# Thread: I want to see if someone can solve this puzzle with algebra(not a homework question)

1. ## I want to see if someone can solve this puzzle with algebra(not a homework question)

Hi,

Funny enough, I just encountered this puzzle in a video game. The game said
this problem could be solved with algebra; I tried but couldn’t solve it like that.
It’s an easy puzzle when you think about it, but I think
using algebra would be very tedious. Nonetheless, I’m curious
to know how it would be solved that way.

A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?

The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.

2. ## Re: I want to see if someone can solve this puzzle with algebra(not a homework questi

\begin{align*}2(b-2) &= s+2 \\ 3(b-3) &= s+3\end{align*}
And solve the simultaneous equations.
Spoiler:
Subtract the first from the second $$b-5=1 \implies b=6$$
and substitute into either of the original equations \begin{aligned}2(6-2) &= s+2 \\8&= s+2 \implies s=6\end{aligned}

3. ## Re: I want to see if someone can solve this puzzle with algebra(not a homework questi

Originally Posted by asquiddi
Hi,

Funny enough, I just encountered this puzzle in a video game. The game said
this problem could be solved with algebra; I tried but couldn’t solve it like that.
It’s an easy puzzle when you think about it, but I think
using algebra would be very tedious. Nonetheless, I’m curious
to know how it would be solved that way.

A boy and his sister. When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age. If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age. What are their real ages?
A little more detail:
Let b be the boy's age and g be the girl's age.
"subtract two from the boy's real age": b- 2
"add two to the girl's age": g+ 2

"his sister becomes twice his NEW (subtracted) age": g+ 2= 2(b- 2)

"subtract three from his real age instead": b- 3
"add three to his sister": g+ 3

"she becomes thrice his NEW (subtracted) age": g+ 3= 3(b- 3)

I would solve these a little differently from the way Archie did.
g+ 2= 2(b- 2)= 2b- 4 so, subtracting 2 from both sides, g= 2b- 6.
g+ 3= 3(b- 3)= 3b- 9. Replace that "g" with 2b- 6 to get 2b- 6+ 3= 2b- 3= 3b- 9.

Subtract 2b from both sides: -3= b- 9.
Add 9 to both sides: 6= b. The boy is 6 years old. g= 2b- 6= 12- 6= 6. The girl is also 6 years old.
Twins!

Check:
"When you subtract two from the boy’s real age and add two to his sister’s age, his sister
becomes twice his NEW (subtracted) age"
Subtracting 2 from 6 gives 4 and adding 2 to 6 gives 8. Yes, 8 is twice 4.

"If you subtract three from his real age instead and add three to his sister then she becomes thrice his NEW (subtracted) age."
Subtracting 3 from 6 gives 3 and adding 3 to 6 gives 9. Yes, 9 is "thrice" 3.

"The answer for both brother and sister is 6 years (real ages). How would you solve
it with algebra.