solve the following equations for real X and Y
1) (2-3i) (x+yi) = 4+i
2) (2-2i) (x+yi) = 2(x=2yi) + 2i - 1
need help :( after getting answer i m gonna write this mathematics in my death note lol
4 answers
6 views!! but still no one answer ...-_-!
impatient much? whining after 10 minutes?
Have you been working on it while waiting for someone to help?
Two complex numbers are equal if their real and imaginary parts are equal
(2-3i) (x+yi) = 4+i
2x - 3xi + 2yi + 3y = 4+i
(2x+3y) + (-3x+2y) = 4+i
That means that
2x+3y = 4
-3x+2y = 1
x = 5/13 and y = 14/13
check: (2-3i)(5+14i)/13 = 4+i yes
assuming a typo, I used
(2-2i) (x+yi) = 2(x+2yi) + 2i - 1
(2x+2y) + (-2x+2y)i = (2x-1) + (4y+2)i
so now just solve
2x+2y = 2x-1
-2x+2y = 4y+2
Have you been working on it while waiting for someone to help?
Two complex numbers are equal if their real and imaginary parts are equal
(2-3i) (x+yi) = 4+i
2x - 3xi + 2yi + 3y = 4+i
(2x+3y) + (-3x+2y) = 4+i
That means that
2x+3y = 4
-3x+2y = 1
x = 5/13 and y = 14/13
check: (2-3i)(5+14i)/13 = 4+i yes
assuming a typo, I used
(2-2i) (x+yi) = 2(x+2yi) + 2i - 1
(2x+2y) + (-2x+2y)i = (2x-1) + (4y+2)i
so now just solve
2x+2y = 2x-1
-2x+2y = 4y+2
No, that's what I was doing
i was trying to solve this and i got right answer
by the way thank you for your help :)
i was trying to solve this and i got right answer
by the way thank you for your help :)
by the way are you teacher?