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 ?
 A: What you do is to subtract or divide each side as a whole by the same thing. Immediately after you do that you have one of
$$ (2x + 1) - 1 = 11 - 1 \qquad\qquad\text{or}\qquad\qquad (2x+1)\div{2} = 11\div 2 $$
So far things look pretty similar. The difference is in what happens next, which is to say the simplification you do after you have dealt with the "same operation on each side" step.
On the left-hand sides the "subtract one" operation combines differently with an addition than "divide by two" does.
In one case you have two piles of things and want to take one thing away from the whole. You can take that from either of the piles, but only take of from one of them.
In the other case you have two piles of things and want to cut the entire configuration in two identical halves. Then you need to cut each pile separately -- otherwise if you put the "halves" together again you'll suddenly have two of the pile you didn't halve.
A: This is not a question about solving equations. What you want to know is why
$$(a+b)+c=a+(b+c)$$
instead of
$$(a+b)+c=(a+c)+(b+c)$$
and why
$$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$$
instead of
$$\frac{a+b}{c}=a+\frac{b}{c}.$$
There are different types of answers to this question. From a mathematically point of view you could argue that the field axioms force it that way (the first one is called associativity of addition, the second one distributivity of multiplication). You could also think about the interpretation of addition and division in real life. Then you would realize that it makes sense.
A: $a(x+y) = (x+y) + (x+y) + (x+y) + \cdots{}$, $\quad a$ times
so regrouping, $x + x + x + \cdots{}$, $\quad a$ times, ${}+ y +y + y \cdots{}$ $\quad a$ times, so $a(x+y) = ax+ay$
so distributive property works, and you can also reverse it
so $(5+15)/5 = 5(1+3)/5$
if you have $12/3$ thats $(4 \times 3)/3$ for example, so if you cancel $3$s, you get $4$, which is the same thing as dividing $12/3$ and getting $4$
so then $5(1+3)/5 = 1 + 3 = 4$
this is also
$5/5 + 15/5 = 1 + 3 = 4$
If $x + 1 = 11$, then that's in language just "what number needs $1$ added to it to get $11$?", and if you were to add $1$ to $10$, you would get $11$, so you're really just subtracting the one from the $11$
A: 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 $(2x-1)+(1-1)$.
