Show two systems of linear equations are equivalent.

The two systems are

\left\{\begin{align*} &2x_1+(-1+i)x_2+x_4=0\\ &3x_2-3ix_3+5x_4=0\; \end{align*}\right.

and

\left\{\begin{align*} &\Big(1+\frac i2\Big)x_1+8x_2-ix_3-x_4=0\\ &\frac 23x_1-\frac12x_2+x_3+7x_4=0\;. \end{align*}\right.

However I cannot find a method of solving them. So far, in class, we have only learned row reduction, nothing else. Of course, the methods of substitution and elimination are assumed knowledge.

What would an effective way of solving each of these be? Every time I end up with an equation for one of the variables, I'm unable to derive an equation for another variable that doesn't result in me getting an equality like $0=0$ after substitution.

Is it possible to combine both systems?

I've been working on this question for hours and I feel that the answer is so simple yet there is something fundamentally important that I am missing.

Any help is greatly appreciated.

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The vector $(-3,0,-10i,6)$ is a solution to the first system of equations, but not to the second system, so the two systems are not equivalent. (I found this solution by using Git Gud's solution to the first system, and checking whether it works in the second system.) –  Michael Joyce Jan 13 '13 at 22:44

Notice that from \left\{\begin{align*} &2x_1+(-1+i)x_2+x_4=0\\ &3x_2-3ix_3+5x_4=0\; \end{align*}\right., you can write $x_1$ as a function of $x_2$ and $x_4$ and the same goes for $x_3$.
You get \left\{\begin{align*} &x_1=-\frac{1}{2}((-1+i)x_2+x_4)\\ &x_3=\frac{1}{3i}(3x_2+5x_4)\; \end{align*}\right.
So your set of solutions is $\{(x_1,x_2,x_3,x_4):x_1=-\frac{1}{2}((-1+i)x_2+x_4) \wedge x_3=\frac{1}{3i}(3x_2+5x_4)\wedge x_2, x_4\in \mathbb{C}\}$=$\{(-\frac{1}{2}((-1+i)x_2+x_4), x_2, \frac{1}{3i}(3x_2+5x_4), x_4): x_2, x_4\in \mathbb{C}\}$.