SOLUTION: The sum of the ages of a man and his wife is six times the sum of the ages of their children. Two years ago the sum of their ages was ten times the age of their children. After six

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Question 1104990: The sum of the ages of a man and his wife is six times the sum of the ages of their children. Two years ago the sum of their ages was ten times the age of their children. After six years the sum of their ages will be 3 times the sum of the ages of their children . How many children do they have ?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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let m = the man's present age
let w = wife's present age
let c = the sum of the children's present age
let n = no. of children
:
The sum of the ages of a man and his wife is six times the sum of the ages of their children.
m + w = 6c
:
Two years ago the sum of their ages was ten times the age of their children.
(m-2) + (w-2) = 10(c - 2n)
m + w - 4 = 10c - 20n
replace (m+w) with 6c
6c - 4 = 10c - 20n
6c - 10c - 4 = - 20n
-4c - 4 = -20n
multiply by -1
4c + 4 = 20n
simplify, divide by 4
c + 1 = 5n
:
After six years the sum of their ages will be 3 times the sum of the ages of their children.
(m+6) + (w+6) = 3(c + 6n)
m + w + 12 = 3c + 18n
replace (m+w) with 6c
6c + 12 = 3c + 18n
6c - 3c + 12 = 18n
3c + 12 = 18n
simplify, divide by 3
c + 4 = 6n
:
How many children do they have?
use elimination on these two simplified equation
c + 4 = 6n
c + 1 = 5n
-----------subtraction eliminated c
0 + 3 = n
they have 3 children