SOLUTION: Wark’s (W) age equals Peter’s (P) age plus the cube root of Ian’s (I) age. P’s age equals I’s age plus the cube root of W’s age, plus 14 years. I’s age equals the cube root of W

Algebra ->  Square-cubic-other-roots -> SOLUTION: Wark’s (W) age equals Peter’s (P) age plus the cube root of Ian’s (I) age. P’s age equals I’s age plus the cube root of W’s age, plus 14 years. I’s age equals the cube root of W      Log On


   

Question 226362: Wark’s (W) age equals Peter’s (P) age plus the cube root of Ian’s (I) age. P’s age equals I’s age plus the cube root of W’s age, plus 14 years.
I’s age equals the cube root of W’s age plus the square root of P’s age.
What is the age of each ???


Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Write an equation for each statement
:
Wark’s (W) age equals Peter’s (P) age plus the cube root of Ian’s (I) age.
W = P + I%5E%281%2F3%29
:
P’s age equals I’s age plus the cube root of W’s age, plus 14 years.
P = I + W%5E%281%2F3%29 +14
;
I’s age equals the cube root of W’s age plus the square root of P’s age.
I = W%5E%281%2F3%29 + sqrt%28P%29

What is the age of each ???
:
This looks horrible at first glance but there are only 3 perfect cubes
in the age range. Ignoring 1, we have 8, 27 & 64, for I and W. And P is
a perfect square.
:
In first equation
W = P + I%5E%281%2F3%29
W = 27, I=8, then P=25
or
W = 8, I=64, P=4; (does not work in the other equations)
:
Test in the 2nd equation using W=27; I=8; P=25
P = I + W%5E%281%2F3%29 +14
25 = 8 + 27%5E%281%2F3%29 + 14; this works,
;
Test in the 3rd equation
I = W%5E%281%2F3%29 + sqrt%28P%29
8 = 27%5E%281%2F3%29 + sqrt%2825%29
8 = 3 + 5
:
The ages; Warks: 27; Peters 25; Ian 8