Problem Statement
Solve for $x$:
$\frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}, \quad [a \neq 0, b \neq 0, x \neq 0, x \neq -(a+b)]$
Verified Solution & Marking Scheme
Transpose 1/x to LHS
$\frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b}$
Taking common denominator on both sides:
$\frac{x - (a + b + x)}{x(a + b + x)} = \frac{b + a}{ab}$
$\frac{-(a + b)}{x(a + b + x)} = \frac{a + b}{ab}$
Divide by (a + b) (Since a + b ≠ 0)
$\frac{-1}{x(a + b + x)} = \frac{1}{ab}$
Cross-multiplying:
$x(a + b + x) = -ab$
$x^2 + (a + b)x + ab = 0$
Factor the Quadratic Equation
$x^2 + ax + bx + ab = 0$
$x(x + a) + b(x + a) = 0$
$(x + a)(x + b) = 0$
$x = -a \quad \text{or} \quad x = -b$.