# Simultaneous equation help

• Feb 18th 2010, 10:06 AM
Magma828
Simultaneous equation help
Okay I'm not even sure if this even is a simultaneous equation, but I don't know how to solve it anyway..

x+y=60

x^2 = 0.25
y^2

Find x and y.

It's part of a massive a level physics question and I've simplified it a lot, so no I'm not a 12 year old asking for the answers to his maths homework :P
• Feb 18th 2010, 10:11 AM
TKHunny
Changing the second around a bit, you get the much simpler

4x^2 = y^2

It is now your responsibility to remember that y = 0 is inappropriate.

If we KNOW that x and y are greater than zero (0), we get:

2x = y

Do we KNOW that x and y are greater than zero (0)?

Simultaneous solution is rather simple after that.
• Feb 18th 2010, 10:15 AM
Magma828
Yes x and y are greater than zero :D

Okay thanks a lot, that did it and got the correct answer. But out of interest, how did you go from 4x^2 = y^2 to 2x = y? Is that just a general rule when x and y are greater than 0?
• Feb 18th 2010, 10:37 AM
masters
Quote:

Originally Posted by Magma828
Yes x and y are greater than zero :D

Okay thanks a lot, that did it and got the correct answer. But out of interest, how did you go from 4x^2 = y^2 to 2x = y? Is that just a general rule when x and y are greater than 0?

Hi Magma828,

TK took the square root of each side of the equation.

$4x^2=y^2$

$\sqrt{4x^2}=\sqrt{y^2}$

$2x=y$
• Feb 18th 2010, 07:29 PM
TKHunny
... and that's why we needed to know things were positive. If things can be negative, that square root is not quite as straight forward.