Re-arranging formula question

• Sep 18th 2008, 02:12 PM
Markus123
Re-arranging formula question
Hi All,

I'm having a slight problem with a re-arranging formula question (not a marked one, the answers are on the sheet, I just don't know how to get to the specified answer) it should be easy and I'm pretty embarrassed that I can't do it.

The question is make 'x' the subject of:
$\displaystyle r^2 = (x - a)^2 + y^2$

According to the sheet the correct answer is
$\displaystyle x = sqrt(r^2 - y^2) + a$

I've tried it various ways though and usually end up with
$\displaystyle x = r - y + a$

If anyone could break down how you get the correct answer that would be great. Thanks in advance.
• Sep 18th 2008, 02:33 PM
11rdc11
$\displaystyle r^2 = (x-a)^2 + y^2$

$\displaystyle r^2 - y^2 = (x-a)^2$

Take square root of both sides

$\displaystyle \sqrt{r^2-y^2} = x-a$

$\displaystyle \sqrt{r^2-y^2} + a =x$
• Sep 18th 2008, 02:45 PM
Markus123
Doh, that was easy, now I feel even more stupid.

Thanks.

I'm having problem with another easy one if you could help (Angel)

Make 'x' the formula of:

$\displaystyle y = c/(x+a)$

$\displaystyle x = (c/y) - a$

I did...

$\displaystyle y = c/(x+a)$
$\displaystyle y(x + a) = c$
$\displaystyle yx + ya = c$
$\displaystyle yx = c - ya$
$\displaystyle x = (c - ya)/y$

but apparently I seem to have wrong somewhere (or just not simplified it enough?). Thanks again.
• Sep 18th 2008, 02:50 PM
11rdc11
$\displaystyle y = \frac{c}{x+a}$

$\displaystyle y(x+a) = c$

$\displaystyle x+a = \frac{c}{y}$

$\displaystyle x = \frac{c}{y} -a$
• Sep 18th 2008, 03:00 PM
Markus123
Thank you, again! That's been a great help.