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To solve the equation \(\sqrt{x-3} - \sqrt{2x-7} = 2x - 8\), we need to isolate the square roots and then square both sides to eliminate them. Here are the steps:

1. **Isolate one of the square roots:**
\[
\sqrt{x-3} = \sqrt{2x-7} + 2x - 8
\]

2. **Square both sides to eliminate the square root on the left:**
\[
(\sqrt{x-3})^2 = (\sqrt{2x-7} + 2x - 8)^2
\]
Simplifying the left side:
\[
x - 3 = (\sqrt{2x-7} + 2x - 8)^2
\]

3. **Expand the right side:**
\[
x - 3 = (\sqrt{2x-7})^2 + 2 \cdot \sqrt{2x-7} \cdot (2x - 8) + (2x - 8)^2
\]
Simplifying each term:
\[
x - 3 = (2x - 7) + 2 \cdot \sqrt{2x-7} \cdot (2x - 8) + (2x - 8)^2
\]

4. **Combine like terms:**
\[
x - 3 = 2x - 7 + 2 \cdot \sqrt{2x-7} \cdot (2x - 8) + (2x - 8)^2
\]

5. **Isolate the square root term:**
\[
x - 3 - 2x + 7 = 2 \cdot \sqrt{2x-7} \cdot (2x - 8) + (2x - 8)^2
\]
Simplifying the left side:
\[
-x + 4 = 2 \cdot \sqrt{2x-7} \cdot (2x - 8) + (2x - 8)^2
\]

6. **Square both sides again to eliminate the remaining square root:**
This step is quite complex, and it might be easier to check for possible solutions by substituting values back into the original equation.

Let's check for possible solutions by substituting values:

- **Check \(x = 3\):**
\[
\sqrt{3-3} - \sqrt{2(3)-7} = 2(3) - 8
\]
Simplifying:
\[
\sqrt{0} - \sqrt{6-7} = 6 - 8
\]
\[
0 - \sqrt{-1} = -2
\]
This is not valid because \(\sqrt{-1}\) is not a real number.

- **Check \(x = 4\):**
\[
\sqrt{4-3} - \sqrt{2(4)-7} = 2(4) - 8
\]
Simplifying:
\[
\sqrt{1} - \sqrt{8-7} = 8 - 8
\]
\[
1 - 1 = 0
\]
This is valid.

Thus, \(x = 4\) is a solution to the equation.