The united ages of a man and his wife are six times to united ages of their children. Two years ago their united ages were ten times the united ages of their children and six years hence their ages will be three times the united ages of their children. How many children have they?
There is no way to tell. The combined ages of the children in no way tells you how many children there are. It does provide an upper limit though, assuming each child is at least 1 year old.
Not only that, but the system of equations using the given conditions has no solution.
m+w = 6c
m-2 + w-2 = 10(c-2)
m+6 + w+6 = 3(c+6)
its 3 ezpz
I'm sorry, but the solution requires more information or clarification about the problem. As stated, it is not possible to determine the number of children they have and the system of equations provided has no solution. Could you please provide more details or constraints to the problem?
To solve this problem, let's break it down into steps:
Step 1: Determine the variables
Let's assume the united ages of the man and his wife as "M" and "W" respectively, and the united ages of their children as "C".
Step 2: Translate the given information into equations
According to the problem statement:
1. The united ages of the man and his wife are six times the united ages of their children:
M + W = 6C
2. Two years ago, their united ages were ten times the united ages of their children:
(M - 2) + (W - 2) = 10(C - 2)
3. Six years hence, their ages will be three times the united ages of their children:
(M + 6) + (W + 6) = 3(C + 6)
Step 3: Simplify and solve the equations
Start by simplifying equations 2 and 3:
2. M + W - 4 = 10C - 20 (expand and simplify)
3. M + W + 12 = 3C + 18 (expand and simplify)
Rearrange equation 2:
M + W = 10C - 16
Now, combine equations 1 and the rearranged equation 2:
6C = 10C - 16
16 = 4C
C = 4
Step 4: Find the number of children
The solution C = 4 represents the number of children they have, which means they have 4 children.
Therefore, they have 4 children.