SOLUTION: A collection of jewelry includes two rings, one is 18 years old and the other 46 years old. In how many years will the older ring be twice as old as the newer ring?

Algebra ->  Equations -> SOLUTION: A collection of jewelry includes two rings, one is 18 years old and the other 46 years old. In how many years will the older ring be twice as old as the newer ring?      Log On


   

Question 30816: A collection of jewelry includes two rings, one is 18 years old and the other 46 years old. In how many years will the older ring be twice as old as the newer ring?
Found 2 solutions by bmauger, Paul:
Answer by bmauger(101) About Me  (Show Source):
You can put this solution on YOUR website!
Let x stand for the number of years from now. So if this year ring 1 is 18, in x years it will be 18+x and the second ring will be 46+x.
The question asks when will the second ring be twice as old. Twice indicates we're multiplying by two. So in essence, the question asks how many years will have passes until twice (2 times) the age of the younger ring (18+x) equals (=)the age of the second (46+x) or:
2%2818%2Bx%29=46%2Bx Distribute on the right:
36%2B2x=46%2Bx Subtract x and 36 from both sides.
x=10
So in ten years ring 1 will be 18+10=28 and the second ring will be 46+10=56. 28 doubled is 56, check.

Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
Let the future age be x
18+x
46+x

2(18+x)=46+x
36+2x=46
x=10

Hence, in two years it will be twice as old.
Paul.