Solve for x and y in the equation: jx(x+3y) = (3x+j4)(1+jy)

The left side becomes:

$$jx^2 + j3xy$$

The right side becomes:

$$3x(1) + 3x(jy) + j4(1) + j4(jy)$$ $$3x + j3xy + j4 + j^24y$$

Since $j^2 = -1$, substitute this into the equation:

$$3x + j3xy + j4 - 4y$$

Group the real and imaginary terms on the right side:

$$(3x - 4y) + j(3xy + 4)$$

The full equation is:

$$jx^2 + j3xy = (3x - 4y) + j(3xy + 4)$$

We can write the left side as $0 + j(x^2 + 3xy)$.
Equating the real parts:

$$0 = 3x - 4y \quad (*)$$

Equating the imaginary parts:

$$x^2 + 3xy = 3xy + 4 \quad (**)$$

From equation $(**)$, we can cancel $3xy$ from both sides:

$$x^2 = 4$$ $$x = \pm \sqrt{4}$$ $$x = \pm 2$$

Now, use equation $(*)$ to find the corresponding values of $y$:

$$3x - 4y = 0$$ $$4y = 3x$$ $$y = \frac{3}{4}x$$

For $x = 2$:

$$y = \frac{3}{4}(2) = \frac{6}{4} = \frac{3}{2}$$

For $x = -2$:

$$y = \frac{3}{4}(-2) = -\frac{6}{4} = -\frac{3}{2}$$

The solutions for $x$ and $y$ are:

$$\boxed{x=2, y=\frac{3}{2} \quad \text{and} \quad x=-2, y=-\frac{3}{2}}$$