Worked Example 1.3.1, 1. Find the set of all real numbers \( x \) which satisfy the inequalities a) \( 3 x+1>2 x \) OR \( 3 x+1 \leq 5 x-2 \). b) \( 3 x+1>2 x \) AND \( 3 x+1 \leq 5 x-2 \). c) Hence solve \( 2 x<3 x+1 \leq 5 x-2 \).

To find the set of all real numbers \( x \) that satisfy the inequalities, let's break each part down. a) For the inequality \( 3x + 1 > 2x \): Subtracting \( 2x \) from both sides gives us \( x + 1 > 0 \) or \( x > -1 \). For \( 3x + 1 \leq 5x - 2 \): Rearranging this gives \( 1 + 2 \leq 5x - 3x \), leading to \( 3 \leq 2x \) or \( x \geq \frac{3}{2} \). Thus, the solution for part (a) is \( x > -1 \) OR \( x \geq \frac{3}{2} \). This implies \( x \in (-1, \frac{3}{2}] \cup (\frac{3}{2}, \infty) \). b) For \( 3x + 1 > 2x \): We already determined \( x > -1 \). For \( 3x + 1 \leq 5x - 2 \): Again, we have \( x \geq \frac{3}{2} \). For \( x > -1 \) AND \( x \geq \frac{3}{2} \), the intersection is simply \( x \geq \frac{3}{2} \). c) To solve \( 2x < 3x + 1 \leq 5x - 2 \), we can split it into two parts. First, \( 2x < 3x + 1 \): This simplifies to \( 0 < x + 1 \), resulting in \( x > -1 \). Next, for \( 3x + 1 \leq 5x - 2 \): As determined earlier, this implies \( x \geq \frac{3}{2} \). Combining these results, we conclude that the solution to the compound inequality is \( x \geq \frac{3}{2} \). Now let’s delve deeper! Though inequalities might feel abstract, they have real-world applications! For instance, when setting budgets, one might use inequalities to determine maximum spends while ensuring costs don’t exceed income or savings. This helps make sound financial decisions using basic algebra! If you’re grappling with inequalities, a common mistake is forgetting to flip the inequality when multiplying or dividing by negative numbers. Remember, it’s like making everyone turn in the opposite direction when the music changes - always a fun twist when you’re trying to solve those tricky inequalities!