Question

if x+y+z=50 yz=400 xy=200 find x,y,z

Answer 1

xy=200 therefore x = 200/y
yz = 400 therefore z = 400/y
keep the values of them in the equation

200/y + y + 400/y = 50 
600 + y² - 50y = 0
y² - 50y + 600 = 0
y² - 20y - 30y + 600=0
y(y-20) - 30(y-20)=0
therefore y = 20 or 30
when y = 20 x = 10 z = 20
when y = 30 x = 200/30   z = 400 / 30

Answer 2

x + y + z = 50    --- (1)
    y z = 400      --- (2)
    x y = 200    -- (3)

(2)  /  (3)    =>    z/x = 2      =>  z = 2 x

substitute that in  (1):    3 x + y = 50 
 from (3) we substitute value of y:  
           3 x + 200 / x = 50
           3 x² - 50 x + 200 = 0
           x  =  [50 + -  √[2500 - 2400] / 6

x = 10 or 20/3
y = 200/x =  20  or  30
z = 2x = 20  or  40/3

there are two solutions.