https://file.aiquickdraw.com/mathbot/file/1726296697941ehr97qlk.jpg

The equation shown in the image is:

\[
|3x + 5| + |2x - 3| = 25
\]

We need to solve for \(x\) and find the sum of the highest and lowest values.

### Step 1: Analyze Breakpoints of the Absolute Values

We have two absolute values:
1. \( |3x + 5| \)
2. \( |2x - 3| \)

These absolute values change based on the points where their respective expressions are zero. Let's find those points:

\[
3x + 5 = 0 \implies x = -\frac{5}{3}
\]

\[
2x - 3 = 0 \implies x = \frac{3}{2}
\]

Thus, the expression changes behavior at \(x = -\frac{5}{3}\) and \(x = \frac{3}{2}\). These are the critical points where we will need to check different cases based on the intervals they create.

### Step 2: Consider the Intervals

The critical points divide the real line into three intervals:
1. \( x < -\frac{5}{3} \)
2. \( -\frac{5}{3} \leq x \leq \frac{3}{2} \)
3. \( x > \frac{3}{2} \)

#### Case 1: \(x < -\frac{5}{3}\)

In this interval, both \( 3x + 5 < 0 \) and \( 2x - 3 < 0 \), so:

\[
|3x + 5| = -(3x + 5) = -3x - 5
\]
\[
|2x - 3| = -(2x - 3) = -2x + 3
\]

Substitute into the equation:

\[
-(3x + 5) + -(2x - 3) = 25
\]
\[
-3x - 5 - 2x + 3 = 25
\]
\[
-5x - 2 = 25
\]
\[
-5x = 27
\]
\[
x = -\frac{27}{5}
\]

Since \(x = -\frac{27}{5} \approx -5.4\) lies in the interval \(x < -\frac{5}{3}\), this is a valid solution.

#### Case 2: \( -\frac{5}{3} \leq x \leq \frac{3}{2} \)

In this interval, \( 3x + 5 \geq 0 \) and \( 2x - 3 < 0 \), so:

\[
|3x + 5| = 3x + 5
\]
\[
|2x - 3| = -(2x - 3) = -2x + 3
\]

Substitute into the equation:

\[
(3x + 5) + -(2x - 3) = 25
\]
\[
3x + 5 - 2x + 3 = 25
\]
\[
x + 8 = 25
\]
\[
x = 17
\]

Since \(x = 17\) does not lie within the interval \( -\frac{5}{3} \leq x \leq \frac{3}{2} \), this is not a valid solution.

#### Case 3: \(x > \frac{3}{2}\)

In this interval, both \( 3x + 5 > 0 \) and \( 2x - 3 > 0 \), so:

\[
|3x + 5| = 3x + 5
\]
\[
|2x - 3| = 2x - 3
\]

Substitute into the equation:

\[
(3x + 5) + (2x - 3) = 25
\]
\[
3x + 5 + 2x - 3 = 25
\]
\[
5x + 2 = 25
\]
\[
5x = 23
\]
\[
x = \frac{23}{5}
\]

Since \(x = \frac{23}{5} \approx 4.6\) lies in the interval \(x > \frac{3}{2}\), this is a valid solution.

### Step 3: Find the Sum of Highest and Lowest Values

The valid solutions are \(x = -\frac{27}{5}\) and \(x = \frac{23}{5}\).

The sum of these two values is:

\[
-\frac{27}{5} + \frac{23}{5} = \frac{-27 + 23}{5} = \frac{-4}{5}
\]

Thus, the sum of the highest and lowest values is \( \frac{-4}{5} \).

Would you like further clarification or additional details?

Here are five related questions:
1. How do absolute value equations work with multiple intervals?
2. What methods can simplify solving absolute value equations?
3. How does solving equations change when inequalities are introduced?
4. How can we graph absolute value functions to find solutions visually?
5. What happens when there are no valid solutions in a certain interval?

**Tip:** When solving absolute value equations, always consider each possible case based on the intervals formed by the points where the expressions inside the absolute values are zero.

|3x + 5| + |2x - 3| = 25, find the sum of the highest and lowest values.

Learn how to solve the absolute value equation |3x + 5| + |2x - 3| = 25. This problem involves algebraic concepts and the use of properties of absolute values. Suitable for students in Grades 9-12.

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