2. Addition and Subtraction II
Often we multiply the numerator and the denominator by β1
to change the denominator to itβs opposite,
3. a
b ο
Addition and Subtraction II
Often we multiply the numerator and the denominator by β1
to change the denominator to itβs opposite, i.e.
βa
βb
4. Example A. Switch the following fractions to the opposite
denominator.
a.
2
3
=
a
b ο
b.
a β b
=
d.
x + y
2x β y
=
Addition and Subtraction II
Often we multiply the numerator and the denominator by β1
to change the denominator to itβs opposite, i.e.
β3
βa
βb
5. Example A. Switch the following fractions to the opposite
denominator.
a.
2
3
=
a
b ο
β2
b.
a β b
=
d.
x + y
2x β y
=
Addition and Subtraction II
Often we multiply the numerator and the denominator by β1
to change the denominator to itβs opposite, i.e.
β3
β3
βa
βb
6. Example A. Switch the following fractions to the opposite
denominator.
a.
2
3
=
a
b ο
β2
b.
a β b
=
β(b β a)
3
d.
x + y
2x β y
=
Addition and Subtraction II
Often we multiply the numerator and the denominator by β1
to change the denominator to itβs opposite, i.e.
β3
β3
βa
βb
=
b β a
3
7. Example A. Switch the following fractions to the opposite
denominator.
a.
2
3
=
a
b ο
β2
b.
a β b
=
β(b β a)
3
d.
x + y
2x β y
=
βx β y
y β 2x
Addition and Subtraction II
Often we multiply the numerator and the denominator by β1
to change the denominator to itβs opposite, i.e.
β3
β3
βa
βb
=
b β a
3
8. Example A. Switch the following fractions to the opposite
denominator.
a.
2
3
=
a
b ο
β2
b.
a β b
=
β(b β a)
3
d.
x + y
2x β y
=
βx β y
y β 2x
Addition and Subtraction II
Often we multiply the numerator and the denominator by β1
to change the denominator to itβs opposite, i.e.
β3
β3
βa
βb
=
b β 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 β y +
x β 3y
y β 2x
Addition and Subtraction II
10. Example B. Switch one of the denominators then combine.
x + y
2x β y +
x β 3y
y β 2x
Opposites
denominators
Addition and Subtraction II
11. Example B. Switch one of the denominators then combine.
x + y
2x β y +
x β 3y
y β 2x
Opposites
denominators
=
x + y
2x β y +
3y β x
2x β y
switched
Addition and Subtraction II
12. Example B. Switch one of the denominators then combine.
x + y
2x β y +
x β 3y
y β 2x
Opposites
denominators
=
x + y
2x β y +
3y β x
2x β y
switched
x + y + 3y β x
2x β y
=
Addition and Subtraction II
13. Example B. Switch one of the denominators then combine.
x + y
2x β y +
x β 3y
y β 2x
Opposites
denominators
=
x + y
2x β y +
3y β x
2x β y
switched
x + y + 3y β x
2x β y
=
4y
2x β y=
Addition and Subtraction II
14. Example B. Switch one of the denominators then combine.
x + y
2x β y +
x β 3y
y β 2x
Opposites
denominators
=
x + y
2x β y +
3y β x
2x β y
switched
x + y + 3y β x
2x β y
=
4y
2x β y=
Addition and Subtraction II
Another way to switch to a denominator to its opposite is to
pull out a βββ and pass it to the numerator.
15. Example B. Switch one of the denominators then combine.
x + y
2x β y +
x β 3y
y β 2x
Opposites
denominators
=
x + y
2x β y +
3y β x
2x β y
switched
x + y + 3y β x
2x β y
=
4y
2x β y=
Addition and Subtraction II
Another way to switch to a denominator to its opposite is to
pull out a βββ 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 β y +
x β 3y
y β 2x
Opposites
denominators
=
x + y
2x β y +
3y β x
2x β y
switched
x + y + 3y β x
2x β y
=
4y
2x β y=
Addition and Subtraction II
Example C. combine x + 3
(4 β x2) +
2x β 1
(x2 β 3x + 2)
Another way to switch to a denominator to its opposite is to
pull out a βββ 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 β y +
x β 3y
y β 2x
Opposites
denominators
=
x + y
2x β y +
3y β x
2x β y
switched
x + y + 3y β x
2x β y
=
4y
2x β y=
Addition and Subtraction II
Example C. combine x + 3
(4 β x2) +
2x β 1
(x2 β 3x + 2)
Another way to switch to a denominator to its opposite is to
pull out a βββ and pass it to the numerator. Specifically, for
polynomials in x, make sure the leading term is positive.
We write the denominator 4 β x2 as β(x2 β 4)
18. Example B. Switch one of the denominators then combine.
x + y
2x β y +
x β 3y
y β 2x
Opposites
denominators
=
x + y
2x β y +
3y β x
2x β y
switched
x + y + 3y β x
2x β y
=
4y
2x β y=
Addition and Subtraction II
Example C. combine x + 3
(4 β x2) +
2x β 1
(x2 β 3x + 2)
as
Another way to switch to a denominator to its opposite is to
pull out a βββ and pass it to the numerator. Specifically, for
polynomials in x, make sure the leading term is positive.
x + 3
(4 β x2)
We write the denominator 4 β x2 as β(x2 β 4) then write
x + 3
β(x2 β 4)
19. Example B. Switch one of the denominators then combine.
x + y
2x β y +
x β 3y
y β 2x
Opposites
denominators
=
x + y
2x β y +
3y β x
2x β y
switched
x + y + 3y β x
2x β y
=
4y
2x β y=
Addition and Subtraction II
Example C. combine x + 3
(4 β x2) +
2x β 1
(x2 β 3x + 2)
as
Another way to switch to a denominator to its opposite is to
pull out a βββ and pass it to the numerator. Specifically, for
polynomials in x, make sure the leading term is positive.
x + 3
(4 β x2)
We write the denominator 4 β x2 as β(x2 β 4) then write
x + 3
β(x2 β 4)
β(x + 3)
x2 β 4
20. Example B. Switch one of the denominators then combine.
x + y
2x β y +
x β 3y
y β 2x
Opposites
denominators
=
x + y
2x β y +
3y β x
2x β y
switched
x + y + 3y β x
2x β y
=
4y
2x β y=
Addition and Subtraction II
Example C. combine x + 3
(4 β x2) +
2x β 1
(x2 β 3x + 2)
as
Another way to switch to a denominator to its opposite is to
pull out a βββ and pass it to the numerator. Specifically, for
polynomials in x, make sure the leading term is positive.
x + 3
(4 β x2)
We write the denominator 4 β x2 as β(x2 β 4) then write
x + 3
β(x2 β 4)
β(x + 3)
x2 β 4
=
βx β 3
x2 β 4
21. x + 3
(4 β x2) +
2x β 1
(x2 β 3x + 2)
Switch to opposite denominator
Addition and Subtraction II
22. x + 3
(4 β x2) +
2x β 1
(x2 β 3x + 2)
Switch to opposite denominator and pass the βββ to the top.
=
βx β 3
(x2 β 4)
+
2x β 1
(x2 β 3x + 2)
Addition and Subtraction II
23. x + 3
(4 β x2) +
2x β 1
(x2 β 3x + 2)
=
βx β 3
(x2 β 4)
+
2x β 1
(x2 β 3x + 2)
=
Addition and Subtraction II
(x β 2)(x + 2)+
2x β 1
(x β 2)(x β 1)
βx β 3
Switch to opposite denominator and pass the βββ to the top.
24. x + 3
(4 β x2) +
2x β 1
(x2 β 3x + 2)
=
βx β 3
(x2 β 4)
+
2x β 1
(x2 β 3x + 2)
=
The LCD is (x β 2)(x + 2)(x β 1),
Addition and Subtraction II
(x β 2)(x + 2)+
2x β 1
(x β 2)(x β 1)
βx β 3
Switch to opposite denominator and pass the βββ to the top.
25. x + 3
(4 β x2) +
2x β 1
(x2 β 3x + 2)
=
βx β 3
(x2 β 4)
+
2x β 1
(x2 β 3x + 2)
=
(x β 2)(x + 2)+
2x β 1
(x β 2)(x β 1)
The LCD is (x β 2)(x + 2)(x β 1), hence
Addition and Subtraction II
βx β 3
(x β 2)(x + 2)+
2x β 1
(x β 2)(x β 1)
βx β 3
Switch to opposite denominator and pass the βββ to the top.
34. Addition and Subtraction II
Weβre less likely to make mistakes in a subtraction problem if
the problem is changed to an addition problem.
35. Addition and Subtraction II
Weβre 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 β x β 2 )
x β 2
(x2 β 2x β 3 )
β
Addition and Subtraction II
Weβre 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 β x β 2 )
x β 2
(x2 β 2x β 3 )
=
β
Addition and Subtraction II
x + 3
(x2 β x β 2 )
x β 2
(x2 β 2x β 3 )
β
x + 3
(x2 β x β 2 )
β(x β 2)
(x2 β 2x β 3 )
+
distribute the subtraction
to the numerator
Weβre 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 β x β 2 )
x β 2
(x2 β 2x β 3 )
=
β
Addition and Subtraction II
x + 3
(x2 β x β 2 )
x β 2
(x2 β 2x β 3 )
β
x + 3
(x2 β x β 2 )
β(x β 2)
(x2 β 2x β 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+
Weβre 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 β x β 2 )
x β 2
(x2 β 2x β 3 )
=
β
LCD = (x + 1)(x β 2)(x β 3)
Addition and Subtraction II
Weβre 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 β x β 2 )
x β 2
(x2 β 2x β 3 )
β
x + 3
(x2 β x β 2 )
β(x β 2)
(x2 β 2x β 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+
40. Example D. Combine x + 3
(x2 β x β 2 )
x β 2
(x2 β 2x β 3 )
=
β
LCD = (x + 1)(x β 2)(x β 3)
Addition and Subtraction II
x + 3
(x2 β x β 2 )
x β 2
(x2 β 2x β 3 )
β
x + 3
(x2 β x β 2 )
β(x β 2)
(x2 β 2x β 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+[ ]*(x + 1)(x β 2)(x β 3) LCD
Weβre 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 β x β 2 )
x β 2
(x2 β 2x β 3 )
=
β
LCD = (x + 1)(x β 2)(x β 3)
Addition and Subtraction II
x + 3
(x2 β x β 2 )
x β 2
(x2 β 2x β 3 )
β
x + 3
(x2 β x β 2 ) (x2 β 2x β 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+[ ]*(x + 1)(x β 2)(x β 3) LCD
(x β 3)
β(x β 2)
Weβre 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 β x β 2 )
x β 2
(x2 β 2x β 3 )
=
β
LCD = (x + 1)(x β 2)(x β 3)
Addition and Subtraction II
x + 3
(x2 β x β 2 )
x β 2
(x2 β 2x β 3 )
β
x + 3
(x2 β x β 2 ) (x2 β 2x β 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+[ ]*(x + 1)(x β 2)(x β 3) LCD
(x β 3) (x β 2)
β(x β 2)
Weβre 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 β x β 2 )
x β 2
(x2 β 2x β 3 )
=
= [(x + 3)(x β 3) + (βx + 2)(x β 2)]
β
LCD = (x + 1)(x β 2)(x β 3)
Addition and Subtraction II
x + 3
(x2 β x β 2 )
x β 2
(x2 β 2x β 3 )
β
x + 3
(x2 β x β 2 ) (x2 β 2x β 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+[ ]*(x + 1)(x β 2)(x β 3) LCD
(x β 3) (x β 2)
LCD
β(x β 2)
Weβre 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 β x β 2 )
x β 2
(x2 β 2x β 3 )
=
= [(x + 3)(x β 3) + (βx + 2)(x β 2)]
= [x2 β 9 β x2 + 4x β 4]
β
LCD = (x + 1)(x β 2)(x β 3)
Addition and Subtraction II
x + 3
(x2 β x β 2 )
x β 2
(x2 β 2x β 3 )
β
x + 3
(x2 β x β 2 ) (x2 β 2x β 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+[ ]*(x + 1)(x β 2)(x β 3) LCD
(x β 3) (x β 2)
LCD
LCD
β(x β 2)
Weβre 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 β x β 2 )
x β 2
(x2 β 2x β 3 )
=
= [(x + 3)(x β 3) + (βx + 2)(x β 2)]
= [x2 β 9 β x2 + 4x β 4]
β
LCD = (x + 1)(x β 2)(x β 3)
Addition and Subtraction II
x + 3
(x2 β x β 2 )
x β 2
(x2 β 2x β 3 )
β
x + 3
(x2 β x β 2 ) (x2 β 2x β 3 )
+
distribute the subtraction
to the numerator
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+
=
x + 3
(x + 1)(x β 2 )
βx + 2
(x + 1)(x β 3 )
+[ ]*(x + 1)(x β 2)(x β 3) LCD
(x β 3) (x β 2)
LCD
LCD
4x β 13
=
(x + 1)(x β 2)(x β 3)
β(x β 2)
Weβre 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 βeasyβ 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 βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ±
49. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
50. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
bd
51. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
bd
Example E. Combine
x + 1
xβ 2
β 3
x + 4
52. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
bd
Example E. Combine
x + 1
xβ 2
β 3
x + 4
x + 1
x β 2 β
3
x + 4Β±
53. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
bd
Example E. Combine
x + 1
xβ 2
β 3
x + 4
x + 1
x β 2 β
3
x + 4 =Β±
(x + 1)(x + 4) β 3(x β 2)
54. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
bd
Example E. Combine
x + 1
xβ 2
β 3
x + 4
x + 1
x β 2 β
3
x + 4 =Β±
(x + 1)(x + 4) β 3(x β 2)
(x β 2)(x + 4)
55. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
bd
Example E. Combine
x + 1
xβ 2
β 3
x + 4
x + 1
x β 2 β
3
x + 4 =Β±
(x + 1)(x + 4) β 3(x β 2)
(x β 2)(x + 4)
No cancellation!
56. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
bd
Example E. Combine
x + 1
xβ 2
β 3
x + 4
x + 1
x β 2 β
3
x + 4 =Β±
(x + 1)(x + 4) β 3(x β 2)
(x β 2)(x + 4)
=
x2 + 5x + 4 β 3x + 6
(x β 2)(x + 4)
No cancellation!
Expand
57. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
bd
Example E. Combine
x + 1
xβ 2
β 3
x + 4
x + 1
x β 2 β
3
x + 4 =Β±
(x + 1)(x + 4) β 3(x β 2)
(x β 2)(x + 4)
=
x2 + 5x + 4 β 3x + 6
(x β 2)(x + 4)
=
x2 + 2x + 10
(x β 2)(x + 4)
No cancellation!
Expand
58. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
bd
Example E. Combine
x + 1
xβ 2
β 3
x + 4
x + 1
x β 2 β
3
x + 4 =Β±
(x + 1)(x + 4) β 3(x β 2)
(x β 2)(x + 4)
=
x2 + 5x + 4 β 3x + 6
(x β 2)(x + 4)
=
x2 + 2x + 10
(x β 2)(x + 4)
This method wonβt work well with example D. Their crossβ
multiplication is messy.
No cancellation!
Expand
59. Addition and Subtraction II
The special case of combining two βeasyβ fractions
When adding or subtracting two easy fractions, we may use
the cross multiplication method. That is,
a
b
c
dΒ± =
ad Β±bc
bd
Example E. Combine
x + 1
xβ 2
β 3
x + 4
x + 1
x β 2 β
3
x + 4 =Β±
(x + 1)(x + 4) β 3(x β 2)
(x β 2)(x + 4)
=
x2 + 5x + 4 β 3x + 6
(x β 2)(x + 4)
=
x2 + 2x + 10
(x β 2)(x + 4)
No cancellation!
This method wonβt work well with example D. Their crossβ
multiplication is messy. Hence this is for two Β± βeasyβ fractions.
Expand