Verify if x=3 and y=1 satisfy the equation (x-j2y) - (y+1/jx) = 2+j.

$$\frac{1}{j} = \frac{1}{j} \times \frac{-j}{-j} = \frac{-j}{-j^2} = \frac{-j}{-(-1)} = \frac{-j}{1} = -j$$

Now substitute this into the equation:

$$(x-j2y) - (y-jx) = 2+j$$

$$(3-j2(1)) - (1-j(3))$$ $$(3-j2) - (1-j3)$$

$$3 - j2 - 1 + j3$$

Group the real and imaginary parts:

$$(3-1) + (-j2+j3)$$ $$2 + j1$$ $$2+j$$

The simplified LHS is $2+j$.
The RHS of the original equation is $2+j$.
Since LHS = RHS ($2+j = 2+j$), the given values are correct.

✅ Correct. The values $x=3$ and $y=1$ satisfy the equation.