Solving $2x + 1 = 11$: Why, when subtracting $1$, do I only do it to a single term on the left but, if dividing by $2$, I must divide both terms? Solving $2x + 1 = 11$ (for example)
Why, when subtracting $1$, do I only do it to a single term on the left but, if dividing by $2$, I must divide both terms ?

The key thing to realise here is that multiplication is distributive over addition. This is essentially a fancy way of saying that you can expand brackets: $(3+x)\times 5=(3 \times 5)+(x\times 5)$, and $(3+7)\times 4=(3\times 4)+(7\times 4)$, and $(a+y)\times 100=(a\times 100)+(y\times 100)$. In general, the distributive property means that for any numbers $a$, $b$, and $c$, the following law is true:
$$(b+c)\times a=(b\times a)+(c\times a)$$
In your example, we are told that $2x+1=11$. If we multiply both sides by $\frac{1}{2}$ (this is the same as dividing both sides by $2$), then we get that
$$
(2x+1)\times\frac{1}{2}=11 \times \frac{1}{2} \, .
$$
Then, we can simplify the left-hand-side by using the distributive property: $(2x+1)\times\frac{1}{2}=\left(2x \times \frac{1}{2}\right)+\left(1\times \frac{1}{2}\right)=x+\frac{1}{2}$. The right-hand-side simplifies to $\frac{11}{2}$. This means that
$$
x+\frac{1}{2}=\frac{11}{2} \, .
$$
Then, if we subtract $\frac{1}{2}$ from both sides, we get that $x=\frac{11}{2}-\frac{1}{2}=\frac{10}{2}=5$. By contrast, subtraction is not distributive over addition. In other words, the following "law" is bogus:
$$
(b+c)-a=(b-a)+(c-a) \, .
$$
For example, $(5+7)-2$ does not equal $(5-2)+(7-2)$, and $(x+y)-5$ does not equal $(x-5)+(y-5)$. So, if we are told that $2x+1=11$, then we are allowed to subtract $1$ from both sides:
$$
(2x+1)-1=11-1 \, .
$$
What we can't do is "simplify" the left-hand-side by writing $(2x+1)-1$ as $(2x-1)+(1-1)$.
