What does something like $b=p(qs+bt)$ mean? (notice $b$ on both sides)

Question

You can say: "$b$ is equal to $p$ times $qs+bt$

But how does that make any sense? $b$ is equal to an expression that contains $b$? How can the definition of $b$ contain $bt$? This doesn't mean you have solved for $b$ right? If the variables are integers, can you say $p$ divides $b$?

If you have another equation:
$p=3s+2b$

And then you "plug in $b$":
$p=3s+2(p(qs+bt))$
then you plug in $b$ again:
$p=3s+2(p(qs+(p(qs+bt))t))$

What?

In general, what does it actually mean when you have just one variable isolated on side of an equation? In $b=p(qs+bt)$ we have $p$ isolated on the leftside but it also on the right side, contained in $bt$. In $p=3s+2b$ we have $p$ isolated on the left side and it is not also on the right side. What is the difference?

Answer 1

You appear to be getting confused between a definition and an equation. While we often write something like "define the number $p$ by $p = 3s+2b$", some people prefer to write $p := 3s+2b$ to emphasize that we define $p$ to be whatever is on the right side of the equality. In this context, $p$ itself CANNOT appear on the right side.

On the other hand, an equality like $b = p(qs+bt)$, where $b$ is a variable (and the other letters are constants), is instead asking "does there exist a $b$ which satisfies this equation?". For some equations (like this one), there is exactly one solution, in which case you might say "$b$ is defined to be the unique solution to the equation $b=p(qs+bt)$", but this does require proof. Some equations might admit no solutions (like $x=x+1$), and some might admit many (like $x^2=1$).

What you're saying with all those nested equalities (like $p=3s+2(p(qs+p(qs+bt)t)$) is: if there is a number $b$ that satisfies the equation $b=p(qs+bt)$, then these equalities are also true. It doesn't say anything about whether such a $b$ exists, and it isn't a definition.

Answer 2

What does it actually mean?

Let's first look at a simpler example: $b=2b-1$. This means that we are looking for a value of $b$ such that, when you multiply it by $2$ and then subtract $1$, you get the same result. In general, something like $b=\text{[something involving b]}$ means that we are looking for a value of $b$ such that plugging it into the thing involving $b$ returns you back $b$.

Going back to your problem, we have $b=p(qs+bt)$. This means that we are looking for a value of $b$ such that, when we multiply it by $t$, add $qs$, and then multiply all that by $p$, we get $b$ back again.

For your last question(what's the difference), consider $p=1+2$ and $p=2p-1$. $1$ and $2$ are clearly not in terms of $p$, so if we want to find $p$, we just have to add $1$ and $2$. However, for $p=2p-1$, we need to do some work. We can only find the value of a variable once it has been fully isolated. So, if we have $p$ appearing on both sides of the equation, all that means is we need to do some work to find $p$.