What Is The Subject? I Need Help!

**phillychum** · March 27th 2008, 12:20 AM

Okay.. it says on my assignment ..

"to solve an equation we sometimes have to change the subject"

first of all.. what is the subject?

thats the first question. and then the second question says.

"Given A = B ( (square root of) P+Q / P - Q ) where B > 0, make P the subject"

ARGH
I NEED HELP AND THIS ASSIGNMENT IS DUE SOON !

why does math have to be so confusing D:

**janvdl** · March 27th 2008, 12:48 AM

Originally Posted by phillychum

Okay.. it says on my assignment ..

"to solve an equation we sometimes have to change the subject"

first of all.. what is the subject?

thats the first question. and then the second question says.

"Given A = B ( (square root of) P+Q / P - Q ) where B > 0, make P the subject"

ARGH
I NEED HELP AND THIS ASSIGNMENT IS DUE SOON !

why does math have to be so confusing D:

The subject will be the variable that is isolated.

So if we have:
$y = x + 5$ and we are asked to change the subject to $x$ then we will say $x = y - 5$

I assume this is how your problem looks:

$A = B \sqrt{ \frac{P+Q}{P-Q} }$

Square both sides.

$A^2 = B^2 \left( \frac{P+Q}{P-Q} \right)$

$A^2 (P-Q) = B^2 (P+Q)$

$A^2 P - A^2 Q = B^2 P + B^2 Q$

$A^2 P - B^2 P = A^2 Q + B^2 Q$

$P(A^2 - B^2) = Q(A^2 + B^2)$

$P = \frac{Q(A^2 + B^2)}{(A^2 - B^2)}$

Keep practising. You'll get the hang of it soon.

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