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?

Question

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 ?

Answer 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.

Answer 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.

Answer 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 $11$

Answer 4

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 $(2x-1)+(1-1)$.

Answer 5

Bear with me.

Consider we start with $\color{red}5$. And we do two things: first we multiply by $2$ and second we add $1$. So $2\times\color{red}5 = \color{blue}{10}$ and $\color{blue}{10} + 1 = 11$. So we end up with with $2\times \color{red}5 + 1 = \color{blue}{10} + 1 = 11$.

Now consider we start with $\color{red}5$. And we do two things but in the opposite order: first we add $1$ and then we multiply by $2$. So $\color{red}5 + 1 = \color{green}6$ and $2\times \color{green}6 = 12$. So we end up with $2\times(\color{red}5 + 1) = 2\times \color{green}6 =12$.

We are the answers different. Well, because the order in which we do things matter and if we do things in a different order the results will be different. That's a no brainer, really. (Just think of shoes and socks, or baking and stirring, or walking through a doorway first and opening the door second)

SO the question becomes if we have something written as $2\times \color{red} 5 + 1$ which order did we do it in? Is it $2\times 5 =10$ and $10 + 1 = 11$ or is it $5+1 = 6$ and $2\times 6$? And the answer is simply: Multiplication first. Then addition. That's just how we mutually had a meeting and we all decided that is how it is going to be.

So if you have something like $2x + 1$ that is "first we had $x$, and then we added $1$". Now suppose we want to do something to it.

I'm going to go on a tangent. Suppose we want to raise it to the third power. $(2x + 1)^3$. That means "first we had $x$, and then we added $1$, and then the third thing is we raised it to the third power". Or we want to multiply by $7$. $7(2x + 1)$; that means "first we had $x$, and then we added $1$, and then the third thing is we is we multiplied by $7$".

Or what if we add $3$. $(2x + 1) + 3$; "first we had $x$, and then we added $1$, and then the third thing is we added $3$. Or we subtract $1$. $(2x+1) -1$; "first we had $x$, and then we added $1$, and then the third thing is we subtracted $1$".

Okay... so what?

Now notice. If we add $3$ that is doing the exact same type of thing. And if we subtract $1$ subtraction is the opposite "undoing" type of same thing as addition.

If we do the same thing or the opposite "undoing" type of the same thing one right after another we can combine them because they are the same type of thing. We can combine the same types of things !!!IF!!!! we do them one after another.

So $(2x + 1) + 3$ if we combine the adding $1$ with the adding $3$ we can combine them to "adding 1+3" is the same as "adding $4$" so: $(2x + 1) + 3 = 2x + 4$.

And same for subtracting $1$. So $(2x+1) -1 = 2x + 0 =2x$.

But we can't combine two of the same actions if the aren't done one after the other. We can't take $x$; multiply by $2$ to get $2x$; add $1$ to get $2x+1$; and then divide by $2$ to get $\frac {2x+1}2$ and expect the dividing by $2$ to cancel the multiplying by $2$. It doesn't work because the multiplication and division aren't done one after the other. There is an "and we add $1$" in between. So $\frac {2x + 1}2 \ne x+1$.

It might be worth noting the same would be true if we added $1$ first and then multiplied by $2$. We couldn't then subtract $1$ and cancel the adding $1$. That wouldn't work because we had an "and then we multiply by $2$" in between. So $2(x+1) - 1 \ne 2x $.

......

Okay. So why can we do something like $(2x + 1)\cdot \frac 12 = 2\cdot \frac 12 x + 1\cdot \frac 12 = x + \frac 12$?

Well, in general we shouldn't be able to undo something if it were done of few steps earlier. For example if we had raised to $x+2$ to the power of $3$ to get $(x+2)^3$ then there is absolutely nothing we can do next to the "$(x+2)^3$" to undo the $+2$ (without undoing the "raise to the third power" first). We just can't.

But we do have another rule. It's called the distributive rule (of multiplication over addition).

If we start with $5$ and then add $3$ to get $5+3$ and then we multiply thing by $2$ to get $2(5+3)$. That will be the exact same thing as if we had multiplied the $5$ by $2$ and the $3$ by $2$ and added them.

$2(5 + 3) = 2\times 5 + 2\times 3$.

In general this is not true $(5+3)^2 \ne 5^2 + 3^2$ and $(5\times 3) + 2 \ne (5+2)\times (3+2)$. But in the case of multiplying over a sum it is true.

$a(b+c) = a\cdot b + a\cdot c$. Always.

So using that rule we can do an "undo" via distribution.

$(2x + 1)\cdot \frac 12 = (2x)\cdot \frac 12 + 1\cdot \frac 12$ happens because of distribution.

......

Okay, that was that... now the rubdown.

In solving a problem such as

$2x + 1 = 11$ we usually want to "unwrap". We consider what we did and the order we did it in.

Start with $x$. Multiply by $2$ to get $2x$. Add $1$ to get $2x + 1$. And we unwrap it backwards.

$2x + 1 = 11$ unwrap by subtracting $1$

$(2x + 1) - 1 = 11-1=10$. As the adding and subtraction were done one after another we can combine them.

$2x + 10$. Now that we have undone the "add $1$", it's as though it had never happened. Now it is as though: We started with $x$. And we multiplied by $2$ to get $2x$. And that's all we did.

Now as it is as though the last (and only) thing we did was multiply by $2$, we can undo it by dividing by $2$.

$2x = 10$

$\frac {2x}2 = \frac{10}{2}$

$x = 5$.

Now, if we wanted to, we could attempt to undo the "multiply" by $2$ first by distribution. (We shouldn't) but we could.

$2x + 1 = 11$ so

$\frac 12(2x + 1) = \frac 12\cdot 11$.

We use the distributive rule $\frac 12(2x + 1) = \frac 12\cdot 2x +\frac 12 \cdot 1$ so

$x + \frac 12 = 5\frac 12$.

Now we undo the $+\frac 12$ by subtracting $\frac 12$

$(x+ \frac 12) -\frac 12 = (5\frac 12) -\frac 12$

$x = 5$.

......

And just to pound it down. If the addition and multiplication were done originally in the other order. We start with $x$. We add $1$: $x + 1$. And we multiply by $2$: $2(x+1)$ then we'd undo in that orde and we would just divide one, not two, terms.

$2(x+1) = 12$ Undo the "multiply by $2$" by dividing by $2$

$\frac {2\color{purple}{(x+1)}}2 = \frac {12} 2$.

$\color{purple}{(x+1)} = 6$. Now it's as though we had never multiplied by $2$. We undo the "add $1$" by subtracting $1$.

$(x+1)-1 = 6-1$

$x = 5$.