The sum of three numbers is 64. One of the numbers is twice the average of the other two. The difference between the other two is 6. What are the three numbers?
Please help me solve this.
x + y + z = 64
z = y + 6 the difference is 6
x = 2((y + z)/2) twice the average
x = y + z
x = y + (y + 6)
x = 2y + 6
x + y + z = 64
(2y + 6) + y + (y + 6) = 64
4y + 12 = 64
4y = 52
y = 13 <<<<<<<<<<<<<<<<<<<<<<
z = y + 6
z = 13 + 6
z = 19 <<<<<<<<<<<<<<<<<<<<<<
x = 2y + 6
x = 2(13) + 6
x = 26 + 6
x = 32 <<<<<<<<<<<<<<<<<<<<<<
Check the answers.
x + y + z = 64
32 + 13 + 19 = 64
64 = 64