# (x/(2+i))-(yi/(1+2i))=i

Solve `x/(2+i)-(yi)/(1+2i)=i` :

First we get the imaginary numbers out of the denominators by rationalizing; multiplying the numerator and denominator by the complex conjugate of the denominator:

`x/(2+i)*(2-i)/(2-i)=(2x-ix)/5`

`(yi)/(1+2i)*(1-2i)/(1-2i)=(yi-2yi^2)/5`  Note that `i^2=-1` so:

`x/(2+i)-(yi)/(1+2i)=(2x-ix)/5-(2y+yi)/5=i` Multiplying by 5:

`2x-2y-(x+y)i=5i`  Equating real terms and imaginary terms we get the following system:

`2x-2y=0`

...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Solve `x/(2+i)-(yi)/(1+2i)=i` :

First we get the imaginary numbers out of the denominators by rationalizing; multiplying the numerator and denominator by the complex conjugate of the denominator:

`x/(2+i)*(2-i)/(2-i)=(2x-ix)/5`

`(yi)/(1+2i)*(1-2i)/(1-2i)=(yi-2yi^2)/5`  Note that `i^2=-1` so:

`x/(2+i)-(yi)/(1+2i)=(2x-ix)/5-(2y+yi)/5=i` Multiplying by 5:

`2x-2y-(x+y)i=5i`  Equating real terms and imaginary terms we get the following system:

`2x-2y=0`

`-x-y=5` Multiplying the second equation by 2 and adding we get:

`-4y=10 ==>y=-5/2` . Then `x=-5/2` .

The solution is `x=y=-5/2` .

Approved by eNotes Editorial Team