20/x - 20/x+2= 1/3

To solve the equation \( \frac{20}{x} - \frac{20}{x+2} = \frac{1}{3} \), we can follow these steps:

1. Find a common denominator: The common denominator for the fractions on the left side is \( x(x + 2) \).

2. Rewrite each term with the common denominator:

   \[ \frac{20(x + 2) - 20x}{x(x + 2)} = \frac{1}{3} \]

   Simplifying the left side:

   \[ \frac{20x + 40 - 20x}{x(x + 2)} = \frac{40}{x(x + 2)} \]

   This gives us the equation:

   \[ \frac{40}{x(x + 2)} = \frac{1}{3} \]

3. Cross-multiply to eliminate the fractions:

   \[ 40 \cdot 3 = 1 \cdot x(x + 2) \]

   This simplifies to:

   \[ 120 = x^2 + 2x \]

4. Rearranging the equation:

   \[ x^2 + 2x - 120 = 0 \]

5. Now we can factor or use the quadratic formula to solve for \(x\). Let's try factoring first:

   We want two numbers that multiply to \(-120\) and add to \(2\). These numbers are \(12\) and \(-10\).

   Thus, we can factor the equation:

   \[ (x + 12)(x - 10) = 0 \]

6. Setting each factor to zero gives:

   \[ x + 12 = 0 \quad \Rightarrow \quad x = -12 \] \[ x - 10 = 0 \quad \Rightarrow \quad x = 10 \]

So the solutions are \( x = -12 \) and \( x = 10 \).

7. Check for extraneous solutions:

   We must ensure neither solution makes any denominator zero:

   • For \( x = -12 \):
     • \( x + 2 = -10 \) (not zero)
   • For \( x = 10 \):
     • \( x + 2 = 12 \) (not zero)

Both solutions are valid. Therefore, the final answers are:

\[ x = -12 \quad \text{and} \quad x = 10 \]