the sum of ages of3 children= 32 the age of the oldest is2x the age ofthe youngestthe 2 older differ by 3yrs?

Question

what iare the ages of the 3 children

Answer 1

Let x = age of the oldest, y = age of middle and z = age of youngest, then
x + y + z = 32
x = 2z
x - y = 3 (so y = x-3)
Replace x with 2z and y with x - 3
2z + x - 3 + z = 32
Now replace x with 2z
2z + 2z - 3 + z = 32
Combine like terms and solve for z.
4z - 3 + z = 32
5z = 35
z = 7
x = 2(7) = 14
y = 14 - 3 = 11
Answer: 14, 11 and 7

Answer 2

14, 11 and 7

Answer 3

Youngest= 7
Next Oldest=11
Oldest=14

Answer 4

let the children be called a, b, and c

a + b + c = 32

2a = c

c - b = 3

2(a + b + c) = 64
c + 2b + 2c = 64
3c + 2b = 64
3(3+b) + 2b = 64
9+5b = 64
11 = b

c - 11 = 3
c = 14

2a = 14
a =7

therefore the age of the children are 7, 11, and 14

Answer 5

x + y + z = 32
x = 2z

x - y = 3
2z - y = 3
-y = -2z + 3
y = 2z - 3

2z + (2z - 3) + z = 32
2z + 2z - 3 + z = 32
5z - 3 = 32
5z = 35
z = 7

x = 2(7) = 14
y = 14 - 3 = 11
z = 7

The ages are 14, 11, and 7

Answer 6

x + y + z = 32
x = 2z
x - y = 3

Solve and you get x = 14, y = 11, and z = 7.

Answer 7

Why are you asking this nerd.