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 Post subject: Please explain #65,71, 76, 81, 83, 86, 91, 93 in Book IVPosted: Sun Mar 04, 2012 5:30 pm

Joined: Thu Jul 28, 2011 1:49 pm
Posts: 1
Please explain #65,71, 76, 81, 83, 86, 91, 93 in Book IV

I looked over the solutions in the back of the book but still have trouble understanding the solutions. Are there alternative ways to explain these?

Thanks

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 Post subject: Re: Please explain #65,71, 76, 81, 83, 86, 91, 93 in Book IVPosted: Mon Mar 05, 2012 5:28 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
#65.
This one's posted in the forums. If you need further explanation, however,
feel free to let us know.
It's actually posted in two different places, with two different
explanations:
"http://www.veritasprep.com/community/viewtopic.php?f=20&t=2747&p=3621&hilit=integer+p+for+10+p+10+p#p3621"
and
"http://www.veritasprep.com/community/viewtopic.php?f=19&t=11974&p=19860&hilit=integer+p+for+10^p#p19860"

#71.
This one's also posted on the forums:
"http://www.veritasprep.com/community/viewtopic.php?f=20&t=11652&p=19513&hilit=maximum+value+of+x+y#p19513"

#76
I can't find this one posted - so here you go:

Let's look at some values here.
We have x/y and x/(y+1)

If x is positive, then as we increase the denominator, we decrease the
value. (think of 2/3 versus 2/4 and 2/5 and 2/100... We keep getting
smaller, though we keep the numerator the same.)
If x is negative, we have
-2/3 and then -2/4 and then -2/5 and then -2/100. These are actually
getting bigger, so this one goes the other direction. (they're -.66, -.5,
-.4, -.02, etc...)

For this reason, we NEED to know whether x is positive or negative in
order to answer this one, and neither statement tells us this, so we don't
have sufficient information to answer this.

#81.
Remember for this one that x^2 - y^2 = (x+y)(x-y)
This means we can evaluate the right half of this instead of the left half.
To know whether the whole thing is positive or negative, we need to know
the sign (positive or negative) of both x+y and x-y.
(1) gives us part of this, and (2) gives us the other part, so together we
have enough information, and our answer is C.

#83.
http://www.veritasprep.com/homework-que ... problem=83
This one is solved on video for you. You'll have to be logged into the
Veritas site to access this, though.

#86.
http://www.veritasprep.com/homework-que ... problem=86
ALso on video - be sure to log in first to access this. (You can get to
this section by going to "Live Instructor Help" and then the last
paragraph on the following screen that says "You can also access our
library of recorded homework help session organized by book and problem
number INSTANTLY. Access Recorded Homework Help Video Library"

#91.
This one is posted in the forum:
viewtopic.php?f=21&t=4088&hilit=absolute+value

#93
This one is a little trickier. We have to combine two inequalities to see
if we can get to the positive or negative sign of both x and y.

We know that neither statement is sufficient alone, because both have two
variables included.

Let's look at them.
x - y > -5
x - 2y< -7

We can't add inequalities if the signs are pointing different ways, so
we'll rearrange.
x - 2y< -7
x < -7 + 2y
x + 7 < 2y
subtract x from both sides
7 < 2y - x
Then turn this one around so that we have the 7 on the right.
2y - x > 7

Now we have
2y - x > 7 and
x - y > -5
We can add these two together:
The x's cancel, and the 2y and -y add together to make just y.
y > 2
This means y is positive.
Then look at this inequality:

2y is "something greater than 4" since y is greater than 2.
so we have
x - 2y< -7
Let's plug in 5 for 2y, just to see what happens
x - 5 < -7
x < -2
In this case, x is negative.
Now let's plug in 10 for 2y to see what happens
x - 2y< -7
x - 10 < -7
x < 3

In this case, x might be positive OR negative.
Since we can have BOTH values for x, we don't have enough information, and

Veritas Help

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