1 divided by 3a = (2 divided by a squared) minus (a-2 divided by a squared)

We are given the equation:

\[
\frac{1}{3a} = \frac{2}{a^2} - \frac{a - 2}{a^2}
\]

### Step 1: Simplify the right-hand side
First, simplify the expression on the right-hand side by combining the fractions. Both terms have a common denominator \(a^2\):

\[
\frac{2}{a^2} - \frac{a - 2}{a^2} = \frac{2 - (a - 2)}{a^2}
\]

Now distribute the minus sign:

\[
2 - (a - 2) = 2 - a + 2 = 4 - a
\]

So, the expression becomes:

\[
\frac{4 - a}{a^2}
\]

Thus, the equation is now:

\[
\frac{1}{3a} = \frac{4 - a}{a^2}
\]

### Step 2: Cross-multiply
We can eliminate the denominators by cross-multiplying:

\[
(1)(a^2) = (3a)(4 - a)
\]

This simplifies to:

\[
a^2 = 12a - 3a^2
\]

### Step 3: Rearrange the equation
Move all terms to one side of the equation:

\[
a^2 + 3a^2 - 12a = 0
\]

Combine like terms:

\[
4a^2 - 12a = 0
\]

### Step 4: Factor the equation
Factor out the common term \(4a\):

\[
4a(a - 3) = 0
\]

### Step 5: Solve for \(a\)
Set each factor equal to zero:

1. \(4a = 0 \Rightarrow a = 0\)
2. \(a - 3 = 0 \Rightarrow a = 3\)

### Conclusion:
The possible solutions are \(a = 0\) and \(a = 3\).

However, we must exclude \(a = 0\) because it would make the original equation undefined due to division by zero. Therefore, the only solution is:

\[
a = 3
\]

Would you like further details on any step?

### Related Questions:
1. How can we verify that \(a = 3\) satisfies the original equation?
2. What are the implications of dividing by zero in an equation?
3. How do we combine fractions with different numerators but the same denominator?
4. How is cross-multiplication used to solve rational equations?
5. Can factoring always be used to solve quadratic equations?

**Tip:** Always check for restrictions in the variable (e.g., division by zero) when solving rational equations.

Learn how to solve the rational equation 1/(3a) = (2/a^2) - ((a-2)/a^2) with a detailed step-by-step solution. This problem involves algebraic concepts such as rational expressions, factoring, and the zero product property. Ideal for students in Grades 9-11.

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