Transcript Document

Unit 13
BASIC ALGEBRAIC OPERATIONS
ADDITION
 Only like terms can be added. The
addition of unlike terms can only be
indicated
 Procedure for adding like terms:
 Add the numerical coefficients, applying the
procedure for addition of signed numbers
 Leave the variables unchanged
6y + (–5y) = 1y = y Ans
–13ab + (–11ab) = –24ab Ans
2
ADDITION
•
Procedure for adding expressions that consist of two or
more terms:
–
–
•
Group like terms in the same column
Add like terms and indicate the addition of the unlike terms
Add: 5y + (–3x) + 6x2y and (–4x) + (–2y) + (–2x2y)
–
–
Group like terms in the same column
Add the like terms and indicate the
addition of the unlike terms
6 x 2 y  3x  5 y
 2 x 2 y  4 x  3 y
4 x 2 y  7 x  3 y Ans
3
SUBTRACTION
 Just as in addition, only like terms can be
subtracted
 Each term of the subtrahend is subtracted
following the procedure for subtraction of
signed numbers
 Subtract: (7x2 + 7xy – 15y2) – (–8x2 + 5xy –
10y2)
7 x 2  7 xy  15 y 2
2
2

8
x

5
xy

10
y
Change the sign of each term in the
subtrahend and follow the procedure
for addition of signed numbers
15 x 2  2 xy  5 y 2
Ans
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MULTIPLICATION
 In multiplication, the exponents of the literal factors do
not have to be the same to multiply the values
 Procedure for multiplying two or more terms:
– Multiply the numerical coefficients, following the
procedure for multiplication of signed numbers
– Add the exponents of the same literal factors
– Show the product as a combination of all numerical and
literal factors
 Multiply: (–4)(5x)(–6x2y)(7xy)(–2y3)
 Multiply all coefficients and
add exponents of the same
literal factors
= (–4)(5)(–6)(7)(–2)(x1 + 2 + 1)(y1 + 1 + 3)
= –1680x4y5 Ans
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MULTIPLICATION
•
Procedure for multiplying expressions that consist
of more than one term within an expression:
–
–
a.
Multiply each term of one expression by each term of the
other expression
Combine like terms
2x(3x2 + 2x – 5)
= 2x(3x2) + 2x(2x) + (2x)(–5)
= 6x3 + 4x2 – 10x Ans
b.
(2a – 3b)(5a + 2b)
= (2a)(5a) + (2a)(2b)
+ (–3b)(5a) + (–3b)(2b)
= 10a2 + 4ab – 15ab – 6b2
= 10a2 – 11ab – 6b2 Ans
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DIVISION
•
Procedure for dividing two terms:
–
Divide the numerical coefficients following the
procedure for division of signed numbers
–
Subtract the exponents of the literal factors of the divisor
from the exponents of the same letter factors of the
dividend
–
Combine numerical and literal factors
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DIVISION
•
Divide: (40a3b4c5)  (–4ab2c3)
40a 3b 4c 5  40  a 3  b 4  c5 
    2  3 
2 3
 4ab c   4  a  b  c 
= (–10)(a3 – 1)(b4 – 2)(c5 – 3)
= –10a2b2c2 Ans
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DIVISION
•
Procedure for dividing when the dividend consists
of more than one term:
–
–
•
Divide each term of the dividend by the divisor,
following the procedure for division of signed numbers
Combine terms
Divide:
24a 5b 4  16a 3b 5  36a 2b3c
 4a 2 b 2
24a 5b 4  16a 3b5 36a 2b3c



2 2
2 2
 4a b
 4a b
 4a 2 b 2
  6a3b 2  4ab3  9bc Ans
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POWERS
•
Procedure for raising a single term to a power:
–
–
–
Raise the numerical coefficients to the indicated power
following the procedure for powers of signed numbers
Multiply each of the literal factor exponents by the exponent
of the power to which it is raised
Combine numerical and literal factors
(2x2y3)2 = 22(x2)2(y3)2 = 4x4y6 Ans
•
Procedure for raising two or more terms to a power:
–
Apply the procedure for multiplying expressions that consist
of more than one term (FOIL)
(x + y)2 = (x + y)(x + y) = x2 + xy + xy + y2 = x2 + 2xy + y2 Ans
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ROOTS
•
Procedures for extracting the root of a term:
–
–
–
•
Determine the root of the numerical coefficient following
the procedure for roots of signed numbers
The roots of the literal factors are determined by dividing
the exponent of each literal factor by the index of the root
Combine the numerical and literal factors
Solve:
3
 8a b c
6 9 3
 3  8 (a 63 )(b 93 )(c 33 )
= –2a2b3c Ans
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REMOVAL OF PARENTHESES
•
•
Procedure for removal of parentheses preceded by a plus sign:
– Remove the parentheses without changing the signs of any terms within
the parentheses
– Combine like terms
7x + (–4x + 3y – 2) = –7x – 4x + 3y – 2 = –11x + 3y – 2 Ans
Procedure for removal of parentheses preceded by a minus sign:
– Remove the parentheses and change the sign of each term within the
parentheses
– Combine like terms
9a – (–4a + 2b – 6) = 9a + 4a – 2b + 6 = 13a – 2b + 6 Ans
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COMBINED OPERATIONS
•
Expressions that consist of two or more different operations are
solved by applying the proper order of operations
•
Simplify: 15x – 4(–2x) + x
= 15x + 8x + x
= 23x + x = 24x Ans
• Simplify: [3x – x + (x2y3)2]2
= [3x – x + x4y6]2
= [2x + x4y6]2
= (2x + x4y6)(2x + x4y6)
= 4x2 + 2x5y6 + 2x5y6 + x8y12
= 4x2 + 4x5y6 + x8y12 Ans
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BINARY NUMERATION SYSTEM
•
•
The binary number system uses only the two digits 0
and 1. These two digits are the building blocks for the
binary code that is used to represent data and program
instructions for computers
Place values for binary numbers are shown below:
25
32
24
16
23
8
22
4
21
2
20
1
·
2-1
0.5
2-2
0.25
2-3
0.125
Remember this can continue as far as needed in either direction
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EXPRESSING BINARY NUMBERS AS
DECIMAL NUMBERS
•
•
We use a subscript 2 to show that a number is binary
Express the binary number 1012 as a base 10 decimal
number:
1012 = 1(22) + 0(21) + 1(20) = 4 + 0 + 1 = 510 Ans
•
Express the decimal number 1910 as a binary number:
–
The largest power of two that will divide into 19 is 24 or 16
19 = 1(24) + 0(23) + 0(22) + 1(21) + 1(20) = 100112 Ans
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PRACTICE PROBLEMS
•
Perform the indicated operations and simplify:
1. 6a + 7a + 9b
2. (–3xy) + 4x + (–5xy2) + 5xy + (–7x)
3. –3.07ab + 7.69c + (–5.76ab) + 9d + (–11.2c)
4. 1/2x + (–2/3y) + 1/4z + (–1/3z) + 2/3x
5. 4x2y + (–5xy2) + 7xy2 + (–2x2y)
6. 7a – 3a
7. –10x – (–20x)
8. (3y2 – 4z) – (–2y2 + 4z)
9. –1 1/2ab – 1 2/3ab
10. (–2.04t2 + 7.6t – 7) – (3t2 – 6.7t – 4)
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PRACTICE PROBLEMS (Cont)
11.(3ab)(–4a2b2)
12.(–1/2x)(–1/3y2)(1/4x3)
13.(a – b)(a – b)
14.(2x2 – 3y)(–3x2 + 2y)
15.16y2  4y
16.1 1/3 a2b3  2/3ab
17.(2.4x3y3 + 4.8x2y2 – 24x)  1.2x
18.(x2y)3
19.(–2.1a2b3)2
20.(2/3 x3y3z2)3
21.(–2m2n + p3)2
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PRACTICE PROBLEMS (Cont)
121a 6b4c 2
22.
23.
24.
25.
26.
27.
28.
29.
3
 64 x12 y 9
4 6 8
m n
9
2x – (x – 2y)
–(x + y – z) – (x2 – y2 + z)
15 – (ab – a2b – b) – 4 + (ab – b)
(18a4b2)  (3a2b) – b3(b2)
5(2x – y)2 – (x2 – y2)
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PROBLEM ANSWER KEY
1. 13a + 9b
2. 2xy + (–3x) + (–5xy2)
3. –8.83ab + (–3.51c) + 9d
4. 7/6x + (–2/3y) + (–1/12z)
5. 2x2y + 2xy2
6. 4a
7. 10x
8. 5y2 – 8z
9. –3 1/6ab
10.–5.04t2 + 14.3t – 3
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PROBLEM ANSWER KEY (Cont)
11.–12a3b3
12.1/24x4y2
13. a2 – 2ab + b2
14.–6x4 + 13x2y – 6y2
15.4y
16.2ab2
17.2x2y3 + 4xy2 – 20
18. x6y3
19.4.41a4b6
20.8/27x9y9z6
21.4m4n2 – 4m2np3 + p6
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PROBLEM ANSWER KEY (Cont)
22.
23.
24.
25.
26.
27.
28.
29.
11a3b2c
– 4x4y3
2 3 4
3m n
x + 2y
–x2 + y2 – x – y
11 + a2b
6a2b – b5
19x2 – 20xy + 6y2
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