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

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

Working on (3)

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

taking y from both sides

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

\$\displaystyle 2p = x + y\$

\$\displaystyle 2p = x + t-p\$

\$\displaystyle 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

those mysterious "symbols" you refer to (Wondering)
• Dec 22nd 2009, 06:15 AM
aquajam
Well, I don't think it's that badly stated. By design, the problem is supposed to be somewhat confusing. I'll try to be clearer next time.

It also possible it cannot be solved by equation, I don't know.

Quote: