To solve the equation \(7x(x-1) + 3y = 2xy\), let's first simplify and rearrange the terms.

1. Distribute \(7x\) in the term \(7x(x-1)\):
\[
7x^2 - 7x + 3y = 2xy
\]

2. Move all terms to one side of the equation to set it to zero:
\[
7x^2 - 7x + 3y - 2xy = 0
\]

3. Rearrange the terms to group \(x\) and \(y\) terms together:
\[
7x^2 - 7x - 2xy + 3y = 0
\]

This is a quadratic equation in terms of \(x\) and \(y\). To solve for \(x\) in terms of \(y\) or vice versa, we can try to isolate one of the variables. However, this equation is not easily factorizable, so we might need to use substitution or other algebraic methods depending on the context or additional constraints.

Let's try to solve for \(y\) in terms of \(x\):

1. Isolate the \(y\) terms on one side:
\[
7x^2 - 7x = 2xy - 3y
\]

2. Factor out \(y\) on the right-hand side:
\[
7x^2 - 7x = y(2x - 3)
\]

3. Solve for \(y\):
\[
y = \frac{7x^2 - 7x}{2x - 3}
\]

So, the solution for \(y\) in terms of \(x\) is:
\[
y = \frac{7x^2 - 7x}{2x - 3}
\]

This is the simplified form of the equation. If you need a specific numerical solution, you would need to substitute a value for \(x\) and solve for \(y\), or vice versa.