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.
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