Chapter 12: Rational Expressions
12.3 ADDING AND
SUBTRACTING RATIONAL
EXPRESSIONS
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Converting Units
ƒ 1 inch=
i h 2.54
2 54 cm
ƒ 1 foot = 12 inches
ƒ 1 mile = 5280 feet
ƒ Can make any
y of these into ‘unit fractions’
ƒ A ratio equal to one
Converting Units
ƒ 1 inch=
i h 2.54
2 54 cm
ƒ 1 foot = 12 inches
ƒ 1 mile = 5280 feet
1 inch
=1
2.54 cm
1 foot
=1
12 inches
1 mile
=1
5280 feet
ƒ And reciprocals also equal one!
Converting Units: inches to cm
ƒ How
H
many centimeters
ti t
iis 46
46.25
25 inches
i h
ƒ 1 inch= 2.54 cm
1 inch
=1
2.54 cm
2.54 cm
=1
1 in.
2.54 cm 46.25 in.
⋅
≅ 117.48 cm
1 in.
1
Converting Units mi./hr to ft./s
5280 ft
=1
1 mi.
mi
ƒ 4 mi./hr
i /h
1 hr
ƒ 1 hour= 60 min.
=1
ƒ 1 min = 60 sec. 60 min.
ƒ 1 mi. = 5280 ft. 1 min.
=1
60 s
4 mi.
mi 5280 ft
1 hr
1 min.
min
feet
⋅
⋅
⋅
≅ 5 .87
hr
1 mi. 60 min. 60 sec.
sec.
Think of each factor of a rational
expression like it is a variable
ƒ Replace any factor with a unique variable
ƒ Write a list,, so you
y can put
p them back in
after simplifying the expression
ƒ Simplify
p y the expression
p
you
y create
ƒ Replace ‘your’ variables with their
meaning
g from your
y
list
Replace FACTORS
2
with variables f ( x) = 81x − 49 ÷ (9 x − 7)
2
ƒ Quotient
ƒ
ƒ
ƒ
ƒ
ffunction
ti
Write as product
Factor
Use a, b, c, d for
binomial factors
Simplify this
ab 6c
=
⋅
cd a
a = (9 x − 7 )
3 x + 16 x + 5
g
18 x + 6
81x − 49 18 x + 6
= 2
⋅
3 x + 16 x + 5 (9 x − 7)
2
(
9 x − 7 )(9 x + 7 ) 6(3 x + 1)
=
⋅
(3x + 1)(x + 5) (9 x − 7)
b 6
= ⋅
d 1
b = (9 x + 7 )
6(9 x − 7 )
=
( x + 5)
c = (3 x + 1)
d = (x + 5 )
Adding Fractions
ƒ Need
N d common d
denominator
i t
ƒ Use lowest value
ƒ Find by comparing prime factors
ƒ Multiply
p y each term byy 1, composed
p
of the
factors missing from the denominator
ƒ Add now that they
y have same denominator
Adding rational expressions
ƒ Need
N d common d
denominators
i t
2 3 5
+ =
7 7 7
(4 x + 3) + 2
x
x
4x + 3 + 2
=
x
Adding rational expressions
ƒ Need
N d common d
denominators
i t
x − 10 x 2 x + 15
+
x −5
x −5
2
x − 8 x + 15
=
x −5
2
x − 10 x + 2 x + 15
=
x −5
2
(
x − 3)( x − 5)
=
( x − 5)
= ( x − 3)
Adding with different
denominators
7
5
+
24 18
ƒ Need
N d tto b
build
ild th
the ffractions,
ti
b
by multiplying
lti l i by
b 1
ƒ 1 is composed of something divided by itself
ƒ Choose the right 1
7 ⎛ 18 ⎞ 5 ⎛ 24 ⎞
⎜ ⎟+ ⎜ ⎟
24 ⎝ 18 ⎠ 18 ⎝ 24 ⎠
126 120 246 41
=
+
=
=
432 432 432 72
ƒ Find lowest common denominator instead
Adding with different
denominators
7
5
+
24 18
ƒ Need
N d tto b
build
ild th
the ffractions,
ti
b
by multiplying
lti l i by
b 1
ƒ Find lowest common denominator instead
ƒ Choose the right 1 using prime factors
24 = 2 ⋅ 2 ⋅ 2 ⋅ 3 = 2 ⋅ 3
3
18 = 2 ⋅ 3 ⋅ 3 = 2 ⋅ 3
2
ƒ Take each unique prime factor
à Raise to g
greatest power
p
it occurs in anyy factor list
ƒ LCD is
= 23 ⋅ 32 = 72
Adding with different
denominators
7
5
+
24 18
ƒ Choose
Ch
the
th right
i ht 1 using
i prime
i
factors
f t
ƒ LCD is = 23 ⋅ 32 = 72
ƒ What does it have that prime factor lists don’t
24 = 2 ⋅ 2 ⋅ 2 ⋅ 3 = 23 ⋅ 3
ƒ Needs a 3
7 ⎛3⎞ 5 ⎛ 4⎞
⎜ ⎟+ ⎜ ⎟
24 ⎝ 3 ⎠ 18 ⎝ 4 ⎠
18 = 2 ⋅ 3 ⋅ 3 = 2 ⋅ 32
needs 2·2
21 20 41
=
+
=
72 72 72
Adding with different 5 + 7
6 2x − 8
denominators
ƒ Need
N d tto b
build
ild th
the ffractions,
ti
b
by multiplying
lti l i by
b 1
ƒ Find lowest common denominator instead
ƒ Choose the right 1 using prime factors
6 = 2⋅3
2 x − 8 = 2( x − 4 )
ƒ Take each unique prime factor
à Raise to g
greatest power
p
it occurs in anyy factor list
ƒ LCD is
= 2 ⋅ 3 ⋅ (x − 4)