2x + 3y = (39 - 6xy)/5
Now by first making 39 - 6xy a multiple of 5 , we get if 39-6xy=30 , or 6xy=9 we get xy=3/2 or 2x+ 3y=6 (x=3/2 , y=1) and rest all cases we are getting either x,y imaginary or 2x+ 3y >6
so minimun value is 6
x, y, z are 3 non zero positive integers such that x+y+z = 8 and xy+yz+zx = 20,
What would be minimum possible value of x*y^2*z^2
x+y+z = 7
2x-y+z = 4
11y-10x = ??
abcde=1 (where a,b,c,d and e are all positive real numbers)
what is the minimum value of a+b+c+d+e?