The given compound inequality is:

$-2 < 1 - 5x \leq 3$

Step 1: Solve the first inequality $-2 < 1 - 5x$

1. Subtract 1 from both sides: $-2 - 1 < -5x \implies -3 < -5x$

2. Divide both sides by $-5$ and reverse the inequality sign (since we are dividing by a negative number): $\frac{-3}{-5} > x \implies \frac{3}{5} > x \implies x < \frac{3}{5}$

Step 2: Solve the second inequality $1 - 5x \leq 3$

1. Subtract 1 from both sides: $1 - 1 - 5x \leq 3 - 1 \implies -5x \leq 2$

2. Divide both sides by $-5$ and reverse the inequality sign: $x \geq \frac{2}{-5} \implies x \geq -\frac{2}{5}$

Step 3: Combine the two results

The solution for $x$ is the intersection of the two inequalities: $-\frac{2}{5} \leq x < \frac{3}{5}$

Thus, the final solution is: $-\frac{2}{5} \leq x < \frac{3}{5}$

Would you like any further details or clarifications?

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Related Questions:

1. How do you graph compound inequalities on a number line?
2. What is the importance of reversing the inequality when dividing by a negative number?
3. How would the solution change if we had strict inequalities on both sides?
4. How can we write the solution in interval notation?
5. How do compound inequalities differ from simple inequalities?

Tip: Always remember to flip the inequality sign when multiplying or dividing by a negative number!