Simplifying Rational Expressions

7.3. SIMPLIFYING RATIONAL EXPRESSIONS
433
49. We are given the distance and the rate. What we want is the time traveled
in days and years. To find this we will use the model D = rt. Let’s solve this
for t and we get t = D/r. Our distance is given to be D = 1.43 × 106 miles
and our rate is r = 65 mph. Putting these values into our formula yields:
1.43 × 106 miles
65 miles per hour
= 0.022 × 106 hours
= 2, 200 hours
t=
We are asked to give our result in days and years.
days
24 hours
days
= 2, 200 hours ×
24 hours
≈ 916.7 days
years
= 916.7 days ×
365 days
× years
= 916.7 days
365 days
≈ 2.5 years
2, 200 hours = 2, 200 hours ×
So, you would have to drive nonstop for about 916.7 days or about 2.5 years
to cover all the paved roads in the USA. Be sure and bring a lot of coffee.
7.3
Simplifying Rational Expressions
1. Multiply numerators and denominators.
12s5
12 s5
=
·
2
s
9
9s2
Now, there several different ways you can reduce this answer to lowest terms,
two of which are shown below.
You can factor numerator and denominator, then cancel common
factors.
Or you can write the answer as a
product, repeat the base and subtract exponents.
2·2·3·s·s·s·s·s
12s5
=
2
9s
3·3·s·s
2·2·
3 · s · s · s · s · s
=
3
· 3 · s · s
4s3
=
3
12 s5
12s5
·
=
2
9s
9 s2
4
= · s3
3
4s3
=
3
Second Edition: 2012-2013
CHAPTER 7. RATIONAL FUNCTIONS
434
3. Multiply numerators and denominators.
12 v 4
12v 4
=
·
3
v 10
10v 3
Now, there several different ways you can reduce this answer to lowest terms,
two of which are shown below.
You can factor numerator and denominator, then cancel common
factors.
Or you can write the answer as a
product, repeat the base and subtract exponents.
2·2·3·v·v·v·v
12v 4
=
10v 3
2·5·v·v·v
2
·
2
· 3 · v · v · v · v
=
2 · 5 · v · v · v
6v
=
5
12 v 4
12v 4
·
=
3
10v
10 v 3
6
= · v1
5
6v
=
5
5. Invert, then multiply.
s5
9s2
s 5 t2
÷
=
·
t4
t2
t4 9s2
5 2
s t
= 2 4
9s t
Now, there several different ways you can reduce this answer to lowest terms,
two of which are shown below.
You can factor numerator and denominator, then cancel common
factors.
Or you can write the answer as a
product, repeat the base and subtract exponents.
s·s·s·s·s·t·t
s 5 t2
=
9s2 t4
3·3·s·s·t·t·t·t
s · s · s · s · s · t · t
= 3 · 3 · s · s · t · t · t · t
s3
= 2
9t
1 s 5 t2
s 5 t2
=
·
·
9s2 t4
9 s 2 t4
1
= · s3 · t−2
9
s3
= 2
9t
7. Invert, then multiply.
b4
9b2
b 4 c2
÷ 2 = 4· 2
4
c
c
c 9b
b 4 c2
= 2 4
9b c
Now, there several different ways you can reduce this answer to lowest terms,
two of which are shown below.
Second Edition: 2012-2013
7.3. SIMPLIFYING RATIONAL EXPRESSIONS
435
You can factor numerator and denominator, then cancel common
factors.
Or you can write the answer as a
product, repeat the base and subtract exponents.
b 4 c2
b·b·b·b·c·c
=
9b2 c4
3·3·b·b·c·c·c·c
b · b · b · b · c · c
=
3 · 3 · b · b · c · c · c · c
b2
= 2
9c
b 4 c2
1 b 4 c2
·
=
·
9b2 c4
9 b 2 c4
1
= · b2 · c−2
9
b2
= 2
9c
9. Becase we have a common denominator, we can simply add the numerators,
placing the answer over the common denominator.
−
−10s + 19s
10s 19s
+
=
18
18
18
9s
18
s
=
2
=
Add the numerators over
the common denominator.
Simplify: −10s + 19s = 9s
Reduce.
11. Becase we have a common denominator, we can simply subtract the numerators, placing the answer over the common denominator.
5
17
5 − 17
−
=
9c 9c
9c
−12
9c
4
=−
3c
=
Subtract the numerators over
the common denominator.
Subtract: 5 − 17 = −12
Reduce.
13. Becase we have a common denominator, we can simply subtract the numerators, placing the answer over the common denominator.
−
8x
16x
−8x − 16x
−
=
15yz 15yz
15yz
−24
15yz
8x
=−
5yz
=
Subtract the numerators over
the common denominator.
Subtract: −8x − 16x = −24x
Reduce.
Second Edition: 2012-2013
CHAPTER 7. RATIONAL FUNCTIONS
436
15. The smallest number divisible by both 10 and 2 is 10; i.e., LCD(10, 2) = 10.
We must first make equivalent fractions with a common denominator of 10.
9z
5z
9z
+
=
10
2
10
9z
=
10
1 5z 5
+
·
1
2 5
25z
+
10
·
Make equivalent fractions
with LCD = 10.
We can now add the numerators and put the result over the common denominator.
34z
10
17z
=
5
=
Add: 9z + 25z = 34z
Reduce.
17. The smallest expression divisible by both 10v and 5v is 10v; i.e., LCD(10v, 5v) =
10v. We must first make equivalent fractions with a common denominator of
10v.
4
3
3
−
=
10v 5v
10v
3
=
10v
1
4 2
−
·
1 5v 2
8
−
10v
·
Make equivalent fractions
with LCD = 10v.
We can now subtract the numerators and put the result over the common
denominator.
−5
10v
1
=−
2v
=
Subtract: 3 − 8 = −5
Reduce.
19. The smallest expression divisible by both 5st and 10st is 10st; i.e.,
LCD(5st, 10st) = 10st. We must first make equivalent fractions with a common
denominator of 10st.
−
9r
8r 2
9r 1
8r
−
=−
· −
·
5st 10st
5st 2 10st 1
9r
16r
−
=−
10st 10st
Make equivalent fractions
with LCD = 10st.
We can now subtract the numerators and put the result over the common
denominator.
−25r
10st
5r
=−
2st
=
Second Edition: 2012-2013
Subtract: −16r − 9r = −25r
Reduce.
7.3. SIMPLIFYING RATIONAL EXPRESSIONS
437
21. Prime factor each denominator, placing the results in exponential form.
18rs2 = 21 · 32 · r1 · s2
24r2 s = 23 · 31 · r2 · s1
To find the LCD, list each factor that appears to the highest power that it
appears.
LCD = 23 · 32 · r2 · s2
Simplify.
LCD = 8 · 9 · r2 · s2
LCD = 72r2 s2
After making equivalent fractions, place the sum of the numerators over this
common denominator.
11
11
5
3s
5
4r
=
+
·
+
·
18rs2
24r2 s
18rs2 4r 24r2 s 3s
44r
15s
=
+
72r2 s2
72r2 s2
44r + 15s
=
72r2 s2
23. Prime factor each denominator, placing the results in exponential form.
24rs2 = 23 · 31 · r1 · s2
36r2 s = 22 · 32 · r2 · s1
To find the LCD, list each factor that appears to the highest power that it
appears.
LCD = 23 · 32 · r2 · s2
Simplify.
LCD = 8 · 9 · r2 · s2
LCD = 72r2 s2
After making equivalent fractions, place the sum of the numerators over this
common denominator.
5
5
17
2s
17
3r
=
+
·
+
·
24rs2
36r2 s
24rs2 3r 36r2 s 2s
15r
34s
=
+
72r2 s2
72r2 s2
15r + 34s
=
72r2 s2
Second Edition: 2012-2013
438
CHAPTER 7. RATIONAL FUNCTIONS
25. Prime factor each denominator, placing the results in exponential form.
36y 3 = 22 · 32 · y 3
48z 3 = 24 · 31 · z 3
To find the LCD, list each factor that appears to the highest power that it
appears.
LCD = 24 · 32 · y 3 · z 3
Simplify.
LCD = 16 · 9 · y 3 · z 3
LCD = 144y 3 z 3
After making equivalent fractions, place the sum of the numerators over this
common denominator.
7
11
7
4z 3
11 3y 3
+
=
·
+
·
36y 3
48z 3
36y 3 4z 3 48z 3 3y 3
28z 3
33y 3
=
+
144y 3z 3
144y 3 z 3
3
3
28z + 33y
=
144y 3 z 3
27. Prime factor each denominator, placing the results in exponential form.
48v 3 = 24 · 31 · v 3
36w3 = 22 · 32 · w3
To find the LCD, list each factor that appears to the highest power that it
appears.
LCD = 24 · 32 · v 3 · w3
Simplify.
LCD = 16 · 9 · v 3 · w3
LCD = 144v 3 w3
After making equivalent fractions, place the sum of the numerators over this
common denominator.
5
13
5
3w3
13
4v 3
+
=
·
+
·
48v 3
36w3
48v 3 3w3
36w3 4v 3
3
15w
52v 3
=
+
3
3
144v w
144v 3 w3
3
3
15w + 52v
=
144v 3 w3
Second Edition: 2012-2013
7.3. SIMPLIFYING RATIONAL EXPRESSIONS
439
29. Prime factor each denominator, placing the results in exponential form.
50xy = 21 · 52 · x1 · y 1
40yz = 23 · 51 · y 1 · z 1
To find the LCD, list each factor that appears to the highest power that it
appears.
LCD = 23 · 52 · x1 · y 1 · z 1
Simplify.
LCD = 8 · 25 · x · y · z
LCD = 200xyz
After making equivalent fractions, place the sum of the numerators over this
common denominator.
11
9
11
4z
9
5x
−
=
·
−
·
50xy 40yz
50xy 4z 40yz 5x
45x
44z
−
=
200xyz 200xyz
44z − 45x
=
200xyz
31. Prime factor each denominator, placing the results in exponential form.
50ab = 21 · 52 · a1 · b1
40bc = 23 · 51 · b1 · c1
To find the LCD, list each factor that appears to the highest power that it
appears.
LCD = 23 · 52 · a1 · b1 · c1
Simplify.
LCD = 8 · 25 · a · b · c
LCD = 200abc
After making equivalent fractions, place the sum of the numerators over this
common denominator.
19
17
19 4c
17 5a
−
=
·
−
·
50ab 40bc
50ab 4c 40bc 5a
76c
85a
=
−
200abc 200abc
76c − 85a
=
200abc
Second Edition: 2012-2013
CHAPTER 7. RATIONAL FUNCTIONS
440
33. We use the distributive property, dividing each term by 3.
6v + 12
6v 12
=
+
3
3
3
= 2v + 4
Distribute 3.
Simplify: 6v/3 = 2v
and 12/3 = 4
35. We use the distributive property, dividing each term by 5.
25u + 45
25u 45
=
+
5
5
5
= 5u + 9
Distribute 5.
Simplify: 25u/5 = 5u
and 45/5 = 9
37. We use the distributive property, dividing each term by s.
2s − 4
2s 4
=
−
s
s
s
4
=2−
s
Distribute s.
Simplify: 2s/s = 2
39. We use the distributive property, dividing each term by r.
3r − 5
3r 5
=
−
r
r
r
5
=3−
r
Distribute r.
Simplify: 3r/r = 3
41. We use the distributive property, dividing each term by x2 .
3x2 − 8x − 9
3x2
8x
9
=
− 2 − 2
x2
x2
x
x
9
8
=3− − 2
x x
Distribute x2 .
Simplify: 3x2 /x2 = 3
and 8x/x2 = 8/x
43. We use the distributive property, dividing each term by x2 .
2x2 − 3x − 6
2x2
3x
6
=
− 2 − 2
x2
x2
x
x
6
3
=2− − 2
x x
Second Edition: 2012-2013
Distribute x2 .
Simplify: 2x2 /x2 = 2
and 3x/x2 = 3/x