x^2+xy+y= 14

y ^ 2+xy+x = 28:

x^2+xy+y+y^2+xy+x=28+ 14

(x^2+2xy+y^2)+(x+y)-42=0

(x+y) 2+(x+y)-42 = 0 (taking (x+y) as a whole, the decomposition factor)

(x+y+7)(x+y-6)=0

Namely: x+y=

-7 or x+y=6

2. Prove:

14*(a^2+b^2+c^2)=(a+2b+3c)^2

14a^2+ 14b^2+ 14c^2=a^2+4ab+4b^2+ 12bc+9c^2+6ac

14a^2+ 14b^2+ 14c^2-a^2-4ab-4b^2- 12bc-9c^2-6ac=0

13a^2+ 10b^2+5c^2-4ab- 12bc-6ac=0

Note that the result to be proved is a:b:c= 1:2:3, that is, c=3a, then the above formula can be simplified as:

(9a^2-6ac+c^2)+(4a^2-4ab+b^2)+(9b^2- 12bc+4c^2)

(3a-c)^2+(2a-b)^2+(3b-2c)^2=0

3a-c=0

2a-b=0

3b-2c=0

Namely: c=3a

b=2a

So: a:b:c=a:2a:3a= 1:2:3.