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Adding algebraic fractions: Level 2
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Problem 7.
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a) |
1 x |
+ |
2 3x |
= |
3 + 2 3x |
= |
5 3x |
|
b) |
3 2x |
− |
4 2x² |
= |
3x − 4 2x² |
c) |
1 (x + 1)² |
+ |
2 (x + 1) |
= |
1 + 2(x + 1) (x + 1)² |
= |
1 + 2x + 2 (x + 1)² |
= |
2x + 3 (x + 1)² |
d) |
6 x(x − 1) |
+ |
2 x(x − 2) |
= |
6(x − 2) + 2(x − 1) x(x − 1)(x − 2) |
|
|
= |
6x − 12 + 2x − 2 x(x − 1)(x − 2) |
|
|
= |
_8x − 14_ x(x − 1)(x − 2) |
Example 7. Add |
4 x² − 25 |
− |
3 x² − 6x + 5 |
Note that we tend to say "add" in algebra, even though the operation is subtraction.
Solution. To construct the LCM of the denominators, we must first factor the denominators.
4 x
² − 25 |
− |
3 x² − 6x + 5 |
= |
_4_ (x + 5)(x − 5) |
− |
_3_ (x − 5)(x − 1) |
|
|
= |
4(x − 1) − 3(x + 5) (x + 5)(x − 5)(x − 1) |
|
|
= |
_4x − 4 − 3x − 15_ (x + 5)(x − 5)(x − 1) |
|
|
= |
__x − 19__ (x + 5)(x − 5)(x − 1) |
Problem 8. Add. First factor each denominator.
a) |
x 2 |
+ |
_5_ 2x + 2 |
= |
x 2 |
+ |
_5_ 2(x + 1) |
|
|
= |
x(x + 1) + 5 2(x + 1) |
|
|
= |
x² + x + 5 2(x + 1) |
b) |
1 x² − x |
+ |
2 x |
= |
_1_ x(x − 1) |
+ |
2 x |
|
|
= |
1 + 2(x − 1) x(x − 1) |
|
|
= |
1 + 2x − 2 x(x − 1) |
|
|
= |
2x − 1 x(x − 1) |
c) |
2 x + 3 |
+ |
12 x² − 9 |
= |
2 x + 3 |
+ |
__12__ (x + 3)(x − 3) |
|
|
= |
2(x − 3) + 12 (x + 3)(x − 3) |
|
|
= |
2x − 6 + 12 (x + 3)(x − 3) |
|
|
= |
___2x + 6___ (x + 3)(x − 3) |
|
|
= |
__ 2(x + 3) __ (x + 3)(x − 3) |
|
|
= |
_2_ x − 3 |
d) |
___6___ x² + 5x + 6 |
+ |
___2___ x² − x − 6 |
= |
___6___ (x + 2)(x + 3 ) |
+ |
___ 2___ (x + 2)(x − 3) |
|
|
= |
6(x − 3) + 2(x + 3) (x + 2)(x + 3)(x − 3) |
|
|
= |
_ 6x − 18 + 2x + 6 _ (x + 2)(x + 3)(x − 3) |
|
|
= |
_____8x − 12_____ (x + 2)(x + 3)(x − 3)
|
|
|
= |
___ 4(2x − 3) ___ (x + 2)(x + 3)(x − 3)
|
Factor the numerator to see if anything might cancel.
e) |
___3___ x² − 7x + 10 |
− |
__2__ x² − 25 |
= |
___3___ (x − 2)(x − 5) |
− |
___ 2___ (x + 5)(x − 5) |
|
|
= |
3(x + 5) − 2(x − 2) (x − 2)(x − 5)(x + 5) |
|
|
= |
_ 3x + 15 − 2x + 4 _ (x − 2)(x − 5)(x + 5) |
|
|
= |
___ __x + 19__ ___ (x − 2)(x − 5)(x + 5) |
f) |
___7___ 3x² − 5x + 2 |
− |
___4___ 3x² + x − 2 |
= |
___7___ (3x − 2)(x − 1) |
− |
___ 4___ (3x − 2)(x + 1) |
|
|
= |
7(x + 1) − 4(x − 1) (3x − 2)(x − 1)(x + 1) |
|
|
= |
_ 7x + 7 − 4x + 4_ (3x − 2)(x − 1)(x + 1) |
|
|
= |
___ __3x + 11__ ___ (3x − 2)(x − 1)(x + 1) |
Example 8. Symmetrically, we can rewrite a fraction as a sum of fractions:
a + b + c d |
= |
a d |
+ |
b d |
+ |
c d |
Problem 9. Rewrite each of the following as a sum of fractions, and simplify.
a) |
1 + 2 + 3 6 |
= |
1 6 |
+ |
2 6 |
+ |
3 6 |
= |
1 6 |
+ |
1 3 |
+ |
1 2 |
b) |
2n² − 4n + 1 n² |
= |
2n² n² |
− |
4n n² |
+ |
1 n² |
= |
2 |
− |
4 n |
+ |
1 n² |
c) |
x³ + 4x² + 2 2x5 |
= |
x³ 2x5 |
+ |
4x² 2x5 |
+ |
2 2x5 |
= |
1 2x² |
+ |
2 x3 |
+ |
1 x5 |
d) |
x − 1 x + 1 |
= |
x x + 1 |
− |
1 x + 1 |
Example 9. Simplify |
|
Problem 10. Simplify.
a) |
|
= |
|
= |
1 6 |
· |
1 10 |
= |
1 60 |
b) |
|
= |
|
= z· |
xy y + x |
= |
zxy y + x |
|
= |
x − (x + h) (x + h)x |
· |
1 h |
|
= |
x − x − h (x + h)x |
· |
1 h |
d) |
|
= |
|
= |
(x + 1)(x − 1) x² |
· |
x x − 1 |
= |
x + 1 x |
Or, on recognizing the numerator as the Difference of Two Squares:
|
|
= |
|
= 1 + |
1 x |
= |
x + 1 x |
e) |
|
= |
|
|
|
= |
(a + b)(a − b) ba |
· |
ba a + b |
|
|
= |
a − b |
Next Lesson: Equations with fractions
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