The question has two fundamental problems. They're both show-stoppers:
1). It asks for the "first" of something that extends infinitely in two directions,
but never tells where to start from.
2). There's no such thing as the "first" one anyway. The ordered pairs that satisfy
the equation can't be listed or numbered. If you give me two of them, then no matter
how close together yours are, I can always put another one in between them.
So if you write down one ordered pair and say it's the "first" one, I can always name
another one that's closer to the starting point than yours is.
(2x)ysquared
Whooop!
4xy + x3y + yx2 + yx + 3yx = x3y + x2y + 8xy = (xy)(x2y + x + 8)
3x3 - x2y + 12x - 4y = x2*(3x - y) + 4*(3x - y) = (x2 + 4)*(3x - y)
(b - x)(ab - xy)
(2x)ysquared
Whooop!
The GCF of 27 x^2 y^3 and 46 x^2 y is xy^2. 27 and 46 are coprime.
It is -2y + x2y - 8, which is an expression that cannot be simplified further, not can it be evaluated.
I do the coefficients first. The GCF of 10 and 15 is 5. I tackle the variables next. The GCF of x2y and x3y2 is x2y I put it all together. The GCF of 15x2y and 10x3y2 is 5x2y.
You will get nonsense. You cannot plug x2y + 4 into anything!
4 (apples) plus 3 (apples) = 7 (apples). 4 (cows) + 3 (cows) = 7 (cows) 4 (x2y) + 3 (x2y) = 7 (x2y)
x2y = k (constant) so 36 x 2 ie 72 is the constant. If x = 3 then y = 72/9 = 8
xy
The GCF is xy
4
4xy + x3y + yx2 + yx + 3yx = x3y + x2y + 8xy = (xy)(x2y + x + 8)