# Thread: Subject of the formula

1. ## Subject of the formula

Make r the subject

$\displaystyle P=\pi r+2r+2a$

I could only get

$\displaystyle \frac{P-2a}{\pi}=r+\frac{2r}{\pi}$

2. Originally Posted by Mukilab
Make r the subject

$\displaystyle P=\pi r+2r+2a$

I could only get

$\displaystyle \frac{P-2a}{\pi}=r+\frac{2r}{\pi}$
Take 2a as normal.

$\displaystyle \pi r + 2r = P-2a$

You can then factor out an r on the LHS

$\displaystyle r(\pi+2) = P-2a$

Can you find $\displaystyle r$ now?

3. Originally Posted by e^(i*pi)
Take 2a as normal.

$\displaystyle \pi r + 2r = P-2a$

You can then factor out an r on the LHS

$\displaystyle r(\pi+2) = P-2a$

Can you find $\displaystyle r$ now?

$\displaystyle r=\frac{P-2a}{\pi +2}?$

4. Originally Posted by Mukilab
$\displaystyle r=\frac{P-2a}{\pi +2}?$
Indeed. $\displaystyle \pi +2$ is a constant so the domain is all the real numbers but you don't need to know that for this question

5. Originally Posted by e^(i*pi)
Indeed. $\displaystyle \pi +2$ is a constant so the domain is all the real numbers but you don't need to know that for this question
But thank you for the supply of extra information ^^