Answer:
Jason is 56; Carolyn is 28; Neil is 32
Step-by-step explanation:
You want the ages of Jason, Carolyn, and Neil if Jason is twice as old as Carolyn and 24 years older than Neil. Half of Jason's age six years ago was 3 less than half the sum of Carolyn's age four years ago and Neil's present age.
Let j, c, n represent Jason's, Carolyn's, Neil's present ages, respectively. The first two relations are ...
j = 2c
j = n +24
The third relation is a bit more tricky. Half of Jason's age 6 years ago is ...
(j -6)/2
The sum of Carolyn's age 4 years ago and Neil's present age is ...
(c -4) +n
The first of these is 3 less than half the sum:
(j -6)/2 = ((c -4) +n)/2 -3
We can simplify this last equation a bit:
j/2 -3 = (c +n -4)/2 -3
Adding 3 and multiplying by 2, we have ...
j = c +n -4
Substituting j-24 for n gives ...
j = c +(j -24) -4)
0 = c -28 . . . . . . . . . subtract j
c = 28 . . . . . . . . . . . add 28
Now we can find j and n:
j = 2c = 2(28) = 56
n = j -24 = 56 -24 = 32
Jason is 56; Carolyn is 28; Neil is 32.
__
Check
Half of Jason's age 6 years ago is (56 -6)/2 = 25. Carolyn's age 4 years ago was 24. Half the sum of that and Neil's present age is (24 +32)/2 = 56/2 = 28. 25 is 3 less than 28, so this answer checks.
