Application of Symbols in Addition and Subtraction：在加法和减法符号的应用

1、Copyright 2013 Michael Bush,1,Session 3 Basic Algebraic Operations,Industrial Application of BasicAlgebraic Principles,Copyright 2013 Michael Bush,2,Basic Algebraic Operations,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,After studying this unit you should be able to add an

2、d subtract single and multiple literal terms. remove parentheses that are preceded by plus or minus signs. simplify combined operations of literal term expressions. solve literal term problems.,Copyright 2013 Michael Bush,3,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,131 D

3、efinitions 331 132 Addition 332 133 Subtraction 334 UNIT EXERCISE AND PROBLEM REVIEW 356,Basic Algebraic Operations,Copyright 2013 Michael Bush,4,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Definitions Page 331,It is important that you understand the following definitions

4、in order to apply procedures that are required for solving problems involving basic operations.,A term of an algebraic expression is that part of the expression that is separated from the rest by a plus or minus sign. There are five terms in the following expression.,Copyright 2013 Michael Bush,5,In

5、dustrial Application of BasicAlgebraic Principles Session 3 In-Class,A factor is one of two or more literal and/or numerical values of a term that are multiplied. For example, 6 and x are each factors of 6x; 2, a2, and b are each factors of 2a2b;,Definitions Page 331,Copyright 2013 Michael Bush,6,In

6、dustrial Application of BasicAlgebraic Principles Session 3 In-Class,A numerical coefficient is the number factor of a term. The letter factors of a term are called literal factors. For example, in the term 7xy, 7 is the numerical coefficient; x and y are the literal factors.,Definitions Page 331,Co

7、pyright 2013 Michael Bush,7,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Like terms are terms that have identical literal factors, including exponents. The numerical coefficients do not have to be the same. For example, 3a and 10a are like terms; 2x2y and 5x2y are like term

8、s.,Definitions Page 332,Copyright 2013 Michael Bush,8,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Unlike terms are terms that have different literal factors or exponents. For example, 8x and 8y are unlike terms. The terms 12xy, 4x2y, and 5xy2 are unlike terms. Although the

9、 literal factors are x and y in each of the terms, they are raised to different powers.,Definitions Page 332,Copyright 2013 Michael Bush,9,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Addition Page 332,Only like terms can be added. The addition of unlike terms can only be i

10、ndicated. As in arithmetic, like things can be added, but unlike things cannot be added. For example, 2 inches and 3 inches = 5 inches. Both values are like units of measure. Both are inches, therefore, they can be added. It can be readily seen that 2 inches and 3 pounds cannot be added. Inches and

11、pounds are unlike units of measure.,Copyright 2013 Michael Bush,10,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Adding Fractional Terms,Denominators must be the same, + 1 =,Both have 8 as denominator,Proceed, + 2 =,Denominators different,Cannot Add,Copyright 2013 Michael Bu

12、sh,11,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Adding Algebraic Terms,Literal factors must be the same,3w + 5w =,Both have literal factor w,Proceed,3w + 5z =,Literal factors different,Cannot Add,Copyright 2013 Michael Bush,12,Industrial Application of BasicAlgebraic Pri

13、nciples Session 3 In-Class,Procedure for Adding Like Terms Page 332,Group all like terms in the expression,Add the numerical coefficients, applying the procedure for addition of signed numbers.,Leave the literal factors unchanged.,NOTE: If a term does not have a numerical coefficient, the coefficien

14、t 1 is understood. For example, x = 1x,Copyright 2013 Michael Bush,13,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Example Page 332,1. Add 5x and 10 x,Both terms have the identical literal factor, x. These terms are like terms.,Add the numerical coefficients.,Leave the lite

15、ral factor unchanged.,5 + 10 =,5x + 10 x = 15x,15,Copyright 2013 Michael Bush,14,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,2. R + (-12R ),Both terms have the identical literal factor, R. These terms are like terms.,Add the numerical coefficients.,Leave the literal factor

16、 unchanged.,1 + (-12) =,R + (-12R ) = -11R,-11,Example Page 332,Copyright 2013 Michael Bush,15,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,3. -7xy + 7xy,Both terms have the identical literal factors, xy. These terms are like terms.,Add the numerical coefficients.,Leave the

17、 literal factor unchanged.,-7 + 7=,-7xy + 7xy = 0 xy or zero,0,Example Page 332,Copyright 2013 Michael Bush,16,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,4. -14a2b3 + (-6a2b3),Both terms have the identical literal factors, a2b3. These terms are like terms.,Add the numeric

18、al coefficients.,Leave the literal factor unchanged.,-14 - 6 =,-14a2b3 + (-6a2b3) =,-20,-20a2b3,Example Page 332,Copyright 2013 Michael Bush,17,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,5. 3CD + (-5CD) + 8CD,Both terms have the identical literal factors, CD. These terms

19、are like terms.,Add the numerical coefficients.,Leave the literal factor unchanged.,3 + (-5) + 8 =,3CD + (-5CD) + 8CD = 6CD,6,Example Page 332,Copyright 2013 Michael Bush,18,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Adding Unlike Terms Page 332,Rules for Adding Unlike Te

20、rms,Group all like terms in the expression,Add the numerical coefficients, applying the procedure for addition of signed numbers.,Do not add any terms that are not like terms.,Leave the unlike terms unchanged.,Copyright 2013 Michael Bush,19,Industrial Application of BasicAlgebraic Principles Session

21、 3 In-Class,1. Add 13 and x.,The literal factors are not identical.,These terms are unlike.,Indicate the addition.,13 + x = 13 + x,Example Page 332,Copyright 2013 Michael Bush,20,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,2. Add 12M and 8P.,The literal factors are not ide

22、ntical.,These terms are unlike.,Indicate the addition.,12M + 8P = 12M + 8P,Example Page 332,Copyright 2013 Michael Bush,21,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,3. Add 4W and 9W2.,The literal factors are not identical.,These terms are unlike.,Indicate the addition.,4

23、W + 9W2 = 4W + 9W2,Example Page 332,Copyright 2013 Michael Bush,22,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,4. Add 7a, 5b, and 10ab.,The literal factors are not identical.,These terms are unlike.,Indicate the addition.,7a + 5b + 10ab = 7a + 5b + 10ab,Example Page 332,Co

24、pyright 2013 Michael Bush,23,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Procedure for Adding Expressions that Consist of Two or More Unlike Terms Page 333,Group all like terms in the expression,Add the numerical coefficients of the like terms, applying the procedure for a

25、ddition of signed numbers.,Indicate the addition of the unlike terms.,Copyright 2013 Michael Bush,24,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Example Page 333,1. Add 7x + (-xy) + 5xy2 and -2x +3xy + (-6xy2),Add numerical coefficients of xs.,7x + (-xy) + 5xy2 + (-2x) + 3

26、xy + (-6xy2) =,Add numerical coefficients of xys.,Add numerical coefficients of xy2s.,7x + (-2x) =,5x,-xy + 3xy =,2xy,5xy2 + (-6xy2) =,-xy2,5x + 2xy - xy2,Copyright 2013 Michael Bush,25,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,2. Add 3c + (-4d), c + 10d + (-4cd), and -1

27、2c + (-5cd) + cd2,Add numerical coefficients of cs.,3c -4d+c + 10d -4cd+-12c -5cd + cd2 =,Add numerical coefficients of ds.,Ind numerical coefficient of cd2.,3c + c +(-12c) =,-8c,-4d + 10d =,6d,cd2 =,cd2,-8c + 6d 9cd + cd2,Add numerical coefficients of cds.,-4cd + (-5cd) =,-9cd,Example Page 333,Copy

28、right 2013 Michael Bush,26,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,EXERCISE 132 Page 333,2. Add 6b and 12b,These expressions consist of groups of single terms. Add these terms.,Like terms,18b,4. Add 7x and 3x,Like terms,10 x,6. Add 15xy and 7xy,Like terms,22xy,8. Add 2

29、5m2 and m2,Like terms,26m2,10. Add 4c3 + 0,Unlike terms,4c3,Copyright 2013 Michael Bush,27,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,EXERCISE 132 Page 333,12. Add 0.4x and 0.8x,These expressions consist of groups of single terms. Add these terms.,Like terms,1.2x,14. Add

30、0.05y and 0.006y,Like terms,0.056y,16. Add 2c2d and -3c2d,Like terms,-c2d,18. Add -50.6abc and 50.5abc,Like terms,-0.1abc,Copyright 2013 Michael Bush,28,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,EXERCISE 132 Page 333,20. Add -0.3dt2, -1.7dt2, -dt2,These expressions consi

31、st of groups of single terms. Add these terms.,Like terms,-3.0dt2,22. Add 20.06D, -19.97D, and -0.9D,Like terms,-0.79D,Copyright 2013 Michael Bush,29,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,26. (2a + 6b + 3c), (a + 5b + 4c),Add numerical coefficients of as.,(2a + 6b +

32、3c) + (a + 5b + 4c) =,Add numerical coefficients of bs.,Add numerical coefficients of cs.,2a + a=,3a,6b + 5b =,11b,3c + 4c =,7c,3a + 11b + 7c,EXERCISE 132 Page 333,These expressions consist of groups of two or more terms. Group like terms and add.,Copyright 2013 Michael Bush,30,Industrial Applicatio

33、n of BasicAlgebraic Principles Session 3 In-Class,28. 6a + (-10ab), (-a) + 12ab,Add numerical coefficients of as.,6a + (-10ab) + (-a) + 12ab =,Add numerical coefficients of abs.,6a + (-a) =,5a,-10ab + 12ab =,2ab,5a + 2ab,EXERCISE 132 Page 333,These expressions consist of groups of two or more terms.

34、 Group like terms and add.,Copyright 2013 Michael Bush,31,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,EXERCISE 132 Page 333,These expressions consist of groups of two or more terms. Group like terms and add.,30. (-8cd) + 7c2d + 14cd2 , 7cd + (-12c2d) + (-17cd2),Add numeric

35、al coefficients of cds.,-8cd + 7cd =,-cd,Add numerical coefficients of c2ds.,7c2d + (-12c2d) =,-5c2d,(-8cd) + 7c2d + 14cd2 + 7cd + (-12c2d) + (-17cd2) =,-cd -5c2d -3cd2,Add numerical coefficients of cd2s.,14cd2 + (-17cd2) =,-3cd2,Copyright 2013 Michael Bush,32,Industrial Application of BasicAlgebrai

36、c Principles Session 3 In-Class,EXERCISE 132 Page 333,These expressions consist of groups of two or more terms. Group like terms and add.,32. 31.3M + (-3N), (-8M) + 0.5N, 20M + (-0.7N),Add numerical coefficients of Ms.,31.3M - 8M + 20M =,43.3M,Add numerical coefficients of Ns.,-3N + 0.5N -0.7N =,-3.

37、2N,31.3M + (-3N) + (-8M) + 0.5N + 20M + (-0.7N) =,43.3M 3.2N,Copyright 2013 Michael Bush,33,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,EXERCISE 132 Page 333,These expressions consist of groups of two or more terms. Group like terms and add.,34. 0.5T + (-2.8T2) + (-T3), 5.

38、5T2 + 0.7T3 ,Add numerical coefficients of Ts.,0.5T =,0.5T,Add numerical coefficients of T3s.,-2.8T2 + 5.5T2 =,2.7T2,0.5T + (-2.8T2) + (-T3) + 5.5T2 + 0.7T3 =,0.5T + 2.7T2 - 0.3T3,Add numerical coefficients of T2s.,-T3 + 0.7T3 =,-0.3T3,Copyright 2013 Michael Bush,34,Industrial Application of BasicAl

39、gebraic Principles Session 3 In-Class,EXERCISE 132 Page 333,These expressions consist of groups of two or more terms. Group like terms and add.,36. 1P + (-V) + (PV) , P + V + (-2PV) + (-P2V),Add numerical coefficients of Ps.,1P + P =,2P,Add numerical coefficients of PVs.,-V+ V =,V,1P + (-V) + (PV) +

40、 P + V + (-2PV) + (-P2V) =,2P + V - 1PV -P2V,Add numerical coefficients of Vs.,PV -2PV =,-1PV,Ind literal term of P2V.,-P2V =,-P2V,Copyright 2013 Michael Bush,35,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,EXERCISE 132 Page 334,0.15x + 0.08x + 0.14x + 0.28x +0.10 x + 0.25x

41、 =,38. The total voltage (Et), in volts, of the circuit shown is represented as x. The amount of voltage taken by each of the six resistors is given as Eg1 through Eg6. What is the sum of the voltage taken by the resistors listed?,R1 and R2 b. R3 and R4 c. R4, R5, and R6 d. All 6 resistors,0.15x + 0

42、.08x =,0.23x,0.14x + 0.28x =,0.42x,0.28x + 0.10 x + 0.25x =,0.63x,x,Copyright 2013 Michael Bush,36,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Subtracting Algebraic Terms,Literal factors must be the same,3w - 5w =,Both have literal factor w,Proceed,3w - 5z =,Literal factor

43、s different,Cannot Subtract,Copyright 2013 Michael Bush,37,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Procedures for Subtracting Like Terms Page 334,Subtract the numerical coefficients applying the procedure for subtraction of signed numbers.,Leave the literal factors unc

44、hanged.,Copyright 2013 Michael Bush,38,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Example Page 334,1. 14xy - (6xy),Both terms have the identical literal factors, xy. These terms are like terms.,Subtract the numerical coefficients.,Leave the literal factor unchanged.,14 -

45、6 =,14xy - (6xy) = 8xy,8,Copyright 2013 Michael Bush,39,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,2. 16H - 13H,Subtract the numerical coefficients.,Leave the literal factor unchanged.,16 - 13 =,16H - 13H = 3H,3,Example Page 334,Copyright 2013 Michael Bush,40,Industrial A

46、pplication of BasicAlgebraic Principles Session 3 In-Class,3. ab - 10ab,Subtract the numerical coefficients.,Leave the literal factor unchanged.,1 - 10 =,ab - 10ab = -9ab,-9,Example Page 334,Copyright 2013 Michael Bush,41,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,4. 4L2

47、- 7L2,Subtract the numerical coefficients.,Leave the literal factor unchanged.,4 - 7 =,4L2 - 7L2 = -3L2,-3,Example Page 334,Copyright 2013 Michael Bush,42,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,5. -15x2y - (-15x2y),Subtract the numerical coefficients.,Leave the litera

48、l factor unchanged.,-15 (-15) =,-15x2y - (-15x2y) = 0,0,Example Page 334,Copyright 2013 Michael Bush,43,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Procedure for Subtracting Unlike Terms Page 335,The subtraction of unlike terms can only be indicated.,Literal factors remain

49、 unchanged.,Copyright 2013 Michael Bush,44,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Examples Page 335,1. Subtract 3b from 4a.,The literal factors are not identical. These terms are unlike.,Indicate the subtraction.,4a 3b = 4a 3b,Copyright 2013 Michael Bush,45,Industrial

50、 Application of BasicAlgebraic Principles Session 3 In-Class,2. Subtract 5xy2 from 16xy.,The literal factors are not identical. These terms are unlike.,Indicate the subtraction.,16xy 5xy2 = 16xy 5xy2,Example Page 335,Copyright 2013 Michael Bush,46,Industrial Application of BasicAlgebraic Principles

51、Session 3 In-Class,3. Subtract 2P from PT.,The literal factors are not identical. These terms are unlike.,Indicate the subtraction.,PT 2P = PT 2P,Examples Page 335,Copyright 2013 Michael Bush,47,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,Procedure for Subtracting Expressi

52、ons that Consist of Two or More Terms Page 335,Group like terms,Subtract like terms and indicate the subtractions of the unlike terms.,NOTE: Each term of the subtrahend is subtracted following the procedure for subtraction of signed numbers.,Copyright 2013 Michael Bush,48,Industrial Application of B

53、asicAlgebraic Principles Session 3 In-Class,1. Subtract 6a + 8b - 5c from 9a - 13b + 7c.,(9a - 13b + 7c) (6a + 8b - 5c) = 3a 21b + 12c,Subt numerical coefficients of as.,9a 6a =,3a,Subt numerical coefficients of cs.,-13b 8b =,-21b,Subt numerical coefficients of bs.,7c (-5c) =,12c,Examples Page 335,C

54、opyright 2013 Michael Bush,49,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,2. Subtract (4x2 + 6x - 15xy) - (9x2 - x - 2y + 5y2),(4x2 + 6x - 15xy) - (9x2 - x - 2y + 5y2) =,Subt numerical coefficients of x2s.,4x2 9x2 =,-5x2,Ind literal term of xy.,6x (-x) =,7x,Subt numerical

55、coefficients of xs.,-15xy =,-15xy,Ind literal term of y.,-(-2y) =,2y,-5y2=,-5y2,Ind literal term of y2.,-5x2 + 7x - 15xy + 2y - 5y2,Examples Page 335,Copyright 2013 Michael Bush,50,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,EXERCISE 133 Page 335,2. 5x - (3x),These express

56、ions consist of groups of single terms. Subtract terms as indicated.,2x,4. 6ab - (-4ab),10ab,6. 16xy - (-16xy),32xy,8. -12c2 - c2,-13c2,10. 0 - 16M,-16M,Copyright 2013 Michael Bush,51,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,EXERCISE 133 Page 335,12. 0.5x2y - 1.2x2y,The

57、se expressions consist of groups of single terms. Subtract terms as indicated.,-0.7x2y,14. -12ax4 - 7ax4,-19ax4,16. 8.12n - 8.82n,-0.7n,18. -c2d - (- c2d ),0,20. H - 1H,- 1H,Copyright 2013 Michael Bush,52,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,26. (5x + 9xy) - (2x 6xy

58、),Subt numerical coefficients of xs.,(5x + 9xy) - (2x 6xy) =,Subt numerical coefficients of xys.,5x 2x =,3x,9xy 6xy =,3xy,3x + 3xy,EXERCISE 133 Page 336,These expressions consist of groups of two or more terms. Group like terms and subtract.,Copyright 2013 Michael Bush,53,Industrial Application of B

59、asicAlgebraic Principles Session 3 In-Class,28. (10y2 - 2) - (10y2 - 2),Subt numerical coefficients of y2s.,(10y2 - 2) - (10y2 - 2) =,Subt numerical coefficients.,10y2 10y2 =,0,-2 (-2) =,0,0,EXERCISE 133 Page 336,These expressions consist of groups of two or more terms. Group like terms and subtract

60、.,Copyright 2013 Michael Bush,54,Industrial Application of BasicAlgebraic Principles Session 3 In-Class,30. (ab - a2b + ab2) - 0,Ind literal term of ab.,(ab - a2b + ab2) - 0 =,Ind literal term of a2b.,Ind literal term of ab2.,ab =,ab,a2b =,a2b,ab2 =,ab2,ab - a2b + ab2,EXERCISE 133 Page 336,These exp