# four variable equation

• Dec 21st 2009, 10:01 PM
aquajam
four variable equation
x + 2y = t + p
2p = x + y
(t - y) + y = p + y

Having trouble with this equation, I need to get it in terms of: t + x + p
and lowest possible values.

I thought with n variables, you needed n equations. This has 4 variables, but only 3 equations.

I gather the idea is to eliminate one variable and get three equations with three variables. I could probably solve it from there, but I keep going in circles trying trying to create new equations.

thanks
• Dec 21st 2009, 10:29 PM
Wilmer
Quote:

Originally Posted by aquajam
x + 2y = t + p
2p = x + y
(t - y) + y = p + y

Re-write the 3 equations:
2y - p = t - x [1]
y - 2p = -x [2]
y + p = t [3]

[2] + [3] : 2y - p = t - x (same as [1])

So you're really given only one equation.
Not much you can do with that...
Have you given us the original problrm IN FULL?
• Dec 21st 2009, 10:35 PM
pickslides
Quote:

Originally Posted by aquajam

I thought with n variables, you needed n equations.

Correct

Quote:

Originally Posted by aquajam

Having trouble with this equation, I need to get it in terms of: t + x + p
and lowest possible values.

What do you mean by this?

Consider the following manipulation

$x + 2y = t + p$ ...(1)
$2p = x + y$ ...(2)
$(t - y) + y = p + y$ ...(3)

Working on (3)

$(t - y) + y = p + y$

taking y from both sides

$(t - y) = p \implies y = t-p$ now using this with (2)

$2p = x + y$

$2p = x + t-p$

$3p = x + t$

This equation is in terms of p,x and t.
• Dec 21st 2009, 11:11 PM
aquajam
Thanks for the replies.

The original problem contains symbols, which I have turned into equations. I'm pretty sure they're accurate.

The question offers sever solutions to the question of what is equal to t + x + p.

2t + y
3x
2p + t
2y + x
3t
p + t + y

Ok, maybe this makes a big difference. I just assumed it could be done without the answers without looking at them properly.
• Dec 22nd 2009, 01:40 AM
aquajam
in the book i have, the values are:

x = 5
y = 1
t = 4
p = 3

And I didn't realise but it stated that the are all different values at the beginning of the puzzle.

I still have no idea how you would arrive at these figures, aside from trial and error.
• Dec 22nd 2009, 06:01 AM
Wilmer
You state that each of these is equal to p + t + x:
2t + y
3x
2p + t
2y + x
3t
p + t + y
...and the values of the 4 variables are:
x = 5
y = 1
t = 4
p = 3

This means p + t + x = 3 + 4 + 5 = 12; BUT only 3t = 12

And if p + t + x = p + t + y
then x = y
But x = 5 and y = 1