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 ?

Solution 1:

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.

Solution 2:

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.

Solution 3:

$a(x+y) = (x+y) + (x+y) + (x+y) + \cdots{}$, $\quad a$ 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 $11$