Chapter 7 Rational Functions - College of the Redwoods

696 Chapter 7 Rational Functions
1
2 + 1 3
1
4 + 2 3
= 5 6 · 12
11
= 5
2 · 3 · 2 · 2 · 3
11
= 5
2 · 3 · 2 · 2 · 3
11
= 10
11
Here is an arrangement of the work, from start to finish, presented without comment.
This is a good template to emulate when doing your homework.
1
2 + 1 3
1
4 + 2 3
=
=
3
6 + 2 6
3
12 + 8 12
5
6
11
12
= 5 6 · 12
11
= 5
2 · 3 · 2 · 2 · 3
11
= 5
2 · 3 · 2 · 2 · 3
11
= 10
11
Now, let’s look at a second approach to the problem. We saw that simplifying the
numerator in (2) required a common denominator of 6. Simplifying the denominator in
(3) required a common denominator of 12. So, let’s choose another common denominator,
this one a common denominator for both numerator and denominator, namely, 12.
Now, multiply top and bottom (numerator and denominator) of the complex fraction
(1) by 12, as follows.
1
2 + 1 3
1
4 + 2 3
=
( 1
2 + 1 3)
12
( 1
4 + 2 3)
12
Distribute the 12 in both numerator and denominator and simplify.
(5)
Version: Fall 2007