20
ALGEBRAIC FRACTIONS
The principle of equivalent fractions
FRACTIONS IN ALGEBRA are typically called rational expressions. (See Topic 18 of Precalculus.) We begin with the principle of equivalent fractions, which appears as follows:
| x y |
= | ax ay |
"Both the numerator and denominator may be
multiplied by the same factor, and the value
of the fraction will not change."
| Both x and y have been multiplied by the factor a. | x y |
and | ax ay |
are |
called equivalent fractions. This principle is the single most important fact about fractions.
Problem 1. Write the missing numerator.
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| 6 n |
= |
18 3n |
The denominator has been multiplied by 3; therefore the numerator will also be multiplied by 3.
Problem 2. Write the missing numerator.
| 4 x |
= |
4x x² |
The denominator has been multiplied by x; therefore the numerator will also be multiplied by x.
Problem 3. Write the missing numerator.
| m x |
= |
8x²m 8x3 |
The denominator has been multiplied by 8x²; therefore the numerator will also be multiplied by 8x².
Problem 4. Write the missing numerator.
("The denominator has been multiplied by _____. Therefore the numerator will also be multiplied by ____.")
| a) | a b |
= |
5a 5b |
b) | 3 x |
= |
6 2x |
c) | 5 y |
= |
5y y² |
||
| d) | 8 x |
= |
8y xy |
e) | a x |
= |
2x²a 2x3 |
f) | b y |
= |
bx²y x²y² |
||
| g) | p q |
= |
prs qrs |
h) | 2 b |
= |
2ac abc |
i) | 4 x |
= |
4(x + 1) x(x + 1) |
||
| Example 1. | a | = | ? b |
| Solution. We may write a as | a 1 |
. Therefore, |
| a 1 |
= | ab b |
Both a and 1 have been multiplied by b.
The numerator ab, however, is simply the product of a times b. It is a kind of cross-multiplying, and the student should not have to write the denominator 1.
| a | = | ab b |
Problem 5. Write the missing numerator.
| a) | x | = |
3x 3 |
b) | 2 | = |
2ab ab |
c) | x | = |
x³ x² |
| d) | 1 | = |
x x |
e) | 2 | = |
2x + 2 x + 1 |
f) | x + 1 | = |
x² − 1 x − 1 |
Part f) is The Difference of Two Squares.
There will be more problems of this type at the 2nd Level.
Reducing to lowest terms
The numerator and denominator of a fraction are called its terms. The symmetric version of the principle of equivalent fractions shows how to reduce to lowest terms, that is, how to relieve the terms of any common factors.
| ax ay |
= | x y |
"If the numerator and denominator have a common factor,
it may be canceled."
Equivalently, we may think of the bar separating ax and ay as the division bar, and we could say that we have divided a in the numerator by a in the denominator.
| Example 2. Reduce | 5x 5y |
. |
| Answer. | 5x 5y |
= | x y |
. |
5 is a common factor.
| Example 3. Reduce | 5 + x 5 + y |
. |
Answer. This can not be reduced. 5 is not a factor. It is a term. We cannot cancel terms!
The word term does double duty in algebra. We speak of the terms of a sum and also the terms of a fraction, which are the numerator and denominator.
Problem 6. Reduce to lowest terms.
| a) | 3a 3b |
= |
a b |
b) | 8xy 12x |
= |
2y 3 |
c) | 56y 77xy |
= |
8 11x |
| Example 4. Reduce | 4x x |
. |
| Answer. | 4x x |
= | 4. |
We may think of this as 4x divided by x.
| Example 5. Reduce | x 4x |
. |
| Answer. | x 4x |
= | 1 4 |
. |
When the numerator cancels completely, we must write 1. For,
x = x· 1.
| x· 1 4x |
= | 1 4 |
. |
| Example 6. Reduce | x − 3 6(x − 3) |
. |
| Answer. | x − 3 6(x − 3) |
= | 1 6 |
. |
We can view x − 3 as a factor of the numerator, because
x − 3 = (x − 3)· 1
Again, when the numerator cancels completely, we must write 1.
Problem 7. Reduce.
| a) | 2a a |
= | 2 | b) | a ab |
= |
1 b |
c) | 2x 8xy |
= |
1 4y |
| d) | 5(x − 2) x − 2 |
= | 5 | e) | x + 1 2(x + 1) |
= |
1 2 |
f) | 3(x + 2)x 6(x + 2)xy |
= |
1 2y |
| Example 7. Reduce | 2x 8x − 10 |
. |
Answer. The denominator in its present form is not composed of factors. But we can make factors:
| 2x 8x − 10 |
= | 2x 2(4x − 5) |
= | x 4x − 5 |
2 is a factor of every term in both the numerator and denominator. Every term can be divided by 2.
Note that there is no more canceling -- we cannot cancel the x's. Why not? Because the denominator is no longer made up of factors. It is made up of two terms.
(Although x is a factor of the first term, in order to cancel, we must be able to factor the entire denominator.)
| Example 8. Reduce | 3a + 6b + 9c 12d |
. |
Answer. Every term in both the numerator and denominator has a factor of 3. Therefore, upon dividing each term by 3, we can write immediately:
| 3a + 6b + 9c 12d |
= | a + 2b + 3c 4d |
The numerator can no longer be composed of factors. There is no more canceling.
| Example 9. Reduce | 3a + 6b + 8c 12d |
. |
Answer. Not possible! The numerator and denominator have no common factor.
| Example 10. Reduce | x² − x − 6 x² − 4x + 3 |
. |
Answer. In its present form, there is no canceling -- because there are no factors. But again, we can make factors:
| x² − x − 6 x² − 4x + 3 |
= | (x − 3)(x + 2) (x − 3)(x − 1) |
= | x + 2 x − 1 |
(x −3) is a common factor. We can cancel it. And when we do cancel it, there are no more factors. The end.
Problem 8. Make factors, and reduce.
| a) | 5x 10x + 15 |
= |
5x 5(2x + 3) |
= |
x 2x + 3 |
| b) | 3x − 12 3x |
= |
3(x − 4) 3x |
= |
x − 4 x |
| c) | 12x − 18y + 21z 6y |
= |
4x − 6y + 7z 2y |
, |
upon dividing every term by their common factor, 3.
| d) | 2m m² − 2m |
= |
2m m(m − 2) |
= |
2 m − 2 |
| e) | x² − x x |
= |
x(x − 1) x |
= | x − 1 |
| f) | 12x² 16x5 − 20x² |
= |
12x² 4x²(4x3 − 5) |
= |
3 4x3 − 5 |
| g) | x + 3 4x + 12 |
= |
x + 3 4(x + 3) |
= |
1 4 |
| h) | 2x − 8 x − 4 |
= |
2(x − 4) x − 4 |
= | 2 |
| i) | 2x − 2y 3x − 3y | = |
2(x − y) 3(x − y) |
= |
2 3 |
Problem 9. Make factors, and reduce.
| a) | x² − 2x − 3 x² − x − 2 |
= |
(x + 1)(x − 3) (x + 1)(x − 2) |
= |
x − 3 x − 2 |
| b) | x² + x − 2 x² − x − 6 |
= |
(x + 2)(x − 1) (x + 2)(x − 3) |
= |
x − 1 x − 3 |
| c) | x² − 2x + 1 x² − 1 |
= |
(x − 1)² (x + 1)(x − 1) |
= |
x − 1 x + 1 |
| d) | x² − 100 x + 10 |
= |
(x + 10)(x − 10) x + 10 |
= | x − 10 |
| e) | x + 3 x² + 6x + 9 |
= |
x + 3 (x + 3)² |
= |
1 x + 3 |
| f) | x³ + 4x² _ x² + x − 12 |
= |
x²(x + 4) (x − 3)(x + 4) |
= |
x² x − 3 |
Problem 10. Simplify by canceling -- if possible.
| a) | 3 + x 3x |
Not possible. The numerator is not made up of factors. |
| b) | 8a + b 2ab |
Not possible. Again, the numerator is not made up of factors. |
| c) | 8a + 2b 2ab |
= |
2(4a + b) 2ab |
= |
4a + b ab |
| d) | 6a + b 3a + b |
Not possible. The numerator and denominator have no common factors. 3 is not a factor of the denominator. It is a factor only of the first term. |
| e) | 6(a + b) 3(a + b) |
= | 2 |
| f) | 2x + 4y + 6z 10 |
= |
x + 2y + 3z 5 |
Divide every term by 2. |
| g) | 2x + 4y + 5z 10 |
Not possible. The numerator and denominator have no common factors. |
| h) | (x + 1) + (x + 2) (x + 1)(x + 3) |
Not possible. The numerator is not made up of factors. |
| i) | (x + 1)(x + 2) (x + 1)(x + 3) |
= |
|
= |
x + 2 x + 3 |
| j) | ab + c abc |
Not possible. The numerator is not made up of factors |
| k) | ab + ac abc |
= |
a(b + c) abc |
= |
b + c bc |
| l) | x² − x − 12 x² + x − 6 |
= |
(x + 3)(x − 4) (x + 3)(x − 2) |
= |
x − 4 x − 2 |
Next Lesson: Negative exponents
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