Getting Rid of Parentheses

In algebra when you have a simplifying problem where there are a number of parentheses, it can look like a difficult problem, but it doesn't have to be, because a complicated problem is just several simple problems if you take it one step at a time. When you have several sets of parentheses inside each other just work from the inside out. Do as much simplifying inside a set of parentheses as you can, and when you can do no more, then get rid of the parentheses by using the distributive property. Then progress to the next set of parentheses out and do the same thing. Continue like this until all the parentheses are gone and all the like terms are combined.

It is also a good idea right from the beginning to change all subtractions to adding the opposite. If you can think of it this way without writing it, that is okay too, but whatever you do you always at least think of all the subtractions you see in algebra as addition of the opposite even if you don't write them that way. This is part of a general principle that you should always follow with respect to leaving out steps in algebra, which is that you should never leave out a step if you are not really doing it in your head. At the end of your problem you should write your addition of opposites back as subtraction, because it is shorter that way, so better form.

Another matter to pay attention to is variables without coefficients, like the first x in x+5x. Some people like to think of this as there being a missing or invisible 1 there, but that isn't really quite right, because it would seem to say that there is something wrong or incomplete about the expression, which is not true. x+5x is a perfectly proper algebraic expression and in simpler form than 1x+5x. The only problem with it is that it doesn't fit well into the pattern needed for combining like terms, because our rule for that requires a coefficient on each term. But this problem is easy to take care of, because 1 times any number is itself, so we can get it into that form by replacing x with 1x, which is legal because 1x=x. So basically the game is that whenever you need to for any purpose, you can write a 1 next to a variable, because 1 times anything is itself. You could even put a 1 next to the 5x, provided you separated it with a dot, so that it was clear that it meant multiplying by it rather than turning the 5 into 15. There just isn't any good reason to do so. At the end of you problem, though, you should always get rid of these 1's that you put in, because otherwise your answer isn't in simplest form. 1x isn't the best answer to any simplifying problem, because it is equal to x, and x is simpler. It doesn't do either to argue that it should matter because they are equal, because in a simplifying problem that could even be said about the original problem. So for example if you are simplifying 6x-5x and get 1x, that's fine, but it isn't finished, because you can simplify it even more by realizing that 1x=x, and if you ask why it should matter, because 1x and x are the same, then why not accept 6x-5x as the final answer, since that is equal to it too? 1x is a better answer than 6x-5x, of course, but it still isn't quite finished.

Another thing to keep in mind in general when doing algebra is that everything you do comes from a really quite small list of properties that must hold for all real numbers. Everything that you do when simplifying expressions of the type that I am talking about here comes from the following properties, and if it can't be justified by one of them, then it can't be done.

Properties of Addition

a+b=b+a
(commutative property)
(a+b)+c=a+(b+c)
(associative property)
a+0=a
a+-a=0

Properties of Multiplication

ab=ba
(commutative property)
(ab)c=a(bc)
(associative property)
1a=a
0a=0
-1a=-a

a

1 a

=1

Distributive Property

a(b+c)=ab+ac

Examples:

The instruction for these problems is to simplify the expression.

Example 1:

2x+3-x-1

Solution:

1. =2x+3+-x+-1
2. =2x+3+-1x+-1
3. =2x+-1x+3+-1
4. =1x+2
5. =x+2

Explanation:

In this problem there are no parentheses to deal with, so all we have to do is combine like terms.

1. We first write all of the subtractions as additions of the opposite so that we can use the properties of addition.
2. Then write -x as -1x to get the expression in the form where we can combine like terms.
3. By using the commutative and associative properties of addition we rearrange the order so that the like terms are together.
4. Now we combine like terms. For the variables this comes from the distributive property. For the numbers it is just arithmetic.
5. Then since 1 times anything is itself we can get rid of the 1.

Example 2:

2(y+3)

Solution:

=2y+6

Explanation:

Here we just get rid of the parentheses by using the distributive property.

Example 3:

z-(1-2z)

Solution:

1. =z+-(1+-2z)
2. =z+-1(1+-2z)
3. =z+-1+-1(-2z)
4. =z+-1+2z
5. =z+2z+-1
6. =1z+2z+-1
7. =3z+-1
8. =3z-1

Explanation:

1. First we write all of the subtractions as addition of the opposite.
2. When you have a minus sign before parentheses you can also use distributive property to get rid of the parentheses, but to do that you have to turn the minus sign into multiplying by -1. This is all right because when you multiply any number by -1 you will always get its opposite.
3. The we use the distributive property to get rid of the parentheses, multiplying the -1 by both the 1 and the -2z.
4. Notice here when we multiplying the -1 by the -2z we don't multiply it by both the -2 and the z, because here we are not using the distributive property, instead we are using the associative property to change the grouping in order to group the -2 with the -1 instead of with the z.
5. Now we use the commutative and associative properties of addition to get the like terms together.
6. Then we combine them,
7. and at the very end change the +-1 to subtraction to write it more compactly.

Example 4:

3(2x+y)-5(3x-2y)

Solution:

1. =3(2x+y)+-5(3x+-2y)
2. =6x+3y+-15x+10y
3. =6x+-15x+3y+-2y
4. =-9x+1y
5. =-9x+y

Explanation:

1. First write all subtractions as additions of the opposite.
2. Then we use the distributive property to get rid of the parentheses. We are also using the associative property of multiplication here in order to change the groupings so that we can get the numbers together to multiply them, changing for example 3(2x) to (3·2)x.
3. Then we rearrange the terms using the commutative and associative properties of addition to get the like terms together.
4. Then we combine like terms
5. and finally get rid of the 1. We can't change the -9x to a subtraction because it is the first term. We could if we rearranged the order and that would give us y-9x, which is in some sense simpler, but here there is a bit of a conflict of customs, because it is also customary to write the terms in alphabetical order, so either way would be acceptable.

Example 5:

3a+2[-a+5(5a+2)]

Solution:

1. =3a+2[-a+25a+10]
2. =3a+2[24a+10]
3. =3a+48a+20
4. =51a+20

Explanation:

This problem is a bit more complicated. We need to follow the instructions that I gave above with it. Remember, a complicated problem like this is just several simpler ones. What we do is work from the inside out, simplifying as much as we can in the innermost parentheses before working outward.

1. Going to the innermost parentheses, we use the distributive property to get rid of them.
2. Then we simplify as much as possible within those parentheses by combining like terms
3. and then get rid of them by using the distributive property.
4. Then combine like terms to get the final answer.

Example 6:

3{-2[(2b-3)-(3-b)]+3[4(b+1)-(2b-1)]}-6b

Solution:

1. =3{-2[(2b+-3)+-(3+-b)]+3[4(b+1)+-(2b+-1)]}+-6b
2. =3{-2[(2b+-3)+-1(3+-b)]+3[4(b+1)+-1(2b+-1)]}+-6b
3. =3{-2[2b+-3+-3+b]+3[4b+4+-2b+1]}+-6b
4. =3{-2[2b+1b+-3+-3]+3[4b+-2b+4+1]}+-6b
5. =3{-2[3b+-6]+3[2b+5]}+-6b
6. =3{-6b+12+6b+15}+-6b
7. =3{-6b+6b+15+12}+-6b
8. =3{27}+-6b
9. =81+-6b
10. =81-6b

Explanation:

This one looks pretty awful, but again it is the same procedure, just over and over again and you don't really know that you understand these sort of problems until you can feel confident that you could get through one no matter how complicated it was provided you had enough time. Just like before all we have to do is go to the innermost parentheses and keep working outward. If you are one of my students, don't panic, I won't give one this hard on an exam, but it is a good test to see if you really understand this stuff.

1. First we change all of the subtractions into additions of the opposite.
2. Then we change the minus signs in before parentheses into multiplication by -1.
3. Then we go to the innermost parentheses, and ignore the rest of the problem while we are working on them. Just copy everything else without paying any attention to what it is. The first set of parentheses isn't really necessary, so we just get rid of it for free. For the next one we use the distributive property to multiply each term by -1. In the next one we multiply each term by 4 and for the last one we multiply each term by -1.
4. Now inside the innermost sets of parentheses there are like terms to combine, so we put a 1 in front of the b and rearrange the terms to get them together
5. and combine them.
6. Then we get rid of those parentheses by using the distributive property
7. and again there are like terms to combine and we need to rearrange for that
8. and when we do that the b's cancel out because they add up to 0, so we are left with just the sum of the numbers, which is 27.
9. Finally we multiply that by 3
10. and change the adding of-6b to subtracting 6b and we are done.

Extra Tip:

All of these simplification are truths about all real numbers, so one interesting way you can check yourself on any of them is to substitute a couple of different numbers in for the variables and see if you get the same thing in the simplified version as in the original. Of course this will be true, because all along the way you are only using properties that are true for all real numbers. One fascinating thing about algebra is that the properties that you use for each step are pretty obvious. For example, it is pretty obvious that the order doesn't matter in multiplication and addition and even distributive property is something you use all of the time in arithmetic, but by combining these properties and applying them several times the way you do when simplifying expressions you come up with truths about all numbers that are not at all obvious.