2 1 addition and subtraction ii

Addition and Subtraction II
Frank Ma © 2011

Often we multiply the numerator and the denominator by –1 to change the denominator to it’s opposite,

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

a
–a

b
–b
Example A. Switch the following fractions to the opposite
denominator.
2
a.
=
3
a – b
b.
=
–3
x + y
d.
=
2x – y

a
–a

b
–b
denominator.
2
–2
a.
=
3
–3
a – b
b.
=
–3
x + y
d.
=
2x – y

a
–a

b
–b
denominator.
2
–2
a.
=
3
–3
–(b – a)
b – a
a – b
=
b.
=
3
3
–3
x + y
d.
=
2x – y

a
–a

b
–b
denominator.
2
–2
a.
=
3
–3
–(b – a)
b – a
a – b
=
b.
=
3
3
–3
x + y
–x – y
d.
=
2x – y
y – 2x

a
–a

b
–b
denominator.
2
–2
a.
=
3
–3
–(b – a)
b – a
a – b
=
b.
=
3
3
–3
x + y
–x – y
d.
=
2x – y
y – 2x
When combining two fractions with opposite denominators, we may switch one of them to make their denominators the same.

Example B. Switch one of the denominators then combine.
x + y
x – 3y
+
2x – y
y – 2x

x + y
x – 3y
+
2x – y
y – 2x
Opposites
denominators

x + y
x + y
x – 3y
3y – x
switched
+
+
=
2x – y
y – 2x
2x – y
2x – y
Opposites
denominators

x + y
x + y
x – 3y
3y – x
switched
+
+
=
2x – y
y – 2x
2x – y
2x – y
x + y + 3y – x
=
Opposites
denominators
2x – y

x + y
x + y
x – 3y
3y – x
switched
+
+
=
2x – y
y – 2x
2x – y
2x – y
x + y + 3y – x
=
Opposites
denominators
2x – y
4y
=
2x – y

x + y
x + y
x – 3y
3y – x
switched
+
+
=
2x – y
y – 2x
2x – y
2x – y
x + y + 3y – x
=
Opposites
denominators
2x – y
4y
=
2x – y
Another way to switch to a denominator to its opposite is to pull out a “–” and pass it to the numerator.

x + y
x + y
x – 3y
3y – x
switched
+
+
=
2x – y
y – 2x
2x – y
2x – y
x + y + 3y – x
=
Opposites
denominators
2x – y
4y
=
2x – y
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 + y
x + y
x – 3y
3y – x
switched
+
+
=
2x – y
y – 2x
2x – y
2x – y
x + y + 3y – x
=
Opposites
denominators
2x – y
4y
=
2x – y
x + 3
2x – 1
Example C. combine
+
(4 – x2)
(x2 – 3x + 2)

x + y
x + y
x – 3y
3y – x
switched
+
+
=
2x – y
y – 2x
2x – y
2x – y
x + y + 3y – x
=
Opposites
denominators
2x – y
4y
=
2x – y
x + 3
2x – 1
Example C. combine
+
(4 – x2)
(x2 – 3x + 2)
We write the denominator 4 – x2 as –(x2 – 4)

x + y
x + y
x – 3y
3y – x
switched
+
+
=
2x – y
y – 2x
2x – y
2x – y
x + y + 3y – x
=
Opposites
denominators
2x – y
4y
=
2x – y
x + 3
2x – 1
Example C. combine
+
(4 – x2)
(x2 – 3x + 2)
We write the denominator 4 – x2 as –(x2 – 4) then write
x + 3
x + 3
as
(4 – x2)
–(x2 – 4)

x + y
x + y
x – 3y
3y – x
switched
+
+
=
2x – y
y – 2x
2x – y
2x – y
x + y + 3y – x
=
Opposites
denominators
2x – y
4y
=
2x – y
x + 3
2x – 1
Example C. combine
+
(4 – x2)
(x2 – 3x + 2)
x + 3
x + 3
–(x + 3)
as
(4 – x2)
–(x2 – 4)
x2 – 4

x + y
x + y
x – 3y
3y – x
switched
+
+
=
2x – y
y – 2x
2x – y
2x – y
x + y + 3y – x
=
Opposites
denominators
2x – y
4y
=
2x – y
x + 3
2x – 1
Example C. combine
+
(4 – x2)
(x2 – 3x + 2)
x + 3
x + 3
–(x + 3)
–x – 3
as
=
(4 – x2)
–(x2 – 4)
x2 – 4
x2 – 4

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
Switch to opposite denominator

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

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1),

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)
The LCD is (x – 2)(x + 2)(x – 1), hence
2x – 1
–x – 3
+
(x – 2)(x + 2)
(x – 2)(x – 1)

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)
2x – 1
–x – 3
[
]
* (x – 2)(x + 2)(x – 1)
LCD
+
(x – 2)(x + 2)
(x – 2)(x – 1)

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)
2x – 1
–x – 3
[
]
* (x – 2)(x + 2)(x – 1)
LCD
+
(x – 2)(x + 2)
(x – 2)(x – 1)
Distribute

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)
(x – 1)
2x – 1
–x – 3
[
]
* (x – 2)(x + 2)(x – 1)
LCD
+
(x – 2)(x + 2)
(x – 2)(x – 1)

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)
(x – 1)
(x + 2)
2x – 1
–x – 3
[
]
* (x – 2)(x + 2)(x – 1)
LCD
+
(x – 2)(x + 2)
(x – 2)(x – 1)

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)
(x – 1)
(x + 2)
2x – 1
–x – 3
[
]
* (x – 2)(x + 2)(x – 1)
LCD
+
(x – 2)(x + 2)
(x – 2)(x – 1)
]
[
=
(–x – 3)(x – 1) + (2x – 1)(x + 2)
LCD

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)
(x – 1)
(x + 2)
2x – 1
–x – 3
[
]
* (x – 2)(x + 2)(x – 1)
LCD
+
(x – 2)(x + 2)
(x – 2)(x – 1)
]
[
=
(–x – 3)(x – 1) + (2x – 1)(x + 2)
LCD
]
[
LCD
= –x2 – 2x + 3 + 2x2 + 3x – 2

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)
(x – 1)
(x + 2)
2x – 1
–x – 3
[
]
* (x – 2)(x + 2)(x – 1)
LCD
+
(x – 2)(x + 2)
(x – 2)(x – 1)
]
[
=
(–x – 3)(x – 1) + (2x – 1)(x + 2)
LCD
]
[
LCD
= –x2 – 2x + 3 + 2x2 + 3x – 2
x2 + x + 1
=
(x + 2)(x – 2)(x – 1)

x + 3
2x – 1
+
(4 – x2)
(x2 – 3x + 2)
–x – 3
2x – 1
+
=
(x2 – 4)
(x2 – 3x + 2)
2x – 1
–x – 3
=
+
(x – 2)(x + 2)
(x – 2)(x – 1)
(x – 1)
(x + 2)
2x – 1
–x – 3
[
]
* (x – 2)(x + 2)(x – 1)
LCD
+
(x – 2)(x + 2)
(x – 2)(x – 1)
]
[
=
(–x – 3)(x – 1) + (2x – 1)(x + 2)
LCD
]
[
LCD
= –x2 – 2x + 3 + 2x2 + 3x – 2
x2 + x + 1
=
This is the simplified answer.
(x + 2)(x – 2)(x – 1)

We’re less likely to make mistakes in a subtraction problem if the problem is changed to an addition problem.

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
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )

x + 3
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
x – 2
–
distribute the subtraction
to the numerator
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–(x – 2)
+
=
(x2 – x – 2 )
(x2 – 2x – 3 )

x + 3
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
x – 2
–
to the numerator
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–(x – 2)
+
=
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–x + 2
+
=
(x+ 1)(x – 2 )
(x + 1)(x – 3 )

x + 3
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
x – 2
–
to the numerator
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–(x – 2)
+
=
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–x + 2
+
LCD = (x + 1)(x – 2)(x – 3)
=
(x+ 1)(x – 2 )
(x + 1)(x – 3 )

x + 3
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
x – 2
–
to the numerator
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–(x – 2)
+
=
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–x + 2
+
LCD = (x + 1)(x – 2)(x – 3)
=
(x+ 1)(x – 2 )
(x + 1)(x – 3 )
[
]
x + 3
–x + 2
+
=
(x+ 1)(x – 2 )
(x + 1)(x – 3 )

x + 3
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
x – 2
–
to the numerator
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–(x – 2)
+
=
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–x + 2
+
LCD = (x + 1)(x – 2)(x – 3)
=
(x+ 1)(x – 2 )
(x + 1)(x – 3 )
[
]
x + 3
–x + 2
+
=
*(x + 1)(x – 2)(x – 3)
LCD
(x+ 1)(x – 2 )
(x + 1)(x – 3 )

x + 3
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
x – 2
–
to the numerator
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–(x – 2)
+
=
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–x + 2
+
LCD = (x + 1)(x – 2)(x – 3)
=
(x+ 1)(x – 2 )
(x + 1)(x – 3 )
(x – 3)
[
]
x + 3
–x + 2
+
=
*(x + 1)(x – 2)(x – 3)
LCD
(x+ 1)(x – 2 )
(x + 1)(x – 3 )

x + 3
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
x – 2
–
to the numerator
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–(x – 2)
+
=
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–x + 2
+
LCD = (x + 1)(x – 2)(x – 3)
=
(x+ 1)(x – 2 )
(x + 1)(x – 3 )
(x – 3)
(x – 2)
[
]
x + 3
–x + 2
+
=
*(x + 1)(x – 2)(x – 3)
LCD
(x+ 1)(x – 2 )
(x + 1)(x – 3 )

x + 3
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
x – 2
–
to the numerator
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–(x – 2)
+
=
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–x + 2
+
LCD = (x + 1)(x – 2)(x – 3)
=
(x+ 1)(x – 2 )
(x + 1)(x – 3 )
(x – 3)
(x – 2)
[
]
x + 3
–x + 2
+
=
*(x + 1)(x – 2)(x – 3)
LCD
(x+ 1)(x – 2 )
(x + 1)(x – 3 )
= [(x + 3)(x – 3) + (–x + 2)(x – 2)]
LCD

x + 3
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
x – 2
–
to the numerator
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–(x – 2)
+
=
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–x + 2
+
LCD = (x + 1)(x – 2)(x – 3)
=
(x+ 1)(x – 2 )
(x + 1)(x – 3 )
(x – 3)
(x – 2)
[
]
x + 3
–x + 2
+
=
*(x + 1)(x – 2)(x – 3)
LCD
(x+ 1)(x – 2 )
(x + 1)(x – 3 )
= [(x + 3)(x – 3) + (–x + 2)(x – 2)]
= [x2 – 9 – x2 + 4x – 4]
LCD
LCD

x + 3
x – 2
–
Example D. Combine
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
x – 2
–
to the numerator
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–(x – 2)
+
=
(x2 – x – 2 )
(x2 – 2x – 3 )
x + 3
–x + 2
+
LCD = (x + 1)(x – 2)(x – 3)
=
(x+ 1)(x – 2 )
(x + 1)(x – 3 )
(x – 3)
(x – 2)
[
]
x + 3
–x + 2
+
=
*(x + 1)(x – 2)(x – 3)
LCD
(x+ 1)(x – 2 )
(x + 1)(x – 3 )
= [(x + 3)(x – 3) + (–x + 2)(x – 2)]
= [x2 – 9 – x2 + 4x – 4]
LCD
LCD
4x – 13
=
(x + 1)(x – 2)(x – 3)

The special case of combining two “easy” fractions

When adding or subtracting two easy fractions, we may use the cross multiplication method.

When adding or subtracting two easy fractions, we may use the cross multiplication method. That is,
c
a
±
d
b

c
ad±bc
a
±
=
d
b

c
ad±bc
a
±
=
d
bd
b

c
ad±bc
a
±
=
d
bd
b
x + 1
3
–
Example E. Combine
x– 2
x + 4

c
ad±bc
a
±
=
d
bd
b
x + 1
3
–
Example E. Combine
x– 2
x + 4
x + 1
3
–
±
x – 2
x + 4

c
ad±bc
a
±
=
d
bd
b
x + 1
3
–
Example E. Combine
x– 2
x + 4
(x + 1)(x + 4) – 3(x – 2)
x + 1
3
–
=
±
x – 2
x + 4

c
ad±bc
a
±
=
d
bd
b
x + 1
3
–
Example E. Combine
x– 2
x + 4
(x + 1)(x + 4) – 3(x – 2)
x + 1
3
–
=
±
x – 2
x + 4
(x – 2)(x + 4)

c
ad±bc
a
±
=
d
bd
b
x + 1
3
–
Example E. Combine
x– 2
x + 4
(x + 1)(x + 4) – 3(x – 2)
x + 1
3
can’t cancel!
–
=
±
x – 2
x + 4
(x – 2)(x + 4)

c
ad±bc
a
±
=
d
bd
b
x + 1
3
–
Example E. Combine
x– 2
x + 4
(x + 1)(x + 4) – 3(x – 2)
x + 1
3
can’t cancel!
–
=
±
x – 2
x + 4
(x – 2)(x + 4)
x2 + 5x + 4 – 3x + 6
=
(x – 2)(x + 4)

c
ad±bc
a
±
=
d
bd
b
x + 1
3
–
Example E. Combine
x– 2
x + 4
(x + 1)(x + 4) – 3(x – 2)
x + 1
3
can’t cancel!
–
=
±
x – 2
x + 4
(x – 2)(x + 4)
x2 + 5x + 4 – 3x + 6
=
(x – 2)(x + 4)
x2 + 2x + 10
=
(x – 2)(x + 4)

c
ad±bc
a
±
=
d
bd
b
x + 1
3
–
Example E. Combine
x– 2
x + 4
(x + 1)(x + 4) – 3(x – 2)
x + 1
3
can’t cancel!
–
=
±
x – 2
x + 4
(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.

c
ad±bc
a
±
=
d
bd
b
x + 1
3
–
Example E. Combine
x– 2
x + 4
(x + 1)(x + 4) – 3(x – 2)
x + 1
3
can’t cancel!
–
=
±
x – 2
x + 4
(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. Hence this is for two ± “easy” fractions.

Ex. Combine and simplify the answers.
–3
2x
2x – 3
3x + 4
+
1.
+
2.
2x – 1
2 – 4x
x – 2
5 – 10x
–3
2x
2x – 3
5x + 4
+
3.
–
4.
x – 3
–6 – 2x
x – 3
5 – 15x
3x + 1
2x + 3
–5x + 7
4x – 3
–
5.
+
6.
6x – 4
2 – 3x
3x – 12
–2x + 8
3x + 5
x + 3
–5x + 7
4x – 3
–
7.
+
8.
3x –2
2 – 3x
3x – 4
–6x + 8
3x + 1
2x + 3
–3x + 2
7x – 2
–
9.
+
10.
2x – 5
5 – 10x
3x – 12
–2x + 8
3x + 1
x + 3
x – 4
x + 5
11.
+
12.
+
4 – x2
x2 – 4x + 4
–x2 + x + 2
x2 – x – 6
3x + 1
2x + 3
3x – 4
2x + 5
13.
+
14.
–
9 – x2
x2 – x – 6
x2 – x – 6
–x2 + 5x + 6
3x + 4
2x – 3
5x – 4
3x – 5
15.
+
16.
–
–x2 – 2x + 3
x2 – x
1 – x2
x2 + 2x – 3

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

2 1 addition and subtraction ii

by math123b on Aug 28, 2011

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2 1 addition and subtraction ii — Presentation Transcript