Which values of x and y would make the following expression represent a real number?

(6 + 3i)(x + yi)
x = 6, y = 0
x = –3, y = 0
x = 6, y = –3
x = 0, y = –3

this expression represent a real number when the imaginer part =0

(6+3i)(x+yi) = 6x+6yi+3ix+3yi² i²=-1

(6+3i)(x+yi) = 6x+6yi+3ix-3y

(6+3i)(x+yi) = 6x-3y +(6y+3x)istandard form

the imaginer part is : 6y+3x=0

for x=6 and y= - 3   : 6(-3)+3(6) =0

answer:

a=9

step-by-step explanation: lmk if you need a explanation