2016/053) Solve the system of equations in real

$4x^2+25y^2 +9z^2 - 10xy -15yz - 6xz = 0\cdots(1)$
$x+y+z=5\cdots(2)$

Solution
from (1) we have
$8x^2+50y^2 +18z^2 - 20xy -30yz - 12xz = 0$
or  $(4x^2 - 20xy + 25y^2) + (25y^2 - 30yz + 9z^2) + (9z^2 - 12xz + 4x^2)= 0$
or $(2x-5y)^2 + (5y-3z)^2 + (3z-2x)^2=0$
above is sum of 3 squares and hence each of them is zero ior $2x = 5y = 3z=k$ (say)
so $x= \frac{k}{2}$, $y= \frac{k}{5}$,$z= \frac{k}{3}$
putting in (2) we get
$\frac{k}{2} + \frac{k}{5}+  \frac{k}{3} = 5$
or $\frac{31k}{30} = 5$
or  $k = \frac{150}{31}$
so $x= \frac{75}{31}$, $y= \frac{30}{31}$,$z= \frac{50}{31}$