# Solve this linear system over the field

#### crypt50

##### New member
$$\displaystyle \mathbb C: (1+i)x + 2y = 3, 3x + (1i)y = 2i$$

#### MarkFL

Staff member
One way to proceed would be to multiply the second equation by $2i$, and then add the two equations:

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

$$\displaystyle 6ix-2y=-4$$

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

#### crypt50

##### New member
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 $y$.
I am not supposed to use Gaussian elimination.

#### MarkFL

Staff member
I thought Gaussian elimination involved an augmented matrix. If you are not to use any type of elimination then use substitution instead.

#### eddybob123

##### Active member
Since this has two equations and two variables, it's easy to apply Cramer's Rule, since the determinant of a 2x2 matrix $\begin{bmatrix}a&b\\c&d \end{bmatrix}$ is just $ad-bc$

#### HallsofIvy

##### Well-known member
MHB Math Helper
I am not supposed to use Gaussian elimination.
It would have helped if you had told us that!

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

If you are not allowed to use the easiest method (Gaussian elimination) you could use Cramer's rule, as eddybob123 suggested:
$$x= \dfrac{\left|\begin{array}{ccc}3 & 2 \\ 2i & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}$$
$$y= \dfrac{\left|\begin{array}{ccc}1+i & 3 \\ 3 & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}$$

(Editted thanks to eddybob123.)

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#### eddybob123

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

#### crypt50

##### New member
It would have helped if you had told us that!

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

If you are not allowed to use the easiest method (Gaussian elimination) you could use Cramer's rule, as eddybob123 suggested:
$$x= \dfrac{\left|\begin{array}{ccc}3 & 2 \\ 2i & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}$$
$$x= \dfrac{\left|\begin{array}{ccc}1+i & 3 \\ 3 & 2i\end{array}\right|}{\left|\begin{array}{ccc}1+i & 2 \\ 3 & i \end{array}\right|}$$
I thought I could figure it out. What if I don't want to use Cramer's rule.

#### MarkFL

$$\displaystyle (1+i)x+2y=3$$
$$\displaystyle 6ix-2y=-4$$
Solve one of these equations for $2y$ and then substitute that into the other equation and solve for $x$.