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.
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.
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