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We're on problem 38.
Which of the following best describes the graph of this
system of equations?
OK, so maybe they're the same line.
Maybe they're parallel.
Maybe they only intersect in one point-- two lines
intersecting in only two points.
Well that's impossible.
Two lines, I mean that can happen with curves, but that's
not going to happen with lines.
So we can already cancel out choice D.
OK, now let's look at these two.
See I have a y here and I have a 5y here.
Let's multiply this top equation times 5 and see what
it looks like.
So if you multiply the left-hand side
by 5, you get 5y.
I'll do it up here.
You get 5y is equal to-- 5 times minus 2 is minus 10x,
plus 5 times 3 is 15.
So if you multiply the top equation-- both sides of it--
by 5 and it really doesn't fundamentally change the line,
the equation might look different, but the equality
will still hold in the same universe, which is
essentially that line.
So if you just multiply both sides by 5, they become the
same equation.
5y is equal to minus 10x plus 15.
So they are the same lines.
So that's A, two identical lines.
Problem 39.
And they want us to simplify 5x to the third over 10x to
the seventh.
So the easiest way to think of this, or at least for me--
well, there's a lot of ways you can do it and
we'll do it both ways.
This is the same thing as 5/10 times x to the third times x
to the minus 7.
1 over x to the 7 is the same thing as x to the minus 7.
And this is equal to-- 5/10 is 1/2.
And then here, we have the same base and we're
multiplying, so we can add the exponents.
3 plus minus 7 is minus 4.
So x to the minus fourth power.
And we could write that as 1/2 times 1 over x to the fourth
or 1 over 2x to the fourth.
And that is choice B.
Now you could have done it other ways.
You could have said, OK, let's see.
Divide the numerator and denominator by 5.
So this would have been 1.
This would have been a 2.
And you say, OK, let's divide the numerator and denominator
by x to the third.
So this will become a 1.
And x to the seventh divided by x to the
third is x to the fourth.
You could have done it that way.
You had 1 over 2x to the fourth.
Either way.
Or you could have even said-- you didn't have
to go to this step.
You could have said, OK, when I'm dividing with the same
base, I can just subtract the exponents.
So 3 minus 7 was minus 4.
Either way.
All of them would have valid ways to approach this problem.
Problem 40.
This looks like a simplification.
They write 4x squared minus 2x plus 8, minus x squared plus
3x minus 2 is equal to.
So the key here is just to realize that this is a minus.
So you could kind of view it as a plus minus 1 times this
whole thing.
So we're just going to have to distribute that out.
So this is equal to 4x squared minus 2x plus 8.
And now we distribute this minus over this whole
expression.
So minus times x squared is minus x squared.
Minus times 3x, positive 3x.
So it's minus 3x.
The minus 1 times negative 2.
Well now they cancel out and you get a plus 2.
We switch the sign on everything here because
they're all being multiplied by this negative 1.
OK, now we can simplify.
So let's take the x squared terms first. So we have a 4x
squared, we have a minus x squared.
So 4x squared minus x squared is 3x squared.
4 minus 1 is 3.
Then let's do our x terms. We have a minus 2x, we
have a minus 3x.
So minus 2 minus 3, that's a minus 5x.
And then last we have our constants.
We have 8 plus 2.
8 plus 2 is 10.
So 3x squared minus 5x plus 10.
And that is choice D.
Problem 41.
OK.
They say the sum of two binomial-- let
me copy this one.
It's interesting.
The sum of two binomials is 5x squared minus 6x.
So a binomial is just a polynomial with two terms. If
one of the binomials is 3x squared minus 2x, what is the
other binomial?
So this binomial is one of them, so they're saying 3x
squared minus 2x, and when you add that to some other
binomial-- and I don't know, let me just write that as A.
I mean there is no constant term here and there is no
constant term here, so I'm assuming that my-- and it has
to be a binomial.
There's only two terms. So I'm assuming my two terms are an x
squared term and x term because that's the only terms
that are involved in both of these.
So let's say my binomial is Ax squared plus Bx.
This is the mystery binomial.
And their sum is equal to this up here.
Is equal to 5x squared minus 6x.
Now let's see what we can do.
Well this is a plus here, so the parenthesis
really don't matter.
We can rearrange this as 3x squared plus Ax squared minus
2x plus Bx is equal to 5x squared minus 6x.
3 plus A.
3x squared plus Ax squared, that's the same thing as 3
plus A, x squared.
And then, minus 2x plus Bx, or we could switch them around.
That's the same thing as plus B minus 2-- I just took the
coefficients and added them together-- x.
I switched them, but we could have written this in the other
order to begin with-- is equal to 5x squared minus 6x.
And now you just compare.
OK, 3 plus A-- if you just look at the x squared terms--
3 plus A has to be equal to 5.
Because that's the coefficient on the x squared term.
So 3 plus A is equal to 5.
Subtract 3 from both sides.
You get A is equal to 2.
And then we have B minus 2 has to be the coefficient on x
here, so it has to be equal to minus 6.
Add 2 to both sides, you get B.
Minus 6 plus 2 is 4.
So the other binomial, just substituting that Ax squared
plus Bx, is 2x squared plus Bx.
Oh, sorry.
This is a minus 4.
Minus 6 plus 2 is minus 4.
So plus Bx.
So minus 4-- that's B-- x.
And that is choice A.
Next problem.
OK, they say, which of the following expressions is equal
to-- this is problem 42.
And they write x plus 2, plus x minus 2, times 2x plus 1.
So we have to simplify this.
And remember, order of operations, multiplication
comes first. So we have to multiply these two expressions
first. So let's do that.
So this is-- I'll rewrite this one over here.
x plus 2 plus-- and now let's multiply this.
When you multiply these two binomials, you're really just
doing the distributive property twice.
And let me show that to you.
We could view this as x minus 2 times 2x plus x
minus 2 plus 1.
So I'm just distributing the x minus 2 times each of these
terms. So I could write this as x minus 2 times 2x, plus x
minus 2 times the 1.
All right, and now we can just simplify that by doing the
distributive property again.
So this is x plus 2 plus-- let's distribute the 2x times
each of these.
2x times x is 2x squared.
2x times minus 2 is minus 4x.
Plus, well, we're disturbing a 1.
1 times anything is just itself.
So plus x minus 2.
And let's see what we can do.
We only have 1x squared terms, so let's write that down.
2x squared.
So 2x squared.
And then our x terms, we have a plus x, a minus
4x, and a plus x.
So we have 1 minus 4 is minus 3.
Plus 1 is minus 2.
So its minus 2x.
And then, let's see.
We have a positive 2 and a minus 2.
They cancel out.
So we're left with 2x squared minus 2x, and that's choice A.
Problem 43, I think we can fit in here.
Let me copy and paste it.
OK, copy it and now pasting it.
OK, it says, a volleyball court is
shaped like a rectangle.
Let me draw that.
Well, I didn't want to draw it filled in like
that, but fair enough.
Shaped like a rectangle.
It has a width of x meters and a length of 2x meters.
So it's width is x.
Let me write, this could be x and this would be 2x.
Because this is longer.
Which expression gives the area of the
court in square meters?
Well the area is just the width times the length.
So it's just x times 2x, which is equal to 2x squared.
That's the same thing as 2 times, x times x, which is the
same thing as 2x squared.
And that's choice B.
Anyway, see you in the next video.