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A sister is as old as the brother will be when the sister is twice the age that the brother was when the sister's age was half the sum of their present ages.

What are the relation between their ages?

They're both younger than 100 years old.

That's a stupid riddle.

the sister was really just a man in a dress.

Since Blaxican is such a buzz-kill I'm going to give it a shot.

Here's a tabular for the relations.

http://i.imgur.com/tama0.png

Since we know that there's always a constant number of days in-between them we can construct (and solve) the following matrix.

http://i.imgur.com/kIUE1.png

Meaning that the relation is such that the sister is 4/3rds as old as her brother.

So the thread is now about checking if that's accurate.

we need to go deeper

Originally posted by Astner
Since Blaxican is such a buzz-kill


You know you love it.

Wait did I miss the funny part?

Originally posted by Miss Fiend
Wait did I miss the funny part?
i think it's supposed to be hidden in all that mathematics drivel

sorry, guess im just too retarded to get the punchline

ahahaa mathematics all the ones and twos just so silly.

Leave him alone, guys. People get off from different things. I get off to big booty'd bitches, Robtard gets off to my cousin, and Astner gets off to mathematical formulae. Be tolerant.

Originally posted by RE: Blaxican
Leave him alone, guys. People get off from different things. I get off to big booty'd bitches, Robtard gets off to my cousin, and Astner gets off to mathematical formulae. Be tolerant.
http://www.wethinkllc.com/blog/wp-content/uploads/2009/08/douchebag.jpg

Oh all right FINE.

Originally posted by Astner
Since Blaxican is such a buzz-kill I'm going to give it a shot.

Here's a tabular for the relations.

http://i.imgur.com/tama0.png

Since we know that there's always a constant number of days in-between them we can construct (and solve) the following matrix.

http://i.imgur.com/kIUE1.png

Meaning that the relation is such that the sister is 4/3rds as old as her brother.

So the thread is now about checking if that's accurate. The future sister would be 2z.

Read it again.

(x+y)/2 - z = x - y = 2z - x = c (constant, which is difference between ages)
Gonna do present n future first.
2z - y - c = 0
Oh shit, I gotta differentiate both y and z and find out the constant, then do the same logic for the past part of the sum.

I gave up on Maths at A Level, and it's 20 past 4. **** that, I'm going to bed, but I think I layed down some of the foundation stuff.

Originally posted by lord xyz
The future sister would be 2z.

Read it again.
A mistyping. You can clearly see in the last row of the first matrix that I wrote: (-1 0 2)

Meaning -1 x, 0 y and 2 z.

The calculation is correct. Learn your linear algebra for ****'s sake.

People need to caaaaaalm down up in here.

Oh man, ALGEBRA?! You're killin me! roflroflrofl

3

Originally posted by lord xyz
(x+y)/2 - z = x - y = 2z - x = c (constant, which is difference between ages)
Gonna do present n future first.
2z - y - c = 0
Oh shit, I gotta differentiate both y and z and find out the constant, then do the same logic for the past part of the sum.

I gave up on Maths at A Level, and it's 20 past 4. **** that, I'm going to bed, but I think I layed down some of the foundation stuff.

I think you "laid" down a gross misspelling.

Well, that's a riddle, I don't see what's funny about it though. It may be a fun riddle to some. Astner is correct though.

Riddle me this???? herb

(x+y)/2 - c = z
x - c = y
2z - c = x

2z - c = y + c
2z - 2c = y
z - c = y/2
{(x+y)/2 - c} - c = y/2
(x+y)/2 - 2c = y/2
x + y - 4c = y
x = 4c

4c - c = y
y = 3c

Yeah, the sister is 4/3s as old as the brother

2z - c = 4c
2z = 5c
z = 5/2c

(x+y)/2 - 5/2c = c
x + y - 5c = 2c
x + y = 7c

Yep, all consistent. c=x/4

x = 16
y = 12

(12+16)/2 - (16-12) = z
14 - 4 = 10
2(10) - 4 = 16
20 - 4 = 16

x = 12
y = 9

(12+9)/2 - 3 = z
10.5 - 3 = 7.5
15 - 3 = 12

I don't understand any of that shit in brackets, but I worked out the difference between their ages.

Originally posted by lord xyz
(x+y)/2 - c = z
x - c = y
2z - c = x

2z - c = y + c
2z - 2c = y
z - c = y/2
{(x+y)/2 - c} - c = y/2
(x+y)/2 - 2c = y/2
x + y - 4c = y
x = 4c

4c - c = y
y = 3c

Yeah, the sister is 4/3s as old as the brother

2z - c = 4c
2z = 5c
z = 5/2c

(x+y)/2 - 5/2c = c
x + y - 5c = 2c
x + y = 7c

Yep, all consistent. c=x/4

x = 16
y = 12

(12+16)/2 - (16-12) = z
14 - 4 = 10
2(10) - 4 = 16
20 - 4 = 16

x = 12
y = 9

(12+9)/2 - 3 = z
10.5 - 3 = 7.5
15 - 3 = 12

I don't understand any of that shit in brackets, but I worked out the difference between their ages.
I can reconstruct another one if you'd like.

A mother is half as old as her son will be when the mother's age is twice the sum of the age of the mother and the son when the mother was as old as the son currently is.

Mother's current age: x
Son's current age: y
Difference between ages: c

Mother's future age: 2x+c
Son's future age: 2x
Mother's past age: z+c
Son's past age: z

z+c=y
2(2z+c)=2x+c
c=x-y

2(2(y-c)+c)=2x+c
2(2y-2c+c)=2x+c
4y-4c+2c=2x+c
4y-2c=2x+c
4y=2x+3c
4y=2x+3(x-y)
4y=2x+3x-3y
7y=5x
y=5/7x

Son is 5/7s the age of his mother.

x-y=c
x-(5/7)x=c
2/7x=c

x=35
y=25
c=10
z=25-10=15

2(2z+c)=2x+c
2(2(15)+10)=2(35)+10
2(30+10)=17+10
2(40)=80

Or simply:

http://i.imgur.com/iEqNi.png

It's virtually the same thing. But it's less messy to solve with matrix algebra and you can even recreate similar problems like I just did without having to spend a lot of time on it.

I do not understand Matrix Algebra.

Although you are right, simultaneous equations are a real *****.

Originally posted by lord xyz
I do not understand Matrix Algebra.

Although you are right, simultaneous equations are a real *****.
It's rather simple actually, it's the same type.

Say you have three equations.

x + y - z
2x - 3z
3y

You'd write that:

1 1 -1
2 0 -3
0 3 0

Then you subtract or add lines with one another until you get something like.

1 0 0
0 1 0
0 0 1

In which case that would be translated to

x =
y =
z =