. 20 R. 3• +9=64 20 X 408 Part I. Or Of Equa- cion. 84 R. 2. III. 9 1+2=16 R. 3. 20+9x=64* R. I. 2055* R. 2. 55 2. Solution of Questions producing fimple Equations. nominator may be taken away, by multiplying all solultion of a variety of useful problems, both in pure mathematics and phyfics, and also in the practical arts founded upon these sciences. In this place, we conThus, if =670 Also, if a sider the application of it to those questions where the quantities are expressed by numbers, and their magnix=ab + ac ax-b-cx tude alone is to be considered. And by trans. ax-x=b When an equation, containing only one unknown b quantity, is deduced from the question by the follow be fimple, it may be resolved by the preceding rules ; be resolved by the same quanity, it remains a true propofition. the rules afterwards to be explained. The examples in this chapter are so contrived, that the final equation may be fimple. equations only, are to be considered as general, and as General Rule. The unknown quantities in the question proposed must be expressed by letters, and the rela- 3b tions of the known and unknown quantities con tained in it, or the conditions of it, as they are call. If then x-4=16. ed, must be expressed by equations. These equa- tions being resolved by the rules of this science, will give the answer of the queftion. For example, if the question is concerning two num- sum of two numbers fought x+y=360 be 6o, that condition is exExamples of fimple Equations resolved by these Rules. prefled thus : If their difference must be 24, then xy=24 If their product is 1640, then xy=1640 4 2, = 2therefore These are some of the relations which are moft ea. fily expreffed. Many others occur which are less ob- 4* vious ; but as they cannot be described in particular rules, the algebraical expression of them is best explain ed by examples, and must be acquired by experience. No it, A 4 16 If their quotient mult be 6, then =6 5x5 2 R. 1. 5* 2 Equa 00. a a 4 a 11 }| 33–100+ 100= 162 162-3700 6 162—3700 of Equa- A diftinct conception of the nature of the question, added a half, a third part, and a fourth part of itself, of Equa. tions. tions, 3' Therefore, zt;tit. 50 3 242 +122 +82 +6z=1200 be found. 502= 1200 Rule. An equation involving the unknown quantity 224 must be deduced from the question (by the general If the operation be more complicated, it may be lowing part of his stock which is not so expended by a Let his firft ftock be there remains 47400 3 3 42-400 42-700 The fecond year be spends L. 100, and there } 41 3 3 He increases the remainder by one-third of 42-700, 42-700 162-2800 5 } 十 9 -2800 remains 9 9 162—3700_643–14800 He increases it by one-third 7 + 27 8 doubled; therefore 27 By R. 3. 9642--14800=542 By R. i. 10 1OZ=14800 112=1480 tion can be derived. Example. 3. Two persons, A and B, were talking of their : says A to B, Seven years ago I was just hence I shall be just twice as old as you will be. I ftion. A value of one of the unknown quantities demand their present ages. must be derived from each of the equations: and Let the ages of A tively 2 x7 and y-7 Seven years hence 3 x+7 and y+7 1. and 2. 4 x–73Xy-y=34-21 and if any third condition were assumed at pleasure, Also by Queft. 2. 5x+7= 2Xy+7=27+14 By 5. and transp. 7 *= 2y+7 9 y=21 101 x=49 of A and B then are 49 and 21, which an ages Vol. I. Part I. The were and 3. 3 F 410 2 2/+*+2 = 34 68-y 2 2 2 a IO 2 I 2 tions. Of Eqau. The operation might have been a little shortened by independent equations may be derived from a question of Equations. fubtracting the 4th from 5th, and thus 14 =-,+35; as there are unknown quantities in it, these quantities and thence g=21. therefore (by 6tb) x= (39–14) may be found by the resolution of equations. Examp. 6. To find three numbers, so that the first, with half the other two, the second with one-third Example. 4. A gentleman distributing money among of the other two, and the third with one-fourth of the other two, may each be equal to 34. Let the numbers be x, y, z, and the equations are ytz = 34 = 34 3 two are employed. There must be, however, two in 32+ ty 4 From the ift 4 x From the 2d 51x=102-3--Z From the 3d 6x=136-42-y 68-y-z =102-34-2 1512=15. 136-Z 8 5 5=6, and reduced 91y= 32-34 Examp. 5. A courier sets out from a certain place, and 32-34_136-Z 136— travels at the rate of 7 miles in 5 hours ; and 8 hours 8 and 9 5 after; another sets out from the same place, and tra Toth reduced 11|152-170=272-22 172=442 or z=26 By 8 and 5 131 before he is overtaken by the second ? y=22 and x=10 Let the number of hours y Examp. 7. To find a number confisting of three places, Then the fecond travelled whose digits are in arithmetical proportion; if this 23-8 The first travelled seven number be didvided by the sum of its digits, the quo 7y 5 tient will be 48 ; and if from the number be subtherefore in y hours tracted 198, the digits will be inverted. In like manner the second 59-40 4 (3:5::y--8: Let the 3 digits be 1 x, y, z travelled in y-8 hours miles 3 Then the number is 2100x+10y + same number of miles; 5 54 — 40 If the digits be 2 therefore by 3. and 4. 3 5 inverted, it is 3 10oz +10y +* Mult. 6257-200= 219 The digits are Transp. 4y= 200 4*+z=2y 8 therfore 100x+10y+z 5 (y-8=) 42 hours. =48 x+y+z By question 6100x+10y+2-198=100Z = . +10y+x From 6 and tranf. 7.99x=99z+198 Case III. When there are three or more unknown Divid. by 99 81x=2+2 quantities. |9f*= 2y--% Rule. When there are three unknown quantities, there Transp. 10|2y-2=2+2 must be three independent equations arifing from the Mult. 5. Ily=zt! queflion; and from each of these a value of one of Transp. 12100xti@y+z=48x+48y+487" 13 52x=38y+472 paring these three values, two equations will arise, for x and y 14 522+104=38z+38+472 . involving only two unknown quantities, which may Transp. therefore be resolved by the rule for Case 2. 115 332= 66 In like manner may the rule be extended to such Divid. 16 By=(z+1=)3 questions as contain four or more unknown quantities; (x=(2+2=)4. and hence it may be inferred, That, when just as many The number then is 432, which succeeds upon trial. 5 It 1 7 in ar. prop. From 4 8 and 9 Band is fublit.} of Equa- It sometimes happens, that all the unknown quan- part, the given quantities (being numbers) disappear Of Equa tions. tions. tities, when there are more than two, are not in all in the last conclusion, so that no general rules for like the equations expressing the conditions, and therefore cases can be deduced from them. But if letters are their original form. Hence also the connection of the quantities will appear in such a manner as to discover and difference are given. Let s be the sum given, and d the given difference. Also, let x and y be the two numbers fought. x+y=s メープd operations by the preceding rules may be much abrid Whence Sx=y ged. This, however, must be left to the skill and practice 2x=dty of the learner. A few examples are the following: d+y=smy 1. It is often easy to employ fewer letters than there 2y=smd sand 2. Sometimes it is convenient to express by letters, Thus, let the given sum be 100, and the difference 24. 76 &y= In the same manner may the canon be applied to a- std and . Let A perform the work in the time x, B in y, and 3 F 2 By y= 2 And x= stad 2 estd 124. 2 sod 2 a 412 tions. bas 11 :1: 200 9=)50. Il to the work performed by By the question: 54:1::-:) a than pso Part I. A L G E B R A. If particular values be inserted for these letters, a Of Involu1: :2:) A in a days particular folution will be obtained for that cafe. Lettion and Evolution. them denote the numbers in Example 5. 2 y:1::a:) Bin a days gra 4 Here it is obvious, that gr must be greater than ps, = else the problem is impoffible ; for then the value of x would either be infinite or negative. This limita- tion appears also from the nature of the question, as the first, in order to overtake him. For the rate of p the first courier is to the rate of the second as 9 += ; and ay tax=xy that is, as ps to qr; and therefore or must be greater Scholium. Sometimes when there are many known quantities. in a general solution, it may fimplify the operation to bc labe abc + abc Mult. 7th by express certain combinations of them by new letters, ftill to be confidered as known. labe abc CHAP. IV. t=ab . 29 Of Involution and Evolution. =batactal у From 13th subt. 2 abc 2abc In order to resolve equations of the higher orders, twice ioth 141Z - 2 abc Evolution. LEMMA. 2abc , 116 The reciprocals of the powers of a quantity may be. , x . Example in numbers. Let a=8 days, b=9 days, of the fame denomination. That is, the series a, 1, 34 23 =177 49 31 m, &c. a3 =a. Also, =1. and so on of others. limitation here mentioned. If a, b, and c, were such, that any of the quantities *, y, or z, became equal to o, it implies that one of the numerator or denominator of a fraction, may be Cor. 1. Hence any quantity which multiplies either : the agents did nothing in the work. "If the values of transposed from the one to the other, by changing the any of these quairtities be negative, the only suppor- fign of its index. . tion which could give them any meaning would be, 43 xy-2. &c. work, either obstructed it, or undid it to a certain exe 3x~ tent. Cor. 2. From this notation, it is evident that thefe negative powers, as they are called, are multiplied by Thus, a_axa_3=1-5. qra gr--ps 1. Of and c= 10; then x=14mm = and z=23,7 a as amo &c . may be expressed by a',ao,a-', a' air |