Question:y - 2x = 20y = x^2 + 5If \(\mathrm{(x, y)}\) is a solution to the system of equations above,...
GMAT Advanced Math : (Adv_Math) Questions
\(\mathrm{y - 2x = 20}\)
\(\mathrm{y = x^2 + 5}\)
If \(\mathrm{(x, y)}\) is a solution to the system of equations above, what is a possible value of x?
\(\mathrm{-5}\)
\(\mathrm{-1}\)
\(\mathrm{3}\)
\(\mathrm{5}\)
1. TRANSLATE the problem information
- Given system:
- Equation 1: \(\mathrm{y - 2x = 20}\)
- Equation 2: \(\mathrm{y = x^2 + 5}\)
- We need to find a possible value of x
2. INFER the solution strategy
- Since Equation 2 already gives us y in terms of x, substitution is the most efficient approach
- We'll substitute the expression for y from Equation 2 into Equation 1
3. SIMPLIFY by substitution and algebraic manipulation
- Substitute \(\mathrm{y = x^2 + 5}\) into \(\mathrm{y - 2x = 20}\):
\(\mathrm{(x^2 + 5) - 2x = 20}\) - Rearrange to standard quadratic form:
\(\mathrm{x^2 + 5 - 2x = 20}\)
\(\mathrm{x^2 - 2x - 15 = 0}\)
4. SIMPLIFY by factoring the quadratic
- We need two numbers that multiply to -15 and add to -2
- These numbers are -5 and 3: \(\mathrm{(-5) \times 3 = -15}\) and \(\mathrm{(-5) + 3 = -2}\)
- Factor: \(\mathrm{(x - 5)(x + 3) = 0}\)
- Using zero product property: \(\mathrm{x = 5}\) or \(\mathrm{x = -3}\)
5. APPLY CONSTRAINTS to select the final answer
- Both \(\mathrm{x = 5}\) and \(\mathrm{x = -3}\) are mathematically valid solutions
- However, checking the answer choices: A. -5, B. -1, C. 3, D. 5
- Only \(\mathrm{x = 5}\) appears among the options
Answer: D. 5
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY skills: Making sign errors during algebraic manipulation, especially when rearranging \(\mathrm{x^2 + 5 - 2x = 20}\) to standard form. Students might incorrectly get \(\mathrm{x^2 - 2x + 15 = 0}\) instead of \(\mathrm{x^2 - 2x - 15 = 0}\).
This leads to attempting to factor an unfactorable expression or getting completely different solutions, causing confusion and random guessing.
Second Most Common Error:
Poor factoring execution: Finding incorrect factor pairs for -15. Students might use factors like (-1, 15) or (1, -15) which don't add to -2, leading to wrong factorizations.
This may lead them to select Choice A (-5) if they confuse the factors with the solutions, or causes them to get stuck and guess.
The Bottom Line:
This problem tests whether students can execute multiple algebraic steps accurately while keeping track of signs and coefficients. The substitution strategy is straightforward, but the algebraic manipulation requires careful attention to detail.
\(\mathrm{-5}\)
\(\mathrm{-1}\)
\(\mathrm{3}\)
\(\mathrm{5}\)