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\(\displaystyle (1+i)x+2y=3\)

\(\displaystyle 6ix-2y=-4\)

Then you will have eliminated $y$ and can solve for $x$, after which you may substitute for $x$ into either equation to determine $y$.

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I am not supposed to use Gaussian elimination.One way to proceed would be to multiply the second equation by $2i$, and then add the two equations:

\(\displaystyle (1+i)x+2y=3\)

\(\displaystyle 6i-2y=-4\)

Then you will have eliminated $y$ and can solve for $x$, after which you may substitute for $x$ into either equation to determine $y$.

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- Aug 18, 2013

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- Jan 29, 2012

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It would have helped if you hadI am not supposed to use Gaussian elimination.

In any case "complex numbers", as far as algebra is concerned, are just

If you are not allowed to use the

[tex]x= \dfrac{\left|\begin{array}{ccc}3 & 2 \\ 2i & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}[/tex]

[tex]y= \dfrac{\left|\begin{array}{ccc}1+i & 3 \\ 3 & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}[/tex]

(Editted thanks to eddybob123.)

Last edited:

- Aug 18, 2013

- 76

I'm sure you meant y on the second equation!

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I thought I could figure it out. What if I don't want to use Cramer's rule.It would have helped if you hadtoldus that!

In any case "complex numbers", as far as algebra is concerned, are justnumbers. Any method that you could use with real coefficients works with complex coefficients.

If you are not allowed to use theeasiestmethod (Gaussian elimination) you could use Cramer's rule, as eddybob123 suggested:

[tex]x= \dfrac{\left|\begin{array}{ccc}3 & 2 \\ 2i & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}[/tex]

[tex]x= \dfrac{\left|\begin{array}{ccc}1+i & 3 \\ 3 & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}[/tex]

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Online service - linear system to complex numbers\(\displaystyle \mathbb C: (1+i)x + 2y = 3, 3x + (1i)y = 2i\)

I don't know how to go about this, am I supposed to do addition of complex number? Please help.

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