SOLUTION: i am having trouble solving: find real numbers x and y if: (x+i)(3-iy)= 1+13i i have done: 3x- xyi +3i +y= 1+13i 3x+y=1 -xy+3=13 x+y=1/3 -xy=10 but i dont know

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: i am having trouble solving: find real numbers x and y if: (x+i)(3-iy)= 1+13i i have done: 3x- xyi +3i +y= 1+13i 3x+y=1 -xy+3=13 x+y=1/3 -xy=10 but i dont know       Log On


   

Question 725092: i am having trouble solving: find real numbers x and y if:
(x+i)(3-iy)= 1+13i
i have done:
3x- xyi +3i +y= 1+13i
3x+y=1 -xy+3=13
x+y=1/3 -xy=10
but i dont know where to go from here
Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
(x+i)(3-iy)= 1+13i
(3x+y)+(3-xy)i=1+13i

(1) 3x+y=1,
(2) 3-xy=13
(2) 3-xy=13, xy=-10, y=-10/x

(1) 3x+y=1, 3x+(-10/x)=1, 3x^2-x-10=0, x=2, x=-5/3

then
(x,y)=(2,-5)
or

(x,y)=(-5/3, 6)