To solve the equation $|3x + 5| + |2x - 3| = 25$, we need to consider different cases based on the values of $x$ that affect the signs inside the absolute values.

Step 1: Break down the cases

1. Case 1: $x \geq \frac{3}{2}$

   • In this case, both expressions inside the absolute values are non-negative. Thus, we can drop the absolute value symbols: $(3x + 5) + (2x - 3) = 25$ Simplifying: $5x + 2 = 25$ Solving for $x$: $5x = 23 \quad \Rightarrow \quad x = \frac{23}{5}$ Since $\frac{23}{5} \approx 4.6 \geq \frac{3}{2}$, this solution is valid.

2. Case 2: $-\frac{5}{3} \leq x < \frac{3}{2}$

   • In this case, $3x + 5 \geq 0$ but $2x - 3 < 0$, so we change the sign of the second term: $(3x + 5) - (2x - 3) = 25$ Simplifying: $x + 8 = 25$ Solving for $x$: $x = 17$ But $x = 17$ does not lie in the range $-\frac{5}{3} \leq x < \frac{3}{2}$, so there is no solution in this case.

3. Case 3: $x < -\frac{5}{3}$

   • In this case, both expressions inside the absolute values are negative, so we change the signs of both terms: $-(3x + 5) - (2x - 3) = 25$ Simplifying: $-5x - 2 = 25$ Solving for $x$: $-5x = 27 \quad \Rightarrow \quad x = -\frac{27}{5}$ Since $-\frac{27}{5} \approx -5.4 < -\frac{5}{3}$, this solution is valid.

Step 2: Final solutions

The valid solutions for $x$ are: $x = \frac{23}{5} \quad \text{and} \quad x = -\frac{27}{5}$

Summary:

The solutions to the equation are: $x = \frac{23}{5} \quad \text{or} \quad x = -\frac{27}{5}$

Would you like more details on any of the steps?

Follow-up questions:

1. How do absolute values impact the number of cases in an equation?
2. Can you explain why case 2 has no valid solution?
3. How can we verify solutions when solving absolute value equations?
4. What are some applications of absolute value in real-life scenarios?
5. Can absolute value equations have infinitely many solutions?

Tip: When solving absolute value equations, always check that each solution fits within the constraints of the case it was derived from.