What are x and y if (8x + yi) / (1 + 4i) = (2x + i)/ (3 + 2i)
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We have to find x and y given that (8x + yi) / (1 + 4i) = (2x + i)/ (3 + 2i)
(8x + yi) / (1 + 4i) = (2x + i)/ (3 + 2i)
=> (8x + yi) (3 + 2i) = (2x +i) (1+ 4i)
=> 24x + 16xi + 3yi + 2yi^2 = 2x + i + 8xi + 4i^2
=> 24x + 16xi + 3yi – 2y = 2x + i + 8xi – 4
=> 24x – 2y + i (16x + 3y) = 2x – 4 + i (8x +1)
Equate the real and complex coefficients
We get 24x – 2y = 2x – 4
22x – 2y = -4
=> 11x – y = -2
=> y = 11x + 2
and 16x + 3y = 8x + 1
=> 8x + 3y – 1 = 0
substitute y = 11x + 2
=> 8x + 33x + 6 – 1 = 0
=> 41x = -5
=> x = -5/41
y = 11*(-5/41) + 2
=> (-55 + 82)/41
=> 27 / 41
Therefore x = -5/41 and y = 27/41
giorgiana1976 | College Teacher | (Level 3) Valedictorian
We'll multiply each ratio by the conjugate of the denominator:
We'll start with the fraction from the left side:
(8x + yi) / (1 + 4i) = (8x + yi)(1 - 4i) / (1 + 4i)(1 - 4i)
(8x + yi) / (1 + 4i) = (8x - 32xi + yi + 4y)/(1+16)
We'll combine real parts and imaginary parts from numerator:
[(8x + 4y) - i(32x - y)/(17) (1)
We'll do the same steps with the fractions from the right side;
(2x + i)/ (3 + 2i) = (2x + i)(3 - 2i)/ (3 + 2i)(3 - 2i)
(2x + i)/ (3 + 2i) = (6x - 4xi + 3i + 2)/(9+4)
We'll combine real parts and imaginary parts from numerator:
(2x + i)/ (3 + 2i) = [(6x+2) - i(4x - 3)]/13 (2)
We'll put (1) = (2):
[(8x + 4y) - i(32x - y)]/(17) = [(6x+2) - i(4x - 3)]/13
We'll cross multiply:
13[(8x + 4y) - i(32x - y)] = 17[(6x+2) - i(4x - 3)]
Comparing, we'll get:
13(8x + 4y) = 17(6x+2)
104x + 52y = 102x + 34
We'll subtract 102x:
2x + 52y = 34
We'll divide by 2:
x + 26y = 17
x = 17 - 26y (3)
13(32x - y) = 17(4x - 3)
416x - 13y = 68x - 51
We'll subtract 68x:
348x - 13y = -51 (4)
We'll substitute (3) in (4):
348(17-26y) - 13y = -51
5916 - 9048y - 13y = -51
9061y = 5967
y = 0.6585
x = 17 - 26y
x = -0.1219
The values of x and y are: x = -0.1219 and y = 0.6585.