# x & yFind x and y if (x-1)/i + (y+1)/2 = (x+2)/3 + (y-1)/i .

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### 2 Answers

We have to solve for x and y given that (x-1)/i + (y+1)/2 = (x+2)/3 + (y-1)/i

(x-1)/i + (y+1)/2 = (x+2)/3 + (y-1)/i

=> [2(x - 1) + i(y +1)]/2i = [(x + 2)i + 3(y - 1)]/3i

=> 6(x - 1) + 3i(y + 1) = 2i(x + 2) + 6(y - 1)

=> 6x - 6 + 3iy + 3i = 2ix + 4i + 6y - 6

equate the real and complex coefficients

=> 6x - 6 = 6y - 6

=> x = y

3y + 3 = 2x + 4

=> 3x + 3 = 2x + 4

=> x = 1

=> y = x =1

**The solution is x = 1 and y = 1**

We'll multiply all over by 6i:

6i*(x-1)/i + 6i*(y+1)/2 = 6i*(x+2)/3 + 6i*(y-1)/i

We'll simplify and we'll get:

6x - 6 + 3iy + 3i = 2ix + 4i + 6y - 6

We'll combine the real parts and the imaginary parts both sides:

(6x - 6) + i(3y + 3) = (6y - 6) + i(2x + 4)

We'll compare and we'll get:

6x - 6 = 6y - 6

6x - 6y = 6 - 6

x - y = 0

x = y (1)

3y + 3 = 2x + 4 (2)

We'll substitute (1) in (2):

3x + 3 = 2x + 4

x = 4 - 3

**x = 1**

**y = 1**