# Solve in Z*Z the system (x-3)*y=2 (x-y)o4=10 x*y=x+y-3 xoy=xy-3(x+y)+12

### 1 Answer | Add Yours

We'll re-write the equations of the system, using the given laws of composition.

(x-3)*y=2, where

x*y=x+y-3

We'll substitute x by (x-3).

(x-3)*y = x - 3 + y - 3

We'll combine like terms:

x + y - 6 = 2

x + y = 6 + 2

x + y = 8 (1)

Now, we'll re-write the second equation of the system using the second law of composition.

(x-y)o4=10

xoy=xy-3(x+y)+12

We'll substitute x by (x-y) and y by 4:

(x-y)o4 = 4(x-y) - 3(x-y+4) + 12

We'll remove the brackets:

4x - 4y - 3x + 3y - 12 + 12 = 10

We'll eliminate and combine like terms:

x - y = 10 (2)

We'll solve the system formed from (1) and (2). We'll add (1) + (2):

x + y + x - y = 8 + 10

We'll eliminate and combine like terms:

2x = 18

**x = 9**

We'll substitute x in (1):

x + y = 8

9 + y = 8

y = 8 - 9

**y = -1**

**Since both values are integers, we'll accept the values resulted for x and y, so the solution of the system is: {9 , -1}.**