⚡ AZUCATION SHORTCUT:
Let common root be $\alpha$.
For Eq 1: $\alpha + \beta_1 = 5/3 \implies \beta_1 = 5/3 - \alpha$.
For Eq 2: $\alpha + \beta_2 = 2/2 = 1 \implies \beta_2 = 1 - \alpha$.
Sum $= (5/3 + 1) - 2\alpha = 8/3 - 2\alpha$.
By elimination: $2(\text{Eq 1}) - 3(\text{Eq 2}) \implies -4\alpha + 2p - 3q = 0 \implies 2\alpha = p - \frac{3}{2}q$.
Substitute: Sum $= 8/3 - (p - \frac{3}{2}q) = \mathbf{\frac{8}{3} - p + \frac{3}{2}q}$.

Step 1: Identify the roots
Let the common root of both equations be $\alpha$.
Let the other root of the first equation ($3x^2 - 5x + p = 0$) be $\beta_1$.
Let the other root of the second equation ($2x^2 - 2x + q = 0$) be $\beta_2$.

Step 2: Use Sum of Roots property
From the first equation: $\alpha + \beta_1 = \frac{-(-5)}{3} = \frac{5}{3} \implies \beta_1 = \frac{5}{3} - \alpha$.
From the second equation: $\alpha + \beta_2 = \frac{-(-2)}{2} = 1 \implies \beta_2 = 1 - \alpha$.
The required sum is $\beta_1 + \beta_2 = (\frac{5}{3} - \alpha) + (1 - \alpha) = \frac{8}{3} - 2\alpha$.

Step 3: Solve for the common root $\alpha$
Since $\alpha$ is a root for both:
1) $3\alpha^2 - 5\alpha + p = 0$ (Multiply by 2 $\to 6\alpha^2 - 10\alpha + 2p = 0$)
2) $2\alpha^2 - 2\alpha + q = 0$ (Multiply by 3 $\to 6\alpha^2 - 6\alpha + 3q = 0$)
Subtracting the two results: $(-10\alpha - (-6\alpha)) + (2p - 3q) = 0$
$-4\alpha + 2p - 3q = 0 \implies 4\alpha = 2p - 3q \implies 2\alpha = p - \frac{3}{2}q$.

Step 4: Final Calculation
Substitute $2\alpha$ back into our sum expression:
Sum $= \frac{8}{3} - (p - \frac{3}{2}q) = \mathbf{\frac{8}{3} - p + \frac{3}{2}q}$.

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