Chapter 6 Solving equations


 Bertram Armstrong
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1 Chpter 6 Solving equtions Defining n eqution 6.1 Up to now we hve looked minly t epressions. An epression is n incomplete sttement nd hs no equl sign. Now we wnt to look t equtions. An eqution hs n = sign nd letter (vrible) in it. It is complete sttement. + is n epression or sometimes we cll it function. But + = 7 is n eqution. It hs n = sign in it nd the unknown vrible,. This sttement is only true for some specific vlue of. In this cse = becuse () + = 7. In other words the LHS will only equl the RHS when is. Now consider + =. In this cse lthough the sttement hs n equl side the LHS = RHS for ll vlues of. We cll this n identity. With the RHS we cn see it is not n identity. So it is lso n eqution but this hs two unknown vribles, nd y in it. + y = is n eqution with two unknowns, nd y lthough the ppers twice in the eqution. E6.1 (i) Is + = + n eqution? How mny vribles? (ii) Is + = 7 n eqution? Eplin. (iii) Is + y + z = 10 n eqution? How mny vribles? (iv) Is 7  = n eqution? How mny vribles? (v) Is 7 n eqution? Eplin. A6.1 (i) + = + :Yes, n eqution with 1 vrible. (ii) + = 7: No, it is n identity. (iii) + y + z = 10:Yes, n eqution with vribles. (iv) 7  = :Yes, n eqution with 1 vrible. (v) 7 : No, it is n epression. 6. Compre + = with +=. + = is n eqution becuse it is only true for =. Now consider + =. This looks like n eqution but is strictly not n eqution. Consider the LHS: + is. This the sme s the RHS. So this eqution is true for ny or ll vlues of. It is n identity rther thn n eqution. So when the LHS cn be shown to equl the RHS, we cll it n 68
2 identity. E6. Is = (+1) n eqution? Consider the LHS nd RHS. A6. It is n identity s the RHS = LHS (epnd the RHS s perfect squre) 6. We will be concerned minly with equtions, rther thn identities. We wnt to lern how to solve n eqution. This mens finding the one (or ) vlue(s) for the unknown vrible tht mkes the sttement true i.e. LHS = RHS. We hve seen tht we cn test for solution by verifiction. For emple: Test if = is solution of the eqution + =. We substitute = in the LHS viz: LHS = () + = 6 + = 10 Now substitute = in the RHS viz: () = 10 = LHS. So LHS = RHS so = is solution. E6. Test if = is solution of the eqution of the eqution: = 7 6. A6. = 7 6. RHS= ( ) = 1 LHS = 7() 6 = 1 Yes = is solution. 6. But how do we solve n eqution when we do not hve vlue to substitute? We re going to lern some techniques to do this. Consider the very simple eqution: + 7 = 10. We cn guess the result =. So one wy is to obtin solution is by inspection. This pplies to esy, convenient cses only. E6. Solve by inspection : 1 = 8. A6. = 0 6. Let s be more introspective: Solve + 1 =. By inspection we know = 9 is the solution. So we cn write: + 1 = Note: LHS hs terms nd the RHS hs 1 term. 69
3 = 1 Note: LHS hs 1 term nd the RHS hs terms. So = 9. Observe tht the term 1 hs moved over the = sign nd hs chnged sign. This shows us tht when we move term over the = sign it chnges sign. E6. Solve: + = 70 by moving the constnt term from the LHS to the RHS. A6. + = 70 = 70 So = Similrly with 1 =  we proceed: = = 10, moving the 1 term. E6.6 Solve. =. A6.6 Solve. =. =. +. = Now consider solving the eqution: = 1 We know the solution by inspection is =. Formlly we write: 1 1 Note the fctor of hs moved from the numertor of the LHS to the denomintor of the RHS i.e. it hs been cross multiplied. E6.7 Solve = 100 by moving the fctor from the LHS to the RHS. A Now solve. 7 We get: = (7) = 1 Another emple: Solve 70
4 7 (7) So when we move fctor over the = sign it is cross multiplied i.e. moves from numertor(denomintor) of one side of the = sign to the denomintor(numertor) of the other side. E6.8 Solve (i) = 1 (ii) 1 (iii) 1 A6.8 (i) 1 1 ( ii) 1 1() 60 (iii) 1 1() () 0 1 Lyout of solution to n eqution 6.9 When we were deling with epressions we joined or connected ech vrint of the epression with n = sign. e.g = + 6. With equtions we must not use the = sign or rther use it with cution i.e. do not introduce new = sign in ddition to the one tht is lredy there. For emple in the eqution of bo 6.6 writing the following is not correct: 1 1 The first = sign of the second line is not pproprite becuse the suggestion is tht 1 =. Insted of = sign we use the therefore symbol, or the impliction rrow or i.e. 71
5 Use ny dditionl, new = signs only between trivil numericl vlues. E6.9 Replce ny incorrect = signs in the following: + 7 = 7 = = 7 7 = 0 A = 7 = 7 7 = 0 Further tools for solving equtions 6.10 We will develop technique tht involves Unpcking n epression until we isolte the unknown vrible. Wht does this men? Lets tke non mthemticl emple: Suppose I (mke) pck present s follows. I tke cricket bll, wrp it in foil then plce it in bubble wrp then plce it in bo nd then sel the bo with tpe. To isolte the bll the present will hve to unpcked in the right order: Remove the tpe, then remove from the bo then remove bubble wrp nd then finlly remove the foil. E6.10 How do you undo: put my socks on then put more shoes on! A6.10 Remove shoes then socks Here is n eqution with mied mode epressions (ddition with multipliction) = 10. To solve this eqution we will need to isolte (UNPACK) the by moving both term nd fctor to the RHS. But in the right order. Wht is the order? The left hnd side shows tht the hs been multiplied by 7 first nd then hs 7 dded to it. So to unpck we first move the + 11 term from the LHS to the RHS nd then move the fctor 7 to the LHS. This is shown below: (move term from LHS to RHS) (move fctor from LHS to RHS) So = is the solution. Check: LHS = 7() + 11 = = 10 = RHS. Are the = signs ok? Additionl = signs in line cn be used to do trivil simplifictions. E6.11 Solve: 1 = 1 7
6 A (move term from LHS to RHS) (move fctor from LHS to RHS) 6.1 Now consider this cse where the unknown ppers more thn once in the eqution: + 7 =. Move terms so tht the terms in the unknown, re on one side of the equl sign nd the constnt terms re on the other side: E6.1 Solve = 7 (simplify both sides) A Trck this emple: It is Ok to hve the unknown on the RHS if it is convenient. E6.1 Solve 8 = by hving the unknown terms on the RHS. A Now consider 8 = 1 8. Note the coefficients of the terms re both negtive. 7
7 We cn chnge the sign of ech term of both sides of the eqution: Chnge the sign of ech term of 8 = 1 8 to get E6.1 Solve 10 = 1 by chnging the sign of ech term first. A Now consider the eqution: ( ) = 0 We hve choice of unpcking s follows: (  ) Or first removing the brckets viz: (  ) E6.1 Solve ( 11) = 10 by unpcking directly nd then by removing brckets first. 7
8 A6.1 ( 11) First removing the brckets: ( 11) Now consider the lgebric emples: i. + b = c ii. ( b) = c b c i. c  b c  b OR c b ( b) c c ii. b c  b OR c  b ( b) c b c repet of ii. c  b c  b c b c OR   b E6.16 Solve : d + ( b) = e by both methods. A6.16 d ( b) e ( b) e  d e  d b e  d e d b OR b Alterntive method: d ( b) e d b e e  b  d e  b  d OR e b d   b e d  7
9 6.17 Consider the eqution: Note the unknown, is in the denomintor. E6.17 A6.17 Trck the following solution: Now tke reciprocls of both sides to get 1. is now in the numertor. Solve Now tke reciprocls of both sides to get 1. is now in the numertor Now consider the lgebric cse: b c c b 1 c  b c  b Note: 1 c 1 b is incorrect. See * below. *Remember the reciprocl of difference is not the difference of the reciprocls. 76
10 E6.18 Solve c b A6.18 c b b 1 c b c b c 6.19 Consider the eqution + = 17. We cn proceed by the unpcking method s follows: nd  Note the unpcking of the squre is by squre rooting. The proper lnguge is the inverse of squring is squre rooting. Note lso the squre root provides two roots, +ve & ve. E6.19 Solve 11 = 7 A nd. 6.0 Now we will use the inverse of rooting is squring to unpck this eqution: ( ) = 1 77
11 (  ) E6.0 Solve = A Zero Product Equtions (ZPE) 6.1 Wht cn you sy bout the? in the following: (?)(?) = 0. We cn sy tht t lest one of the? is zero. If product is zero then t lest one fctor of product must be zero. Here is n emple of ZPE: y = 0.The LHS is product nd the RHS is zero. So in the ZPE eqution, y = 0 we hve two solutions: = 0; y =0 Now (+) = 0 is lso ZPE. By setting ech fctor to zero we get the two solutions: = 0; And +=0 i.e. = . E6.1 Solve ( ) = 0 A6.1 ( ) = 0 Setting ech fctor = 0 we get. = 0 ; = 0 =. 6. Consider the emple: (+)() = 0 This is ZPE. So + = 0 i.e. =  And  = 0 i.e. = i.e. = E6. Solve ()() = 0 78
12 A6. In the ZPE, ()() = 0 set ech fctor to zero to get: = 0 =. And = 0 =. Qudrtic equtions 6. Here is qudrtic eqution: + = 0 The is the qudrtic term. In its present form it is not ZPE. To solve it, mke it into ZPE by fctorising the LHS. Note the RHS is lredy zero. + = 0 i.e. (+) = 0 So = 0 nd = . E6. Solve:  = 0 A6.  = 0 ( ) = 0 It is now ZPE. = 0 nd =. 6. Now consider the qudrtic eqution: + = . Note the RHS is not zero. To mke this into ZPE we first mke the RHS = 0 by moving the term,  from the LHS to the RHS viz: + + = 0. Now mke LHS into product. So let s fctorise the LHS using our qudrtic pir rule. ( + )( + 1) = 0 Note: Now we hve ZPE. So solve to get =  nd = 1. E6. Solve the qudrtic eqution, + =  A6. + = . Mke the RHS = 0: + + = 0. Fctorise the LHS using our qudrtic pir rule: ( + )( + ) = 0 79
13 = . Note only 1 (repeted) root You my spotted tht the LHS is perfect squre. (+) = 0 So = Here is nother qudrtic eqution in which the coefficient of the qudrtic term is not 1. Solve + = 0 To mke this into ZPE we need to fctorise the LHS. We will find the qudrtic pir nd then use grouping (  ) 1(  ) 0 ( 1)(  ) And  0 E6. Solve 9 = 0 Hint: (?)(?) = 10? +? = 9 A6. 9 = 0 Now: (10)(1) = = (  ) 1(  ) 0 ( 1)(  ) And 80
14 81
15 Eercise 6 1. Solve by unpcking i. ( + ) = 1 ii. ( ) =  iii. ( ) =10 iv. ( ) = 1. Solve the equtions by converting to ZPE: (i) + 7 = 0 (ii) 8 = 0 (iii) X + = 9. Solve the eqution 8 = 0 8
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