Algebra
Review of Algebra
2
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REVIEW OF ALGEBRA
Review of Algebra
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Here we review the basic rules and procedures of algebra that you need to know in order to be successful in calculus.
Arithmetic Operations
The real numbers have the following properties: a b b a ab a b c a b ab c ab ac In particular, putting a b and so b c b c ba c (Commutative Law) (Associative Law) (Distributive law)
ab c
a bc
1 in the Distributive Law, we get c 1 b c 1b 1c
EXAMPLE 1
(a) 3xy 4x 3 4 x 2y 12x 2y (b) 2t 7x 2tx 11 14tx 4t 2x 22t (c) 4 3 x 2 4 3x 6 10 3x If we use the Distributive Law three times, we get a b c d a bc a bd ac bc ad bd
This says that we multiply two factors by multiplying each term in one factor by each term in the other factor and adding the products. Schematically, we have a In the case where c or
1
b c
d
a and d a b
b, we have
2
a2
ba
ab
b2
a
b
2
a2
2ab
b2
Similarly, we obtain
2
a
b
2
a2
2ab
b2
REVIEW OF ALGEBRA
x
3
EXAMPLE 2
6x 2 3x (a) 2x 1 3x 5 (b) x 6 2 x 2 12x 36 2x 6 (c) 3 x 1 4x 3
10x 3 4x 2 12x 2 12x 2
5 x 3x 5x
6x 2
7x
5 12 12
3 2x 9 2x 21
Fractions
To add two fractions with the same denominator, we use the Distributive Law: a b Thus, it is true that a b c a b c b c b 1 b a 1 b c 1 a b c a b c
But remember to avoid the following common error:
| b a c
a b
a c
(For instance, take a b c 1 to see the error.) To add two fractions with different denominators, we use a common denominator: a b c d ad bd bc
We multiply such fractions as follows: a b In particular, it is true that a b a b a b c d ac bd
To divide two fractions, we invert and multiply: a b c d
a b
d c
ad bc
4
s
REVIEW OF ALGEBRA
EXAMPLE 3
3 x x 3 x 3x 2 xx 1 3x 6 x 2 x (b) x 1 x 2 x 1 x 2 x2 x 2 2 x 2x 6 2 x x 2 s2t ut s 2 t 2u s2t 2 (c) u 2 2u 2 x
2
s
REVIEW OF ALGEBRA
Review of Algebra
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
Here we review the basic rules and procedures of algebra that you need to know in order to be successful in calculus.
Arithmetic Operations
The real numbers have the following properties: a b b a ab a b c a b ab c ab ac In particular, putting a b and so b c b c ba c (Commutative Law) (Associative Law) (Distributive law)
ab c
a bc
1 in the Distributive Law, we get c 1 b c 1b 1c
EXAMPLE 1
(a) 3xy 4x 3 4 x 2y 12x 2y (b) 2t 7x 2tx 11 14tx 4t 2x 22t (c) 4 3 x 2 4 3x 6 10 3x If we use the Distributive Law three times, we get a b c d a bc a bd ac bc ad bd
This says that we multiply two factors by multiplying each term in one factor by each term in the other factor and adding the products. Schematically, we have a In the case where c or
1
b c
d
a and d a b
b, we have
2
a2
ba
ab
b2
a
b
2
a2
2ab
b2
Similarly, we obtain
2
a
b
2
a2
2ab
b2
REVIEW OF ALGEBRA
x
3
EXAMPLE 2
6x 2 3x (a) 2x 1 3x 5 (b) x 6 2 x 2 12x 36 2x 6 (c) 3 x 1 4x 3
10x 3 4x 2 12x 2 12x 2
5 x 3x 5x
6x 2
7x
5 12 12
3 2x 9 2x 21
Fractions
To add two fractions with the same denominator, we use the Distributive Law: a b Thus, it is true that a b c a b c b c b 1 b a 1 b c 1 a b c a b c
But remember to avoid the following common error:
| b a c
a b
a c
(For instance, take a b c 1 to see the error.) To add two fractions with different denominators, we use a common denominator: a b c d ad bd bc
We multiply such fractions as follows: a b In particular, it is true that a b a b a b c d ac bd
To divide two fractions, we invert and multiply: a b c d
a b
d c
ad bc
4
s
REVIEW OF ALGEBRA
EXAMPLE 3
3 x x 3 x 3x 2 xx 1 3x 6 x 2 x (b) x 1 x 2 x 1 x 2 x2 x 2 2 x 2x 6 2 x x 2 s2t ut s 2 t 2u s2t 2 (c) u 2 2u 2 x