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SAT Prep: Test 4 Section 2 Part 2

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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
Title:
SAT Prep: Test 4 Section 2 Part 2
Description:

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Video Language:
English
Team:
Khan Academy
Duration:
08:52

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