1. **State the problem:** Find a number such that one-fifth of the sum of the number and -3 is twice the number. 2. **Set up the equation:** Let the number be $x$. The sum of the number and -3 is $x + (-3) = x - 3$. 3. One-fifth of this sum is \frac{x - 3}{5}. 4. According to the problem, this equals twice the number, which is $2x$. 5. So, the equation is: $$\frac{x - 3}{5} = 2x$$ 6. **Solve the equation:** Multiply both sides by 5 to eliminate the denominator: $$\cancel{5} \times \frac{x - 3}{\cancel{5}} = 2x \times 5$$ $$x - 3 = 10x$$ 7. Subtract $x$ from both sides: $$x - 3 - x = 10x - x$$ $$-3 = 9x$$ 8. Divide both sides by 9: $$\frac{-3}{9} = \frac{9x}{9}$$ $$-\frac{1}{3} = x$$ 9. **Final answer:** The number is $x = -\frac{1}{3}$.