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The equation provided is:

\[
P = \frac{1}{3} \rho v^2
\]

To solve for the variable \(\rho\) (denoted as lowercase "p" in your request):

1. **Multiply both sides by 3** to isolate the term involving \(\rho\) on one side:
   \[
   3P = \rho v^2
   \]

2. **Divide both sides by \(v^2\)** to solve for \(\rho\):
   \[
   \rho = \frac{3P}{v^2}
   \]

So the equation solved for \(\rho\) is:

\[
\rho = \frac{3P}{v^2}
\]

Would you like more details or have any questions? 

Here are some related questions you might consider:
1. How does the density (\(\rho\)) affect the pressure (P) in this equation?
2. What are the typical units for pressure, density, and velocity in this context?
3. How would this equation change if you included a different factor, such as temperature?
4. How is this equation applied in practical scenarios, like fluid dynamics or kinetic theory of gases?
5. Could you derive a related equation to solve for velocity \(v\) instead?

**Tip:** Always check the units of your variables when solving or rearranging equations, especially when dealing with physical quantities, to ensure consistency and accuracy.

Learn how to rearrange the equation P = (1/3) * ρ * v^2 to solve for density (ρ). This step-by-step guide covers the algebraic manipulation required to isolate ρ, useful for students in Grades 9-12 studying physics or algebra.

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