<Question>
If $\frac{\sqrt{2x}}{\sqrt{3x-1}}=\frac32$, solve for $x$. Express your answer in simplest fractional form.
</Question>

<think>
<Plan_1>
Okay, so I need to solve the equation √(2x) divided by √(3x - 1) equals 3/2. Hmm, let's see. Alright, first, I should probably get rid of the square roots to make the equation easier to handle. The standard way to do that is by squaring both sides of the equation, right? Because if two square roots are equal, then their squares should be equal too. Let me try that.

None

<prompt_1.1> Square both sides of the equation √(2x)/√(3x - 1) = 3/2 and simplify. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Starting with the original equation:
√(2x) / √(3x - 1) = 3/2

If I square both sides, that would eliminate the square roots. Let's do that:

(√(2x) / √(3x - 1))² = (3/2)²

Which simplifies to:

(2x) / (3x - 1) = 9/4
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Okay, so now I have a rational equation. To solve this, I can cross-multiply.
Multiply both sides by (3x - 1) and by 4 to get rid of the denominators. Let me write that out:

<prompt_2.1> Cross-multiply to solve the rational equation (2x)/(3x - 1) = 9/4. </prompt_2.1>
-----

But I need to check if this solution is valid in the original equation because sometimes squaring both sides can introduce extraneous solutions.

Let me verify by plugging x = 6 back into the original equation:

<prompt_2.2> Verify the solution x = 6 by substituting it back into the original equation √(2x)/√(3x - 1) = 3/2. </prompt_2.2>

Wait, but just to be thorough, did I consider all steps correctly? Let me retrace my steps.

<prompt_2.3> Retrace the steps of solving the equation to ensure all steps are correct and valid. </prompt_2.3>

Another thing to consider: when squaring both sides, sometimes we can introduce solutions that don't satisfy the original equation's domain. Let's check the domain first.

<prompt_2.4> Check the domain of the original equation √(2x)/√(3x - 1) = 3/2 to ensure the solution x = 6 is valid. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
4 * 2x = 9 * (3x - 1)

Simplify both sides:

8x = 27x - 9

Hmm, now I need to get all the x terms on one side. Let me subtract 27x from both sides:

8x - 27x = -9

Which simplifies to:

-19x = -9
</execution_2.1>

<execution_2.2>
Left side: √(2*6) / √(3*6 - 1) = √12 / √(18 - 1) = √12 / √17

Simplify √12: √12 = 2√3, so it's 2√3 / √17

Multiply numerator and denominator by √17 to rationalize the denominator:

(2√3 * √17) / 17 = 2√51 / 17

Now, let's compute 3/2:

3/2 = 2.5

Compute 2√51 / 17:

√51 ≈ 7.1414, so 2*7.1414 ≈ 14.2828

14.2828 / 17 ≈ 0.8401

But 3/2 is 1.5, so 0.8401 is not equal to 1.5. Wait, that's not right. Did I make a mistake here?
</execution_2.2>

<execution_2.3>
Original equation: √(2x)/√(3x -1)=3/2. Squared both sides to get (2x)/(3x -1)=9/4. Cross-multiplied: 8x=27x -9. Then 8x -27x= -9 => -19x= -9 => x=9/19.
</execution_2.3>

<execution_2.4>
Original equation: √(2x) / √(3x - 1) = 3/2

The expressions under the square roots must be non-negative, and the denominator can't be zero. So:

1. 2x ≥ 0 ⇒ x ≥ 0

2. 3x - 1 > 0 ⇒ 3x > 1 ⇒ x > 1/3

Therefore, the domain is x > 1/3. Since x = 6 is greater than 1/3, it's within the domain.
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Divide both sides by x? Wait, no, actually, let's solve for x. So, if -19x = -9, then x = (-9)/(-19) = 9/19.

Wait, so x is 9/19? Let me check that. Let me substitute x = 9/19 back into the original equation to see if it works.

<prompt_3.1> Substitute x = 9/19 back into the original equation √(2x)/√(3x - 1) = 3/2 to check if it works. </prompt_3.1>
-----

Based on execution_2.2:
Hmm, that's confusing. Let me check my calculations again.

<prompt_3.2> Re-evaluate the verification process to identify any mistakes. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
First, check the domain. x = 9/19 ≈ 0.473, which is greater than 1/3 ≈ 0.333, so that's okay. No division by zero in the original equation, since 3x - 1 when x = 9/19 is 3*(9/19) - 1 = 27/19 - 19/19 = 8/19, which is positive. So sqrt(2x) is sqrt(2*(9/19)) = sqrt(18/19), which is defined. Similarly, sqrt(3x -1) is sqrt(8/19). So both square roots are defined and the denominator isn't zero. So x = 9/19 is in the domain.
</execution_3.1>

<execution_3.2>
Wait, when I substituted x = 6, I think I miscalculated. Let me recalculate:

Wait, x is 9/19, not 6. Wait, hold on! Wait, maybe I messed up when solving for x? Let me go back. Wait, in step 2.1, when I had -19x = -9, solving for x would be x = (-9)/(-19) = 9/19. So x is 9/19, not 6. Wait, so I must have made a mistake when plugging back in. Let me correct that.

Wait, so x = 9/19. Let's substitute that into the original equation. Let me compute numerator and denominator separately.

Numerator: √(2x) = √(2*(9/19)) = √(18/19) = √18 / √19 = (3√2)/√19

Denominator: √(3x - 1) = √(3*(9/19) - 1) = √(27/19 - 1) = √(27/19 - 19/19) = √(8/19) = √8 / √19 = (2√2)/√19

So the entire expression is [(3√2)/√19] / [(2√2)/√19] = (3√2)/√19 * √19/(2√2) = (3√2 * √19) / (2√2 * √19) = 3/2. Because √2 and √19 cancel out from numerator and denominator. So that's 3/2, which matches the right side.
</execution_3.2>
</Execution_3>

<Final_answer>
Therefore, the solution to the equation, expressed in its simplest fractional form, is $\boxed{\frac{9}{19}}$.
</Final_answer>
</think>