The equations $3x^2-5x+p=0$ and $2x^2-2x+q=0$ have one common root. The sum of the other roots of this equa

The correct option is $\frac{8}{3}-p+\frac{3}{2}q$.

Let's find the sum of the other roots step-by-step. First, let the common root of the two given quadratic equations be $\alpha$. Let the other root of the first equation be $\beta$ and the other root of the second equation be $\gamma$.

For the first quadratic equation:

$3x^2-5x+p=0$

Using Vieta's formulas, the sum of its roots is given by the negative coefficient of $x$ divided by the coefficient of $x^2$:

$\alpha +\beta =\frac{5}{3}$

For the second quadratic equation:

$2x^2-2x+q=0$

The sum of its roots is:

$\alpha +\gamma =\frac{2}{2}=1$

We are asked to find the sum of the other roots, which is $\beta +\gamma$. From the sum of roots equations, we can express $\beta$ and $\gamma$ in terms of $\alpha$:

$\beta =\frac{5}{3}-\alpha$

$\gamma =1-\alpha$

Adding these together gives us the expression we want to evaluate:

$\beta +\gamma =(\frac{5}{3}-\alpha )+(1-\alpha )=\frac{8}{3}-2\alpha$

Now, we need to find the value of $-2\alpha$ in terms of $p$ and $q$. Since $\alpha$ is a common root, it must satisfy both original equations:

$3{\alpha }^2-5\alpha +p=0$    (Equation 1)

$2{\alpha }^2-2\alpha +q=0$    (Equation 2)

To eliminate the ${\alpha }^2$ term, we can multiply Equation 2 by $\frac{3}{2}$:

$\frac{3}{2}(2{\alpha }^2-2\alpha +q)=0$

$3{\alpha }^2-3\alpha +\frac{3q}{2}=0$    (Equation 3)

Next, we subtract Equation 3 from Equation 1:

$(3{\alpha }^2-5\alpha +p)-(3{\alpha }^2-3\alpha +\frac{3q}{2})=0$

$-2\alpha +p-\frac{3q}{2}=0$

Rearranging this to solve for $-2\alpha$:

$-2\alpha =-p+\frac{3q}{2}$

Finally, we substitute this back into our expression for $\beta +\gamma$:

$\beta +\gamma =\frac{8}{3}-2\alpha$

$\beta +\gamma =\frac{8}{3}-p+\frac{3}{2}q$

This matches our correct option perfectly.