To solve the equation \(3x + 1 = \sqrt{8x + 1} + \sqrt{2x - 1}\), let's proceed step by step.

1. **Isolate one of the square roots:**

Let's isolate \(\sqrt{8x + 1}\):
\[
\sqrt{8x + 1} = 3x + 1 - \sqrt{2x - 1}
\]

2. **Square both sides to eliminate the square root:**

Squaring both sides, we get:
\[
(\sqrt{8x + 1})^2 = (3x + 1 - \sqrt{2x - 1})^2
\]
Simplifying the left side:
\[
8x + 1 = (3x + 1)^2 - 2(3x + 1)\sqrt{2x - 1} + (\sqrt{2x - 1})^2
\]
Simplifying the right side:
\[
8x + 1 = 9x^2 + 6x + 1 - 6x\sqrt{2x - 1} - 2\sqrt{2x - 1} + 2x - 1
\]
Combining like terms:
\[
8x + 1 = 9x^2 + 8x + 1 - 6x\sqrt{2x - 1} - 2\sqrt{2x - 1}
\]

3. **Isolate the remaining square root term:**

Subtract \(8x + 1\) from both sides:
\[
0 = 9x^2 - 6x\sqrt{2x - 1} - 2\sqrt{2x - 1}
\]

4. **Isolate the square root term:**

Factor out \(\sqrt{2x - 1}\):
\[
0 = 9x^2 - \sqrt{2x - 1}(6x + 2)
\]

5. **Solve for \(x\):**

For the equation to hold, either \(9x^2 = 0\) or \(\sqrt{2x - 1}(6x + 2) = 0\).

- \(9x^2 = 0\) implies \(x = 0\).

- \(\sqrt{2x - 1}(6x + 2) = 0\) implies either \(\sqrt{2x - 1} = 0\) or \(6x + 2 = 0\).

- \(\sqrt{2x - 1} = 0\) implies \(2x - 1 = 0\), so \(x = \frac{1}{2}\).

- \(6x + 2 = 0\) implies \(6x = -2\), so \(x = -\frac{1}{3}\).

6. **Check the solutions:**

- For \(x = 0\):
\[
3(0) + 1 = \sqrt{8(0) + 1} + \sqrt{2(0) - 1}
\]
\[
1 = \sqrt{1} + \sqrt{-1}
\]
This is not valid because \(\sqrt{-1}\) is not a real number.

- For \(x = \frac{1}{2}\):
\[
3\left(\frac{1}{2}\right) + 1 = \sqrt{8\left(\frac{1}{2}\right) + 1} + \sqrt{2\left(\frac{1}{2}\right) - 1}
\]
\[
\frac{3}{2} + 1 = \sqrt{4 + 1} + \sqrt{1 - 1}
\]
\[
\frac{5}{2} = \sqrt{5} + 0
\]
This is not valid because \(\frac{5}{2} \neq \sqrt{5}\).

- For \(x = -\frac{1}{3}\):
\[
3\left(-\frac{1}{3}\right) + 1 = \sqrt{8\left(-\frac{1}{3}\right) + 1} + \sqrt{2\left(-\frac{1}{3}\right) - 1}
\]
\[
-1 + 1 = \sqrt{-\frac{8}{3} + 1} + \sqrt{-\frac{2}{3} - 1}
\]
\[
0 = \sqrt{-\frac{5}{3}} + \sqrt{-\frac{5}{3}}
\]
This is not valid because \(\sqrt{-\frac{5}{3}}\) is not a real number.

Therefore, there are no real solutions to the equation \(3x + 1 = \sqrt{8x + 1} + \sqrt{2x - 1}\).