1. **State the problem:** Solve the equation $$\frac{x - 1}{6} = \frac{x + 5}{5} = -50$$ which means both fractions equal $$-50$$. 2. **Set each fraction equal to $$-50$$:** $$\frac{x - 1}{6} = -50$$ $$\frac{x + 5}{5} = -50$$ 3. **Solve the first equation:** Multiply both sides by 6: $$\cancel{6} \times \frac{x - 1}{\cancel{6}} = -50 \times 6$$ $$x - 1 = -300$$ Add 1 to both sides: $$x = -300 + 1$$ $$x = -299$$ 4. **Solve the second equation:** Multiply both sides by 5: $$\cancel{5} \times \frac{x + 5}{\cancel{5}} = -50 \times 5$$ $$x + 5 = -250$$ Subtract 5 from both sides: $$x = -250 - 5$$ $$x = -255$$ 5. **Check for consistency:** The two values for $$x$$ are different ($$-299$$ and $$-255$$), so the original equation with both fractions equal to the same value $$-50$$ cannot be true simultaneously unless the problem means something else. 6. **Interpretation:** If the problem means $$\frac{x - 1}{6} = \frac{x + 5}{5}$$ and both equal $$-50$$, then the two fractions are equal and equal to $$-50$$, but the values of $$x$$ found are different, so no $$x$$ satisfies both simultaneously. 7. **Solve the equality of fractions:** Set $$\frac{x - 1}{6} = \frac{x + 5}{5}$$ Cross multiply: $$5(x - 1) = 6(x + 5)$$ $$5x - 5 = 6x + 30$$ Subtract $$5x$$ from both sides: $$-5 = x + 30$$ Subtract 30 from both sides: $$-5 - 30 = x$$ $$x = -35$$ 8. **Check if $$x = -35$$ satisfies $$\frac{x - 1}{6} = -50$$:** $$\frac{-35 - 1}{6} = \frac{-36}{6} = -6 \neq -50$$ So $$x = -35$$ does not satisfy the fractions equal to $$-50$$. **Final conclusion:** The equation $$\frac{x - 1}{6} = \frac{x + 5}{5} = -50$$ has no solution where both fractions equal $$-50$$ simultaneously. The fractions are equal when $$x = -35$$ but then they do not equal $$-50$$. **Answer:** No solution for $$x$$ such that both fractions equal $$-50$$ simultaneously.