Download MAT 1275: Introduction to Mathematical Analysis Dr

MAT 1275: Introduction to
Mathematical Analysis
Dr. A. Rozenblyum
I. Rational Expressions
A. Adding and Subtracting Rational Expressions
Adding and Subtracting Rational Expressions
Recall that a rational number is a number that can be written as a fraction (having a
numerator on the top and a denominator on the bottom). Usually, we write a fraction in
m
the form
, where m and n are two integers (m is the numerator, and n is the
n
denominator). We treat a fraction as a ratio of its numerator to denominator. We will
always assume that denominator is not equal to zero.
We will consider here rational expressions. These are also fractions. However, their
numerators and denominators are not simply numbers. Instead, they are another
expressions that are called polynomials. Non-formally speaking, a polynomial is an
expression (or a function) that contains a variable, say x, together with the operations of
addition, subtraction, and multiplication (but not division) of x by numbers and by itself.
2
Here are examples of polynomials: 2x –1, 5 x  3 x  2 . The following example is not a
1
polynomial: 3  .
x
Note. In an example of polynomials, x 2 means x  x , and it is called the exponent of the
second power (also, we read it as “x square”). We will discuss exponents in greater
details in section II.A.
Let’s return to rational expressions. As we mentioned, a rational expression is a ratio of
two polynomials (a fraction whose numerator and denominator are polynomials). Here
are several examples of rational expressions:
5 x 2  3x  2 1
,
x
2x 1
, 
3
3x  2
, 3
.
x  1 x  5x 2
2
In this section we will consider examples on how to add and subtract rational
expressions. Mostly, these operations can be done in a manner, similar to rational
numbers (hopefully, you remember how to add and subtract fractions). Where possible,
we will point out the similarity between rational numbers and rational expressions.
Adding and Subtracting Rational Expressions
5x
3

.
2x 1 2x 1
Solution. Recall that it is very easy to add (or subtract) two fractions if they have the
same denominator: just add (or subtract) numerators, and copy (do not add or subtract)
2 3 5 5 4 1
their common denominator. For example,   ,   . The same thing with
7 7 7 9 9 9
5x
3
5x  3


the rational expressions, so
.
2x 1 2x 1 2x 1
Example 1. Add
Example 2. Subtract
3a  4 a  2

.
8
6
Solution. This time the denominators are different. Recall the situation with the rational
7 5
numbers. Suppose, we need to subtract  . The first step is to find LCD (Least
8 6
Common Denominator). This is the smallest number that is divisible by both
denominators. For denominators 8 and 6, LCD = 24. We put LCD in the denominator of
the resulting fraction. The second step is to find complements to each denominator: we
divide LCD by given denominator. For the denominator 8, the complement is 3
( 24  8  3 ), and for the denominator 6, the complement is 4 (24  6  4 ). The third step
is to get the numerator of the resulting fraction: we multiply numerators of given
fractions by the corresponding complements. So, we multiply numerator 7 by 3
(complement to denominator 8), and numerator 5 by 4 (complement to denominator 6).
We have
7 5 7 3  5 4 1
 

.
8 6
24
24
To subtract given rational expressions, we do the same thing:
3a  4 a  2 (3a  4)3  (a  2)4 9a  12  4a  8 5a  20




.
8
6
24
24
24
Adding and Subtracting Rational Expressions
Now consider an example when denominators are also different, but contain variables.
3
5

Example 3. Add
.
10 x 15 y
Solution. First, we construct LCD. Denominators of given fractions contain both
numbers and letters (variables). For numbers 10 and 15, the “numerical” part of LCD is
30. Letters x and y do not have common factors. Therefore, the “letter” part of LCD is
their product xy. The entire LCD = 30xy. Next, we find complements for each
denominator. The complement for the denominator 10x is 30xy/10x = 3y, and the
complement for 15y is 30xy/15y = 2x. From here we have
3
5
3  3 y  5  2 x 9 y  10x 10x  9 y




.
10x 15 y
30xy
30xy
30xy
Example 4. Combine
5 11 9


.
a 2 6a 14
Solution. To construct LCD, consider separately its numerical and letter parts. For
numbers 6 and 14, the “numerical” part of LCD is 42. For lettersa 2 and a, the letter part
2
2
of LCD is a . The entire LCD = 42a . Now, let’s find complements for denominators
a 2 , 6a and 14.
a 2 : the complement is 42 ( 42a 2 / a 2  42 ).
6a: the complement is 7a
(42a 2 / 6a  7a ).
14: the complement is 3a 2 ( 42a 2 / 14  3a 2 ).
From here,
5 11 9 5  42  11 7a  9  3a 2 210  77a  27a 2 27a 2  77a  210

 


.
a 2 6a 14
42a 2
42a 2
42a 2
Adding and Subtracting Rational Expressions
Example 5. Subtract
4
2

.
5 x  3 3x  5
Solution. Denominators 5x – 3 and 3x – 5 do not have common factors, therefore, LCD is
simply their product: LCD = (5x – 3)(3x – 5). The denominators 5 x  3 and 3 x  5 are
complement to each other, therefore
4
2
4(3 x  5)  2(5 x  3) 12 x  20  10 x  6
2 x  14




.
5 x  3 3x  5
(5 x  3)( 3 x  5)
(5 x  3)( 3 x  5)
(5 x  3)( 3 x  5)
Example 6. Add
7
4 .
x 3
4
. From
1
here, LCD of the denominators x – 3 and 1 is x – 3, and these denominators are
complement to each other. Therefore,
Solution. We can treat the integer 4 as a fraction with the denominator 1: 4 
7
7 1  4( x  3) 7  4 x  12 4 x  5
4


.
x 3
x 3
x3
x 3
End of the Topic