We're now on problem
number seven.

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

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

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

what's x minus 1?

What's 1/2 minus 1?

What's minus 1/2, right?

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

And you could evaluate it.

Or you could immediately see
that these would cancel out,

because you could take this
minus and put it right here

and make this a plus.

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

minus 2 and that equals 0.

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

you can do it.

How much time do you
waste on this?

OK, problem number eight.

I'll do it in green.

Let's see, let me draw that.

So we've got a coordinate
axis.

Oh, my god.

Edit, undo.

It's a coordinate axis.

I didn't have my line tool on.

And then that's the x-axis.

This is the y-axis.

Then let me draw some
points here.

I have the point right here.

They say this is 1, 0.

Let's see, x-axis.

That's the y-axis.

So then I still use
the line tool.

They have this going straight
up like this.

It goes like that.

Right angle there.

Right angle there.

This is point s.

This is point t.

This is point r.

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

That's just saying that their
lengths are equal.

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

So what does that tell us?

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

right here is k.

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

That's not t, 3.

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

What is the value of k?

Well we know that this length
and this length

are the same, right?

What is this length?

Actually, that was a very
ugly curly bracket.

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

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

we know that distance is 3.

We know that this
distance is 3.

That's also 3.

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

that this distance is 3.

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

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

We've gone 3 units in the
negative direction.

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

negative 3 because we
went 3 to the left.

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

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

Next question, number nine.

OK, let's see.

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

way they did it.

So they did x and f of x.

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

The table above gives the
values of the quadratic

function f for the selected
values of x.

Which of the following
defines f?

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

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

well it could be ax squared
plus bx plus c.

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

you-- simple.

So let's see.

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

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

and this is actually the biggest
clue right here,

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

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

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

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

And we still have
it equal to 1.

So we know c is 1.

So we know that already,
that's ax squared

plus bx plus 1.

And then let's see.

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

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

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

a very confusing way.

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

this plus 1, I get 2.

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

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

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

squared plus 1.

If your brain doesn't
just stumble on

that-- although it should.

You should probably just say
well it's probably just

something simple like I
squared and I add 1 or

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

something really complicated.

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

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

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

And then you could have
said well, when x is

1, f of x is 2.

So you could have substituted
here.

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

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

plus 1, right?

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

is equal to 2.

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

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

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

crazy fractional thing.

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

direction, because
I think it's just

overcomplicating the thing.

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

then get a system of equations,

et cetera, et cetera.

And you would have wasted
a lot of time.

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

going to be really
simple equations.

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

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

equations that they give you.

And it's lucky that the
first one works.

Actually, you should just
try out all of them and

see which ones work.

Next problem.

Write me a message if you found
that last explanation a

little confusing.

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

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

the correct way to
do it on the SAT.

Problem number ten.

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

years ago, the person
was y years old.

So this is just something
to confuse you.

So x years ago the person
was y years old.

So let's put it this way.

Let me think of the best
way to write this.

So let's write their
current age.

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

So we know a couple of things.

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

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

This is their current age.

And what do we want
to figure out?

We want to figure out their
age 1 year ago.

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

So the current age minus 1?

Well their current
age is x plus y.

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

minus 1 is how old they
were 1 year ago.

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

Next problem.

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

problem in the next video.

I'll see