Question 1104990
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