Complex numbers problem no.2

[mathwannabe]

Hello everybody

I'm short on time, so I'm gonna get straight to the point.

1) If i is imaginary unit and x and y are real numbers for which stands that (2+3i)x+(3+2i)y=1, then x-y=?

I tried and I tried... I failed and I failed. Hint?

 

[pka]

1) If i is imaginary unit and x and y are real numbers for which stands that (2+3i)x+(3+2i)y=1, then x-y=?

You know that
\(\displaystyle \begin{align*}2x+3y&=1 \\3x+2y&=0\end{align*}\)
Solve for x & y.

 

[mathwannabe]

You know that
\(\displaystyle \begin{align*}2x+3y&=1 \\3x+2y&=0\end{align*}\)
Solve for x & y.

Yes... It is completely clear to me now. So simple when you see the answer. I knew I needed to construct a system of two linear equations with two unknowns, I was just getting so confused with those imaginary units. But if the result is real, imaginary parts of complexes must be 0.

Thank you pka

 

[HallsofIvy]

Yes... It is completely clear to me now. So simple when you see the answer. I knew I needed to construct a system of two linear equations with two unknowns, I was just getting so confused with those imaginary units. But if the result is real, imaginary parts of complexes must be 0.

Thank you pka

What you should have known from the start is that a+ bi= c+ di if and only if a= c and b= d.

Interestingly, it is not necessary to actual find x and y. Since 2x+ 3y= 1 and 3x +2y= 0, subtracting the first equation from the second immediately gives x- y= -1.