Question

Solve the following equation: \[ \frac{1}{x} - \frac{1}{x-2} = 3, \quad x \neq 0, 2. \]

Hint:

To solve equations involving fractions, find a common denominator, simplify, and then solve the resulting equation.

Answer 1

The given equation is: \[ \frac{1}{x} - \frac{1}{x - 2} = 3. \]
Step 1: Find the LHS with a common denominator. The common denominator is \( x(x - 2) \), so we rewrite the left-hand side: \[ \frac{1}{x} - \frac{1}{x - 2} = \frac{(x - 2) - x}{x(x - 2)} = \frac{-2}{x(x - 2)}. \] Thus, the equation becomes: \[ \frac{-2}{x(x - 2)} = 3. \]
Step 2: Multiply both sides by \( x(x - 2) \) to eliminate the denominator: \[ -2 = 3x(x - 2). \]
Step 3: Expand the right-hand side: \[ -2 = 3x^2 - 6x. \]
Step 4: Move all terms to one side of the equation: \[ 3x^2 - 6x + 2 = 0. \]
Step 5: Solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \( a = 3 \), \( b = -6 \), and \( c = 2 \). Substitute the values: \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm \sqrt{12}}{6}. \] Simplify \( \sqrt{12} \): \[ x = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}. \] Thus, the two possible values of \( x \) are: \[ x = 1 + \frac{\sqrt{3}}{3} \quad \text{or} \quad x = 1 - \frac{\sqrt{3}}{3}. \]
Conclusion: The solutions are \( x = 1 + \frac{\sqrt{3}}{3} \) and \( x = 1 - \frac{\sqrt{3}}{3} \).