To solve the equation $|3x + 5| + |2x - 3| = 25$, we need to break it down by considering the cases where the expressions inside the absolute value signs change.

Step 1: Identify the points where the absolute values change

The absolute value expressions will change depending on when the terms inside them are zero:

• $3x + 5 = 0 \Rightarrow x = -\frac{5}{3}$
• $2x - 3 = 0 \Rightarrow x = \frac{3}{2}$

This gives us three intervals to consider:

1. $x < -\frac{5}{3}$
2. $-\frac{5}{3} \leq x \leq \frac{3}{2}$
3. $x > \frac{3}{2}$

Step 2: Solve the equation in each interval

Case 1: $x < -\frac{5}{3}$

In this interval, both expressions inside the absolute values are negative: $|3x + 5| = -(3x + 5) = -3x - 5$ $|2x - 3| = -(2x - 3) = -2x + 3$ The equation becomes: $-(3x + 5) + -(2x - 3) = 25$ $-3x - 5 - 2x + 3 = 25$ $-5x - 2 = 25$ $-5x = 27$ $x = -\frac{27}{5}$ Since $x = -\frac{27}{5}$ is less than $-\frac{5}{3}$, it satisfies the condition for this case.

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

In this interval, $3x + 5 \geq 0$ and $2x - 3 \leq 0$, so: $|3x + 5| = 3x + 5$ $|2x - 3| = -(2x - 3) = -2x + 3$ The equation becomes: $(3x + 5) + (-2x + 3) = 25$ $3x - 2x + 5 + 3 = 25$ $x + 8 = 25$ $x = 17$ But $x = 17$ is outside the interval $-\frac{5}{3} \leq x \leq \frac{3}{2}$, so this is not a valid solution.

Case 3: $x > \frac{3}{2}$

In this interval, both expressions inside the absolute values are positive: $|3x + 5| = 3x + 5$ $|2x - 3| = 2x - 3$ The equation becomes: $(3x + 5) + (2x - 3) = 25$ $3x + 2x + 5 - 3 = 25$ $5x + 2 = 25$ $5x = 23$ $x = \frac{23}{5}$ Since $\frac{23}{5} > \frac{3}{2}$, this solution is valid.

Final Solutions

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

Would you like to see more details or have any questions?

Here are 5 related questions:

1. How do you handle absolute value inequalities?
2. What happens if the equation contains three absolute values?
3. Can the method be applied to other nonlinear functions?
4. How do we graph absolute value functions with multiple breakpoints?
5. What real-world situations can be modeled using absolute value equations?

Tip: Always check if your solutions satisfy the original equation, especially when working with absolute values.