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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

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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.**

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