traditional method

[prithibi]

1. if x^2+xy+1=16 and x-y=1 then x+y=?
answer: 5 or -6
2. if x+y=4, y+z=8 and z+x=6; then 2y-z-x=?
answer: 0
3. if 5/x=3 and y/6=2 then (3+y)/(x+5)=?
answer: 9/4
these are solved by traditionally. is there other method to solve these?

[grexam]

well the second problem can be solved by trial n error method

x+y=4
y+z=8
z+x=6

Assume values x=1,y=3,z=5
substitute them in each
1+3=4
3+5=8
5+1=6
and finally 6-5-1=0

hope this works out what do u think?????if any other way plzz do post

[prithibi]

i've solved the 2nd problem in this way
2y-z-x
=y+y-z-z+z-x-x+x
=(y+z)+(y+x)-2(x+z)
=8+4-2*6
=0

but it costs me 2 valuable minutes