# Given the law of composition x*y= xy+4mx+2ny, determine m and n if the law is commutative.

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### 2 Answers

If an operation % is commutative it implies that x%y = y%x.

As * is commutative for x*y = xy + 4mx + 2ny, we have:

xy + 4mx + 2ny = yx + 4my + 2nx

=> 4mx + 2ny = 4my + 2nx

=> x(2m - n) + y(n - 2m) = 0

Equating the coefficients of x and y to 0

=> 2m - n = 0 and n - 2m = 0

=> n = 2m

**We cannot find an exact value for m and n, we can only say that they are related by the relation n = 2m**

We'll write the commutative property of a law of composition:

x*y = y*x, for any value of x and y.

We'll substitute x*y and y*x by the given expression:

x*y = xy + 4mx + 2ny (1)

y*x = yx + 4my + 2nx (2)

We'll put (1) = (2) and we'll get:

xy + 4mx + 2ny = yx + 4my + 2nx

We'll remove like terms:

4mx + 2ny = 4my + 2nx

We'll move the terms in "m" to the left side and the terms in "n" to the right side:

4mx - 4my = 2nx - 2ny

We'll factorize and we'll get:

4m(x-y) = 2n(x-y)

We'll divide by x - y:

4m = 2n

m = 2n/4

m=n/2

**So, for the law to be commutative, we find m = n/2, for any real value of m and n.**