Simplifying using the Commutative Property of Addition

[coughsyrup78]

Hello,

I'm having trouble understanding how to use the commutative property of addition to simplify this expression.

4(x + 3) - 2(x - 1)

This is how the book solves the problem:

4(x + 3) - 2(x - 1)
= 4x + 12 - 2x + 2 <-- I don't understand why the sign changed from minus to plus during the distribution of 2
= 4x - 2x + 12 + 2 <-- how on earth did they reorder the expression when there is a subtraction in there? subtraction is not commutative or associative
= 2x + 14

This is how I work this problem:

4(x + 3) - 2(x - 1)
= 4(x) + 4(3) - 2(x) - 2(1) <-- use the distrubutive property
= 4x + 12 - 2x - 2
= 4x + 12 + (-2x) + (-2) <-- I don't know how to deal with the subtraction unless I convert it to addition
= 4x + (-2x) + 12 + (-2) <-- reorder using commutative property of addition
= 2x + 10 <-- my answer is incorrect

I tried converting the subtraction to addition as the first step but I was no nearer to getting the correct result.

Thanks for your help.

 

[Denis]

4(x + 3) - 2(x - 1)
= 4(x) + 4(3) - 2(x) - 2(1) <-- use the distrubutive property

Your last term should be -2(-1):
4(x) + 4(3) - 2(x) - 2(-1)

and -2(-1) = +2 : quite basic stuff!

 

[JeffM]

Hello,

I'm having trouble understanding how to use the commutative property of addition to simplify this expression.

4(x + 3) - 2(x - 1)

This is how the book solves the problem:

4(x + 3) - 2(x - 1)
= 4x + 12 - 2x + 2 <-- I don't understand why the sign changed from minus to plus during the distribution of 2 Because you are not distributing 2 but (-2). And (-2) * (-1) = 2.
= 4x - 2x + 12 + 2 <-- how on earth did they reorder the expression when there is a subtraction in there? subtraction is not commutative or associative
Do you really not see that that 12 - 2x = 12 + (-2x) = (-2x) + 12 = - 2x + 12? If that is where your problem is, let's discuss further.

= 2x + 14

This is how I work this problem:

4(x + 3) - 2(x - 1)
= 4(x) + 4(3) - 2(x) - 2(1) <-- use the distrubutive property And how did (-1) become (+1)?
= 4x + 12 - 2x - 2
= 4x + 12 + (-2x) + (-2) <-- I don't know how to deal with the subtraction unless I convert it to addition
= 4x + (-2x) + 12 + (-2) <-- reorder using commutative property of addition
= 2x + 10 <-- my answer is incorrect

I tried converting the subtraction to addition as the first step but I was no nearer to getting the correct result. That is because the conversion from subtraction to addition was not an error. Your error is in assigning a negative sign to the product of two negatives.
Thanks for your help.

Do my comments help? If not, we can take it step by step.

PS Congratulations on showing your work. You will get MUCH better help at this site if you continue to do it.

 

[coughsyrup78]

4(x + 3) - 2(x - 1)

My problem is that I can't convert the subtraction to addition in my head without getting my signs all jumbled up.

If I remove all of the subtraction at the very beginning, it makes it easier for me, although there are more steps. Something like this:

4(x + 3) - 2(x - 1) =
4(x + 3) + (-2)[x + (-1)] = <-- remove all subtraction signs 1st
4(x) + 4(3) + (-2)(x) + (-2)(-1) = <-- distribute
4x + 12 + (-2x) + 2 =
4x + (-2x) + 12 + 2 = <-- reorder
2x + 14 <-- seems to be correct

Assuming I did that right, can someone explain to me why the x above didn't also become negative when I converted to addition?
For example, why isn't it: 4(x + 3) - 2(x - 1) = 4(x + 3) + (-2)[-x + (-1)]

 

[Denis]

Assuming I did that right, can someone explain to me why the x above didn't also become negative when I converted to addition?
For example, why isn't it: 4(x + 3) - 2(x - 1) = 4(x + 3) + (-2)[-x + (-1)]

You seem to be "inventing" questions; are you a student attending math classes?

 

[JeffM]

4(x + 3) - 2(x - 1)

My problem is that I can't convert the subtraction to addition in my head without getting my signs all jumbled up.

If I remove all of the subtraction at the very beginning, it makes it easier for me, although there are more steps. Something like this:

4(x + 3) - 2(x - 1) =
4(x + 3) + (-2)[x + (-1)] = <-- remove all subtraction signs 1st
4(x) + 4(3) + (-2)(x) + (-2)(-1) = <-- distribute
4x + 12 + (-2x) + 2 =
4x + (-2x) + 12 + 2 = <-- reorder
2x + 14 <-- seems to be correct

Assuming I did that right, can someone explain to me why the x above didn't also become negative when I converted to addition?
For example, why isn't it: 4(x + 3) - 2(x - 1) = 4(x + 3) + (-2)[-x + (-1)]

You did it right.

The different uses of the + sign and - sign are a source of confusion to the beginner. + and - are used in at least three different senses in math. At a very fundamental level these senses are all coherent, but that coherence is not intuitive at first to the beginner. So be patient.

Additive inverses: For a given number a (which may be positive or negative) there is an ADDITIVE INVERSE indicated as (- a) such that a + (- a) = 0.

Operations: + means addition and - means subtraction.

The rule is that (a - b) = [a + (- b)] where the first minus sign means the operation of subtraction and the second minus sign means additive inverse. So your "removing the subtraction" is CORRECT. It may add a step but if it prevents mistakes, SO WHAT?

Now consider an expression like [a - (c * d)]. That is equal to {a + [-(c * d)]}, where [-(c * d)] is the additive inverse of (c * d). Now, in general, the additive inverse of (c * d) is NOT [(-c) * (-d)] because (c * d) + [(-c) * (-d)] = (c * d) + (c * d) = 2(c * d). The additive inverse of (c * d) = [(-c) * d] = [c * (-d)] = - (c * d).

So when "you remove the subtraction" from 4(x + 3) - 2(x - 1) = 4(x + 3) + {- [2(x - 1)]} you have a choice.
You can compute {- [2(x - 1)]} = (- 2) * (x - 1) = - 2x + 2, or you can compute {- [2(x - 1)]} = 2[- (x - 1)] = 2(- x + 1) = - 2x + 2.

Clear now?

 

[coughsyrup78]

Assuming I did that right, can someone explain to me why the x above didn't also become negative when I converted to addition?
For example, why isn't it: 4(x + 3) - 2(x - 1) = 4(x + 3) + (-2)[-x + (-1)]

You seem to be "inventing" questions; are you a student attending math classes?

Is that your way of saying it was a stupid question? :mrgreen:

 

[coughsyrup78]

You did it right.

The different uses of the + sign and - sign are a source of confusion to the beginner. + and - are used in at least three different senses in math. At a very fundamental level these senses are all coherent, but that coherence is not intuitive at first to the beginner. So be patient.

Additive inverses: For a given number a (which may be positive or negative) there is an ADDITIVE INVERSE indicated as (- a) such that a + (- a) = 0.

Operations: + means addition and - means subtraction.

The rule is that (a - b) = [a + (- b)] where the first minus sign means the operation of subtraction and the second minus sign means additive inverse. So your "removing the subtraction" is CORRECT. It may add a step but if it prevents mistakes, SO WHAT?

Now consider an expression like [a - (c * d)]. That is equal to {a + [-(c * d)]}, where [-(c * d)] is the additive inverse of (c * d). Now, in general, the additive inverse of (c * d) is NOT [(-c) * (-d)] because (c * d) + [(-c) * (-d)] = (c * d) + (c * d) = 2(c * d). The additive inverse of (c * d) = [(-c) * d] = [c * (-d)] = - (c * d).

So when "you remove the subtraction" from 4(x + 3) - 2(x - 1) = 4(x + 3) + {- [2(x - 1)]} you have a choice.
You can compute {- [2(x - 1)]} = (- 2) * (x - 1) = - 2x + 2, or you can compute {- [2(x - 1)]} = 2[- (x - 1)] = 2(- x + 1) = - 2x + 2.

Clear now?

Thanks JeffM for your thorough explanation. The more I do this kind of problem the more I can kind of do it in my head.

4(x  +  3)  -  2(x  -  1)
 |       |       |      |
 1       2       3      4           <-- I look at it in four steps
 |       |       |      |
4(x)   4(3)   -2(x)  -2(-1)        <-- 1, 2, 3, 4
 |       |       |      |
 4x   + 12  +  (-2x)   + 2          <-- stick a bunch of + signs in there

 2x + 14                             <-- reorder and simplify

 

[Denis]

Is that your way of saying it was a stupid question? :mrgreen:

Yes and no...hard for us to help "properly" if we don't know if the "student" is learning on his/her own,
or attending math classes with a math teacher...

 

[coughsyrup78]

Is that your way of saying it was a stupid question? :mrgreen:

Yes and no...hard for us to help "properly" if we don't know if the "student" is learning on his/her own,
or attending math classes with a math teacher...

In that case, I'm planning on taking algebra I and I need to brush up on my pre-algebra before I do (it's been a couple years since I took pre-algebra so I've forgotten most of it).

 

[JeffM]

The more I do this kind of problem the more I can kind of do it in my head.

This is generally true. The way to get a mathematical concept into your head is to do many problems that utilize the concept. After a while, it becomes automatic, and you wonder why it ever seemed so hard.

HOWEVER, doing stuff (particularly lots of stuff) in your head increases the probability of error, which is not zero even if you use pencil and paper. So it is a good idea to check your work if you have time to do so. In fact, I'd say that doing the blindingly obvious stuff in your head, using pencil and paper for all the rest, and then checking the answer is probably the most efficient use of time.

How do you check your work when simplifying expressions in one variable? Simple, just substitute a small prime number for the variable and evaluate all expressions in parentheses first, starting with inner parentheses and working outward. This extends to expressions in multiple variables, but use a different prime number for each variable.

4(x + 3) - 2(x - 1) = 2x + 14.
I like using 5 because it's a prime with an easy multiplication table.
4(5 + 3) - 2(5 - 1) = 4 * (8) - 2 * (4 ) = 32 - 8 = 24 = 10 + 14 = (2 * 5) + 14.

Please remember when using this site ALWAYS to show your work, identify where you are stuck, and give some clue as to where you are in your math education so that you will get the most precise, pertinent, and prompt help that people at this site can provide.