SOLUTION: the combined age of 2 brothers is 35 years old.7 years ago one brother was twice the age of the other. how many years apart were they born

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Question 1204081: the combined age of 2 brothers is 35 years old.7 years ago one brother was twice the age of the other. how many years apart were they born

Found 3 solutions by mananth, greenestamps, josgarithmetic:
Answer by mananth(16946)   (Show Source):
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the combined age of 2 brothers is 35 years
let the brother's ages be x & y
x+y=35.................................................(1)
7 years ago one brother was twice the age of the other.
7 years ago
One brother was x-7 years
The other was y-7 years
x-7 = 2(y-7)
x-7 =2y-14
x-2y = -7............................................(2)
subtract (2) from (1)
x+y=35
x-2y = -7
3y =42
y = 14
x+y =35
x+14=35
x= 21
Since we know their present ages we can find how many years apart were they born

Answer by greenestamps(13200)   (Show Source):
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The solution from tutor @mananth is a very inefficient solution using two variables, starting from the given information that the sum of their current ages is 35 years.

"7 years ago one brother was twice the age of the other" gives us a much easier way to set up the problem for solving, using a single variable.

Let x = age of younger brother 7 years ago
then 2x = age of older brother 7 years ago

The difference in their ages at any time is the difference in their ages 7 years ago, which is 2x-x = x.

The sum of their current ages is 35:

(x+7)+(2x+7)=35
3x+14=35
3x=21
x=7

ANSWER: the difference in their ages is 7 years -- i.e., they were born 7 years apart

Answer by josgarithmetic(39617)   (Show Source):
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OR maybe this way:
v and 35-v

and

Those two ages differ by 7 years.