In a $1 \leftarrow 10$ machine the place-value of boxes were the powers of ten: $1, 10, 100, 1000, ...$.

In a $1 \leftarrow 2$ machine the place-value of boxes were the powers of two: $1, 2, 4, 8, 16, ...$.

In a $1 \leftarrow x$ machine the place-values of boxes must be the powers of $x$: $1,x,x{2},x{3}, ...$.

As a check:

If I tell you $x$ is ten in my mind, then we really are getting $1, 10, 100, 1000, ...$ and if I tell you instead $x$ really is two, then we are getting $1, 2, 4, 8, 16, ...$, and so on. So the $1 \leftarrow x$ machine is indeed representing ALL machines all at once.

Okay … without any warning, here’s a high-school algebra problem:

Compute $\left( 2x^{2}+7x+6\right) \div \left( x+2\right)$.