1. BASIC ALGEBRAIC OPERATIONS
Unit 13
BASIC ALGEBRAIC OPERATIONS

2. 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

3. 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)
Add the like terms and indicate the addition of the unlike terms

4. 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) Change the sign of each term in the subtrahend and follow the procedure for addition of signed numbers

5. 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)(x )(y )
= –1680x4y5 Ans

6. MULTIPLICATION
Procedure for multiplying expressions that consist of more than one term within an expression:
Multiply each term of one expression by each term of the other expression
Combine like terms
b (2a – 3b)(5a + 2b)
= (2a)(5a) + (2a)(2b)
+ (–3b)(5a) + (–3b)(2b)
a x(3x2 + 2x – 5)
= 2x(3x2) + 2x(2x) + (2x)(–5)
= 10a2 + 4ab – 15ab – 6b2
= 6x3 + 4x2 – 10x Ans
= 10a2 – 11ab – 6b2 Ans

7. 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

8. DIVISION Divide: (40a3b4c5)  (–4ab2c3)
= –10a2b2c2 Ans

9. 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:

10. (x + y)2 = (x + y)(x + y) = x2 + xy + xy + y2
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
Procedure for raising two or more terms to a power:
Apply the procedure for multiplying expressions that consist of more than one term (FOIL)
(2x2y3)2 = 22(x2)2(y3)2 = 4x4y6 Ans
(x + y)2 = (x + y)(x + y) = x2 + xy + xy + y2
= x2 + 2xy + y2 Ans

11. ROOTS Procedures for extracting the root of a term: Solve:
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:
= –2a2b3c Ans

12. 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
9a – (–4a + 2b – 6) = 9a + 4a – 2b + 6 = 13a – 2b + 6 Ans

13. COMBINED OPERATIONS Simplify: [3x – x + (x2y3)2]2
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

14. 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:
Remember this can continue as far as needed in either direction

15. EXPRESSING BINARY NUMBERS AS DECIMAL NUMBERS
We use a subscript 2 to show that a number is binary
Express the binary number as a base 10 decimal number:
1012 = 1(22) + 0(21) + 1(20) = = 510 Ans
Express the decimal number 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) = Ans

16. PRACTICE PROBLEMS Perform the indicated operations and simplify:
6a + 7a + 9b
(–3xy) + 4x + (–5xy2) + 5xy + (–7x)
–3.07ab c + (–5.76ab) + 9d + (–11.2c)
1/2x + (–2/3y) + 1/4z + (–1/3z) + 2/3x
4x2y + (–5xy2) + 7xy2 + (–2x2y)
7a – 3a
–10x – (–20x)
(3y2 – 4z) – (–2y2 + 4z)
–1 1/2ab – 1 2/3ab
(–2.04t t – 7) – (3t2 – 6.7t – 4)

17. PRACTICE PROBLEMS (Cont)
(3ab)(–4a2b2)
(–1/2x)(–1/3y2)(1/4x3)
(a – b)(a – b)
(2x2 – 3y)(–3x2 + 2y)
16y2  4y
1 1/3 a2b3  2/3ab
(2.4x3y x2y2 – 24x)  1.2x
(x2y)3
(–2.1a2b3)2
(2/3 x3y3z2)3
(–2m2n + p3)2

18. PRACTICE PROBLEMS (Cont)
22.
23.
24.
x – (x – 2y)
–(x + y – z) – (x2 – y2 + z)
– (ab – a2b – b) – 4 + (ab – b)
(18a4b2)  (3a2b) – b3(b2)
(2x – y)2 – (x2 – y2)

19. PROBLEM ANSWER KEY 13a + 9b 2xy + (–3x) + (–5xy2)
–8.83ab + (–3.51c) + 9d
7/6x + (–2/3y) + (–1/12z)
2x2y + 2xy2 4a
10x
5y2 – 8z
–3 1/6ab
–5.04t t – 3

20. PROBLEM ANSWER KEY (Cont)
–12a3b3
1/24x4y2
a2 – 2ab + b2
–6x4 + 13x2y – 6y2 4y
2ab2
2x2y3 + 4xy2 – 20
x6y3
4.41a4b6
8/27x9y9z6
4m4n2 – 4m2np3 + p6

21. PROBLEM ANSWER KEY (Cont)
11a3b2c
– 4x4y3
x + 2y
–x2 + y2 – x – y
11 + a2b
6a2b – b5
19x2 – 20xy + 6y2