Q34
ArithmeticAlgebra, Equations & Operations
$\frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}$, $x \ne 0$, $x = ?$
- a.$a$
- b.$b$
- c.$-a, -b$
- d.$a, b$
Answer: (C) $-a, -b$
Move $\frac{1}{x}$ to the left: $\frac{1}{a + b + x} - \frac{1}{x} = \frac{a + b}{ab}$, which gives $\frac{-(a + b)}{x(a + b + x)} = \frac{a + b}{ab}$. So $x^2 + (a + b)x + ab = 0$, i.e. $(x + a)(x + b) = 0$ and $x = -a$ or $-b$. Positive $a$ or $b$ do not satisfy the equation.