# The rule of equation, commonly

The rule of equation, commonly called Algebers Rule.

<side: The rule of equation.> HEtherto haue I taughte you, the common formes of worke, in nombers Denominate. Whiche rules are vsed also in nôbers Abstracte, & likewaies in Surde nombers. Although the formes of these workes be seueralle, in eche kinde of nomber. But now will I teache you that rule, that is the principall in Coßike woorkes: and for whiche all the other dooe serue.

This Rule is called the Rule of Algeber, after the name of the inuentoure, as some men thinke: or by a name of singular excellencie, as other iudge. But of his vse it is rightly called, the rule of equation: bicause that by equation of nombers, it doeth dissolue doubtefull questions: And vnfolde intricate ridles. And this is the order of it.

The somme of the rule of equation:

WHen any question is propoûded, apperteinyng to this rule, you shall imagin a name for the nomber, that is to bee soughte, as you remember, that you learned in the rule of false position. And with that nomber shall you procede, accordyng to the question, vntil you finde a Coßike nomber, equalle to that nomber, that the question expresseth, whiche you shal <page: Ff. i.> reduce euer more to the leaste nombers. And then diuide the nomber of the lesser denomination, by the nomber of the greateste denomination, and the quotient doeth aunswere to the question. Except the greater denominatiô, doe beare the signe of some rooted nôber. for then must you extract the roote of that quotiente, accordyng to that signe of denomination.

Scholar. It semeth that this rule is all one, with the rule of false position: and therefore mighte so bee called: seyng it taketh a false nôber, to worke with al.

Master. This rule doeth farre excell that other. And dooeth not take a false nomber, but a true nomber for his position, as it shall bee declared anon. Wherby it maie bee thoughte, to bee a rule of wonderfull inuention, that teacheth a manne at the firste worde, to name a true nomber, before he knoweth resolutely, what he hath named.

But bicause that name is common to many nombers (although not in one question) and therefore the name is obscure, till the worke doe detectt it, I thinke this rule might well bee called, the rule of darke position, or of straunge position: but not of false position.
and for the more easie and apte worke in this arte wee dooe commonly name that darke position. 1.x. And with it doe we worke, as the question intendeth, till we come to the equation.

<side: The partes of the rule.> This rule of equation, is diuided by some men, into diuerse partes. As namely Schuebelius dooeth make. 3. rules of it. And in the seconde rule, he putteth. 3. seueralle cannôs. some other men make a greater nôber of distinctiôs in this rule. But I intende (as I thinke beste for this treatice, whiche maie serue as farre <page> as their workes doe extende) to distincte it onely into twoo partes. Whereof the firste is, when one nomber is equalle vnto one other. And the seconde is, when one nomber is compared as equalle vnto. 2. other nombers.

Alwaies willyng you to remêber, that you reduce your nombers, to their leaste denominations, and smalleste formes, before you procede any farther.

And again, if your equation be soche, that the greateste denomination Coßike, be ioined to any parte of a compounde nomber, you shall tourne it so, that the nomber of the greateste signe alone, maie stande as equalle to the reste.

Howbeit, for easie alteratiô of equations. I will propunde a fewe exâples, bicause the extraction of their rootes, maie the more aptly bee wroughte. And to auiode the tediouse repetition of these woordes: is equalle to: I will sette as I doe often in woorke vse, a paire of paralleles, or Gemowe lines of one lengthe, thus: =, bicause noe. 2. thynges, can be moare equalle. And now marke these nombers.

1. 14.x.+.15.p=71.p.
2. 20.x.-.18.p=.102.p.
3. 26.z+10x=9.z-10x+213.p.
4. 19.x+192.p=10z+108p-19x
5. 18.x+24.p.=8.z.+2.x.
6. 34z-12x=40x+480p-9.z

<side: 1.> In the firste there appeareth. 2. nombers, that is <page: FF. ii.> 14.x.+15.p. equalle to one nomber, whiche is 71.p. But if you marke them well, you maie see one denominatiô, on bothe sides of the equation, which neuer ought to stand. Wherfore abating the lesser, that is. 15.p. out of bothe the nombers, there will remain .14.x=56.p. that is, by reduction, 1x=4.p.

Scholar. I see, you abate. 15.p. from them bothe. And then are thei equalle still, seyng thei wer equalle before. Accordyng to the thirde common sentence, in the patthewaie:

If you abate euen portions, from thynges that bee equalle, the partes that remain shall be equall also.

Master. You doe well remêber, the firste groundes of this arte. For all springeth of those principles Geometricalle. Wherfore call to your minde likewaies the secende common sentence, in the same booke, and then haue you another reason, whiche will helpe you not onely, in the other formes of woorke here, but also very often in the practise of this arte.

Scholar. That is this.

If you adde equalle portions, to thynges that bee equalle, what so amounteth of them shall be equalle.

Master. These twoo sentences doe instructe you that wen you see on bothe the sides of the equation, any one denominatiô Coßike, you shall marke the signe that is annexed to the lesser of them bothe: and if it be the signe of addition. +. then shall you abate that lesser nomber, from bothe the partes of the equation. As I did in this firste example. But if the signe be of abatemente -, then shall you adde that lesser nôber, to bothe partes. And so shall you doe, till there be noe one denominiation on bothe partes, but diuerse and distincte.

<side: 2.> So the seconde nomber will be. 20.x=120p and in the leaste termes. 1.x=.6.p.

<side: 3.> Scholar. I see that you adde .18.p. to bothe partes <page> of the equation. But by that reason, I doubte in the thirde somme, bicause. 10.x. is in bothe partes of the equation: in the firste parte with +, and in the seconde parte with -, whether I shall adde 10.x, or abate them.

Master. In soche a case, you maie dooe either of bothe, at your libertie: and all will be as to one eande.

Scholar. If I adde. 10.x. then will it be. 26.z.+20.x.=.9z.+213.p.

Master. And doe you not see. z. on bothe sides of the equation?

Scholar. I did loke but for one alteration onely.

Master. If there were twentie like denominations, you should alter them all. For that is the principalle and peculiare reduction, that belongeth to equations.

Scholar. Then must I abate. 9.z. on bothe partes, and so will there remaine. 17.z.+.20.x.=.213.p.

Master. Now reduce it by abatyng. 10.x.

Scholar. So it will bee. 17.z.=.213.p.-.20.x.

And now I remêber, that this is the better forme of reduction. Bicause the greater denomination, that is. z, is alone with his nomber on the one side of the equation, and the. 2. lesser denominations, on the other side.

Master. How doe you reduce the other equatiôs, to their smalleste formes?

<side: 4.> Scholar. In the fourthe example, there is noe denominatiô, before the signe of equation, or in the first parte, but the like is in the seconde parte also, after the signe of equation. Wherfore firste, bicause I see 19.x. on bothe sides, I will abate it on bothe sides. And then will it be thus.

192.p.=10.z.+108.p.-.38.x.

<page: F. iii.> But bicause I see p. yet remainyng on bothe partes, I abate the lesser, that is. 108p. from bothe partes, and it will be. 84.p.=.10.z-.38.x.

Master. This equation would be better, if the greater denomination, did stande as one parte of the equation alone. Whiche thyng you maie easily doe, by addyng. 38.x. to bothe partes: bicause so moche foloweth -, on the one parte.

<side: Translation of nombers.> And euermore when occasion serueth, to translate nombers compounde, – on the one side is equalle to + on the other side.

Scholar. Then it will be thus. 84.p.+38.x.=.10.z.

Master. It were better thus. 10.z.=.38.x.+.84.p.
And in smaller termes. 5.z.=.19.x.+.42.p.
But now procede with the examples.

<side: 5.> Scholar the fifthe is easily reduced, by abatyng 2.x. on bothe sides: For so will it bee. 8.z.=.16.x.+.24.p.

<side: 6.> The sixthe equation will be, by addyng. 12.x. on bothe sides. 34.z=52.x+480.p-9z. But yet I must reduce it farther, by addyng. 9.z. on bothe sides. And then it will stande thus. 43.z.=.52.x.+.480.p.

<side: Varieties of equations.> Master. Now will I shewe you the varieties of equatiôs, taught by Scheubelius, bicause you maie perceiue, how thei bee contained in those. 2. formes, named by me. As for the manyfolde varieties, that some other doe teache, I accoumpte it but an idle bablyng, or (to speake moare fauourably of them) an vnnessary <page> distinction.

<side: The firste equation.> The first equatiô after Scheubelius, & after my meanyng also, is, when one nomber is equall to an other: meanyng that thei bothe must be simple nombers Cossike, and vncompounde. As. 6.x. equalle to. 18.p:

4.z.=.12.x.
14.c.=.70.z:
15.sz.=.90.zz:
20.z c=180.sz:
26.z sz.=.117.c c.

In all these examples, as you see but one nomber, compared to an other: so to finde the quantitie of one roote, you shall diuide the nomber of the lesser Character, by the nomber of the greater Character, and so shall the quotiente bryng forthe the quantitie of. 1.x.

Scholar. It semeth at the firste vewe, that it is against reason, to diuide the nomber of the lesser signe, by the nomber of the greater. But when I consider, that if I compare a nomber of crounes, or any like denomination, to a nomber of shillynges in equalitie, the nomber of crounes, or other soche like, must neades be lesser, then the nôber of shillinges. And so diuiding the nôber of the shillinges (or other lesser name) by the nomber of crounes (or other greater name) the quotiente will shewe, how many shillynges make a croune: and generally, how many of the lesser, dooe make one of the greater.

As if. 20. crounes bee equalle to. 100. shillynges, then. 5. shillynges dooeth make a croune. So when 6.x. bee equall to. 18.p. then. 3.p. doeth make. 1.x. And. 4.z.=.12.x. dooeth cause that. 3.p. must be a roote.

Master. As your examplarie profe is good, so reduction will be a sufficiente proofe in this.

Scholar. I see it manifestly. For it. 14.c. bee equall to. 70.z. then. 1.c. is equalle to. 5.z. by that <page> reduction in nombers. And again by reduction in signes. 1.x. is equalle to. 5.p.

Likewaies. 15.sz. beyng equalle to. 90.z z. reduction by signes and nombers also, will make 1.x=6.p. So shall. 20.z c.=.180.sz. be reduced to. 1.x.=.9.p. And. 26.z sz.=.104.c c. will make. 1.x.=.4.p.

Master. And so generally, when there is noe denomination omitted, betwene those. 2. that bee compared in equalitie, still the diuision of the nomber, of the lesser denomination, by the nomber of the greater denomination, will bryng forthe in the quotiente, the quantitie of. 1.x.

<side: The seconde forme of the firste equatiô> But if there bee any denominations omitted, betwene those. 2. whiche be compared together in equalitie: loke how many denominations are omitted, and so many in order is the rooted quantitie, whose roote you must extract, for the aunswere to the questiô. For in soche a case, euer more you shall extracte the roote of your laste nomber.

As for example, when. 6.c. be equalle to. 24.x. by the former rule, you shall finde. 4. in the quotiente. But here that. 4. is not the quantitie of a roote, but is a rooted nomber, whose roote I shall extracte. And seyng betwene. c. and. x. there is no quantitie omitted, but one, that is. z. Therefore I shall accoumpte. 4. the first quantitie, that is to saie, a Square nomber, and so take his Square roote, beyng. 2. for the quantitie of a roote.

Again if. 7.sz. be equalle to. 567.x. the quotinete will be. 81. and declareth a zenzizenzike nomber, bicause there are omitted betwene. sz. and. x. three nombers: and zenzizenzike is the thirde quantitie: as you did learne in the beginnyng of this treatice, of nombers denominate.

Scholar. I perceiue it. And therfore I must take <page> the zenzizenzike roote of. 81. whiche is. 3. and that is the true roote, where. 7.sz. be equalle to. 567.x.

Master. And if those. 7sz. were accôpted equalle to. 56.z. the quotiente will be. 8. And bicause there are omitted. 2. quantities, that is. c. and. z z. therfore you shall accompte that. 8. to be 1c. or a seconde quantitie. And his roote Cubike is. 2. whiche standeth as the valewe of a roote, in the former equation.

And it is not possible that any other nomber, maie be placed as a roote, in that equation: or in any other forme of this firste kinde. Howbeit in one sorte of equation, of the seconde kinde, there maie be. 2. diuerse rootes, when one nomber hath. 2. rootes in valewe. As I taught you before in the extraction of rootes.

<side: The seconde kinde of equation.> The secôd kinde of equatiô, after Scheubelius minde and myne also, is, when one simple nomber Coßike, is compared as equalle to. 2. other simple nôbers Coßike, of seueralle denominations, and like distaunce.

And in soche equation, beynd reduced as is taught before, the roote of those. 2. nombers compounded, as in one (or rather the valewe thereof) shal be extracted: As I haue before taughte also. And that roote doeth aunswere to the question.

<side: The seconde forme of the second kinde> Howbeeit, here is the like obseruation, as was in the seconde forme of the firste kinde. For if those. 3. denominations be not immediate, but doe omit some other betwene them, then shall you extracte the roote of that laste nomber, in all poinctes, as you did in the firste equation.

Examples of the firste sorte.

4.z.=.6.x+4.p. whiche beyng reduced, will bee: 1.z.=.3/2.x.+.1.p. And the roote wil be .2

And. 6.sz.=.12.z z.+.18.c. <page: Gg. i.> That is by reduction. 1.sz.=2.z z.+3.c. or 1.z.=2.x.+.3.p. And the roote. 3.

5.sz=25.z z-30.c. Or by reduction. 1.sz.=5.z z.-6.c. Or. 1.z.=.5.x-.6.p. whose roote is. 3. or. 2.

Likewaies. 2.z=120.p.-8.x Or by reductiô. 1.z=60.p-4.x. whose roote is. 6.

Examples of the seconde sorte.

5.z z=60.z+320.p. That maketh by reduction. 1.z z.=12.z.+.64.p. And the square roote. 4.

Likewaies. 8z c.=.40.c.+.30208.p. Or by the orderly reduction. 1.z c=5.c.+3776.p. whose Cubike roote is. 4.

Again in residualles. 8.z c=864.z.-24.z z. That maketh by reduction. 1.z c=108.z.-3.z z. Or els. 1.z z=108.p-3z. whose roote is. 3.

So. 9.bsz=90.z z.-144.x. Or by reduction. 1.z c=10.c.-16.p. whose roote is. 8. or. 2.

But now bicause Scheubelius dooeth make. 2. seueralle equations of these. 2. formes: And giueth. 3. diuerse rules, or canons for eche of them, I will declare his. 6. canons to be all contained in this seconde kind of equation.

<page> He maketh his diuision thus. When. 1. nomber is compared as equalle to. 2. other, other that one nôber is of the smalleste denomination. And then is it of the firste Canon. As. 1.z+8.x=65.p. or els that one nomber, is of the greateste denominatiô: As. 3.x.+4.p=1.z. And then is it of the seconde Canon: Or els thirdely, the alone nomber is of the middle denominatiô: and then is it of the thirde Canon. As. 1.z+.12.p.=8.x.

The like forme he vseth, for the nombers of denominations distaunte.

Wherby you maie perceiue, that in my rule there is noe forme of nombers, like thê of the firste Canon, norther yet of the thirde: but onely of the seconde. But then again in my rule, there are. 2. sortes of examples whiche he hath not. And if you compare them well together, you shall pereiue, that thei bee agreable together.

As for exâple. In his firste canon, this is the forme 1.z+6.z.=27.p. whiche equation in my rule, by translation, is expresses thus.

1.z=27.p-6.x. bicause I doe still set the greateste denomination alone.

Again in his thirde Canon, this is an example.

1.z+15.p=8x and that nomber doe I translate into this forme 1z.=8.x-15p.

Now where as he giueth seueralle rules, for euery Cannon, I saie for them all: extracte the roote of that compounde nomber. For all his rules doe teache nothyng els.

Scholar. I doe vnderstande the diuersitie, and agremente of your rules and his. But for my exercise, I dooe couette some apte questions, appertainyng to these equations.

<side: A question of ages.> Master. Take this for the firste question.

Alexander beyng asked how olde he was, I am. 2. <page: G. ii.> yeres elder (quod he) then Ephestio. Yea, saied Ephestio. Any my father was as olde as we bothe, and. 4. yeres moare. And my father hauyng all those yeres, saied Alexander, was. 96. yeres of age. I demaunde now of you, how olde was eche of them.

Scholar. I praie you aunswere the question your self, to teache me the forme.

<side: 1x. is the common supposition.> Master. I will begin with the yongeste mannes age, and that will I call 1.x. whiche is the common supposition in all soche questions. Then is Alexanders age. 2. yeres moare, that is. 1.x+2.p. And those bothe together dooe make. 2.x.+.2.p. whereunto if you put. 4. more, then haue you the age of Ephestio his father, that will be. 2.x.+.6.p. And all these put together, that is, that is. 4.x+.8.p. will make 96 whiche is the equation that shall open the question.

Wherfore I set doune the equation thus. 4.x.+8.p=96.p. And bicause I see on bothe sides, one denomination of. p. I doe abate. 8.p. frô bothe sides: & then there remaineth. 4x=88p And by reduction or diuision, 1.x.=22.p.

<side: The proofe.> Scholar. Then maie I easily saie, that Ephestio was. 22. yeres olde, seyng you did putte. 1.x. for his age: and now. 1x. is founde to be .22. And therby all the other yeres be manifeste. For Alexander beyng. 2 yeres elder, must be. 24. And Ephestio his father had in age. 22. and. 24: and. 4. more, that is. 50. yeres. All whiche make. 96. So is that question fully aunswered.

22
24
50

96

<side: A question of debte.> Master. An other question is this. I had a somme of money owing vnto me: whereof I did receiue at one tyme 1/4 and afterward I receiued 2/5 of that residue, whiche remained vnpaied. And so remained the reste of the debte 27.li. I would knowe what was the firste debte, & what wer the. 2. seuerall paiementes

<page> Scholar. This muste I obserue still, to name the firste doubtfull thyng. 1.x. wherefore I saie that the firste debte was 1.x. whereof I receiued 1/4 And so did there remain. 3/4.x. of whiche reste, againse I receiued 2/5, that is 6/20. of the whole somme, or 6/20 x. And that beyng abated also, then did there remaine 9/20 x. whiche you named to be. 27.li. Then if 9/20.x. bee equalle to 27.li, diuide. 27.li. by 9/20, and the quotiente will bee 520/9, that is. 60. whiche was the whole debte: And then is it plaine, that 1/4. of it is. 15. and 2/5. of the residue is. 18. whiche maketh. 33. and then remaineth. 27.

Master. There is nothyng better then exercise, in attainyng any kynde of knowlege: And therfore I will proue you with diuerse questions, to make you the moare experte in this rule. And this is one.

<side: A question of pauyng.> There is a floore Paued with Square Bricke, the lengthe of that floore beyng longer then the breadth, by 1/7, and the whole Pauemente containeth. 3584. brickes: I require to knowe the bredthe and lengthe.

Scholar. The lesser quâtitie, whiche is the bredth I doe name. 1.x. And then the lengthe will bee, by your proportion. 1 1/7.x. Now must I multiplie the bredthe by the lengthe (for that is the orderly worke in all flatte formes, to finde out the whole platte) that is here. 1.x. by. 8/7.x. and there will amounte the whole platte. 8/7.z. whiche by your supposition is equalle to. 3584.

Wherfore accordyng to your rule, I diuide. 3584. by 8/7, and the quotiente will be. 3136. whiche is a Square nomber, bicause there is one denomination omitted in this equatio. For betwene z and p. there is omitted. x. And therfore must I extracte the square roote of. 3136. and it will bee the quantitie of. 1.x. that I woorke in my tables, and finde it. 56. whiche must be the bredthe: for that I named. 1.x. Then the length must be moare by 1/7. of it: and so shall it be. 64.

<page: Gg. iii.> Now for to confirme my woorke, I multiplie. 56. by. 64 and it will make. 3584. whiche is the nomber that you did name.

<side: An other woorke of that question> Master. That question is well aunswered: And if you had put. 1.x. for the lengthe, as you might do, then the bredthe will be 7/8.x. and the square 7/8z. and so. 1.x. would bee. 64. as you maie proue at leiser: but in the meane time, what saie you to this questiô?

<side: A questian of an armie.> There is a capitain, whiche hath a greate armie, & would gladly Marshall thê, into a square battaile, as large as mighte bee. Wherefore in his firste proofe of square forme, he had remainyng. 284. to many. And prouyng again by puttyng. 1. moare in the fronte, he founde wante of. 25. men. How many souldiars had he, as you gesse?

Scholar. I call the firste fronte. 1.x. and then multipliyng it Squarely: I shall haue for the whole battaile. 1.z. and so by your saiyng, there was lesse 284. men, wherefore the whole nomber of men, was 1.z.+.284.p.

Now for the seconde proofe, when the fronte was increased by. 1. man: I shall set the former fronte, and 1. manne moare, that is 1.x+1.p. And multipliynge that nomber, squarely: there will arise for the whole armie. 1.z+2x+1p. out of whiche I muste abate 25 that, you saie, did wante, to make up that square battaile. And then it will be. 1.z.+2.x.-24.p.

1.x. + .1.p.
1.x. + .1.p.
——————-
1.z. + .1.x.
1.x + 1.p
——————-
1.z + 2.x. + 1.p.

Now haue I one nomber of menne, expressed by. 2 Coßike nombers: Of necessitie therefore must these. 2. nombers be equalle: seyng thei represente one armie.

Wherfore I set them thus.

<page> 1.z.+284.p.=1.z+2.x-24.p.

And findyng. 1.z. on bothe partes of the equation, I doe abate it, & then standeth. 284p=2x-24p. Yet again I see. p. on bothe sides of the equation, and therfore, seing the lesser of them hath the signe of subtraction, I doe adde. 24. to bothe nombers, and then will there be. 308.=.2.x. that is. 154=1.x

So that seing 1x was set for the first fronte: the same front must be. 154. whose Square is. 23716. vnto whiche I muste adde the. 284. that did abounde, and then will the whole nomber be. 24000.

154.
154.
——
616.
770
154
——
23716.
284.
——
24000.

For farther trialle whereof, I take the seconde fronte to be. 155. that is. 1. moare then the firste: and his Square will be 24025. And so is there. 25. moare then the iuste nomber of the armie, as the question supposed.

<side: An other woorke of that questiô.> Master: That question maie be wrought also by namyng the seconde fronte. 1.x. and then will his square bee. 1.z. but seyng there wanteth. 25. menne, to make that Square battaile, the nomber shall bee 1.z.-25.p.

Then for the firste front, you must set. 1. man lesse, as the question importeth, & that will be. 1.x-1.p whose square will be 1.z+1.p.-2.x.

1.x. – .1.p.
1.x. – .1.p.
——————–
1.z. – .1.x.
– .1.x. + .1.p.
——————–
1.z. + .1.p. – .2.x.

vnto whiche I must adde the. 284. menne that did abounde, whê that battaile was framed, and then will <page> the nomber be. 1.z.+.285.p-.2.x. And it must bee equalle to. 1.z.-.25.p. Wherfore to reduce that equation, firste I adde on bothe sides 25.p & then resteth. 1.z. equalle to. 1.z+310-2.x Then I adde. 2.x. bicause I will haue noe – in the equation: and it will be, 1.z+2.x.=.1.z.+310.p. Thirdely I abate. 1.z. on bothe sides of the equation: and then remaineth. 2.x.=.310.p. that is. 1.x.=155.p. wherby it appeareth that the seconde fronte was. 155 and the firste fronte. 154. & so forthe, as you wrought it before.

An other question is this.

<side: A question of an armie.> There is a kyng with a greate armie: And his aduersarie corrupteth one of his heraultes with giftes, and maketh hym swere, that he will tell hym, how many Dukes, Erles and other souldiars there are in that armie. The Heraulte lothe to lease those giftes, and as lothe to bee vntrue to his Prince, diuiseth his aunswere, whiche was true, but yet not so plain, that the aduersarie could therby vnderstande that, whiche he desired. And that aunswere was this.

Looke how many Dukes there are, and for eche of them, there are twise so many Erles. And vnder euery Erle, there are fower tymes so many soldiars, as there be Dukes in the fielde. And when the muster of the soldiars was taken, the. 200. parte of them, was 9. tymes so many as the nomber of the Dukes.

This is a true declaratiô of eche nomber, quod the Heraute: and I haue discharged my othe. Now gesse you how many of eche sorte there was.

Scholar. Although the question seme harde, I see many tymes, that diligence maketh harde thynges easie, and therfore I will attempte the worke of it.

And firste for the nomber of Dukes, I sette. 1.x. then will the nomber of Erles bee. 2.z. that is. 1.x <page> by. 1.x multiplied twise: And the nomber of soldiars are. 8.c. that is. 2.z. multiplied by. 1.x. fower tymes, but bicause the. 200. parte of the soldiars is. 9. tymes so moche as the nomber of the Dukes, therfore must the. 200. parte of. 8.c. be equalle to. 9.x. And so consequently. 8.c=1800.x and 1.c=225.x. or. 1.z=225.p.

For if I set 8/200 and. 9. as equalle together, & would by the arte of fractions, brynge the same proportion in whole nombers, I shall haue for. 9. this fraction 1800/200. And seyng the denominators, be all one in 8/200 and in 1800/200 the proportion consisteth betwene the numerators.

Then to procede, if. 225. be equalle to. 1.z. I shall take the square roote of. 225. for. 1.x. and that is. 15 whiche must be the nomber of Dukes.

And so haue I the firste nomber, wherefore the seconde nomber, that is the nomber of Erles, must bee 15. tymes. 15. twise: that is. 450. And the nomber of soldiars shall be. 4. tymes. 15. multiplied by. 450. that is. 27000. And for iuste trialle of this woorke, I take the. 200. parte of the soldiars that is. 1350. and I finde it to bee. 9. tymes. 15. that is. 9. tymes so moche as the nomber of the Dukes. And so is that question solued, and tried.

450.
60
——
27000

<side: An other question of walles.> Master. This is an other question. There is a grounde inclosed with. 4. walles, beyng like iambes and of one heigthe. The longest. 2. walles are in proportion to the shorteste, as. 5. to. 3 And vnto the height thei bee double Sesquialter. Now multipliyng the longeste by the shorteste, and that totalle by the heighte, there will rise. 39930. foote. I demaunde then, what is the lengthe and the heighte of eche walle?

Scholar. The least quâtitie is the heighte, whiche I call. 1.x. and vnto it the longeste walle is double <page: Hh. i.> Sesquialter: that is. 2 1/2.x. Now that same longeste is in proportion Superbipartiente quintas, to the shorteste walle. So must the shorter wall be 1 1/2 x. Then must I multiplie all those. 3. nôbers together, that is 1.x. by. 1 1/2 x. whereof doeth come. 3/2 z. then shall I multiplie that totalle, by 5/2 x. and it will be 15/4 c. or 3 3/4 c whiche must be equalle, by the woordes of the question, to. 39930.

So by reducyng them to one denomination. 15/4 c. shall be equalle to 159720/4 that is. 15.c=159720.p. and. 1.c.=.10648. wherfore I shall extracte the Cubike roote out of .10648. and that is the quantitie of. 1.x. or the heighte of the walle.

In my Tables I woorke that extraction of Cubike roote, and finde it to be. 22. And therfore must the lôgeste walle bee double Sesquialter to it, that is. 55. And the shorteste walle will be. 33.

<side: The proofe.> For proofe whereof I dooe multiplie. 22. with. 55. and it maketh. 1210. whiche nomber I shall multiplie by. 33. and it will be. 39930. according to the supposition of the question.

Master. You doe chose still the leaste nomber, to be equalle to. 1x. as the easieste forme. Howbeit you maie put. 1.x. for the lengthe of any of the walles.

<side: An other forme of woorke.> And if you sette it for the longeste walle, then the shorteste walle will be 3/5 x. and the heighte 2/5.x and all those. 3. nombers will make, by multiplication together 6/25 c. equalle to. 39930. And so will. 6.c. be equalle to. 998250.p. and. 1.c.=.166375.p. whereof the Cubike roote is 55. and aunswereth to the quantitie of. 1x.

<side: The thirde forme of woorke.> But if. 1.x. be set for the measure of the shorteste walle, then the longeste walle will be 5/3.x. And the heighte 2/3.x. And so all. 3. nombers multiplied together will make 10/9 c.=.39930. So shall. 10.c. be equall to. 359370. And. 1.c.=.35937. whereof <page> the Cubike roote is. 33. and is the value of. 1.x. in this position.

Scholar. This varietie of woorke, is not onely pleasaunte, but it maketh the reason of the woorke to appeare moare plainly. So that I could neuer be werie to heare soche questions.

Master. Then will I propounde one or 2. moare before we passe from this kinde of equation. Whereof one shall be somewhat like that last. And this it is.

<side: A question of Bricke.> A Brickeleiar had a pile of Bricke, whiche he sold by the yarde. The lengthe of it was 7/2 to the bredthe, that is Triplasesquialtera. And the heighte was fiue tymes so moche as the lêgthe. This pile the owner sold for. 980. crounes. By soche rate that he had for euery yarde so many Crounes, as the Pile had yardes in bredthe. Now is the question, what was the lengthe, bredthe, and heighte of this pile?

Scholar. I suppose the bredthe of bee. 1.x. then was the length 3 1/2 x. the the heighte. 17 1/2 x. These 3. sommes dooe I multiplie together, and thei make 245/4 c. whiche standeth as equalle to all the yardes in the whole pile. But yet what that is. I knowe not.

Wherfore to procede farther, I consider that euery yarde coste as many crounes, as the bredthe contained yardes. Now the bredthe being 1.x I must saie, that euery yarde did coste. 1.x. of crounes. And then by the Golden rule: if. 1. yarde coste. 1.x. of Crounes, what shall 245/4 c. coste?

1. —- 1.x.
/
/
245/4 c —- 245/4 z z.

Woorkyng by the rule, I finde that it shall cost 245/4 z z. And the question doeth suppose that it coste. 980. crounes. Wherfore I must saie, that. 980. crounes, are equalle to 245/4 z z. And consequently. 245.z z.=.3920.p. wherfore diuidynge the nomber of the lesser name, by the other, the quotiente will be 16. whose zenzizenzike roote is 2 <page: Hh. ii.> And that therfore must be the value of a roote, and the bredthe of the pile. So shall the lengthe be. 7. yardes, and the heighte. 35. yardes.

<side: The proofe.> For trialle of it, I multiplie the lengthe, by the bredthe, and that totalle by the heighte, and so haue I 490. for all the yardes of Bricke. Then consideryng that euery yarde coste. 2. crounes, bicause. 2. yardes is the bredthe of the pile: the nomber of crounes must be twise. 490. that is. 980. And so is the woorke good.

<side: An other forme of woorke.> Scholar. If the lengthe be. 1.x. the bredthe must be 2/7.z. that is Subtriplasesquialtera to. 1.x. And the heighte must be. 5.x. All whiche somes make by multiplication 10/7 c.

Then farther, if 1. yarde coste 2/7 x. 10/7 c shall coste 20/49.z z, whiche is equalle to 980. And so is. 20.z z. equal to. 48020. and by diuision 1.z z.=.2401. whose zenzizenzike roote is. 7. And that is the lengthe of the walle, and is the value of. 1.x.

1. —- 2/7 x.
/
/
10/7 c. —- 20/49 z z.

The reste of this worke, is like as before.

Master. Yet proue the thirde waie.

<side: A thirde forme of woorke.> Scholar. The heighte beeyng. 1.z the lengthe must be the firft part of it, that is 1/5 x. And the bredth 2/35 x. All these make by multiplication 2/175 c. Then for the price, if. 1. yard coste 2/35 x what shall 2/175 c. coste? By the Golden Rule there is founde, 4/6125.z z, whiche is equalle to 980. And so shall. 4.z z. be equalle to. 6002400. And. 1.z z=1500625. whose zenzizenzike roote is. 35. And that is the value of. 1.x. and the heights of the Pile.

1. —- 2/35 x.
/
/
2/175 c —- 4/6125 z z

Master. One question moare will I propounde, <page> and so eande with this equation.

<side: A question of a Testament.> A poore man died, whiche had fower children, and all his goodes came to. 72. crounes: whiche he would haue parted so, that the seconde & thirde childe should haue. 7. times so moche as the firste. And that the portions of the thirde and fourthe childe should bee. 5. tymes so moche as the secondes parte: And that the first and thr fourthe, should haue twise as moche as the thirde. If you worke the solution wel, you maie seme worthy to be master of those wardes.

Scholar. I trust to obtaine moare benefite by the question, then by that office. Wherefore I will giue good hede vnto it. And for the first nôber, I set. 1.x then muste the seconde and thirde portions make together. 7.x. And the fourthe must bee all the reste of the. 72. that is. 72-8x. Now the thirde must be halfe the firste & the fourthe, that is. 36-3 1/2 x. And the third & fourthe, is. 5. tymes the second, wherfore the seconde shall be the. 5. part of. 108-11 1/2 x that is. 21 3/5 – 2 3/10 x, whiche nomber I shall set in order with Letters, as here I haue dooen for my owne ease, and aide of memory. And then shal I adde them all together. Whereof there commeth. 129 3/5 – 12 4/5 x, whiche is equalle to 72. First therfore I do adde all that foloweth – to bothe partes of the equatiô. And so haue I 129 3/5=12 4/5 x+72. But bicause there are nombers Absolute on bothe sides, I shall abate the lesser somme, that is. 72. from bothe partes, and then will there bee lefte, 57.3/5=12 4/5.x. that is. 288.=.64.x. An by diuision 4 1/2.=.1.x.

A 1.x.
B 21 3/5 – 2 3/10 x.
C 36. – .3 1/2 x.
D 27. – .8.x.
———————
129. 3/5 – 12 4/5 x.

<side: The proofe.> So shall the firste mannes portion bee 4 1/2. And the seconde and thirde mannes portion. 7. times so moche <page> that is. 31 1/2. Whereby it followeth, that the fourthe manne, shall haue the reste of 72. that is. 36.

A 4 1/2
B 11 1/4 }
C 20 1/4 } 31 1/2
D 36.
——–
72.

Then seeyng the thirde manne, hath halfe so moche as the first and the fourthe, his portiô shall be 20 1/4. And then by diuerse reasons, the seconde mânes part shall bee. 11 1/4. And all these partes added together, doe make iuste 72. Wherfore the woorke is good.

<side: An other forme of woorke.> Master. You haue wroughte it well. And yet maie you woorke it thus. Firste sette doune. 1.x. for the firste mannes parte. And then for the seconde and thirde ioyntly. 7.x. so shall the fourthe manne haue 72.p.-.8.x. And bicause the seconde mannes parte is 1/5. of the thirde and fourthe mannes portion, if you ioyne all their. 3. partes together, the seconde mannes portion will be 1/6 of that totalle. Put therfore 7.x, whiche is the partes of the second and the third vnto. 70-8x, whiche is the fourthe mannes parte, and the totalle will be. 72.p-1.x. whose sixte parte is 12.p.-1/6 x, for the seconde mannes share. Whiche somme if you abate out of. 7.x. there wil remain for the thirde mannes parte 7 1/6 x-12.p.

A 1.x.
B 12.p. – 1/6 x
C 7 1/6 x – 12.p
D 72. – .8.x.
———————
72.

And so haue you euery mannes portiô allotted to hym duely. As I haue here set it forthe for you. And all their added together, doe make. 72.

Scholar. But here is noe equatiô yet, though the partes be diuided iustly.

Master. Now shall you see it.

The question saieth, that the thirde mannes portion is halfe the portions of the firste and fourthe man. wherefore seyng the firste and fourthe mannes portions doe make. 72-7.x. the thirde mannes portion <page> beeyng doubled, shall make as moche. But the double of the thirde mânes parte, is 14 1/3 x-24p. and therfore I saie, that those. 2. nombers be equalle, that is. 72.p.-7.x.=14 1/3 x.-24.p. Firste adde. 7.x. to eche parte, and it will bee 72.p.=21 1/3 x.-24.p. Then adde. 24.p. on bothe sides, and there will be. 96.p.=21 1/3 x. that is by reduction. 288.=.64.x. as you made it. And then all agreeth.

Likewaies for the equation, you maie set the third mannes portion, with the halfe of the firste & fourthe mennes partes. And so will. 7 1/6.x.-.12.p. be equalle to. 36.p.-3 1/2 x. And by reduction, 10 2/3 x.=.48.p. That is in other termes of whole nomber. 32x.=.144. And by diuision it will bee 1.x.=.4 1/2. And thus will we eande the examples of the firste equation, for this tyme. And will shewe you some questions of the seconde equation.

Examples of the seconde equation, by questions propounded.

<side: A question of silkes.> There are two men that haue silke to sell. The one hath. 40. elnes, and the other. 90. And the firste man his silke is not so fine as the seconde man his silke. So that he selleth in euery angell, price more by 1/3 of an elne, then the seconde mâ doeth. And at the eande, bothe their moneis made but 42. angelles. Now I demaunde of you, how moche eche man solde for an angell?

Scholar. I will folowe my olde forme, in putting 1.x. for the leaste quantitie, whiche is the seconde mannes somme, and then shall the firste mannes somme be. 1 1/3 x.

Master. You are deceiued all readie. For you set <page> 1.x. for an elne. Seyng you name 1/3 of an elne, to be 1/3.x. And so were the position neadelesse, and likewaies all the woorke.

Scholar. I see my faulte: but I knowe not how to amende it. For that. 1.x. maie bee a parte or partes of an elne: and so maie it be moare then. 1. or. 2. elnes so that I ought not to haue set 1/3 (whiche is certainly referred, in this question, to an elne) as the parte of a doubtfull quantitie, but rahter as the parte of a quantitie certaine. Whereas. 1.x. is euer put for a nomber vnknowen.

Master. To helpe you herein, I will set the firste nombers, as you began them. The seconde man his nombers of elnes, shall be. 1.x. as you did name it, and then shal the firste mannes portion be as moche, and 1/3 of an elne moare: whiche 1/3 I maie beste call 1/3 p. And so shall it bee distaunte from 1.x. clerely in all woorke Arithmeticall.

A 1.x+1/3 p.
B 1.x.

But now to proceade, I shall diuide eche mannes nomber of elnes, whiche he had, by the nomber of elnes, whiche he solde for an angelle, and the quotiente will declare how many angelles eche man had receiued. So that the firste mannes nomber of elnes, beeyng. 40. shall bee the nomberator, and the somme of measure, whiche he solde for an Angelle, that is 1x+1/3 p. shall bee the denominator. And so is the diuision eanded. And that fraction is the quotiente.

40.
————
1.x + 1/3 p.

90.
————
1.x.

Scholar. Now I perceiue the woorke. And by like reason: the seconde mannes somme of elnes beyng. 90. shall bee the numerator, and. 1.x. beyng the somme of measure, solde for one Angelle, shall be the denominator, that is in one fraction 90/{1x}: accordingly as I haue sette bothe nombers <page> here distinctly.

Master: It were moare ease for you in workyng, if you did tourne that fractiô of 1/3 into an intere vnitis.

Scholar. That wil easily be doen, by multipliyng euery nomber, of that whole fraction by. 3. And then will it be 120/{3x+1p}, whiche is all one in value with 40./{1.x+1/3 p.} And this I consider farther, that as these. 2. fractions, seuerally dooe expresse the sommes of angelles, that eche of them receiued, so ioynctly bothe together, dooe declare the full somme, of all their angelles. Wherefore I shall adde theim bothe together. And thei will make. {390x+90p}/{3z+1x} As here in woorke I haue expressed.

390.x.+.90.p.
———————–
120. 90.
———- ———
3.x.+.1.p. 1.x.
———————–
3.z.+.1.x.

And by your supposition, their bothe sommes of Angelles made. 42. So that those. 2. sommes are equalle: and therefore am I come to an equation. In whiche I see a nomber absolute, equalle to a fraction Coßike compounde.

Master. When so euer that, or the like dooeth chaunce, you shall reduce the whole nôber, to the like denomination: and then their numerators will bee equalle.

Scholar. Then shall I multiplie. 42. by the denominator 3z+1x & it wil be 126z+42x whiche must bee equalle to. 390.x.+.90.p. That is in lesser termes. 21z+7x.=.65.x.+.15.p. Where firste I dooe abate. 7.x. on bothe sides: and there remaineth then. 21z.=.58.x.+.15.p.

<page: Ii. i.> But now I remêber your admonitie, that bicause the nomber annexed to the greateste signe, is moare then. 1. I shall diuide all the nombers by it, and sette their quotientes in their stede, with their signes. And so will the nomber of the greateste signe, euermore be 1. And this equation will be 1.z.=58/21 x.+15/21 p. Where I must extracte the square roote of the later part, accordying to your doctrine, and it will be. 3. As it appereth in this worke folowing, whiche I did frame in my tables.

29/21. in square doeth make 841/441, vnto whiche I muste adde 315/441, whiche is all one with 15/21, by reduction to one denominatiô. So is the full additiô 1156/441. whose square roote is 34/21. vnto whiche I shall adde 29/21, and it will bee 63/21, that is. 3.

Master. This is well doen. Now worke the same questiô, as it was proponed, and you shall easily finde all the other nombers to bee true, and agreable to the question.

<side: The proofe.> Scholar. Seyng the seconde manne solde. 3. elnes for an angell, the firste manne did sell. 3. elnes and 1/3. So. 40 (whiche is the somme of elnes of the first man his silke) diuided by. 3 1/3. doeth yelde. 12. and sheweth how many angelles that man receiued.

Again for the seconde man, whiche had. 90. elnes, diuide that. 90. by. 3. and so shall you finde. 30. for the nomber of his Angelles. And that. 30. and. 12. dooe make. 42, it neadeth not to be proued.

<side: An other forme of woorke.> Master. Now againe for your exercise, suppose the firste mannes somme to be. 1.x.

Scholar. Then muste the seconde manne sell for an angelle. 1.x-1/3.p. And their nombers of elnes, diuided by those nombers will make. 40/{1x}. and 90/{3x-1p}. whiche bothe added toghether, will bee {390x-40p}/{3z-1x} equalle to. 42.p. That is by reduction. 390.x.-.40p.=.126.z.-.42.x. <page> And by addition of. 42.x. on bothe partes. 432.x-40.p.=.126.z. And by diuision it will be. 24/7 x-20/63 p.=1.z.

So that now I must extracte the roote of that compounde Coßike fraction, thus. 12/7 squarely, dooe make 144/49 out of whiche I shall abate 20/63. And therfore, firste of all I doe reduce thê to one denomination, & thei make 9072/3087. and 980/3087. wherefore if I abate the lesser out of the greater: there will remaine 8092/3087. that is in lesser termes 1156/441 and is a square nomber, whose roote is. 34/21 vnto whiche if I adde 12/7 that is 36/21. it will make 70/21, or 10/3. that is the valewe of. 1.x. And is the firste mannes nomber of elnes, agreably as I tried it before. And so doe bothe workes agree.

But now commeth to my remembraûce, that this nomber, whose roote I did extract, in this laste worke is of that sorte, where. x.-p. is equalle to. z. And therfore hath in it. 2. rootes: thone by addition, as this, whiche I now founde: And the other by subtraction, whiche in this example, by abatyng 34/21 out of 36/21, will bee 2/21. But how I maie frame that roote, to agree to this question, I doe not see.

Master. That varietie of rootes dooeth declare, that one equation in nomber, maie serue for. 2. seueralle questions. But the forme of the question, maie easily instract you, whiche of those. 2. rootes, you shall take for your purpose. Howbeit sometymes you shall take bothe. As for example again, marke this question.

<side: A question of money.> A gentilman, willyng to proue the cunnyng, of a braggyng Arithmetician, saied thus: I haue in bothe my handes. 8. crounes: But and if I accoumpte the somme of eche hande by it self seuerally, and put therto the squares and the Cubes of thê bothe, it will make in nomber. 194. Now tell me (quod he) what is in eche hande: and I will giue you all for your laboure.

<page: Ii. ii.> Scholar. Soche incoragementes, would make me studie harde, and trauell very willyngly in learned exercises: though learnyng bee moste to be loued, for knowledges sake. But for to finde the true aunswere thus I doe proceade.

Firste I suppose the one nomber in one hand, to be 1.x. And then must the other nedes be 8.p-.1.x Then doe I make theim bothe Squares. And for the firste I haue. 1z. and for the seconde. 1.z+64p-16.x. Thirdely I multiplie them bothe Cubikely: and so haue I for the firste. 1.c and for the other 24.z.+.512.p.-1.c.-192.x. Then must I adde bothe the nôbers, with their squares, and their Cubes, into one somme. As here in work it is set forthe.

1.x.+ .1.z. + .1.c.
8.p. – .1.x.
1.z.+ .64.p. – .16.x.
24.z + .512p – 1c – 192.x.
————————–
26.z + 584.p. – .208.x.

Where for ease I haue set. 1.x, 1.z. and. 1.c (whiche is the Roote, the Square and the Cube of one nomber) all in one line: and the other Roote, Square, and Cube, I haue set seuerally. And so all thei doe make. 26z+584p-208x whiche is equalle to. 194. by the intente of the question. Wherefore I adde firste. 208.x. to bothe partes, and there remaineth. 26.z.+584.p=208.x+194.p. Then I abate. 194. from bothe sides, and so restethe the equatiô thus. 26.z+390.p=208x That is by diuision. 1.z.+.15.p.=8.x. And by translation of. 15.p. to sette. 1.z. alone, it wil be. 1.z=8x-15.p. And now haue I the exacte and complete equation, where I must seke for <page> the value of. 1.x. by extractyng the roote. Therefore firste I take halfe of. 8 and that is. 4. whose square is. 16. out of whiche I abate. 15. and the remainer is 1. whiche I maie either adde to .4. and so haue I. 5. other, I maie abate it from 4 and so haue I 3. Whiche nombers also according to the same rule, beynd added together dooe make. 8. that is the nomber of the middell denomination. And beyng multiplied together, thei dooe make. 15. that is the other parte of the same compounde Coßike nomber.

Master. And if you had marked that firste, you might easily haue found bothe those nombers, by the partes of. 15. whiche can be none other, but. 5. and 3.

And farther, seyng thei. 2. doe make. 8. and. 8. is the nomber (named in the questiô) that thei should make, therfore you shall take them bothe. And name whiche of them you liste to be. 1.x. And the other shall be of necessitie, the reste of. 8.

<side: The proofe.> Scholar. To examine theim, by the order of the question, I doe proceade thus. 3. with his Square. 9. and his Cube, 27. dooeth make. 39. And. 5. with his square 25 and his Cube. 125. doe yelde 155. And bothe thei together doe bryng forthe. 194. accordyng to the saiyng of the question: and therfore it is certain, that the woorke is good.

<side: An other woorke for equations.> Master. Before you passte any farther, I will admonishe you of one waie, whiche I oftê vse in reduction of soche equations, as this is, when there is noe denomination on the one side, but the like is on the other side, with a greater nomber annexed to it. Then maie you abate all the lesser nôbers, out of their greaters, and the reste shall bee accoumpted equalle to nothyng. Whiche chaunce can neuer happen: excepte there bee some nombers on the greater side, with the signe of abatemente. -.

As here you had. <page: Ii. iii> 26z.+584p.-208x.=194.p. Bicause on the one side, there is noe nôber but 194p and on the other side, there is. 584.p. beeyng a greater nomber, and of the same denomination: therefore maie you abate. 194. from bothe sides, and then remaineth. 26z+390p-208x=0 Wherfore you maie well consider, that the nombers whiche be ioined wich +. are equalle to the nombers that bee set with -. And therfore the one abatyng the other iustly, dooe remaine together as equalle to nothyng.

Wherefore it is reasonable, that seeyng the nombers with – bee equalle to the nombers with + that I maie translate the nombers with – from that side of the equation, and set them on the côtrary side, with the signe of +. And so in this exâple it will bee. 26z.+.390p.=208.x. And this forme shall ease you moche, in reducynge of equations.

Scholar. I thanke you moche. And I will not forget to vse it, as occasiô shall happen. But I praie you propounde yet some moare questions, that I maie see their diuerse varieties.

<side: A question of money.> Master. There were twoo seueralle men, which had certaine sommes of angelles, in soche rate, that the seconde manne his somme, was triple sesquiquarta to the firste: and if their. 2. sommes were multiplied together, and to that totall the 2 firste sommed added, there would amounte. 142 1/2. I demaunde of you, what was eche of their sommes in angelles?

Scholar. The firste mannes somme I call. 1.x. And the seconde mannes sôme shall be. 3 1/4 x. whiche 2. sommes beeyng multiplied together, dooe make 3 1/4 z. vnto whiche I must adde bothe the firste nombers, that is 4 1/4 x. And it will be 3 1/4 z+4 1/4 x equalle to. 142 1/2. All whiche nombers, I shal bring <page> into whole nombers, if I multiplie theim by. 4. And so will it be. 13.z.+17.x=570. And by reducyng the greateste denomination Coßike, to an vnitie. 1.z.+.17/13 x.=.43 11/13. And laste of all, by translatyng the nomber of. x. to set. 1.z. alone, on one side of the equation, it will be. 1.z=43 11/13 p-17/13.x. where I must extracte the value of the roote thus. 17/26. squarely dooe make 289/676. vnto whiche I shall adde the. 43 11/13 (it beeyng firste multiplied by. 52. to bryng it to the denomination of. 676. And so makyng 29640/676) And it will be 29929/676 whiche is a square nomber (as I haue proued in my Tables) and his roote is 173/26. from whiche I must abate 17/26, and then wil there remain 156/26, that is. 6.

And that. 6. is the value of. 1.x, and standeth for the firste mannes nomber. So that the seconde mannes nôber must be as. 13/4 to it: that is triple sesquiquarta. And so shall it be. 19 1/2.

<side: The proofe.> Master. Now proue those nombers, accordying to the question.

Scholar. 19 1/2 multiplied by. 6. doeth make. 117. vnto whiche I shall adde. 25 1/2. amountyng of their. 2 additiôs, and all will be. 142 1/2, accordyng to the purporte of the question.

<side: An other worke of the same questiô.> Master. So is your woorke good. Yet woorke it again, by chaungyng the position.

Scholar. I maie put. 1.x. to betoken the seconde manne his somme. And then shall the firste mannes somme bee 4/13.x. whiche bothe multiplied together doe make 4/13 z. And then addyng the. 2. firste sommes to it, it wil bee 4/13 z+1 4/13 x. And that is equalle to. 142 1/2. All whiche nombers will bee reduced to whole nombers, by multiplication conueniente. And so will it be. 8.z+34.x. equalle to. 3705: that is by reduction, 1.z.+.4 1/4 x=463 1/8 p. and by translation of the termes. <page> 1z.=463 1/8.p.-4 1/4.x. out of whiche nomber I shall extract the value of the roote, in this sorte.

Firste I saie 17/8 multiplied Square, doeth make 289/64, vnto whiche nomber I must adde. 463 1/8, reduced as it ought, and it will bee in all 29929/64, whiche is a square nomber, and hath for his roote 173/8. from whiche I must abate 17/8. And then will there remain 156/8, that is 19 1/2, for the value of. 1.x. And so consequently for the second mannes nôber: whiche was named in this position, 1.x. And this maie bee proued as the other was.

<side: A questian of iorneyng.> Master. What saie you then to this question? There is a straunge iorneye appoincted to a manne. The firste daie he must goe 1 1/2 mile, and euery daie after the firste, he must still augemente his iorney, by 1/6 of a mile. So that his iorney shall procede by an Arithmeticalle progression. And he hath to trauell for his whole iorney. 2955. miles. I demaunde in what nôber of daies, shall he eande his iorney?

Scholar. I knowe not how to proceade in this question.

Master. Doe you not heare me name it, an Arithmeticalle progression? Wherby you might be adsured, that it doeth appertaine to that rule. And accordyng to the canons of that rule, must you woorke this question. But for your better instruction, I will helpe you in this woorke.

Firste aunswere to the question, by the common position: and saie that the tyme of his iorney is. 1.x. of daies. And then shall all the excesses (whiche maie also be called the nomber of the spaces) be. 1.x-1p The common excesse was supposed to bee. 1/6. of a mile. And therefore the somme of all the excesses muste bee {1x-1p}/6 that is to saie, the nomber of all the excesses multiplied by 1/6, that is here, the sixte parte of the <page> nomber of the excesses.

And bicause that the firste nomber is. 1 1/2. I must adde it vnto the somme of the excesses, and so haue I the laste nomber of that progression. Wherefore addyng. 1 1/2. (whiche is 3/2, or in like denomination with the other, 9/6) with {1x-1p}/6 it will make {1x.+.8p}/6. And that is the laste nomber of the progression.

Now you remember, that in progression Arithmeticall, if you adde the firste nomber to the laste: and multiplie that totalle, by the nomber of halfe the places, there doeth amounte the somme totalle of that progression.

And therfore in this exâple, if you adde. 1 1/2 (whiche is the firste nôber in the progression) vnto {1x.+8p}/6 (that is the laste nomber of the progression) there wil amounte {1x+17p}/6, whiche beeyng multiplied by the nomber of halfe the places, that is 1/2 x. (For all the nomber of places is. 1.x) there will rise, {1z+17x}/12, whiche is the totalle somme of all the miles: and therfore is equalle to. 2955.

Scholar. All the reste, and this againe can I dooe now. Seyng {1z+.17x}/12. is equalle to. 2955. I will firste bryng the whole nomber to the like denomination, with the fraction, and it will bee. {1z+17x}/12=35460/12. And then omittyng the like denominations. 1.z+17.x.=35460.p. That is by translation 1.z=35460.p.-17x. whose roote in value I shall finde out thus. I multiplie 17/2 squarely, and it will be 289/4. vnto whiche I shal adde. 35460. & it will make 142129/4, whiche is a square nomber, and hath for his roote 377/2, frô whiche I shall abate 17/2, and then remaineth 360/2, that is. 180. whiche is the value of. 1x. And expresseth the nomber of dayes, whiche the question requireth.

<side: The proofe.> Master. The proofe in this, and the like questions, is, to set foorthe the progression with all his termes. <page: Kk. i.> Excepte you will for shortnesse, sette doune the firste terme, whiche in this example is. 1 1/2: and then by the nomber of the excesses, or distaunces (whiche is euer one lesse then the nôber of places) multiplie the quantitie of one excesse: and put to it the firste terme: and so haue you the laste terme. Then hauyng the firste terme and the laste, with the nomber of excesses you knowe how to finde the totalle.

As in this example, the nomber of excesses beeyng 179. And the quantitie of one excesse beyng 1/6. their multiplcation giueth 179/6. vnto whiche if you adde the firste nomber, that is 1 1/2, it will be 188/6. And that is the laste nomber of that progression. Then to trie the totalle somme of the miles, adde the firste nomber. 1 1/2 to the laste, and thei will make 197/6, that you shall multiplie by halfe the nomber of the places, whiche in our example are. 90 (sith the whole nomber is. 180) and there will amounte. 2955. accordyng as the question saieth.

Scholar. This is sufficient for this question. And at some idle time, I will not sticke to trie it out, by settyng the progression foorthe at large. In the meane tyme I praie you for better exercise, giue me some moare questions.

<side: An other question.> Master. There is a nomber, whiche I haue forgotten: and it is diuided into. 2. partes, whereof the one I haue forgotten also, but the other was. 4. And yet this I remember, that if the parte, whiche I haue forgotten, be multiplied by it self, and then also with 4. those. 2. sommes will make. 117. Now would I knowe what was the whole nomber, and also what is the parte, whiche I haue forgotten.

Scholar. I suppose this whole nomber to be. 1x. And bicause. 4. is his one parte, the other parte must neades bee. 1.x.-4. Then doe I accordyng to the question, multiplie. 1.x.-4. firste by it self, <page> and it will make. 1z+16.p-8.x. Secondarily, I doe multiplie it (that is. 1.x.-4) by. 4 And it giueth. 4.x.-16.

Then adde I bothe those nombers together, and it will be. 1z-4.x. whiche by the question shall be equalle to. 117.

1.z.+.16.p.-.8.x.
4.x -.16.p.
—————–
1.z.-.4.x.

But then must I vse the accustomed translation, to bryng the greateste quantitie in denomination, to stande alone. And so will it bee. 1.z.=.4.x.+.117.p.

Where I must serche for the value of a roote. And therfore I multiplie. 2. by it self squarely, and so haue I. 4. vnto whiche I adde. 117. and it maketh. 121. whose roote is. 11. vnto whiche I muste adde. 2. and there commeth. 13. as the value of. 1x and the quantitie of the whole nomber.

<side: The proofe.> For proofe of this worke, I abate. 4. out of. 13. and there resteth. 9. as the other parte. Then doe I multiplie. 9. by it self, and therof riseth. 81. Also I doe multiplie. 9. by. 4. and it maketh. 36, whiche bothe together, doe make. 117. as the question would.

<side: An other woorke.> Master. Set. 1.x. for the unknowen parte, and then woorke it, to see the diuersitie of the woorkes.

Scholar. If. 4. bee one parte, and. 1.x. the other parte, then will the whole nomber be. 1.x+4p Wherefore firste I multiplie. 1.x. by it self, and it yeldeth. 1.z. Then dooe I multiplie. 1.x. by. 4. and it giueth. 4.x. whiche bothe sommes together, dooe make. 1.z.+.4.x. whiche is equalle to. 117 And by translatiô. 1.z.=.117.p-4.x.

Wherefore I doe multiplie. 2. squarely, and it giueth. <page: KK. ii.> 4, whiche added to. 117. maketh. 121. and the roote of that is. 11. from whiche I shall abate. 2. and there will reste. 9. as the other parte of the nomber. This is verie plain, & the profe of it as it was before.

Master. Then answere to this question.

<side: A question of proporti> There are 3 nôbers in proportion Geometricall. And one of the extremes is. 20 1/4. the other extreme, with the double of the middell terme, doeth make 22. Now would I knowe of you, what those. 2. nombers bee?

Scholar. For trialle, I name the other extreme, 1.x. And bicause it, with the double of the middle terme dooeth make. 22. the middell terme shall bee 11.-.1/2.x. for his double is. 22.-1x. whiche with. 1.x. doeth make. 22.

Then to proceade, I knowe the propertie of those nombers in proportion Geometricall to bee soche, that the multiplication of bothe the extremes, is equalle to the square of the middell terme, wherefore I multiplie the. 2. extremes together, and there will rise. 81/4 x. Then dooe I multiplie. 11-1/2 x. by it self in Square, and it will bee. 121.p.+.1/4 z.-.11x. whiche must bee equalle to 81/4 x. or. 20 1/4 x. Then to reduce it, I adde. 11.x. on bothe sides, and it will be. 31 1/4 x.=1/4 z.+.121p. and by translation. 1/4 z=31 1/4 x.-.121.p. That is 1.z.=125.x.-484.p.

Now resteth nothyng, but to searche the value of 1.x. Wherfore I take 125/2 and multiplie it Square, and so haue 15625/4. from whiche I must abate. 484. that is 1936/4. And there will remain 13689/4 whose roote is 117/2, whiche I shall abate from 125/2, and there will remain 8/2, that is. 4. for the other extreme.

<side: The proofe.> Then for the middell terme, thus shall I doe. Multiplie. 4. and. 20 1/4 together, and there will rise. 81. whose roote is. 9. and is the middell nomber. That 9 doubled will make. 18. and 4. ioined therto, giueth 22 <page> So are those. 3. termes in progression Geometricall, accordyng to the conditions limited in the question.

Master. Proue the worke now, how it wil frame if. 1.x. be set for the middell nomber. For it wer follie, to crie whether this question, would admitte addition of the. 2. laste nombers. Although the rule doe declare that in soche sorte of equations, there is double valuation to eche roote.

Scholar. Yet I beseke you, let me examine it a little, to see the cause, why I maie not adde them, and so take the roote.

Master. I must bere with you so moche. By addition you see, there will rise 242/2, that is 121. And then the middell nomber will be. 49 1/2. And so the proportion is 22/9. that is Dupla superquadripartiens nonas. Where as in the other. 3. nombers. 4. 9. 20 1/4. the proportion is Dupla sesquiquarta.

But in the question is one côdition, that secludeth the roote, that riseth by additiô. For the double of the middell terme, with the other vnknowen extreme, should make. 22. As in. 4. and. 9. it doeth. But in 49 1/2 and 121. it would be 220. that is 10. tymes so moche.

Scholar. And if you had saied in the question, that the double of the middell nomber, with the other extreme, would haue made. 220. then I should haue taken this later roote by additiô, and not the firste roote by subtraction.

And so I perceiue the varietie of conditions in the question dooeth limite, whiche of the. 2. rootes I shall of necessitie take, and leaue the other.

<side: An other woorke.> But now to varie that worke, I will set. 1.x. for the middell terme. And then the double of it, with the other terme, will make. 22. The double of. 1.x. is. 2.x. So must the other terme be 22p-2.x.

Then to seke out an equation, I multiplie the. 2. extremes together, that is. 22.p-2x. by 20 1/4. <page: Kk. iii.> And there riseth. 445 1/2.-40 1/2.x. And the square of. 1.x. beyng the middell terme, is sone perceiued to be. 1.z. And so the firste equation is, 1z=445 1/2 p.-.40 1/2 x.

Wherefore I take halfe. 40 1/2, that is .81/4, whose Square is 6561/16. And vnto it I putte. 445 1/2. whereby there commeth 13689/16. whose roote is 117/4. from whiche roote I must abate 81/4, and so remaineth 36/4. that is. 9. As the value of. 1.x. And for the middle nomber.

<side: The proofe.> Then for the proofe: if. 9. bee the middell nomber, the square of it, whiche is. 81. shall bee equalle to the multiplications of the extremes. Wherefore if I diuide. 81. by 20 1/4, the quotiente beyng. 4. declareth the other extreme.

Master. You seme experte inough in this forme of woorke. Therfore I will procede to other questiôs, that differ somewhat from these.

<side: A double question.> There are. 2. menne telkyng together of their monies, and nother of theim willyng to expresse plainly his somme, but in this sorte. The nomber of angelles in my purse, saieth the firste manne, maie bee parted into soche 2. nombers, whiche beyng multiplied together, will make. 24. And their Cubes beeyng added together, will make. 280. Then, quod the other man. And the like maie I saie of my money, saue that the Cubes of the. 2. partes, will make. 539. Now I desire to knowe, what monie eche of them had.

Scholar. The firste mannes sôme, I set to be 1x whiche I must parte into twoo soche partes, that thei bothe multiplied together, maie make. 24.

Master. You erre verie moche. For it is not possible, that the partes of any Coßike nomber multiplied together, can make an absolute nomber. Wherefore in soche cases, where you perceiue that there is required, after the firste position, any multiplication to make an absolute nomber, you shall call the firste nôbers, <page> by some other name of pleasure. As here you maie call the firste mannes somme. A. And the second mannes somme. B. and then in their partition, vse the name of. 1.x.

And as thei are twoo questions in one, so shall you make seueralle woorkes for them.

Scholar. Then shal I saie, that the firste mannes somme is. A. and it is diuided as he declared. Wherefore for one nomber of that diuision, I set. 1.x. And then the other shall be {24p}/{1x}. for as the one nomber multiplied by the other, doeth make. 24. So. 24.p. diuided by the one of them, must neades bryng forthe the other.

Master. That is well remembred of you. For as 4. and. 5. by multiplication, doe make. 20. So. 20. diuided by. 5. bringeth forthe 4. and diuided by. 4. it yeldeth. 5.

Scholar. So 20/5 is but. 4. and 20/4 is. 5.

Master. Go forth then with the rest of the worke.

Scholar. The Cube of. 1.x. is. 1.c. and the Cube of {24p}/{1x} is {13824p}/{1c} whiche. 2. nombers I maie not adde together, vntill I haue reduced theim vnto one denomination: whiche thyng I shall doe, by settyng. 1.c. as a fraction thus {1c}/{1p}. And then woorkyng after the rate of fractiôs, in the firste reduction thei will stande thus. {1z c}/{1c} + {13824p}/{1c}. And by farther addition thus. {1z c+13824p}/{1c}

And hetherto the woorke of bothe these. 2. mennes sommes, are indifferente and agreynge. So that this one woorke serueth for theim bothe. But now thei will differ. For in the firste mannes woordes, and so in the worke for him {1z c+13824p}/{1c} is equalle to 280: but in the seconde mannes woorke, it must be accompted equalle to. 539.

But firste to goe forward with the firste man. Seyng {1z c+13824p}/{1c} is equalle to. 280. Therefore by <page> to one denomination {1z c+13824p}/{1c} is equalle to {280c}/{1c}. And remouyng the common denominator, the numerators shal kepe the same proportion: and therfore. 1z c+13824.p. shall be equalle to. 280c. And by translation, to haue the greateste denomination alone, 1z c=280c-13824p Where I shall seke the value of. 1.x. whiche shall not be here accoumpted the square roote, but the zenzicubike roote, or the Cubike roote of the square roote, accordyng to the greateste denomination.

Wherfore. 140. in square, maketh 19600. from whiche I must abate 13824. And there doeth remain 5776 whose square roote is. 76. whiche beyng added vnto. 140. dooeth giue. 216. and beyng abated from it, it leaueth. 64. of whiche bothe I must extracte the Cubike roote, bicause in the equation there are. 2. quâtities omitted. So that of. 216. the Cubike roote is 6. And of. 64. the Cubike roote is. 4. Were I see bothe rootes serue so my purpose, that I shall take thê both.

Master. And good reason. For as in settyng 1.x for your position, you could not tell whether it were the greater parte, or the lesser, so maie you not now applie it to either of theim bothe, but take bothe rootes for the. 2. partes of your nomber.

<side: The proofe.> Scholar. So doeth the firste mannes nomber appeare to be. 10. seyng the partes bee. 4. and. 6. whiche I maie examine thus. That thei make. 24. by multiplication, it is easily seen. And that their Cubes added together, doe make. 280. is sone perceiued: seyng the Cube of. 4. is. 64: and the Cube of. 6. is. 216. whiche. 2. nombers by addition, doe make. 280.

<side: The worke of the second parte.> Master. Now proue the seconde mannes worke.

Scholar. In his woorke {1z c+13824p}/{1c} is equalle to 539. And by reduction to one denomination, it is equalle to {539c}/{1c}. So that. 1.z c+13824.p. is equalle to. 539.c. and by translation. <page> 1.z c.=.539.c.-.13824.p. whose zenzicubike roote I seke, thus: 539/2 doeth make in square 290521/4, from whiche I must abate 55296/4, and then remaineth 235225/4, whose roote is 485/2 vnto whiche I maie adde 539/2. and then will it bee 1024/2, that is. 512. whose Cubike roote is. 8. And is one parte of the seconde mannes nomber. And for the other parte, I shall abate 485/2 out of 539/2, and there remaineth 52/2. that is, 27. whose Cubike roote is. 3. And is the other parte of the seconde mannes nomber. <side: The proofe.> As it maie sone be tried thus. For. 3. tymes. 8. maketh. 24. and. 27. whiche is the Cube to. 3. added with. 512. whiche is the Cube to. 8, dooeth make 539, as the question intendeth.

<side: A question of an armie.> Master. One other question I will propounde, of. 2. armie beyng bothe square, and of like nomber. And if you abate. 4. from the one armie, and adde. 10. to the other armie, and then multiplie them bothe together, there will amounte. 9853272. I demaunde of you, what is the fronte of those square battailes.

Scholar. I call the fronte 1x. And then must the battaile bee. 1.z. Now abatyng. 4. from the one, it will bee. 1.z. Now abatyng. 4. from the one, it will bee. 1.z.-.4.p. Then addyng. 10. to the other, it wil make. 1.z.+.10.p. And if you multiplie those. 2. nombers together, there will amounte by it. 1.z z+6.z.-40.p. whiche somme must be equalle to. 9853272.

1.z. +.10.p.
1.z. -.4.p.
———————
1.z z.+.10.z.-.4.z.
-4.z.-.40.p.
———————
1.z z.+.6.z.-.40.p.

And if you adde. 40.p. to bothe partes of the equation, it will be. 1z z+6z. equalle to. 9853312 <page: Ll. i.> And by translation. 1.z z=9853312.-6.z. out of whiche laste equation, I shall searche for the value of. 1.x. by multipliyng first. 3. squarely, whereof commeth. 9. and then addyng it to. 9853312. And so commeth. 9853312. whose roote is. 3139. from whiche I must abate. 3. And then remaineth. 3136. whiche is the full nomber and Square of the one armie. And hath for his roote. 56.

For as here is one onely quantitie omitted, so the firste nomber, whiche in other questiôs of immediate equations, was the verie roote, in these interrupte equations is a rooted nomber, and is here a square nôber: whose roote therfore, I haue drawen accordyngly. <side: The proofe.> And for triall of this woorke. 56. in square maketh 3136. from whiche if you abate. 4. there will reste 3132. Again if you adde. 10. there will rise. 3146. And those. 2. nombers multiplied together, doe make 9853272, as the question intendeth.

Master. This you see, what vse is in these equations, yet are there many other equatiôs, whiche here be not spoken of: but here after you shall haue moare largely declared, if you shewe your self diligente in this parte.

<side: A question of straunge equation.> And one question I will propounde, & assoyle with out woorke for brefernesse, that you maie see there is moare behinde. There is a nomber whose Square abated by. 16. and the firste nomber augemented by 8. and then bothe thei multiplied together, will bryng forthe. 2560.

Scholar. I will proue the woorke of it. And therefore suppose the firste nomber to be. 1.x. Then is his square 1z whiche abated by 16. leueth. 1z-16p. and the nôber augemêted by. 8. yeldeth 1x+8p. These. 2. nombers multiplied together, will make 1c.+.8.z.-16.x.-128.p. beyng equalle to. 2560.

<page>
1.z.-.16.p.
1.x.+.8.p.
————————–
1.c.-.16.x.
8.z.-.128.p.
————————–
1.c.+.8.z.-.16.x.-.128.p.

And addyng 128.p. on bothe sides of the equation, it will be. 1c+8z-16x=2688p Againe addynge. 16.x. on bothe sides, it will bee 1.c+8z=16x+2688p.

Master. Where at staie you now?

Scholar. I see you shifte, but other to leaue it, as it, 2. nombers equalle to. 2: other els to make. 1. nomber equalle to. 3. And all that is aboue my cunnyng. For hetherto I haue learned now rule for any of them bothe. So that I can not gesse, what the firste nomber might bee.

Master. The nomber is. 12. And his Square is 144. from whiche if you abate. 16. it will bee. 128. And if you adde. 8. to. 12. it will yelde. 20. Then multipliyng. 128. by. 20. the somme will will be. 2560. as the question declared.

<side: Of other equations.> But to put you out of doubte, this equation is but a trifle, to other that bee vntouched. And yet I will tourne this equation a litle, to giue some light in it, and other soche. As here. 1.c.=.16.x.+.2688.p.-8z. where you see. 1.c. equalle to. 3. other nombers. And is it not certaine to you, that this equation is true?

Scholar. Yes, I am adsured thereof.

Master. And yet to auoide doubtfulnes the more trie it by resolution, accoumptyng. 12. for. 1.x.

Scholar. Where. 12. is. 1.x, there. 1.z. is. 144. and. 1.c. is. 1728. whiche. 1728. must bee equalle to <page: Ll. ii.> 16.x (that is. 192) and to. 2688. saue that you must abate. 8.z, that is 1152. Now if I adde 192. to 2688 it will make. 2880. out of whiche abatynge. 1152. there will remaine. 1728. wherby I see the equation is iuste.

Master. Then you see that the equation is true. And can you doubte, that any nomber, whiche is equalle to a Cubike nomber, hath in it a Cubike roote?

Scholar. It must neades be a Cubike nomber, that is equalle to a Cubike nomber: and therefore muste neades haue a Cubike roote: although I knowe not how to extracte that roote.

<side: Here the roote is. 2.> Master. Likewaies, when I saie: 8.z c.=.12.sz.+.128.p. it is certaine, not onely that. 12.sz.+.128.p. containeth in it as moche as. 8.z c. but that the. 8. parte of it is a z c. nomber, and hath a zenzicubike roote.

And farther it is manifeste, that as euery. z c. nomber, dooeth containe in it certaine. sz. nombers exactly, so if any nomber be annexed with those Sursolides (as here in this example are set 128) it is of necessitie, that that. 128. must containe in it certaine Sursolides exactly.

<side: The roote is 5> So if. 8.z c. bee equalle to 10.sz+20.z z+400.c+31250.p. it must neades be that the. 8. parte of this compounde nomber shall bee a. z c. nomber. And also that the z z. with the other nombers folowyng dooeth containe a certain nomber of sz. nombers. And the. c. in like sorte includeth a nomber of. z z. nombers. And laste of all. 31250.p. doeth comprehende certain Cubike nombers exactly.

<side: The roote here is. 3.> In like sorte, when we saie, that. 1.sz. is equalle to 6.c.+.8.z.+.9.p. All this compounde nomber is a Sursolide, and hath a. sz. roote. And 8.z.+.9.p. includeth certaine Cubes. And so <page> doeth. 9.p. containe exactly. 1.z. or moare.

But of these and many other verie excellente and wonderfulle woorkes of equation, at an other tyme I will instructe you farther, if I see your diligence applied well in this, that I haue taughte you. And therefore here will I make an eande of Coßike nombers, for this tyme.