute \(\mathrm{x = 7}\) into equation (2): \(\mathrm{7 + 3y = 10}\) → \(\mathrm{y = 1}\)
- Check in equation (1): \(\mathrm{5(7) + 3(1) = 35 + 3 = 38}\) ✓
Answer: 7
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY execution: Students make arithmetic errors during the elimination process, particularly when distributing the negative sign or combining like terms.
For example, they might write: \(\mathrm{(5x + 3y) - (x + 3y) = 5x + 3y - x + 3y = 8x + 6y = 28}\), forgetting to distribute the negative to both x and \(\mathrm{3y}\) terms. This leads to an incorrect equation that cannot be solved for x alone.
This leads to confusion and incorrect answers.
Second Most Common Error:
Poor INFER reasoning: Students choose substitution over elimination, leading to unnecessary algebraic complexity.
They might solve \(\mathrm{x + 3y = 10}\) for x to get \(\mathrm{x = 10 - 3y}\), then substitute into the first equation: \(\mathrm{5(10 - 3y) + 3y = 38}\). While this works, it creates more opportunities for algebraic errors and is less efficient than the direct elimination approach.
The Bottom Line:
This problem rewards students who recognize patterns (identical coefficients) and choose the most efficient solution method, while penalizing those who rush through algebraic manipulations without careful attention to signs and arithmetic.