>> Hello, again.
This is Professor Shannon Gracie [phonetic]
on section 1.6 from the Blitzer Introductory
and Intermediate Algebra text.
Now we'll be discussing subtraction of real
numbers.
And the beauty of subtraction of real numbers
is that it's all the same as addition of real
numbers, except what you're doing is you're
changing that subtraction symbol to adding
the opposite.
So, when you're done with your homework, you
should be able to subtract real numbers, simplify
additions and subtractions, and then use the
definition of subtraction to identify terms,
and use subtraction definition to simplify
algebraic expressions, and we want to solve
problems involving subtraction.
So here is the warm-up.
Go ahead and pause the movie, and do the warm-up.
[ Silence ]
Alright, so let's see how you did.
So this will yield a result of one-half times
2x minus one-half times 7.
That's using which law?
Good, the distributive law or distributive
property.
And then we'll be adding to that three xs.
So the one-half times 2x just gives us 1x,
and then we'll have minus seven-halves plus
three xs.
And notice that we have like terms here, so
at the end of the day we will get four xs
minus seven-halves.
How did you do?
Good.
Okay, so for the second problem we will have
the opposite of negative x plus the opposite
of 5 -- that's distributing that negative
throughout.
And then I'm going to go ahead and do this
3 times 2 to get 6, and we will have the 5x
minus 1 remaining in the parentheses.
And notice I just distributed this minus throughout
the parentheses.
So the opposite of -- the opposite of x is
positive x, we will have minus 5, and then
distributing the 6 throughout we will get
plus 30 xs minus 6.
Now we've got like terms x, and then positive
30x, and remember that constants are always
like terms.
So we will have a result of 31 xs, and then
negative 5 minus 6 is really just like negative
5 plus negative 6, which we learned last time.
And that'll yield a result of 31x minus 11,
and we're all done.
Put your happy box, and we go from there.
Okay.
So for all real numbers a and b, a minus b
is equal to a plus the opposite or negative
b.
And like I said, we learned all those rules
for addition last time, so the beauty of this
section is you just get to hone your addition
of real numbers skills by -- and just, you
know, there's one little additional element
of changing a subtraction symbol to -- to
adding the opposite.
Okay?
So, to subtract b from a, add the opposite,
or additive inverse, of b to a.
The result of subtraction is called the difference.
[ Silence ]
Okay.
So there's a procedure one can use to subtract
real numbers.
So this procedure consists of changing the
subtraction operation to addition.
And remember, nothing in math is free, so
you have to change the sign of the number
being subtracted in order to pay for that.
And then your third step is just to add, using
the rules we learned in section 1.5.
So here we go.
[ Silence ]
Using the guidelines, this is equivalent to
negative 16 plus the opposite of negative
9.
So this minus was changed by adding the opposite.
Now this will yield a result of negative 16
plus positive 9, because the opposite of negative
9 -- this stuff here is positive 9.
And now using what we learned from before,
we'll have -- this is equivalent to negative
7.
Okay?
Okay, so next up -- and at -- if you want
to, actually you should pause the movie and
see if -- how you do on the next three problems.
[ Silence ]
Okay, let's see how you did.
So, this will be 16 plus negative 20.
So again, this minus was changed to plus a
negative.
So this would be negative 4, 'cause the absolute
value of negative 20 is 20, and we'll be subtracting
16 from that to get 4, and you attach the
negative sign, since that was attached to
the number with the larger absolute value.
Then number 3, we would get negative 6 plus
negative 32.
They've got the same sign, and again, you
know, this minus was changed into plus a negative.
So we are going to get 6 plus 32 is 38, they're
both negative so the negative stays there.
[ Silence ]
Alright.
Number 4.
[ Silence ]
Now actually if you think about it, 10 -- 10.2
minus 0.2, we can just do that regularly -- you
know how to do that regular subtraction.
So do you see that's going to be 10.0, and
then we'll do plus the opposite of negative
5.1.
So 10.2 minus 0.2 gave us 10.0, and then this
minus was changed into plus a negative.
So we'll have 10.0 plus a positive 5.1, which
yields a result of 15.1.
Okay.
So now the next example gets a little more
in-depth.
We're going to use all of our mad addition
and subtraction skills to simplify these problems,
okay?
So go ahead and pause the movie and do the
next two examples, and see -- or next two
items from example 2.
See how you do.
[ Silence ]
Okay, let's see how you did.
We will get -- so 3 times 4 gives us 12 xs.
I'm going to distribute this minus through,
so we will get minus 2 xs plus 21.
So the minus distributed because you've got
your finding -- do you see that minus a minus
is positive.
So now 12 xs minus 2 xs, so these here are
like terms, right?
So we're going to get 10 xs plus 21.
There you go.
[ Silence ]
Okay.
Next up we have -- notice these parentheses
around the x plus 5, they're not really doing
anything, except for kind of grouping this
x plus 5.
But this positive won't change anything.
So let's go ahead and remove those parentheses,
since they're not really doing us any -- any
favors there.
Now, I'm going to go ahead and distribute
this minus 2 throughout.
So we'll have minus 2 times x minus 2 times
negative 11, and then we'll have a minus 2
times 3y.
So this is our distributive law, where that
one is headed there.
So we've got -- actually we can combine these
two terms that have xs. 9x plus 1x is 10 xs
plus 5.
We've got minus 2x -- minus a minus is going
to be positive, so we're going to have plus
22, and then we'll have minus 6y.
Now do you see we have like terms for the
10x and the negative 2x.
So this will yield a result of 8 xs.
The 5 -- positive 5 and positive 22 are like
terms, so we'll have plus 27, and then minus
6y for our last term.
We're all set.
Okay?
So this section is a nice short section.
I bet you're going to be very, very sad that
you finished so quickly.
Okay.
So applications.
This line graph illustrates the value of Sarah's
[phonetic] car in dollars from the year 2001
to the year 2007.
So --
[ Silence ]
This graph, if you notice that in year 2001
-- so let me get the little hand going -- from
the year 2001 right here, she -- her -- her
car was valued at 24,000.
Well the next year she was at 22.5, etcetera.
So it's decreasing in value.
Alright?
So how much was Sarah's car worth in 2005?
So here, if you just kind of look, okay 2005,
let's see where I hit the graph.
Okay?
And so you -- here it is.
Her car -- actually I'm blocking the number
-- was valued at 14,500 dollars.
And then how much more was Sarah's car worth
in 2002?
So here in 2002, let's see what it was worth.
We said that in 2002, remember her car -- her
car -- in 2002 her car was worth -- we verbalized
this a minute ago.
Her car was worth 22,500.
So basically, to figure out how much more
her car was worth, we would say 22,500 minus
14,500.
And then we'll get, you know, 12 minus 4 is
8.
So do you see that answering the question,
we would say Sarah's car was worth 8,000 dollars
more in 2002.
And we're all set folks, have a fabulous day.
