Changing the subject - more than one answer? 
Make $x$ the subject of
$5(x+2y)=p(2x-3)$

My textbook says that it doesn't matter which side you move the $x$ terms to when solving the problem, but if I put the $x$ terms over to the right-hand side, the answer becomes:
$\frac{10y+3p}{2p-5} = x$
Conversely, if I move them over to the left-hand side, the answer becomes:
$x = \frac{3p-10y}{5-2p}$
Are both answers correct?
 A: You have $$5(x+2y)=p(2x-3)$$
$x$ to LHS we get:
$$5x+10y=2px-3p$$ $$5x-2px=-3p-10y$$ $$x(5-2p)=-3p-10y$$ $$x=\frac{-3p-10y}{5-2p}$$ $$\boxed{x=\frac{10y+3p}{2p-5}}$$
$x$ to RHS we get:
$$5x+10y=2px-3p$$ $$10y+3p=2px-5x$$ $$10y+3p=x(2p-5)$$ $$\boxed{\frac{10y+3p}{2p-5}=x}$$
Be careful with the signs.
A: Method 1
$5(x+2y)=p(2x-3)$
$\iff 5x+10y=2px-3p$
$\iff 2px-5x=10y+3p$
$\iff x(2p-5)=10y+3p$
$\iff x=\frac{10y+3p}{2p-5}$
Method 2
$5(x+2y)=p(2x-3)$
$\iff 5x+10y=2px-3p$
$\iff 5x-2px=-3p-10y$
$\iff x=\frac{-3p-10y}{5-2p}$
Key Observation
These two answers are the same. Notice that:
$\frac{-3p-10y}{5-2p}$ = $\frac{-1}{-1}$($\frac{10y+3p}{2p-5})$=$\frac{10y+3p}{2p-5}$
Therefore, answer (1) is the same as answer (2), and so this is consistent with you being taught that we are able to rearrange for x in any way that you want.
A: It shouldn't really matter what side $x$ is on. $x=\frac{-3p-10y}{5-2p}$ is the correct answer, so you should probably check your working. Be careful when working with signs - it can lead to wild answers! Hope this helps.
