20
ALGEBRAIC FRACTIONS
The principle of equivalent fractions
FRACTIONS IN ALGEBRA are often called rational expressions. (See Topic 18 of Precalculus.) We begin with the principle of equivalent fractions, which appears as follows:
| x y |
= | ax ay |
"We may multiply both the numerator and denominator
by the same factor."
| 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.
The student will see that the original denominator on the left will be a factor of the new denominator on the right. It must be a factor, because to produce that new denominator it was multiplied 
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. To explain the solution, we will write a as | a 1 |
. |
| 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. Since we may multiply both terms, then, symmetrically, we may divide both terms.
| ax ay |
= | x y |
"We may divide both the numerator and denominator
by a common factor."
When we do that, we say that we have reduced the fraction to its lowest terms.
| Example 2. Reduce | 5x 5y |
. |
| Answer. | 5x 5y |
= | x y |
. |
5 is a common factor of the numerator and denominator. Therefore we may divide each of them by 5. We say that we have "canceled" the 5's.
| Example 3. Reduce | 5 + x 5 + y |
. |
Answer. This can not be reduced. 5 is not a factor of either the numerator or the denominator. It is a term. You cannot cancel terms. You can divide both the numerator and denominator only when they have a common factor.
*
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. A fraction is in its lowest terms when the numerator and denominator have no common factors.
Problem 6. Reduce to lowest terms.
| a) | 3a 3b |
= |
a b |
b) | 8xy 12x |
= |
2y 3 |
c) | 56y 77xy |
= |
8 11x |
| d) | 2x + 3 4x + 9 |
= | No canceling! The numerator and denominator are made up of terms. You cannot cancel terms. And those terms have no common factors. |
| See Example 7 below. | |||
| 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 | 3a + 6b + 9c 12d |
. |
Answer. When the numerator or denominator is made up of terms, then if every term has a common factor, we may divide every term by it.
In this example, every term in both the numerator and denominator has a factor 3. Therefore, upon dividing every term by 3, we can write immediately:
| 3a + 6b + 9c 12d |
= | a + 2b + 3c 4d |
There is no more canceling. The numerator and denominator no longer have a common factor.
This example illustrates the following principle:
To divide a sum -- such as 3a + 6b + 9c -- by a number,
we must be able to divide every term by that number.
| Example 8. Reduce | 3a + 6b + 8c 12d |
. |
Answer. Not possible The numerator and denominator have no common factor. That fraction is in its lowest terms.
| Example 9. Reduce | 8x 8x + 10 |
. |
Answer. 2 is a factor of every term in both the numerator and denominator. Therefore,
| 8x 8x + 10 |
= | 4x 4x + 5 |
. |
There is no more canceling. We cannot cancel the 4x's, because 4x is not a common factor of the denominator. 4x appears only as the first term.
| Example 10. Reduce | 15x 5x − 3 |
. |
Answer. Not possible The numerator and denominator have no common factor.
| Example 11. Reduce | x² − x − 6 x² − 4x + 3 |
. |
Answer. In its present form, there is no canceling -- because there are no common factors. But 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 shown to be a common factor. We can cancel it. And when we do, the numerator and denominator no longer have a common factor. The end.
| Example 12. Reduce: | 4x³ − 9x² 4x³ + 6x² |
. |
Answer. The only common factor is x². And we could display it by factoring both the numerator and denominator:
| 4x³ − 9x² 4x³ + 6x² |
= | x²(4x − 9) 2x²(2x + 3) |
= | 4x − 9 2(2x + 3) |
The fraction is now in its lowest terms. No common factors.
Problem 8. 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 and denominator have no common factors. |
| b) | 8a + b 2ab |
Not possible. Again, no common 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 either the numerator or 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 factor. |
| 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 and denominator have no common 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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