The sum of three numbers is 50. one number is 10 more than the second number it's also twice the third find the number

X + y + z = 50

y + 10 = x, solving for y, y = x - 10

x = 2z, solving for z, z = x/2

Substituting for y and z,

x + x - 10 + x/2 = 50

(5/2) x - 10 = 50

(5/2) x = 60

x = (2/5) * 60 = 24

y = x - 10 = 24 - 10 = 14

Z = x/2 = 24/2 = 12

Checking:

x + y + z = 50, 24 + 14 + 12 = 50, 50 = 50. Solution meets the initial constraint.