Using the theory of simultaneous equations

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SUMMARY

The ages of A, B, C, and D add up to 100 years, with specific relationships defined among their ages. The equations established are: a + b + c + d = 100, a + d = b + c, a - d = 2c, and c + (a - d) = 2b. Solving these equations yields the ages: A is 45, B is 30, C is 20, and D is 5. The solution confirms that A is older than D, aligning with the problem's constraints.

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adam640
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Homework Statement


A, B, C and D are celebrating their joint birthdays and find
that their ages add up to exactly 100 years. The sum of A's and
D's ages equals the sum of B and C, while the di fference between
the ages of A and D is twice C's age. Finally when D is as old
as A is now, C will be twice as old as B is now.
Given that A is older than D, how old are the four of them?


Homework Equations


I constructed the following but I am unsure if they are correct, and where I would go from there?
a+b+c+d = 100
a+d = b+c
a-d = 2c
a > d
when... d = a
2c = b

Any help greatly appreciated, thanks
 
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adam640 said:

Homework Statement


A, B, C and D are celebrating their joint birthdays and find
that their ages add up to exactly 100 years. The sum of A's and
D's ages equals the sum of B and C, while the difference between
the ages of A and D is twice C's age. Finally when D is as old
as A is now, C will be twice as old as B is now.
Given that A is older than D, how old are the four of them?

Homework Equations


I constructed the following but I am unsure if they are correct, and where I would go from there?
a+b+c+d = 100
a+d = b+c
a-d = 2c
a > d
when... d = a
2c = b

The last part (which I bolded) isn't right. Let's look at the statement: "Finally when D is as old as A is now, C will be twice as old as B is now."

Suppose D is 15 and A is 21. "When D is as old as A is now" would mean 6 years (21 minus 15) from now. In general, this would be the difference in their ages (A-D). So 'after this number of years, C's age' would mean that you would add this difference to C's age, and the sum would equal twice B's age. So the remaining equation would be
c + (a - d) = 2b

Now you have four equations and four unknowns.
a + b + c + d = 100
a + d = b + c
a - d = 2c
c + (a - d) = 2b(Hope I'm not giving too much away -- the OP did attempt at the set up.)
 
Solved! Thanks for your help. I got a = 45, b = 30, c = 20, d = 5!
 

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