Hot X: Algebra Exposed
Solution Guide for Chapter 5
Here are the solutions for the “Doing the Math” exercises in Hot X: Algebra Exposed!
DTM from p.59
(assume that all denominators ≠ 0)
2. Since they have the same denominator, we can just add these fractions by adding
across the top:
Answer:
x 2x
x + 2x
3x
. Since it’s already reduced, we’re done!
+
= !!
=
5 5
5
5
3x
5
3. Again, we can just add across the top, since the denominators are the same:
x 2x
x + 2x
3x
+
= !!
=
. Since it’s already reduced, we’re done!
y y
y
y
Answer:
3x
y
4. Since they have the same denominator, we can just subtract these fractions by
subtracting across the top, but let’s use parentheses so we distribute that negative
correctly. Remember, the numerator or denominators of fractions are understood to be
grouped together,, even though they don’t usually have parentheses surrounding them.
1 a +1
1 ! (a + 1)
1! a !1
1 ! a + (!1)
!a
!
=
!!=
=
=
b
b
b
b
b
b
Done! Answer: –
a
b
DTM from p.65
(assume that all denominators ≠ 0)
2. We want to rewrite these so they have the same denominator. What’s the LCD of these
fractions? In other words, what’s the LCM of their denominators, 2a and a? We could
use the birthday cake method, but it’s pretty easy to eyeball this one – the LCM is just 2a.
So what copycat should we multiply times
3
to rewrite it as an equivalent fraction, but
a
with the denominator of 2a? We should use
2
3
2 3
6
, and we get:
. Great,
= ! =
2
a
2 a
2a
now we can rewrite our original problem and solve:
1 3
1
6
1+ 6
7
+
=
+
(samedenominator, yay!) =
=
2a a
2a 2a
2a
2a
Answer:
7
2a
3. In this case, since the denominators have no common factors, the LCD will just be
their product: 3v. So how can we rewrite each of the fractions to have this denominator?
For the first fraction, we’ll use the copycat
v
1
v 1
v
, and we get: = ! =
. For the
v
3
v 3
3v
second fraction, we’ll rewrite the second fraction by multiplying by the copycat
get:
3
, so we
3
2
3 2
6
. Fabulous! Now we can rewrite our original problem with
= ! =
v
3 v
3v
fractions whose denominators are the same, and do the addition:
1 2
v
6
v+6
+
=
+
(samedenominator, yay!) =
3 v
3v 3v
3v
There are no common factors on the top and bottom, so it’s reduced. Done!
Answer:
v+6
3v
4. It might seem at first that these denominators both have a factor of “n” in them, but
1– n is a very different value from n! In fact, n and 1– n have no common factors, so the
LCD we’ll use will just be their product: n(1– n). That means for the first fraction, we’ll
use the copycat
1! n
in order to get that denominator:
1! n
1
1! n 1
(1 ! n) "1
1–n
=
" =
=
n
1! n n
(1 ! n) " n
n(1 – n)
And for the second fraction in our original problem, we’ll use the copycat
1
n 1
=
"
1! n
n 1! n
=
n
, and we get:
n
n "1
n
=
n(1 ! n)
n(1 – n)
Read those slowly and make sure you follow what’s going on!
Great, now we have a common denominator, so we can rewrite our problem, and add the
two fractions together:
1
1
1! n
n
1! n + n
1
+
=
+
(samedenominator, yay!)=
=
n 1! n
n(1 ! n) n(1 ! n)
n(1 ! n)
n(1 – n)
We can also multiply the bottom with distribution to get
1
, but it’s not necessary.
n ! n2
Both answers are fine!
Answer:
1
1
or
n(1 – n)
n – n2
5. Let’s take the hint and rewrite x as a fraction, so our problem becomes:
x 1
! . So
1 y
what’s the LCD here? Well, the LCM of 1 and y is y, so that will be our denominator for
both fractions! We’ll use the copycat fraction
y
on the first fraction, and it becomes:
y
x
y x
xy
=
!
=
. Great! Now our fractions have a common denominator, and we can
1
y 1
y
1
xy 1
xy – 1
finish the problem: x ! !!!=!!! ! !!!!=!!!
y
y y
y
We can’t reduce this (don’t go thinking we could cancel those y’s – we can’t because
there is a separate term, 1, which doesn’t share a factor of y), so we’re done!
Answer:
xy – 1
y
6. Hm, monster problem. Okay, we’ll take this one step at a time. First job? To find the
LCD, which will be the LCM of 6b, 2a, and 4ab. As I talked about on p.25, we can do
this with the birthday cake method in two steps: First, we find the LCM of 6b and 2a,
take that answer, and find the LCM between that and the final, unused term, 4ab. So for
the 6b and 2a, we’ll write the common factor of 2 on the outside and we’re left with 3b
and a on the bottom, resulting in the LCM 6ab.
As you can see above, we’ll then take 6ab and our third, unused term, 4ab, and find their
LCM in a second “cake”, and the LCM is the stuff along the big “L” of this second cake:
ab ! 2 ! 3 ! 2 = 12ab.
Okay, now our challenge is to write each of the original fractions with this new
denominator, 12ab. Let’s figure out which copycats we need to use for each of them!
For the first fraction,
a
2a
, we should use the copycat
. Check it out:
6b
2a
a
2a a
2a 2
=
!
=
6b
2a 6b
12ab
That’s the denominator we wanted! Okay, for the second fraction,
denominator, we should use the copycat
1
, to have that same
2a
6b
1
6b 1
6b
. And we get:
.
=
!
=
6b
2a
6b 2a
12ab
Thirdly, to get the denominator 12ab on the fraction
c
, we’ll need to use the copycat
4ab
3
c
3 c
3c
, and we get:
.
= !
=
3
4ab
3 4ab
12ab
Phew! Now we get to do the easy part. Since they all have the same denominator, we just
get to add across. Here’s what our problem has transformed into:
a
1
c
2a 2
6b
3c
2a 2 + 6b + 3c
+
+
=
+
+
=
6b 2a 4ab
12ab 12ab 12ab
12ab
The terms in the numerator have no common factors, so there’s no way to factor it, and as
a result, there’s no way to reduce this fraction, so we’re done!
Answer:
2a 2 + 6b + 3c
12ab
7. Okay, let’s not get confused by these variables. What’s the LCM of 2(y – 1) and 2y?
Let’s see, the factors of 2(y – 1) are 2 & (y – 1). And the factors of 2y are 2 & y.
Make sense? Now, we could use the birthday cake method on this, but it would look a
little strange. Better to just eyeball it in this case. But with these variables, if you don’t
have much experience with finding LCMs, it can be confusing!
The important thing is that we rewrite the fractions of our problem so that they have the
same denominator, and then we’ll be allowed to add them together. That’s our goal. The
truth is, if we end up using a denominator that isn’t exactly the LCD, but if we are able to
write equivalent fractions and attain our goal, then we’ll still get the right answer; we’ll
just end up needing to reduce the answer afterwards. Not the end of the world!
Eyeballing it, LCM of 2(y – 1) and 2y is 2y(y – 1), but let’s say that we don’t see that. We
can always just take the two original denominators and multiply them together for our
new denominator. That’s what works when the original denominators don’t have any
common factors (as on p.63), and it also works any other time! So let’s just do that.
Our new denominator will be: 2y ! 2(y – 1) = 4y(y – 1). Secretly, you and I know that
we could use 2y(y – 1) as the common denominator, but let’s see how this goes. So, to
rewrite our first fraction,
7
, with the denominator 4y(y – 1), we’ll use the copycat
2(y ! 1)
2y
7
2y
7
14 y
=
"
=
, and we get:
.
2y
2(y ! 1)
2y 2(y ! 1)
4 y(y – 1)
Ok, for the second fraction,
1
2(y ! 1)
, we’ll use the copycat
, and we’ll get:
2y
2(y ! 1)
1
2(y ! 1) 1
2(y – 1)
=
"
=
.
2y
2(y ! 1) 2y
4 y(y – 1)
And now we add them together! Our original problem has become:
7
1
14y
2(y ! 1)
14y + 2(y ! 1)
+
=
+
(same denominator, yay!) =
2(y ! 1) 2y
4y(y ! 1) 4y(y ! 1)
4y(y ! 1)
Let’s distribute that 2, and continue:
14y + 2y ! 2
4y(y ! 1)
=
16y ! 2
4y(y ! 1)
Notice that we can factor the top by pulling out a factor of 2, and then wouldn’t you
know it, it will cancel with a factor of 2 on the bottom!
16y ! 2
2(8y ! 1)
=
=
4y(y ! 1)
2 " 2y(y ! 1)
8y – 1
2 y(y – 1)
Although it seems like we should be able to factor another 2 out of the top and bottom,
we can’t, because of that 1! We’re now done, but if we wanted, we could distribute the 2y
in the denominator. Both answers are fine.
Answer:
8y – 1
8y – 1
or
2 y2 – 2 y
2 y(y – 1)
DTM from p.68-9
(assume that all denominators ≠ 0)
2. In order to simplify this complex fraction, we’ll need to first make sure that both the
numerator and denominator are written as single fractions. So let’s simplify the
numerator, and pretend nothing else exists. So, to add
denominator of 3x, and we’ll use the copycats
3 x
+ , we’ll want a common
x 3
3
x
and , respectively, so we get:
3
x
3 x
3 3 x x
9 x2
9 + x2
+
=
! + !
=
+
(same denominator, yay!) =
x 3
3 x x 3
3x 3x
3x
9 + x2
Now our problem looks like this: 3x , and it’s ready for the means and extremes! (see
1
3x
9 + x2
p. 52 to review this method) Using the means and extremes shortcut, we get: 3x
=
1
3x
3x(9 + x 2 )
3x(9 + x 2 )
=
. Before we go distributing that 3x, let’s notice that we can
1! 3x
3x
cancel a factor of 3x from the top and bottom! Yep, we discovered a hiding kitty cat,
3x
.
3x
3x(9 + x 2 )
(9 + x 2 )
So now our answer has become:
=
= 9 + x2. Done!
3x
1
Answer: 9 + x2
3. Let’s simply the denominator by adding those little fractions together:
1 1
+ . We’ll
c d
want a common denominator of cd, so let’s use some copycats to make that happen:
1 1
d 1 c 1
d+c
+
=
! + !
=
c d
d c c d
cd
Great, and also taking the hint and writing the original numerator, 1, as
1
, our original
1
1
1! cd
cd
=
problem has become: 1 . Using means and extremes, we get:
. And
d+c
1! (d + c) d + c
cd
just because it’s nicer to write things in alphabetical order (and easier to check our
answers against the answer key!), we get
Answer:
cd
.
c+d
cd
c+d
4. We’re not even going to think about that final
3
until we simplify the big fraction!
4
Okay, let’s just look at just the numerator of the big fraction first, and let’s do the
subtraction:
3b b
. We’ll use a common denominator of 4a, so that means we need to
!
4a 2a
use the copycat fraction
2
on the second fraction to make this happen, so the subtraction
2
becomes:
3b b
3b 2 b
!
=
! "
4a 2a
4a 2 2a
=
3b 2b
3b ! 2b b
!
(samedenominator, yay!) =
=
4a 4a
4a
4a
Okay, that was the numerator of the original big fraction. The big fraction has now
b
become: 4a . That’s looking a bit better. Now we can use the means and extremes on it,
b
3a
and we get:
b ! 3a
3ab
3
=
= .
4a ! b
4ab
4
3b b
!
4a
2a ! 3
So now our entire problem has transformed like this:
b
4
3a
How nice!
Answer: 0
=
3 3
= 0.
!
4 4