Real Solutions for a Complex Equation: (4+2i)x + (5-3i)y = 13+i

[NewtonianAlch]

Homework Statement

Find the real solutions of

\left( 4+2\,i \right) x+ \left( 5-3\,i \right) y=13+i

The Attempt at a Solution

4x + 2xi + 5y - 3yi = 13 + i

4x + 5y - 13 + i(2x -3y - 1) = 0

I am not really sure if my method is correct here for starters.

Would I then only consider the real part and solve that?

 

[SammyS]

Homework Statement

Find the real solutions of

\left( 4+2\,i \right) x+ \left( 5-3\,i \right) y=13+i

The Attempt at a Solution

4x + 2xi + 5y - 3yi = 13 + i

4x + 5y - 13 + i(2x -3y - 1) = 0

I am not really sure if my method is correct here for starters.

The above is correct.

Would I then only consider the real part and solve that?

No. you are looking for the real solutions for x and y, which make the complex equation true.

In order for a complex equation to be true, the real part of the right side must equal the real part of the left side AND the imaginary part of the right side must equal the imaginary part of the left side

 

[NewtonianAlch]

I'm not entirely sure what you mean, but I think

4x + 5y = 13 and 2x -3y = 1 ?

So it's just a simultaneous equation, and solving for x and y gives us real solutions. Is that it?

 

[NewtonianAlch]

Anyone know if this is right?

 

[SammyS]

I'm not entirely sure what you mean, but I think

4x + 5y = 13 and 2x -3y = 1 ?

So it's just a simultaneous equation, and solving for x and y gives us real solutions. Is that it?

This is correct.

From your equation:

4x + 5y - 13 + i(2x -3y - 1) = 0 :​

Equating the real parts gives:

4x + 5y - 13 = 0 .​

Equating the imaginary parts gives:

2x -3y - 1 = 0 .​

That's what I mean.

 

[NewtonianAlch]

Damn...thanks.