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
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:
- =2x+3+-x+-1
- =2x+3+-1x+-1
- =2x+-1x+3+-1
- =1x+2
- =x+2
Explanation:
In this problem there are no parentheses
to deal with, so all we have to do is
combine like terms.
- We first write all of the subtractions
as additions of the opposite so
that we can use the properties of
addition.
- Then write -x as -1x to get the
expression in the form where we
can combine like terms.
- By using the commutative and associative
properties of addition we rearrange
the order so that the like terms
are together.
- Now we combine like terms. For
the variables this comes from the
distributive property. For the numbers
it is just arithmetic.
- 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:
- =z+-(1+-2z)
- =z+-1(1+-2z)
- =z+-1+-1(-2z)
- =z+-1+2z
- =z+2z+-1
- =1z+2z+-1
- =3z+-1
- =3z-1
Explanation:
- First we write all of the subtractions
as addition of the opposite.
- 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.
- The we use the distributive property
to get rid of the parentheses, multiplying
the -1 by both the 1 and the -2z.
- 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.
- Now we use the commutative and
associative properties of addition
to get the like terms together.
- Then we combine them,
- and at the very end change the
+-1 to subtraction to write it more
compactly.
Example 4:
3(2x+y)-5(3x-2y)
Solution:
- =3(2x+y)+-5(3x+-2y)
- =6x+3y+-15x+10y
- =6x+-15x+3y+-2y
- =-9x+1y
- =-9x+y
Explanation:
- First write all subtractions as
additions of the opposite.
- 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.
- Then we rearrange the terms using
the commutative and associative
properties of addition to get the
like terms together.
- Then we combine like terms
- 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:
- =3a+2[-a+25a+10]
- =3a+2[24a+10]
- =3a+48a+20
- =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.
- Going to the innermost parentheses,
we use the distributive property
to get rid of them.
- Then we simplify as much as possible
within those parentheses by combining
like terms
- and then get rid of them by using
the distributive property.
- 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:
- =3{-2[(2b+-3)+-(3+-b)]+3[4(b+1)+-(2b+-1)]}+-6b
- =3{-2[(2b+-3)+-1(3+-b)]+3[4(b+1)+-1(2b+-1)]}+-6b
- =3{-2[2b+-3+-3+b]+3[4b+4+-2b+1]}+-6b
- =3{-2[2b+1b+-3+-3]+3[4b+-2b+4+1]}+-6b
- =3{-2[3b+-6]+3[2b+5]}+-6b
- =3{-6b+12+6b+15}+-6b
- =3{-6b+6b+15+12}+-6b
- =3{27}+-6b
- =81+-6b
- =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.
- First we change all of the subtractions
into additions of the opposite.
- Then we change the minus signs
in before parentheses into multiplication
by -1.
- 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.
- 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
- and combine them.
- Then we get rid of those parentheses
by using the distributive property
- and again there are like terms
to combine and we need to rearrange
for that
- 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.
- Finally we multiply that by 3
- 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.
