WEBVTT

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We're now on problem
number seven.

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Problem number seven says x
equals 1/2, or if x equals

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1/2, what is the value of 1 over
z plus 1 over x minus 1?

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So 1 over x, well that's just
1 over 1/2 plus 1 over--

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what's x minus 1?

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What's 1/2 minus 1?

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What's minus 1/2, right?

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So this is x minus
1 is minus 1/2.

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And you could evaluate it.

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Or you could immediately see
that these would cancel out,

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because you could take this
minus and put it right here

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and make this a plus.

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Or you could evaluate it and say
well, this is equal to 2

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minus 2 and that equals 0.

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So this is just kind of a speed
question to see how fast

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you can do it.

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How much time do you
waste on this?

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OK, problem number eight.

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I'll do it in green.

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Let's see, let me draw that.

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So we've got a coordinate
axis.

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Oh, my god.

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Edit, undo.

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It's a coordinate axis.

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I didn't have my line tool on.

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And then that's the x-axis.

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This is the y-axis.

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Then let me draw some
points here.

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I have the point right here.

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They say this is 1, 0.

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Let's see, x-axis.

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That's the y-axis.

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So then I still use
the line tool.

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They have this going straight
up like this.

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It goes like that.

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Right angle there.

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Right angle there.

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This is point s.

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This is point t.

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This is point r.

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They say in the figure above
r, s is equal to s, t.

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That's just saying that their
lengths are equal.

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And the coordinate of s,
right here, is k, 3.

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So what does that tell us?

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That means that the x value,
right here, this point

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right here is k.

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And that this y value, right
here, is 3, right? k, 3.

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That's not t, 3.

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I'm just saying the y value
right here is 3.

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What is the value of k?

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Well we know that this length
and this length

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are the same, right?

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What is this length?

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Actually, that was a very
ugly curly bracket.

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But we know the y value
here is 3, right?

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This intersects the y,
intercepts at 3 right here, so

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we know that distance is 3.

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We know that this
distance is 3.

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That's also 3.

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And we also know that this is
equal to this, so we also know

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that this distance is 3.

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And if that distance is 3, we
know that this distance is 3.

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But we're 3 to the left
of the y-axis, right?

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We've gone 3 units in the
negative direction.

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So the value of k, or the
x-coordinate here would be

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negative 3 because we
went 3 to the left.

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If it was out here it'd be
positive 3, but since it's

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here it's negative 3,
and that's choice A.

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Next question, number nine.

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OK, let's see.

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So they drew a table and
I'll just write it the

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way they did it.

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So they did x and f of x.

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And then they say 0, 1,
2, 3, 1, 2, 5, 10.

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The table above gives the
values of the quadratic

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function f for the selected
values of x.

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Which of the following
defines f?

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So it's a quadratic function, so
we know it's going to have

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the form-- f of x is going to be
something like x squared--

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well it could be ax squared
plus bx plus c.

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My suspicion though is it's
going to be something fairly

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you-- simple.

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So let's see.

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The way I would do it is I would
play around with it.

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So if I squared this x value and
I still end up with a 1--

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and this is actually the biggest
clue right here,

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because when x is
0, f of x is 1.

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So we know that this constant
term is going to be 1.

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It's going to be something plus
1 because when x is 0,

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these things are equal to 0,
this ax squared plus bx.

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And we still have
it equal to 1.

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So we know c is 1.

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So we know that already,
that's ax squared

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plus bx plus 1.

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And then let's see.

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We know that we have a
plus 1 here, right?

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So we know that this whole term
is-- you can put it in

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here, but when you get a plus
1-- well I'm explaining it in

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a very confusing way.

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But your brain might just say
hey, if I square this and add

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this plus 1, I get 2.

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If I square this and
I add 1, I get 5.

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If I square 3 and I
add 1, I get 10.

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And you would say well, f
of x must just equal x

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squared plus 1.

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If your brain doesn't
just stumble on

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that-- although it should.

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You should probably just say
well it's probably just

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something simple like I
squared and I add 1 or

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subtract one, because they're
never going to give you

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something really complicated.

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But if they did, or I guess if
you're not doing this on the

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SAT, hopefully you see how I
immediately got that the y

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intercept is 1, because when
x is 0, f of x is 1.

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And then you could have
said well, when x is

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1, f of x is 2.

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So you could have substituted
here.

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You could have said 2 is
equal to ax-- sorry.

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You could say 2 is equal to a
times 1 squared plus b times 1

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plus 1, right?

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Because I just substituted 1 for
x, and then I put f of x

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is equal to 2.

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And then you would get-- if
you subtract 1 from both

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sides, you get 1 is equal
to a plus d, right?

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And you know on the SAT they're
not going to give some

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crazy fractional thing.

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And I'm actually hesitant
to even go in this whole

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direction, because
I think it's just

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overcomplicating the thing.

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Because you could then do the
same thing with 2 and 5, and

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then get a system of equations,

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et cetera, et cetera.

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And you would have wasted
a lot of time.

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The big discovery here is that
most of these on the SAT are

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going to be really
simple equations.

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And frankly, more than doing
this-- this is just multiple

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choice, I just realized-- you
could just try out the

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equations that they give you.

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And it's lucky that the
first one works.

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Actually, you should just
try out all of them and

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see which ones work.

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Next problem.

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Write me a message if you found
that last explanation a

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little confusing.

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I just didn't want to show you
the correct way to do it,

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because that would take you
forever and it's not really

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the correct way to
do it on the SAT.

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Problem number ten.

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How old was a person exactly
one year ago if, exactly x

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years ago, the person
was y years old.

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So this is just something
to confuse you.

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So x years ago the person
was y years old.

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So let's put it this way.

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Let me think of the best
way to write this.

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So let's write their
current age.

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Well let's just write
it as equal to a.

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So we know a couple of things.

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We know that a minus x is equal
to y, or we know that

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their current age is equal
to x plus y, right?

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This is their current age.

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And what do we want
to figure out?

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We want to figure out their
age 1 year ago.

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So we want to figure out their
current age minus 1.

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So the current age minus 1?

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Well their current
age is x plus y.

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So you just substitute that back
in, and you get x plus y

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minus 1 is how old they
were 1 year ago.

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So x plus y minus 1,
that's choice E.

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Next problem.

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Actually I only have a minute
left, so I'll do the next

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problem in the next video.

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I'll see