To find the maximum value of $a + b$ given the equation:

$\frac{a-2}{b} + \frac{b-2}{a} = 1,$

where $a$ and $b$ are both positive rational numbers, let's solve step by step.

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Step 1: Rewrite the equation

Multiply through by $ab$ (since $a \neq 0$ and $b \neq 0$) to eliminate the fractions:

$a(a - 2) + b(b - 2) = ab.$

Expanding both terms:

$a^2 - 2a + b^2 - 2b = ab.$

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Step 2: Rearrange terms

Rearrange to group terms into a quadratic equation:

$a^2 - ab + b^2 - 2a - 2b = 0.$

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Step 3: Symmetry and maximum $a + b$

To maximize $a + b$, observe the symmetry in the equation. Setting $a = b$ simplifies the equation, as symmetry often helps find extreme values.

If $a = b$, substitute $b = a$ into the original equation:

$\frac{a - 2}{a} + \frac{a - 2}{a} = 1.$

This simplifies to:

$2 \cdot \frac{a - 2}{a} = 1.$

Simplify further:

$\frac{a - 2}{a} = \frac{1}{2}.$

Multiply through by $a$:

$a - 2 = \frac{a}{2}.$

Solve for $a$:

$2a - 4 = a \implies a = 4.$

Since $a = b$, we also have $b = 4$.

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Step 4: Compute $a + b$

If $a = b = 4$, then:

$a + b = 4 + 4 = 8.$

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Final Answer:

The maximum value of $a + b$ is:

$\boxed{8}.$

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Would you like a detailed explanation of any step? 😊

Related Questions:

1. Can the value of $a + b$ be less than 8? Under what conditions?
2. How would the solution change if $a$ and $b$ were not rational numbers?
3. Could a geometric interpretation help in solving this problem?
4. What happens if $a \neq b$? How do we proceed to find the general solution?
5. Can we verify this result by graphing the equation $\frac{a-2}{b} + \frac{b-2}{a} = 1$?

Tip:

Symmetry often simplifies problems in algebra. Always consider setting variables equal or exploring symmetry to find extrema.