# What are x and y if (x-1)/i + (y+1)/2 = (x+2)/3 + (y-1)/i ?

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We have to find 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

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

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

=> x + yi/2 + i/2 = xi/3 + 2i/3 + y

Equate the real and imaginary terms

=> x = y and (y +1)/2 = (x +2)/3

substitute x = y in (y +1)/2 = (x +2)/3

=> (x + 1)/2 = (x +2)/3

=> 3x + 3 = 2x + 4

=> x = 1

**Therefore x = y= 1.**

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

First we will rewrite and simplify the equation.

We will multiply by 6i.

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

Now we will open the brackets:

==> 6x - 6 + 3yi + 3i = 2xi + 4i + 6y - 6

Now we will combine like terms.

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

==> 6x -6 = 6y -6

==> 6x = 6y ==> x= y ..........(1)

==> 3y+3 = 2x + 4

==> 3y -2x = 1

But x = y

==> 3y -2y = 1

**==> y= 1 ==> x= 1**

What are x and y if (x-1)/i + (y+1)/2 = (x+2)/3 + (y-1)/i

(x-1)/i +(y+1)/2 = (x+2)/3 +(y-i)/i. We bring both sides to x+yi form.

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

(y+1)/2 - (x-1)i = (x+2)/3 - (y-1)i , as i^2 = 1.

We equate real parts on both sides and then equate imaginary parts on both sides:

Real parts: (y+1)/2 = (x+2)/3

=>3(y+1) = 2(x+2).

=> 3y-2x = 1....(1).

Imaginary parts: x-1 = y-1.

=> x= y.....(2).

So putting x= y in (1), we get: 3x-22x = 1. So x= 1.

Therefore x= 1 . From (2), y = x= 1.

So 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**