The document provides instructions for simplifying algebraic fractions. It states that one should always factorize the expression first before cancelling terms. Several worked examples are provided that show the steps to (1) create a common denominator, (2) identify the difference between the old and new denominators, and (3) multiply the numerator by this difference when factorizing.

1. Algebraic Fractions

2. Algebraic Fractions
1) Always FACTORISE FIRST
2) Only cancel ( ), not parts of them

3. Algebraic Fractions
1) Always FACTORISE FIRST
2) Only cancel ( ), not parts of them
16a 3b 2 c
e.g. (i )
24ab3c 3

4. Algebraic Fractions
1) Always FACTORISE FIRST
2) Only cancel ( ), not parts of them
16a 3b 2 c 2a 2
e.g. (i ) 3 3

24ab c 3bc

5. Algebraic Fractions
1) Always FACTORISE FIRST
2) Only cancel ( ), not parts of them
16a 3b 2 c 2a 2
e.g. (i ) 3 3

24ab c 3bc
ap 2  aq 2
(ii )
2ap  2aq

6. Algebraic Fractions
1) Always FACTORISE FIRST
2) Only cancel ( ), not parts of them
16a 3b 2 c 2a 2
e.g. (i ) 3 3

24ab c 3bc
ap  aq
2 2 a  p2  q2 
(ii ) 
2ap  2aq 2a  p  q 
a  p  q  p  q 

2a  p  q 

7. Algebraic Fractions
1) Always FACTORISE FIRST
2) Only cancel ( ), not parts of them
16a 3b 2 c 2a 2
e.g. (i ) 3 3

24ab c 3bc
ap  aq
2 2 a  p2  q2 
(ii ) 
2ap  2aq 2a  p  q 
a  p  q  p  q 

2a  p  q 

 p  q
2

8. 9 x2 1 x 1
(iii )  2
3x  1 3x  2 x  1

9. 9 x2 1 x 1
(iii )  2
3x  1 3x  2 x  1
 3x  1 3x  1  x  1
 
 3x  1  3x  1 x  1

10. 9 x2 1 x 1
(iii )  2
3x  1 3x  2 x  1
 3x  1 3x  1  x  1
 
 3x  1  3x  1 x  1
1

11. 9 x2 1 x 1
(iii )  2
3x  1 3x  2 x  1
 3x  1 3x  1  x  1
 
 3x  1  3x  1 x  1
1
ab  2b 2 a 2  4ab  4b 2
(iv) 2

6a b 3a

12. 9 x2 1 x 1
(iii )  2
3x  1 3x  2 x  1
 3x  1 3x  1  x  1
 
 3x  1  3x  1 x  1
1
ab  2b 2 a 2  4ab  4b 2
(iv) 2

6a b 3a
b  a  2b  3a
= 
 a  2b 
2 2
6a b

13. 9 x2 1 x 1
(iii )  2
3x  1 3x  2 x  1
 3x  1 3x  1  x  1
 
 3x  1  3x  1 x  1
1
ab  2b 2 a 2  4ab  4b 2
(iv) 2

6a b 3a
b  a  2b  3a
= 
 a  2b 
2 2
6a b
1
=
2a  a  2b 

14. a 1 a 1
(v )  2
a2  a a  a

15. a 1 a 1
(v )  2
a2  a a  a

 a  1   a  1
a  a  1 a  a  1

16. a 1 a 1
(v )  2
a2  a a  a

 a  1   a  1 1. create a common denominator
a  a  1 a  a  1 “if its not there, put it down”

a  a  1 a  1

17. a 1 a 1
(v )  2
a2  a a  a

 a  1   a  1 1. create a common denominator
a  a  1 a  a  1 “if its not there, put it down”

a  a  1 a  1
2. compare the old denominator with the
new denominator, and ask yourself;
“what’s different?”

18. a 1 a 1
(v )  2
a2  a a  a

 a  1   a  1 1. create a common denominator
a  a  1 a  a  1 “if its not there, put it down”

a  a  1 a  1
2. compare the old denominator with the
new denominator, and ask yourself;
“what’s different?”
3. that is what you multiply the
numerator by

19. a 1 a 1
(v )  2
a2  a a  a

 a  1   a  1 1. create a common denominator
a  a  1 a  a  1 “if its not there, put it down”
 a  1
2

a  a  1 a  1
2. compare the old denominator with the
new denominator, and ask yourself;
“what’s different?”
3. that is what you multiply the
numerator by

20. a 1 a 1
(v )  2
a2  a a  a

 a  1   a  1 1. create a common denominator
a  a  1 a  a  1 “if its not there, put it down”
 a  1   a  1
2 2

a  a  1 a  1
2. compare the old denominator with the
new denominator, and ask yourself;
“what’s different?”
3. that is what you multiply the
numerator by

21. a 1 a 1
(v )  2
a2  a a  a

 a  1   a  1 1. create a common denominator
a  a  1 a  a  1 “if its not there, put it down”
 a  1   a  1
2 2

a  a  1 a  1
2. compare the old denominator with the
a 2  2a  1  a 2  2a  1 new denominator, and ask yourself;

a  a  1 a  1 “what’s different?”
3. that is what you multiply the
numerator by

22. a 1 a 1
(v )  2
a2  a a  a

 a  1   a  1 1. create a common denominator
a  a  1 a  a  1 “if its not there, put it down”
 a  1   a  1
2 2

a  a  1 a  1
2. compare the old denominator with the
a 2  2a  1  a 2  2a  1 new denominator, and ask yourself;

a  a  1 a  1 “what’s different?”
4a

a  a  1 a  1 3. that is what you multiply the
numerator by

23. a 1 a 1
(v )  2
a2  a a  a

 a  1   a  1 1. create a common denominator
a  a  1 a  a  1 “if its not there, put it down”
 a  1   a  1
2 2

a  a  1 a  1
2. compare the old denominator with the
a 2  2a  1  a 2  2a  1 new denominator, and ask yourself;

a  a  1 a  1 “what’s different?”
4a

a  a  1 a  1 3. that is what you multiply the
4 numerator by

 a  1 a  1

24. 1 1 2
(vi )   2
 x  y  x  y x  y
2 2 2

25. 1 1 2
(vi )   2
 x  y  x  y x  y
2 2 2
1 1 2
  
 x  y   x  y   x  y  x  y 
2 2

26. 1 1 2
(vi )   2
 x  y  x  y x  y
2 2 2
1 1 2
  
 x  y   x  y   x  y  x  y 
2 2

 x  y  x  y
2 2

27. 1 1 2
(vi )   2
 x  y  x  y x  y
2 2 2
1 1 2
  
 x  y   x  y   x  y  x  y 
2 2
 x  y
2

 x  y  x  y
2 2

28. 1 1 2
(vi )   2
 x  y  x  y x  y
2 2 2
1 1 2
  
 x  y   x  y   x  y  x  y 
2 2
 x  y  x  y
2 2

 x  y  x  y
2 2

29. 1 1 2
(vi )   2
 x  y  x  y x  y
2 2 2
1 1 2
  
 x  y   x  y   x  y  x  y 
2 2
 x  y    x  y  2  x  y  x  y 
2 2

 x  y  x  y
2 2

30. 1 1 2
(vi )   2
 x  y  x  y x  y
2 2 2
1 1 2
  
 x  y   x  y   x  y  x  y 
2 2
 x  y    x  y  2  x  y  x  y 
2 2

 x  y  x  y
2 2
x 2  2 xy  y 2  x 2  2 xy  y 2  2 x 2  2 y 2

 x  y  x  y
2 2

31. 1 1 2
(vi )   2
 x  y  x  y x  y
2 2 2
1 1 2
  
 x  y   x  y   x  y  x  y 
2 2
 x  y    x  y  2  x  y  x  y 
2 2

 x  y  x  y
2 2
x 2  2 xy  y 2  x 2  2 xy  y 2  2 x 2  2 y 2

 x  y  x  y
2 2
4 y2

 x  y  x  y
2 2

32. 1 1 2
(vi )   2
 x  y  x  y x  y
2 2 2
1 1 2
  
 x  y   x  y   x  y  x  y 
2 2
 x  y    x  y  2  x  y  x  y 
2 2

 x  y  x  y
2 2
x 2  2 xy  y 2  x 2  2 xy  y 2  2 x 2  2 y 2

 x  y  x  y
2 2
4 y2

 x  y  x  y
2 2
Exercise 1D; 1cf, 2cfil, 3cfil, 4cgk, 5cfi, 6ace, 7bdf, 8ace,
9ace etc, 11aceg, 12, 13*bc