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The image shows a graph of the absolute value function \( y = |3 - 2x| \), and you are asked to solve the equation \( |3 - 2x| = 7 + x \).

### Steps to solve \( |3 - 2x| = 7 + x \):

1. **Handle the absolute value function**:
   The equation involves an absolute value, so we need to consider two cases:
   
   - Case 1: \( 3 - 2x \geq 0 \), which gives \( |3 - 2x| = 3 - 2x \).
   - Case 2: \( 3 - 2x < 0 \), which gives \( |3 - 2x| = -(3 - 2x) = -3 + 2x \).

### Case 1: \( 3 - 2x \geq 0 \) (i.e., \( x \leq \frac{3}{2} \))

The equation becomes:
\[
3 - 2x = 7 + x
\]
Solve for \( x \):
\[
3 - 2x = 7 + x
\]
Subtract \( x \) from both sides:
\[
3 - 3x = 7
\]
Subtract 3 from both sides:
\[
-3x = 4
\]
Divide by \( -3 \):
\[
x = -\frac{4}{3}
\]
Since \( -\frac{4}{3} \leq \frac{3}{2} \), this solution is valid for this case.

### Case 2: \( 3 - 2x < 0 \) (i.e., \( x > \frac{3}{2} \))

The equation becomes:
\[
-3 + 2x = 7 + x
\]
Solve for \( x \):
\[
-3 + 2x = 7 + x
\]
Subtract \( x \) from both sides:
\[
-3 + x = 7
\]
Add 3 to both sides:
\[
x = 10
\]
Since \( 10 > \frac{3}{2} \), this solution is valid for this case.

### Final Solution:
The solutions are:
\[
x = -\frac{4}{3} \quad \text{or} \quad x = 10
\]

Would you like further details on the steps, or have any additional questions?

### Related Questions:
1. How do you graph \( y = |3 - 2x| \) from the equation?
2. Can you solve a similar equation like \( |x - 4| = 2x + 1 \)?
3. How do you handle absolute value inequalities like \( |3 - 2x| \geq 7 \)?
4. How does the slope of the lines change when we break the absolute value equation?
5. How would the solution change if the right-hand side of the equation were \( 7 - x \) instead of \( 7 + x \)?

### Tip:
When solving absolute value equations, always remember to check if the solutions fall within the correct range for each case to avoid extraneous solutions.

This guide provides a detailed step-by-step approach to solving the absolute value equation |3 - 2x| = 7 + x. Learn the method to break down absolute value functions into two cases and solve linear equations. Ideal for students in Grades 10-12.

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