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)– Group like terms in the same column– Add the like terms and indicate the

addition of the unlike terms

Ansyxyx

yxyxyxyx

374

342536

2

2

2

4

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

SUBTRACTION

22

22

10581577

yxyxyxyx

Ansyxyx 22 5215

5

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

MULTIPLICATION

= (–4)(5)(–6)(7)(–2)(x1 + 2 + 1)(y1 + 1 + 3)

= –1680x4y5 Ans

6

• 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

MULTIPLICATION

a. 2x(3x2 + 2x – 5)

b. (2a – 3b)(5a + 2b)

= 10a2 + 4ab – 15ab – 6b2

= 10a2 – 11ab – 6b2 Ans

= 2x(3x2) + 2x(2x) + (2x)(–5)

= 6x3 + 4x2 – 10x Ans

= (2a)(5a) + (2a)(2b) + (–3b)(5a) + (–3b)(2b)

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)

3

5

2

43

32

543

4

40

4

40

c

c

b

b

a

a

cab

cba

= (–10)(a3 – 1)(b4 – 2)(c5 – 3)

= –10a2b2c2 Ans

9

DIVISION

22

325345

4

361624

ba

cbababa

22

32

22

53

22

45

4

36

4

16

4

24

ba

cba

ba

ba

ba

ba

Ansbcabba 946 323

• 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

POWERS

(x + y)2 = (x + y)(x + y) = x2 + xy + xy + y2 = x2 + 2xy + y2 Ans

• 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

11

• 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:

ROOTS

3 3968 cba ))()((8 3339363 cba

= –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

– Combine like terms

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

13

• Expressions that consist of two or more different operations are solved by applying the proper order of operations

• Simplify: 15x – 4(–2x) + x

COMBINED OPERATIONS

= 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

25 24 23 22 21 20 · 2-1 2-2 2-3

32 16 8 4 2 1 0.5 0.25 0.125

• 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 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

16

PRACTICE PROBLEMS

• Perform the indicated operations and simplify:1. 6a + 7a + 9b2. (–3xy) + 4x + (–5xy2) + 5xy + (–7x)3. –3.07ab + 7.69c + (–5.76ab) + 9d + (–11.2c)4. 1/2x + (–2/3y) + 1/4z + (–1/3z) + 2/3x5. 4x2y + (–5xy2) + 7xy2 + (–2x2y)6. 7a – 3a7. –10x – (–20x)8. (3y2 – 4z) – (–2y2 + 4z)9. –1 1/2ab – 1 2/3ab10. (–2.04t2 + 7.6t – 7) – (3t2 – 6.7t – 4)

17

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 4y16. 1 1/3 a2b3 2/3ab17. (2.4x3y3 + 4.8x2y2 – 24x) 1.2x18. (x2y)3

19. (–2.1a2b3)2

20. (2/3 x3y3z2)3

21. (–2m2n + p3)2

18

22.

23.

24.25. 2x – (x – 2y)26. –(x + y – z) – (x2 – y2 + z)27. 15 – (ab – a2b – b) – 4 + (ab – b)28. (18a4b2) (3a2b) – b3(b2)29. 5(2x – y)2 – (x2 – y2)

PRACTICE PROBLEMS (Cont)

246121 cba

3 91264 yx

86

9

4nm

19

PROBLEM ANSWER KEY

1. 13a + 9b2. 2xy + (–3x) + (–5xy2)3. –8.83ab + (–3.51c) + 9d4. 7/6x + (–2/3y) + (–1/12z)5. 2x2y + 2xy2

6. 4a7. 10x8. 5y2 – 8z9. –3 1/6ab10. –5.04t2 + 14.3t – 3

20

PROBLEM ANSWER KEY (Cont)

11. –12a3b3

12. 1/24x4y2

13. a2 – 2ab + b2

14. –6x4 + 13x2y – 6y2

15. 4y16. 2ab2

17. 2x2y3 + 4xy2 – 2018. x6y3

19. 4.41a4b6

20. 8/27x9y9z6 21. 4m4n2 – 4m2np3 + p6

21

PROBLEM ANSWER KEY (Cont)

22. 11a3b2c23. – 4x4y3

24. 25. x + 2y26. –x2 + y2 – x – y27. 11 + a2b28. 6a2b – b5

29. 19x2 – 20xy + 6y2

23 m3n4