# Prove 1o(2*3)=(1o2)*(1o3)xoy=xy-3x-3y+12 x*y=x+y-3

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You need to evaluate the expression `1o(2*3)` using the given laws `xoy=xy-3x-3y+12 = xy - 3(x+y)+ 12` and `x*y=x+y-3` such that:

`1o(2*3) = 1(2*3) - 3(1+(2*3)) + 12`

`2*3 = 2+3-3 = 2`

Substituting 2 for `2*3` yields:

`1o(2*3) = 2- 3(1+2) + 12`

`1o(2*3) = 2 - 6 + 12`

`1o(2*3) = 8`

Hence, evaluating `1o(2*3)` yields 8 and you need to check if the expression to the right gives 8 also.

By the law`"*" ` yields:

(1o2)*(1o3) = (1o2) + (1o3) - 3

You need to evaluate (1o2) and (1o3) using the law "o" such that:

`(1o2) = 2 - 3(1+2) + 12`

`(1o2) = 2 - 6 + 12`

`(1o2) = 8`

`(1o3) = 3 - 3(1+3) +12`

`(1o3) = 3 - 12 + 12`

`(1o3) = 3`

You need to substitute 8 for `(1o2) ` and 3 for `(1o3)` such that:

`(1o2)*(1o3) = 8 + 3 - 3`

`(1o2)*(1o3) = 8`

**Hence, evaluating both sides using the given laws, yields `1o(2*3) = (1o2)*(1o3) = 8.` **

The function o and * are defined as:

xoy=xy-3x-3y+12 and x*y=x+y-3

1o(2*3)

1o(2 + 3 - 3)

= 1o2

=1x2 - 3x1 - 3x2 + 12

= 2 - 3 - 3 + 12

=14 - 6

= 8...(1)

(1o2)*(1o3)

(1x2 - 3x1 - 3x2 + 12)*(1x3 - 3x1 - 3x3 + 12)

= 14*3

= 14 - 3 - 3

= 14 - 6

= 8...(2)

(1) = (2)

**This proves that 1o(2*3)=(1o2)*(1o3)**