The document discusses techniques for combining fractions with opposite denominators. It explains that we can multiply the numerator and denominator by -1 to change the denominator to its opposite. It provides examples of switching fractions to their opposite denominators and combining fractions with opposite denominators by first switching one denominator so they are the same. It also discusses an alternative approach of pulling out a "-" from the denominator and passing it to the numerator when switching denominators, ensuring the leading term is positive for polynomial denominators.

1. Addition and Subtraction II

2. Addition and Subtraction II
Often we multiply the numerator and the denominator by –1
to change the denominator to it’s opposite,

3. a
b 
Addition and Subtraction II
Often we multiply the numerator and the denominator by –1
to change the denominator to it’s opposite, i.e.
–a
–b

4. Example A. Switch the following fractions to the opposite
denominator.
a.
2
3
=
a
b 
b.
a – b
=
d.
x + y
2x – y
=
Addition and Subtraction II
Often we multiply the numerator and the denominator by –1
to change the denominator to it’s opposite, i.e.
–3
–a
–b

5. Example A. Switch the following fractions to the opposite
denominator.
a.
2
3
=
a
b 
–2
b.
a – b
=
d.
x + y
2x – y
=
Addition and Subtraction II
Often we multiply the numerator and the denominator by –1
to change the denominator to it’s opposite, i.e.
–3
–3
–a
–b

6. Example A. Switch the following fractions to the opposite
denominator.
a.
2
3
=
a
b 
–2
b.
a – b
=
–(b – a)
3
d.
x + y
2x – y
=
Addition and Subtraction II
Often we multiply the numerator and the denominator by –1
to change the denominator to it’s opposite, i.e.
–3
–3
–a
–b
=
b – a
3

7. Example A. Switch the following fractions to the opposite
denominator.
a.
2
3
=
a
b 
–2
b.
a – b
=
–(b – a)
3
d.
x + y
2x – y
=
–x – y
y – 2x
Addition and Subtraction II
Often we multiply the numerator and the denominator by –1
to change the denominator to it’s opposite, i.e.
–3
–3
–a
–b
=
b – a
3

8. Example A. Switch the following fractions to the opposite
denominator.
a.
2
3
=
a
b 
–2
b.
a – b
=
–(b – a)
3
d.
x + y
2x – y
=
–x – y
y – 2x
Addition and Subtraction II
Often we multiply the numerator and the denominator by –1
to change the denominator to it’s opposite, i.e.
–3
–3
–a
–b
=
b – a
3
When combining two fractions with opposite denominators,
we may switch one of them to make their denominators the
same.

9. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Addition and Subtraction II

10. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
Addition and Subtraction II

11. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
=
x + y
2x – y +
3y – x
2x – y
switched
Addition and Subtraction II

12. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
=
x + y
2x – y +
3y – x
2x – y
switched
x + y + 3y – x
2x – y
=
Addition and Subtraction II

13. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
=
x + y
2x – y +
3y – x
2x – y
switched
x + y + 3y – x
2x – y
=
4y
2x – y=
Addition and Subtraction II

14. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
=
x + y
2x – y +
3y – x
2x – y
switched
x + y + 3y – x
2x – y
=
4y
2x – y=
Addition and Subtraction II
Another way to switch to a denominator to its opposite is to
pull out a “–” and pass it to the numerator.

15. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
=
x + y
2x – y +
3y – x
2x – y
switched
x + y + 3y – x
2x – y
=
4y
2x – y=
Addition and Subtraction II
Another way to switch to a denominator to its opposite is to
pull out a “–” and pass it to the numerator. Specifically, for
polynomials in x, make sure the leading term is positive.

16. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
=
x + y
2x – y +
3y – x
2x – y
switched
x + y + 3y – x
2x – y
=
4y
2x – y=
Addition and Subtraction II
Example C. combine x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
Another way to switch to a denominator to its opposite is to
pull out a “–” and pass it to the numerator. Specifically, for
polynomials in x, make sure the leading term is positive.

17. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
=
x + y
2x – y +
3y – x
2x – y
switched
x + y + 3y – x
2x – y
=
4y
2x – y=
Addition and Subtraction II
Example C. combine x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
Another way to switch to a denominator to its opposite is to
pull out a “–” and pass it to the numerator. Specifically, for
polynomials in x, make sure the leading term is positive.
We write the denominator 4 – x2 as –(x2 – 4)

18. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
=
x + y
2x – y +
3y – x
2x – y
switched
x + y + 3y – x
2x – y
=
4y
2x – y=
Addition and Subtraction II
Example C. combine x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
as
Another way to switch to a denominator to its opposite is to
pull out a “–” and pass it to the numerator. Specifically, for
polynomials in x, make sure the leading term is positive.
x + 3
(4 – x2)
We write the denominator 4 – x2 as –(x2 – 4) then write
x + 3
–(x2 – 4)

19. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
=
x + y
2x – y +
3y – x
2x – y
switched
x + y + 3y – x
2x – y
=
4y
2x – y=
Addition and Subtraction II
Example C. combine x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
as
Another way to switch to a denominator to its opposite is to
pull out a “–” and pass it to the numerator. Specifically, for
polynomials in x, make sure the leading term is positive.
x + 3
(4 – x2)
We write the denominator 4 – x2 as –(x2 – 4) then write
x + 3
–(x2 – 4)
–(x + 3)
x2 – 4

20. Example B. Switch one of the denominators then combine.
x + y
2x – y +
x – 3y
y – 2x
Opposites
denominators
=
x + y
2x – y +
3y – x
2x – y
switched
x + y + 3y – x
2x – y
=
4y
2x – y=
Addition and Subtraction II
Example C. combine x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
as
Another way to switch to a denominator to its opposite is to
pull out a “–” and pass it to the numerator. Specifically, for
polynomials in x, make sure the leading term is positive.
x + 3
(4 – x2)
We write the denominator 4 – x2 as –(x2 – 4) then write
x + 3
–(x2 – 4)
–(x + 3)
x2 – 4
=
–x – 3
x2 – 4

21. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
Switch to opposite denominator
Addition and Subtraction II

22. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
Switch to opposite denominator and pass the “–” to the top.
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
Addition and Subtraction II

23. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
Addition and Subtraction II
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
Switch to opposite denominator and pass the “–” to the top.

24. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
The LCD is (x – 2)(x + 2)(x – 1),
Addition and Subtraction II
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
Switch to opposite denominator and pass the “–” to the top.

25. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1), hence
Addition and Subtraction II
–x – 3
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
Switch to opposite denominator and pass the “–” to the top.

26. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1), hence
Addition and Subtraction II
–x – 3
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
[ ] * (x – 2)(x + 2)(x – 1) LCD
Switch to opposite denominator and pass the “–” to the top.

27. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1), hence
Addition and Subtraction II
–x – 3
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
[ ] * (x – 2)(x + 2)(x – 1) LCD
Switch to opposite denominator and pass the “–” to the top.
Distribute

28. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1), hence
Addition and Subtraction II
–x – 3
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
[ ]* (x – 2)(x + 2)(x – 1) LCD
(x – 1)
Switch to opposite denominator and pass the “–” to the top.

29. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1), hence
Addition and Subtraction II
–x – 3
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
[ ]* (x – 2)(x + 2)(x – 1) LCD
(x – 1) (x + 2)
Switch to opposite denominator and pass the “–” to the top.

30. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1), hence
Addition and Subtraction II
–x – 3
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
[ ]* (x – 2)(x + 2)(x – 1) LCD
(x – 1) (x + 2)
= [(–x – 3)(x – 1) + (2x – 1)(x + 2)] LCD
Switch to opposite denominator and pass the “–” to the top.

31. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1), hence
Addition and Subtraction II
–x – 3
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
[ ]* (x – 2)(x + 2)(x – 1) LCD
(x – 1) (x + 2)
= [(–x – 3)(x – 1) + (2x – 1)(x + 2)] LCD
= –x2 – 2x + 3 + 2x2 + 3x – 2[ ] LCD
Switch to opposite denominator and pass the “–” to the top.

32. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1), hence
Addition and Subtraction II
–x – 3
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
[ ]* (x – 2)(x + 2)(x – 1) LCD
(x – 1) (x + 2)
= [(–x – 3)(x – 1) + (2x – 1)(x + 2)] LCD
= –x2 – 2x + 3 + 2x2 + 3x – 2[ ] LCD
=
x2 + x + 1
(x + 2)(x – 2)(x – 1)
Switch to opposite denominator and pass the “–” to the top.

33. x + 3
(4 – x2) +
2x – 1
(x2 – 3x + 2)
=
–x – 3
(x2 – 4)
+
2x – 1
(x2 – 3x + 2)
=
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1), hence
Addition and Subtraction II
–x – 3
(x – 2)(x + 2)+
2x – 1
(x – 2)(x – 1)
–x – 3
[ ]* (x – 2)(x + 2)(x – 1) LCD
(x – 1) (x + 2)
= [(–x – 3)(x – 1) + (2x – 1)(x + 2)] LCD
= –x2 – 2x + 3 + 2x2 + 3x – 2[ ] LCD
=
x2 + x + 1
(x + 2)(x – 2)(x – 1)
This is the simplified answer.
Switch to opposite denominator and pass the “–” to the top.

34. Addition and Subtraction II
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem.

35. Addition and Subtraction II
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.

36. Example D. Combine x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
–
Addition and Subtraction II
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.

37. Example D. Combine x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
=
–
Addition and Subtraction II
x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
–
x + 3
(x2 – x – 2 )
–(x – 2)
(x2 – 2x – 3 )
+
distribute the subtraction
to the numerator
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.

38. Example D. Combine x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
=
–
Addition and Subtraction II
x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
–
x + 3
(x2 – x – 2 )
–(x – 2)
(x2 – 2x – 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.

39. Example D. Combine x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
=
–
LCD = (x + 1)(x – 2)(x – 3)
Addition and Subtraction II
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.
x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
–
x + 3
(x2 – x – 2 )
–(x – 2)
(x2 – 2x – 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+

40. Example D. Combine x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
=
–
LCD = (x + 1)(x – 2)(x – 3)
Addition and Subtraction II
x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
–
x + 3
(x2 – x – 2 )
–(x – 2)
(x2 – 2x – 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+[ ]*(x + 1)(x – 2)(x – 3) LCD
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.

41. Example D. Combine x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
=
–
LCD = (x + 1)(x – 2)(x – 3)
Addition and Subtraction II
x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
–
x + 3
(x2 – x – 2 ) (x2 – 2x – 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+[ ]*(x + 1)(x – 2)(x – 3) LCD
(x – 3)
–(x – 2)
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.

42. Example D. Combine x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
=
–
LCD = (x + 1)(x – 2)(x – 3)
Addition and Subtraction II
x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
–
x + 3
(x2 – x – 2 ) (x2 – 2x – 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+[ ]*(x + 1)(x – 2)(x – 3) LCD
(x – 3) (x – 2)
–(x – 2)
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.

43. Example D. Combine x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
=
= [(x + 3)(x – 3) + (–x + 2)(x – 2)]
–
LCD = (x + 1)(x – 2)(x – 3)
Addition and Subtraction II
x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
–
x + 3
(x2 – x – 2 ) (x2 – 2x – 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+[ ]*(x + 1)(x – 2)(x – 3) LCD
(x – 3) (x – 2)
LCD
–(x – 2)
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.

44. Example D. Combine x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
=
= [(x + 3)(x – 3) + (–x + 2)(x – 2)]
= [x2 – 9 – x2 + 4x – 4]
–
LCD = (x + 1)(x – 2)(x – 3)
Addition and Subtraction II
x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
–
x + 3
(x2 – x – 2 ) (x2 – 2x – 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+[ ]*(x + 1)(x – 2)(x – 3) LCD
(x – 3) (x – 2)
LCD
LCD
–(x – 2)
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.

45. Example D. Combine x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
=
= [(x + 3)(x – 3) + (–x + 2)(x – 2)]
= [x2 – 9 – x2 + 4x – 4]
–
LCD = (x + 1)(x – 2)(x – 3)
Addition and Subtraction II
x + 3
(x2 – x – 2 )
x – 2
(x2 – 2x – 3 )
–
x + 3
(x2 – x – 2 ) (x2 – 2x – 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+
=
x + 3
(x + 1)(x – 2 )
–x + 2
(x + 1)(x – 3 )
+[ ]*(x + 1)(x – 2)(x – 3) LCD
(x – 3) (x – 2)
LCD
LCD
4x – 13
=
(x + 1)(x – 2)(x – 3)
–(x – 2)
We’re less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem. We do this by
distributing the subtraction as a negative sign to the numerator.

46. Addition and Subtraction II
The special case of combining two “easy” fractions

47. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method.

48. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d±

49. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc

50. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc
bd

51. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc
bd
Example E. Combine
x + 1
x– 2
– 3
x + 4

52. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc
bd
Example E. Combine
x + 1
x– 2
– 3
x + 4
x + 1
x – 2 –
3
x + 4±

53. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc
bd
Example E. Combine
x + 1
x– 2
– 3
x + 4
x + 1
x – 2 –
3
x + 4 =±
(x + 1)(x + 4) – 3(x – 2)

54. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc
bd
Example E. Combine
x + 1
x– 2
– 3
x + 4
x + 1
x – 2 –
3
x + 4 =±
(x + 1)(x + 4) – 3(x – 2)
(x – 2)(x + 4)

55. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc
bd
Example E. Combine
x + 1
x– 2
– 3
x + 4
x + 1
x – 2 –
3
x + 4 =±
(x + 1)(x + 4) – 3(x – 2)
(x – 2)(x + 4)
No cancellation!

56. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc
bd
Example E. Combine
x + 1
x– 2
– 3
x + 4
x + 1
x – 2 –
3
x + 4 =±
(x + 1)(x + 4) – 3(x – 2)
(x – 2)(x + 4)
=
x2 + 5x + 4 – 3x + 6
(x – 2)(x + 4)
No cancellation!
Expand

57. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc
bd
Example E. Combine
x + 1
x– 2
– 3
x + 4
x + 1
x – 2 –
3
x + 4 =±
(x + 1)(x + 4) – 3(x – 2)
(x – 2)(x + 4)
=
x2 + 5x + 4 – 3x + 6
(x – 2)(x + 4)
=
x2 + 2x + 10
(x – 2)(x + 4)
No cancellation!
Expand

58. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc
bd
Example E. Combine
x + 1
x– 2
– 3
x + 4
x + 1
x – 2 –
3
x + 4 =±
(x + 1)(x + 4) – 3(x – 2)
(x – 2)(x + 4)
=
x2 + 5x + 4 – 3x + 6
(x – 2)(x + 4)
=
x2 + 2x + 10
(x – 2)(x + 4)
This method won’t work well with example D. Their cross–
multiplication is messy.
No cancellation!
Expand

59. Addition and Subtraction II
The special case of combining two “easy” fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
d± =
ad ±bc
bd
Example E. Combine
x + 1
x– 2
– 3
x + 4
x + 1
x – 2 –
3
x + 4 =±
(x + 1)(x + 4) – 3(x – 2)
(x – 2)(x + 4)
=
x2 + 5x + 4 – 3x + 6
(x – 2)(x + 4)
=
x2 + 2x + 10
(x – 2)(x + 4)
No cancellation!
This method won’t work well with example D. Their cross–
multiplication is messy. Hence this is for two ± “easy” fractions.
Expand

60. Ex. Combine and simplify the answers.
–3
x – 3
+ 2x
–6 – 2x
3. 2x – 3
x – 3
– 5x + 4
5 – 15x
4.
3x + 1
6x – 4
– 2x + 3
2 – 3x5.
–5x + 7
3x – 12+
4x – 3
–2x + 86.
3x + 1
+
x + 3
4 – x211.
x2 – 4x + 4
x – 4
+
x + 5
–x2 + x + 2
12.
x2 – x – 6
3x + 1
+
2x + 3
9 – x213.
x2 – x – 6
3x – 4
–
2x + 5
x2 – x – 6
14.
–x2 + 5x + 6
3x + 4
+
2x – 3
–x2 – 2x + 3
15.
x2 – x
5x – 4
–
3x – 5
1 – x216.
x2 + 2x – 3
–3
2x – 1
+ 2x
2 – 4x
1.
2x – 3
x – 2
+
3x + 4
5 – 10x
2.
3x + 1
2x – 5
– 2x + 3
5 – 10x
9.
–3x + 2
3x – 12
+
7x – 2
–2x + 8
10.
3x + 5
3x –2
– x + 3
2 – 3x7. –5x + 7
3x – 4 + 4x – 3
–6x + 88.
Addition and Subtraction II

61. 3x + 1
–
x – 2
x2 – 4x +4
17.
4 – x2
x + 1
–
2x – 1
x2 + 3x + 3
18.
x2 + 2x – 4
2x – 3
–
x – 3
x2 – 6x + 8
19.
x2 – 4
4x – 1
–
2x +3
x2 + 3x – 4
20.
1 – x2
2x
–
x + 4
x2 – 5x + 4
21.
16 – x2
2x + 1
–
x +3
x2 + 25
22.
x2 – 3x – 10
Addition and Subtraction II