The denominator has been multiplied by 3; therefore the numerator will also be multiplied by 3.
("The denominator has been multiplied by _____. Therefore the numerator will also be multiplied by ____.")
| 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.
"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.
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. |
We may think of this as 4x divided by x.
When the numerator cancels completely, we must write 1. For,
x = x· 1.
|
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,
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. |
| 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 |
2nd Level
Next Lesson: Negative exponents
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