Math 60-Algebraic Properties

Instructor: Hi class.
Today we're going to take a look at the properties of algebra
and how to use those in order to simplify expressions.
There are three main properties of algebra
that are going to be very important for us,
so we're going to take a look at them.
They are actually things that you may have noticed before
but didn't realize they had a fancy name for them.
So we're going to give them that name.
The first one is the 'commutative property'
And with the commutative property
you want to think of having a community of addition
or you could have a community of multiplication.
Both of the addition and the multiplication operations
do allow this property.
And if you were in the community of addition
and all you have is addition, commutative property says
that you can move around wherever you would like.
For instance, if you had 5 plus 3 [ 5 + 3 ]
that would be the same
as if you chose to do 3 plus 5. [ 3 + 5 ]
So we were able to change the order and move those numbers around
within the community of addition and still get the same answer.
Either way you'll get the answer of 8.
Similarly if you were multiplying
it doesn't matter what order you multiply in,
so you can move around wherever you want
as long as the only thing you have is multiplication.
So for instance if I have 4 times 6 [ 4 · 6 ]
I'm going to get the same answer as if I do 6 times 4 [ 6 · 4 ].
Both of those are the commutative property.
The associative property
is the next one we're going to take a look at,
and that is the one
where you want to think of re-associating parentheses.
And again, this property right here will only work
if you are in only addition, or only multiplication.
For instance, if I have 5 plus 3 in parentheses, ( 5 + 3 )
and then I'm going to add 4 to it. [(5 + 3) + 4]
I will get the same answer as if I re-associate those parentheses
and say, don't want to add 3 and 5 first. [ not: (5 + 3) + 4 ]
So I decide to add 3 and 4 first. [ 5 + (3 + 4) ]
So I don't change the order of the numbers,
but I do change where my parentheses are,
and I re-associate it with two of the other numbers.
[5 + (3 + 4)] And I can only do this
because the only operation I have is addition.
Notice, that if we take a look over here on the left-hand side
and we first add our 5 and 3, I get 8, [ ( 5 + 3 ) = 8]
and when I add that to 4, I get 12. [ 8 + 4 = 12 ]
If I chose to do this over here on the right-hand side instead,
I should get the same answer.
So for instance here I would do
3 plus 4 first, to get 7, [ ( 3 + 4 ) = 7 ]
and then when I add 5 to that I get 12. [ 5 + 7 = 12]
Notice that either way we got the same answer.
Now this property also works
if I have multiplication instead of addition.
It only works as long as you're in one of those two operations.
So for instance let's say that I had
2 times 3 in parentheses, ( 2 · 3 )
and I was going to multiply that answer by 4.
[ ( 2 · 3 ) · 4 ] I'll get the same thing
as if I chose to re-associate my parentheses
and not change the order of the numbers,
but instead do 2 times 3 times 4. [ 2 · ( 3 · 4 ) ]
Having the 3 and the 4 in parentheses [ ( 3 · 4 ) ]
means that I would do those first.
So let's look at the left-hand side
and calculate what we get and show that it matches
what we're going to get on the right-hand side.
So over here, 2 times 3 is 6. [ ( 2 · 3 ) = 6 ]
When I multiply that by 4, I get 24. [ 6 · 4 = 24 ]
What you'll notice is when we do the right-hand side
we'll get the same answer.
We just went about it a different way.
So here we chose to multiply 3 and 4 together,
which gives us 12. [ ( 3 · 4 ) = 12 ]
When we multiply 2 and 12 [ 2 · 12 = 24] we still get 24.
So as long as all you have
is either addition or only multiplication
you can re-associate your parentheses and it's completely fine.
Now let's take a look at our third property.
This is probably one of our most important properties.
It's called the distributive property. [Video jumps.]
For instance, let's say that we have
2 times 5x plus 3. [ 2( 5x + 3 ) ]
In this situation when you have the number 5 here,
right next to the x, what you have is 5 times x [ 5 · x ]
And the x is waiting to be replaced by some value,
and we don't know what it is yet.
So because we don't know what it is,
order of operations would require us
to do this multiplication here first.
But since we don't know what x is,
we can't do this multiplication.
Since we can't do this multiplication,
we can't add the answer of 5 times whatever x is to the 3.
Which means we can't complete the inside of the parentheses.
Which would make you think you can't do anything to this problem.
But because of the distributive property, we can.
We can actually make this expression here
much simpler to calculate once somebody tells us what x is.
We do that by taking the value out front here, number 2,
Distribute your 2 times your 5x 2( 5x )
To get 2 times 5x we multiply the 2 times the 5
and end up with 10x. [ 2( 5x ) = 10x ]
Now we distribute our 2 times our 3 [ 2 ( 3 ) ]
and 2 times 3 is 6. [ 2 ( 3 ) = 6 ]
What we end up with, right here [ 10x + 6 ]
is an expression that is much easier to calculate
once you know what x is.
If you know what x is, you plug it in here for x,
then you multiply by 10,
and after you do that, you add 6,
which is a lot less work,
than if you were to plug it into this expression.
Because plugging it in to the original expression,
you have to multiply that value that we find out x is going to be,
times 5, then you add the 3, and finally you multiply by the 2.
Which is a lot more work
than plugging it in to this more simplified form.
in the cash only checkout line
but all you have is checks.
And you decide if your check is written out
by the time you get up to the counter
that no one behind you in the line will get mad
and maybe the cashier
will be just fine with the fact that you have a check.
And you happen to be buying 7 items.
And all of those items are selling for 5 dollars and 95 cents.
So in your head, since you don't have a calculator,
you need to figure out what 5 times,
7 times 5 dollars and 95 cents is. [ 7 ( 5.95 ) = ? ]
Well, let's think about the 5 dollars and 95 cents [ $5.95 ]
5 dollars and 95 cents [ $5.95 ] is the same thing
as 6 dollars minus a nickel. [ $6.00 - $0.05 ]
If you take that 7, since you have 7 items,
times the 6 dollars minus the nickel, [ 7( $6.00 - $0.05 ) ]
it's going to be much easier
than doing 7 times your $5.95. [ 7 · $5.95 ]
So, we're going to do 7 times 6, which is 42 dollars.
[ 7 · $6 = $42 ]
And then 7 times the nickel [ 7 · $0.05 = $0.35 ]
happens to be 35 cents.
So if you take 42 dollars and subtract 35 cents,
[ $42.00 - $0.35 = $41.65 ] you get 41 dollars and 65 cents.
So that's what you would write your check out for
and have it ready for the cashier when you get up to the counter.
And it's much easier to do the multiplication in this manner,
by breaking the $5.95 into 6 dollars minus a nickel,
than it is to try and just do the straight multiplication.
Many of you may have already done this in your head
when you're out shopping and not realized what you were doing
was using the distributive property.
Now let's take our algebraic properties
and use them for what we call 'simplifying expressions.'
Now, when we simplify expressions
one of the things that we're going to do
is we're going to apply a definition of subtraction,
and one thing that you may have noticed is,
when we were talking about adding and subtracting
positive and negative numbers,
if we had something like 5 minus 7, [ 5 - 7 ]
what we did was we changed this
so that it was the same thing
as 5 plus a negative 7. [ 5 + -7 ]
Changing the subtraction here so that it's adding the opposite
of whatever the value after it is.
That's going to be extremely important
when we simplify expressions.
Once you do this, you can just think about money and say,
well, this is like having 5 dollars and spending 7 dollars;
we're going to owe somebody 2 dollars, so the answer was -2.
And that's a process that we talked about in our video
on adding and subtracting positive and negative numbers.
Now we're going to be using that little fact
when we do what's called 'simplifying expressions.'
And we have three steps for simplifying expressions.
The first step is to change all of our subtraction
to adding the opposite.
And then, once we do that, we distribute if necessary.
So that's why it was so important for us
to learn about the distributive property.
And finally, our third thing we're going to do
is combine like terms.
Now let's take a look at a problem
where we might have to do these three steps.
For instance, we might have something like 3 times 2x minus 5,
plus 4 times 6x plus 1. [ 3( 2x - 5 ) + 4( 6x + 1 ) ]
The first thing we want to do is go through
and change our subtraction to adding the opposite.
Doing this one little step will help prevent many mistakes
that people make with positive and negative numbers.
So we're going to change this
to adding the opposite of positive 5, which is negative 5.
We now know that this 5 is a negative number.
So when we do our 3 times it, [ 3(... + -5) ]
we will end up with a negative number,
because positive times negative is negative.
So now that we've done this first step here,
let's go to the next step which is distribute.
And we have some distribution that has to happen here.
We also have some that has to happen right here.
So let's go ahead
and do those two distributive property situations.
So we have 3 times 2x which is 6x [ 3( 2x ) = 6x ]
3 times negative 5, which is a negative 15. [ 3(...+ -5 = -15 ]
We hit the end of the parentheses,
so we're done with distributing our 3.
So we're going to move over here and distribute our 4.
4 times 6x, [ 4( 6x ) ]
we multiply the numbers in front
and we get a positive 24x. [ 4( 6x ) = 24x ]
[ 4(...+ 1) = 4 ] 4 times 1 is 4.
Now, our distributive property is completely finished
and the parentheses are all gone.
So we have to look at what's left here.
We want to combine the like terms.
The like terms would be your x's are alike,
and then we also have our numbers which are alike.
And notice right now the way things are,
they're not next to each other.
So we can use the commutative property
to rearrange these two items in the middle,
which will make it so the x's are together,
and the numbers are together.
So we're going to just reverse the order of these two
using the commutative property
and we get 6x plus 24x [ 6x + 24x..]
plus negative 15 plus 4. [...+ (-15) + 4 ]
Now we can combine our like terms 6x and 24x.
We add the numbers in front; those are both x's.
So when you add 6 x's to 24 more of them,
you end up with 30 of them. [ 6x + 24x = 30x ]
So that would be 30x.
And then over here if you owe somebody 15 dollars,
and then you only have 4, you're still going to owe them 11.
And at the end here, rather than adding
doing plus a negative 11,
we go back to our subtraction and write it as minus 11.
And this here [ 30x - 11 ] would be our final answer.
We're not solving for anything because there's no equal sign.
We will be solving in the next chapter,
so there's no need to find out what x is.
No one has s actually told us what x is.
Let's take a look at another problem
where this first step up here
of changing the subtraction to adding the opposite
will prevent us from making a very common error.
So our next problem is going to be 5 times 2x plus 3
minus a 3 times 3x minus 4. [ 5( 2x + 3 ) - 3( 3x - 4 ) ]
For this particular problem,
since we have a subtraction right here,
it's going to be extremely important to do this first step.
Without changing this to adding a negative 3,
instead of minus 3 here,
many people will actually make a mistake
of distributing a positive 3 instead of a negative 3.
And that will make the sign of the value we get right here
the wrong sign, which will throw the rest of the problem off.
So it's very important to go through and change the subtraction,
which this is the first one,
to adding the opposite of the number after.
And this is a positive 3, so we make it a negative 3 [ + -3 ]
And then go over here and this is subtract 4,
so we're going to say plus,
and the opposite of positive 4 is negative 4.
Now we know which numbers are positive,
which numbers are negative.
When we distribute and we rearrange our numbers
the chances of losing our positive and negative signs
is going to be lessened for us.
So let's go back and distribute.
Here we're going to distribute 5.
5 times 2x is 10x. [ 5( 2x) = 10x ]
5 times 3 is 15. [ 5( ...+ 3) = 15 ]
We hit the end of the parentheses so we're done.
And so we're going to go over to the next one,
and negative 3 times 3x is a negative 9x. [ -3( 3x ) = -9x ]
Negative 3 times the negative 4 [ -3(...+ -4) = 12 ]
is a positive 12,
because negative times negative is positive.
In this particular problem most errors will happen right here
with the sign on this 12.
Now since everything is addition
we can rearrange items if we like,
and so let's use the commutative property
to reverse the order right here
so that the like terms are next to each other.
So these will get reversed, and we end up with 10x,
plus a negative 9x, plus 15, plus 12. [ 10x + -9x + 15 + 12 ]
Finally, combine our like terms.
These two are alike. [ 10x + -9x ]
10x and negative 9x makes 1x. [ 10x + -9x = 1x ]
We don't actually have to write the 1.
That's just an x.
And then 15 and 12 makes 27. [ 15 + 12 = 27 ]
Now let's take a look at one more problem.
It's a problem where there's a little trick
to making it so that you don't make
any little errors with your negatives.
This one is going to be 3 times 2x minus 7,
minus 4x plus 5. [ 3( 2x - 7 ) - ( 4x + 5 ) ]
Now, when we go through and change our subtraction,
the first one right here is minus 7.
That one's fairly easy; we're going to just do add,
and the opposite of positive 7 is negative 7. [ (...+ -7) ]
But when we get to this one right here,
when we change this to addition
there's no number right after it to change the sign of.
Whenever there's no number in front of a letter
or in front of a parentheses,
there's actually an invisible 1 there.
If it helps to insert the invisible 1,
by all means go ahead, and do that. [ ... -1( 4x + 5 ) ]
That will make it so you can go,
plus negative one [ ...+ -1( 4x + 5 ) ]
and you actually have a value there to change.
And that's a little trick
that will help you with problems like that
where there's no number in front of the parentheses,
or no number in front of a variable.
Now let's go back and distribute.
3 times 2x is 6x. [ 3( 2x ) = 6x ]
3 times negative 7 is negative 21. [ 3(.. -7) = -21 ]
Negative 1 times 4x is negative 4x. [ -1( 4x ) = -4x ]
And negative 1 times 5 is negative 5. [ -1(... -5) + -5 ]
Now we have finished distributing.
Let's rearrange things so our like terms are next to each other.
We'll do that by switching the order of these two
since we have just addition,
and we can use the commutative property.
This gives us 6x plus negative 4x, [ 6x + -4x ]
plus negative 21, plus negative 5. [.... + -21 + -5 ]
Combining the 6x and the negative 4x, [ 6x + -4x = 2x ]
we get 2x.
and then over here
negative 21 and negative 5 is negative 26. [...-21 + -5 = -26 ]
And we don't actually write plus negative 26.
We go back to subtraction and write minus 26.
And this here would be the final simplified expression.
[ 2x - 26 ]