Algebra
System of equations via substitution
MJAT_TS1_P2
Grade 12
Question:
If $\dfrac{5}{x}\left(1 + \dfrac{1}{x^2+y^2}\right) = 12$ and $\dfrac{5y}{4}\left(1 - \dfrac{1}{x^2+y^2}\right) = \dfrac{4}{5}$, with $x, y \in \mathbb{R}$, then correct option(s) is/are
A) $x + y = 3$
B) $x + y = 15$
C) $x + y = 35$
D) $x^2 + y^2 = 5$
Step-by-Step Solution
Key Concept: Let $u = x^2 + y^2$. From the equations: $x = \frac{12}{15}\left(1 + \frac{1}{u}\right)^{-1}\cdot\frac{5}{12}$... Use substitution: $\frac{5}{x} = 12 - \frac{12}{u}$ and $\frac{5y}{4} = \frac{4}{5}\cdot\frac{u}{u-1}$. Then form $x^2+y^2 = u$ and solve.
Put $u = x^2+y^2$. From equations: $x = \frac{5(u+1)}{12u}$ ... After algebraic manipulation leading to $25(u^2-2+\frac{1}{u^2}) = 32(\frac{5}{4}+\frac{1}{4}-8)$... solutions give $u = 4$ or $u = 5$. Corresponding $(x,y)$: one gives $x+y=3$, $x^2+y^2=5$; another gives $x+y=15$ or $x+y=35$. All four options A, B, C, D are satisfied across the solutions.
Correct Answer: ABCD