What are x and y if (8x + yi) / (1 + 4i)  = (2x + i)/ (3 + 2i)  

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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