The equation you're referring to is:

$\frac{1}{3} = 2x - 10$

Let's solve for $x$:

1. Add 10 to both sides: $\frac{1}{3} + 10 = 2x$ $\frac{1}{3} + \frac{30}{3} = 2x$ $\frac{31}{3} = 2x$

2. Divide both sides by 2: $x = \frac{\frac{31}{3}}{2} = \frac{31}{6}$

Thus, the solution to the equation is:

$x = \frac{31}{6}$

Do you have any other equations you'd like to solve or need further clarification?

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Here are some related questions to explore:

1. How would the solution change if the equation had been $\frac{1}{4} = 2x - 10$?
2. What would happen if you had to solve $3x - 5 = 2x + 7$?
3. What methods can you use to solve linear equations like this one without fractions?
4. How would you check your solution after solving for $x$?
5. What is the general process to solve for $x$ in linear equations involving fractions?

Tip: When solving equations with fractions, it's often helpful to clear the fractions by multiplying both sides of the equation by the least common denominator (LCD).