Arithmetic, Part 4

Arithmetic · Part 4

The fourth Part: BEING AN ABRIDGEMENT OF THE PRECEPTS OF ALGEBRA

Written in French By JAMES de BILLY: And now tranſlated into Engliſh.

With divers Queſtions added, which were not in the Original. LONDON, Printed Anno, MDCLIX.

p. 323ALGEBRA BREVIS.

I Have been alwayes of opinion, that the practice of Algebra ſhould not be entangled with a great number of precepts: This Science is of it ſelf dark enough, without adding unto it new obſcurity, by the confuſion of many different operations. You have here an Abridgement which hath pleaſed many of good judgements, and I hope, ſuch as will with attention read it, ſhall from thence receive both ſatisfaction and profit.

I ſhall in the firſt place ſet forth a Table of three ranks: In the firſt of which there is a progreſſion natural, of which the terms are diſpoſed in that order, that immediately under them you have the Coſſick Characters, of which they are Exponents, and in the loweſt Cell, a progreſſron Geometrical, which beginning with an Unite, may be doubled, trebled, quadrupled, &c. we have for the greater eaſe onelyonly doubled them. Obſerve then that R is a Coffick Character ſignifying a Root, and that the Exponent thereof is marked above in the uppermoſt rank, and C is p. 324the Coſſick Character of a Cube, whoſe Exponent is 3, and ſo of the reſt. I call thoſe terms, which are in the upper Cell, Exponents; becauſe they expound the Coſſick Characters, and the Numbers of the Geometrical Progreſſion which are below. Mark well this manner of ſpeaking, conſider diligently this Table,

and for the preſent content your ſelf with this.

You may continue this Table, if you pleaſe, infinitely in this manner. Take two numbers, which if you multiply, they will produce ſome Exponent, you ſhal preſently ſee what a character is to be put under that Exponent. For example, If you would know the character of the Exponent 6, take the numbers 2 and 3 (becauſe theſe multiplied together make 6) after that adde their characters which are Q and C, and you ſhall have Q C for the character of the Exponent 6. In like manner the character p. 325of the Exponent 8 is QQQ, becauſe under the exponents 2 and 4, which multiplied together produce 8, are contained the characters Q and QQ. Alſo the character of the exponent 12 will be Q Q C, becauſe 12 is a number produced by the multiplication of 2 and 6, or of 3 and 4.

But if the exponent be a firſt number (that is to ſay, a number not produced by the multiplication of any two other) mark in what order it is after the exponent 5, and call it ſurſolide ſecond, third, or fourth, &c. according to its rank. The character of 5 is ſolide, of 7 ſolide ſecond, of 11 ſolide 3; and ſo conſequently under ſuch expoponents as are firſt numbers, under which onelyonly are found ſuch ſolides.

CHAP. I. The Alegorithm of Coſſick Numbers, ſimple, compounded, or diminiſhed.

BY the word Alegorithm, I mean all the operations comprehended under theſe four kinds, Addition, Subtraction, Multiplication, and Diviſion. By the word Coſſick ſimple, I underſtand ſuch as have not this + (which ſignifies plus, nor this − ſignifying minus, expreſſed before them. On the other ſide, by Numbers compoſed, are meant ſuch as have the ſigne +, and by diminiſhed, ſuch as have the ſigne −. Note ſuch p. 326numbers as have no ſigne expreſſed, are ſuppoſed to have this of +.

Sect. 1. Addition of ſimple Coſſick numbers.

All ſimple coſſick numbers are of the ſame denomination (that is to ſay, have the ſame character) or of different: If they be of the ſame denomination, the Addition is as in common Arithmetick. Example, 5Q added with 3Q, makes 8Q

If they be of different denominations, they muſt be added by the interpoſition of the figure +, as 6R added to 4Q makes 6R+4Q, in like manner 3 added to 4R makes 3+4R.

Sect. 2. Subtraction of ſimple Coſsick numbers.

Either they are of the ſame, or different denominations: if of the ſame, you muſt ſubtract as in ordinary Arithmetick; for example, 3Q ſubtracted from 8Q, there reſts 5Q.

If of different, you muſt ſubtract by the interpoſition of the ſigne − as 6R ſubtracted from 4Q there reſts 4Q−6R, ſo 3C ſubtracted from 66, there remains 66−3C.

Sect. 3. Multiplication of ſimple Coſsick numbers.

You muſt here have regard both to the abſolute number, and to the Coſſick characters: If therefore a Coſſick number be to be multiplied p. 327by an abſolute, you muſt multiply the abſolute numbers, and unto the product give the ſame character, as 5R multiplied by 12 produce 60R.

But if you multiply Coſſick numbers by Coſſick, you muſt multiply the abſolute numbers together, and to the product give the character of that exponent, which is made by the addition of the exponents belonging unto the foreſaid Coſſick characters. For example, 2R multiplied by 3Q make 6C, becauſe the exponent of Q is 2 added to one the exponent of R, make 3 the exponent of C, which in this reſpect ought to be given to the product. In like manner 5R multiplied 4C it makes 20QQ for the reaſon above given.

Sect. 4. Diviſion of ſimple Coſſick numbers.

The ſpeculation of this is marvellous. But the practice of it is by putting the Diviſor under the number to be divided, drawing between them a little ſmall line in manner of common fractions. For example, 13Q divided by 7R, the quotient is 13q / 7R and 6QQ divided by 5C the quotient is 699/5C.

Sect. 5. Addition of numbers compoſed and diminiſhed.

Some order is to be obſerved in this, in which the numbers are to be diſpoſed in ſuch manner, that thoſe that are of the ſame denomination muſt be put right under one another. After you p. 328have done this, if they have the ſame ſigne, they are added as in common Arithmetick, and to the product give the ſame ſigne. As for example, 7Q−4C added to 3Q−2C, the Sum is 10Q−6C.

But if the numbers be of different ſignes, the leſſer muſt be ſubtracted from the greater, and to the reſidue you muſt give the ſigne of the greater number as 6Q+7R added with 7Q−12R, give for the Sum total 13Q−5R.

Sect. 6. Subtraction of numbers compoſed and diminiſhed.

There is nothing more intricate to beginners than the precepts commonly given for ſubtraction. You have here an order plain, ſure, and very caſie to practiſe.

Change the ſign of the particulars of that number you deſire to ſubtract, and after this change adde them with the number from which the ſubtraction is to be made, and you ſhall have the reſidue: As if from 6Q−10R you would ſubtract 18Q−15R by the 5 Sect. the reſidue will be 5R−12Q. So likewiſe if you would ſubtract −8R−9Q from 16R+6Q, the reſidue will be 24R+15Q

Sect. 7. Multiplication of numbers compoſed and diminiſhed.

Mark what I ſaid of Coſſick ſimple in the 3 Sect. and remember that the ſame ſignes have alwayes the ſigne + in the product and different − & there is no difficulty in multiplication, ſo as you multiply every particular of the mulplicand by every particular of the multiplicator, as in common Arithmetick, as if you multiply 3Q−2R by 8R, multiply 2R by 8R, it makes 16Q, and becauſe the multiplicator and the multiplicand have different ſignes, the product muſt have the ſigne of −; and therefore that ſhall be 16Q, farther 3Q by 8R make 24C, to which you muſt give the ſigne +, becauſe the multiplicator and the multiplicand have the ſame ſign, ſo that the product of this multiplication will be 24C−16Q. So 2R+4Q multiplied by 3Q−5 the product is 6C+12QQ−10R−20Q

6C+12QQ/6C+12QQ−10R−20Q

p. 330Sect. 8. Diviſion of numbers compoſed and diminiſhed.

There is no great difficulty in this, onelyonly put a line between the Dividend and the Diviſor, and you have the quotient, as 4C−3Q+2R divided by 5R−4C make for this quotient 4C−3Q+2R / 5R−4C

Sect. 9. Algorithm of Fractions.

I ſhall not give any particular precepts, becauſe if a man underſtand the Fractions of common Arithmetick, and practiſe according to what is before ſaid, there will be no need of them.

CHAP. II. The rule of Algebra, with the explication thereof.

IT was meerly neceſſary by the precedent rules, to trace out inſenſibly the way to Algebra, which cannot be practiſed without Addition, Subtraction, Multiplication and Diviſion. Having therefore made plain theſe difficulties, we will proceed in the propoſition of the rule of Algebra, and the explication of every part of it briefly and plainly.

p. 331Sect. 1. The rule of Algebra.

You muſt firſt for the number unknown put 1R, and after examine this root according to the tenour of the queſtion, until you come to an Equation. Secondly, this Equation is to be reduced, if need require. Thirdly, you muſt divide every part of the Equation by the number of the greateſt coſſick character. After which, either the quotient, or ſome root of the quotient, will give the root unknown: This is the rule of Algebra, let us now explain it.

Sect. 2. How the Equation muſt be found.

The rule ſaith, this is done by examining the queſtion propounded, according to the tenour of the ſame. That is to ſay, you muſt well obſerve all the conditions of the queſtion propounded, to the end you may fully accompliſh it. For after you have gone thorow it, you ſhall find an equation between two numbers. As if I ſearch a number, which added with its ſquare ſhall make 20, I ſuppoſe this number unknown to be 1R, the ſquare thereof is 1Q. (becauſe every number multiplied by it ſelf, makes its ſquare) then 1Q+1R is equal to 20. See thus an equation found between 1Q+1R and 20.

p. 332Sect. 3. How your Equation muſt be reduced.

Your Equation being found, it is reduced by adding the ſame number to both the terms of the equation, or by ſubtracting from them the ſame number. So is it performed by multiplying, or dividing both the terms by the ſame number. For by this means your equation ſhall remain the ſame after theſe things done. As for example, 1R+1Q=20, adding throughout 2C, you ſhall have alſo an equation between 1R+1Q+2C and 20+2C. So ſubtracting 1R from the termes, you have 1Q=20−1R. Likewiſe multiplying, or dividing both your terms by 3, you ſhall have by the multiplication 3R+3 Qequal to 60, and by the diviſion the equation with be between on ⅓R+⅓Q and 20/3.

Now to make your reduction judiciouſly and profitably, you muſt take care alwayes that your greateſt character remain alone on one ſide of your equation. As of all the reductions before made, there is none uſeful but the ſecond, becauſe in that onelyonly you finde on the one ſide alone 1Q=20 − 1R which is the onelyonly end of your reduction.

I ſaid in the rule of Algebra, that your equation muſt be reduced, if it be neceſſary; becauſe it ſometimes happens, that there is no need of it: as when your equation fals out between two ſimple collateral numbers, I call thoſe numbers collateral, whoſe exponents do not ſurpaſſe one another by more than an unite.

p. 333Sect. 4. When you muſt extract the root.

When your coſſick numbers are ſimple and collateral, you muſt not extract any root; but if you divide by the number of the greateſt coſſick character, the quotient ſhall ſhew you the value of the root, which is all you ſeek for in Algebra. For example, If you finde an equation between 2R and 28, dividing ſimply 28 by 2 the quotient, ſhall be the value of 1R.

But when the terms of your equation are not collateral, you muſt extract ſome root; either ſquare, cube, ſquared ſquare, &c. according to the coſſick character which remains after your hypobibaſme.

Now hypobibaſme is nothing elſe but an abatement, or depreſſion of the character, and is done by ſubtraction of the letter exponent from the greater. As if you finde an equation between 10QC and 90 QQ take notice of the exponent of QC in the Table inſerted at the beginning, the which exponent is 6; afterward look to the exponent of QQ which is 4, ſubtract 4 out of 6 there reſts 2, of which the coſſick character is Q. From hence I conclude, that 10Q are equal to 90, after dividing 90 by 10, and finding 9 in the quotient, I conclude, that the ſquare root of 9 muſt be extracted by reaſon of the character Q.

p. 334Sect. 5. How to extract the ſquare root of numbers, compound and diminiſhed.

No man hath yet perfectly found out the way to extract the root of numbers compound and diminiſhed, unleſſe the exponents of three terms of the equation keep between them in ſome ſituation or other an Arithmetical proportion, that is to ſay, the ſame diſtance. As if the equation be between 1Q and 20−1R, you may now extract the root of 20−1R, becauſe the exponents of the three numbers which make the equation, are 2, 0, 1, which thus place, 0 1, 2, keep the ſame diſtance.

The greateſt coſſick character left after the Hypobibaſme ſhews the root to be extracted; as in this example, before the ſquare root is to be extracted, becauſe the greateſt character is Q. The method to be followed in the extraction, ſee here propounded in general terms.

Take firſt the half of the number of roots. Secondly, to the ſquare of this half adde, or from it ſubduct your abſolute number according to the figne of + or −. Thirdly, extract the ſquare root of this ſum, or of the reſidue. Fourthly, to this root adde, or from it ſubtract half the number of roots, and this laſt ſum or reſidue ſhews you the value of the root unknown. For example, I would find a number, the double whereof added to its ſquare, ſhould be equal to 24, I ſhall find an equation between 2R+1Q and 24 by p. 335the ſecond Section. Moreover; I ſhall reduce this equation after this manner 1Q equal to 24−2R by the third Section. Then if I divide 24−2R by 1, the number of the greateſt coſſick character there ſtill remains 24−2R, becauſe an unite doth neither multiply nor divide. Then in as much as the three terms of the equation do keep an Arithmetical proportion, I extract the ſquare root of 24−2R in this manner. I take firſt half the number of roots, which is 1. Secondly, the ſquare of 1 is 1, to which I adde the abſolute number, which is 24, becauſe of the ſigne + before it, that makes 25. Thirdly, I extract the ſquare root of 25, which is 5. Fourthly, from this root I extract the moity of the number of roots which is 1, becauſe of the ſigne − the reſidue will be 4; whence I conclude that to be the value of one root, and that the number ſought is 4, whoſe double is 8, added to the ſquare (of 4,) 16, makes 24.

Here note that numbers diminiſhed, where the abſolute number hath the ſigne —, have two roots. The greater is extracted, as before we have taught; the leſſer is found out by ſubtracting the ſquare root of the reſidue from the half ſum of the roots. As if I ſeek a number whoſe octuple diminiſhed by 12, ſhall be equal to its ſquare, you will find an equation between 1Q and 8R−12. The greateſt root is 6, the leſſer is 2, here both the roots anſwer the queſtion. But this happens not often.

But if you be to extract the biquadrate root: p. 336Firſt, extract the ſquare root, as is taught, and again extract the ſquare root of this, and this ſhall be your biquadrate root. As if the equation be between 1QQ and 2Q+8, you may finde the ſquare root to be 4 by the method taught, firſt, taking half the number of the ſquares, &c. and afterward you muſt extract the ſquare root of 4, which is 2, this ſhall be the value of the root. In like manner, if your equation be between 1QC and 2C+48, firſt, I extract the ſquare root of 2C+28, which is 8, of which extract the cube root 2, becauſe the root to be extracted is the ſquare cube, as the character QC, which is one of the terms of the equation denoteth.

Sect. 6. How to know if the queſtion be impoſſible, vain, or ill propounded.

You may know the queſtion to be impoſſible, if you come to an equation impoſſible: As if following the conditions of the Probleme, you meet with an equation between 6R and 24R, or between 3Q+5 and 4+…Q.

Secondly, the queſtion is vain, when the equation is between two equal numbers of the ſame denomination, as between 6Q and 6Q.

Thirdly, the queſtion is ill propounded, when without any difficulty many numbers will anſwer the probleme propounded.

p. 337CHAP. III. Algorithme of ſecond roots with their uſe.

ALgebraiſts ſometimes uſe more than one root to finde out divers numbers propounded, and then to the end they may proceed with leſſe confuſion they uſually help themſelves with ſecond roots, which they expreſſe thus, 1A, 1B, &c.

Sect. 1. Addition of ſecond roots.

If your ſecond roots be of the ſame denomination, adde the numbers, and to the ſum give the ſame denomination, as 5A added to 4A make 9A, if they be of different denomination, adde them with the ſigne of + as 5 A added to 6B make 5A+6B.

Sect. 2. Subtraction of ſecond roots.

If they be of the ſame denomination, ſubtract one number from the other, and to the reſidue give the ſame denomination, as 5A taken from 9A, there remains 4A, if different, they are ſubtracted with the ſigne − as 6B taken from 8A, there reſts 8A−6B.

p. 338Sect. 3. Multiplication of ſecond roots.

If they be of the ſame denomination, do as you do with the firſt roots, as 4A multiplied by 7A make 28AQ, if of different, both denominations are reteined in the product, as 3R multiplied by 5A make 15RA.

Sect. 4. Diviſion of ſecond roots.

Diviſion is onelyonly performed by the interpoſition of a little line, as is before taught, notwithſtanding it is to be obſerved, that if for example 3AR be to be divided by the Diviſor 1R the quotient ſhall be 3A, becauſe in ſuch caſe there needs nothing elſe but to take from the Dividend the character of the Diviſor.

Sect. 5. The extraction and uſe of ſecond roots.

After you have found and reduced your equation, according to the manner of working in ſecond roots, you muſt extract the root after the manner taught in the precedent Chapter. As if 1AQ be equal to 25, I ſay that 5 is the value of the ſecond root, and if 1AQ be equal to 4A+12, you muſt take the moity of the number of roots, &c. As is ſaid in the 5. Sect. of the precedent Chapter, and you ſhall find 6 to be the value of 1A.

Now ſince the end of the ſecond roots is to be p. 339reduced to firſt, you muſt not forget after you have found the value to begin again your work, and to put in firſt roots that which you have found to be the value of the ſecond, as I ſhall ſhew you in ſome example in a Chapter following.

CHAP. IV. The Algorithme and extraction of the roots of ſurd & irrational numbers.

SUrd roots are thoſe that have a radical ſigne before them, and which in propriety of ſpeech ought to be called abſolute numbers, notwithſtanding they cannot be expreſſed by any common number, neither whole, nor broken, we will hereafter expreſſe the radical ſigne by this character ℛ.

There are many ſorts of ſurd roots, ſome are ſimple, as ℛQ5, that is to ſay, the root ſquare of 5, others are compound, as ℛQ5+ℛC6, that is to ſay, the root ſquare of 5, plus the root cube of 6, ſome are univerſal, whoſe radical character, extends to all the particulars following it, and for that end are encloſed in a Parentheſis in this manner ℛQ (14+ℛQ4) the root univerſal of 14 joyned with the root ſquare of 4, all which number is 4 for 14+, the root ſquare of 4 which is 2, maketh 16 whoſe root is 4.

p. 340Sect. 1. Reduction of ſurd roots ſimple to the ſame denomination.

Firſt, you muſt put the radical ſignes under the numbers to which they belong. Secondly, you muſt multiply the numbers by the ſignes acroſſe, for to get new ones. Thirdly, you muſt adde the ſignes together, which is done by multiplying their exponents, & give the character of the product common to the two new products, as if you would reduce to the ſame denomination ℛQ5 and ℛC4, you muſt firſt place them as followeth. … Secondly, you muſt multiply the numbers 4 and 5 by their ſignes acroſſe, that is to ſay, you muſt take the ſquare of 4, and the cube of 5, which are 16, and 125. Thirdly, the exponents of the ſignes ℛQ and ℛC, which are 2 and 3, ought to be multiplied together, the product is 6. I look then in the table what coſſick character is under the exponent 6, and finding QC, I take that for my common denominator, and in ſtead of my two firſt ſurd roots, which were of different denomination, that is to ſay ℛQ5, and ℛC4. I have two new ones of the ſame denomination, that is to ſay, ℛQC of 125, and ℛQC of 16.

p. 341Sect. 2. Multiplication and Diviſion of ſurd ſimple roots.

If the roots be of the ſame denomination you muſt onelyonly multiply and divide the numbers by themſelves, and to the product and quotient givse the ſame radical ſigne as ℛQ7 by ℛQ2 give for the product ℛQ14. In like manner ℛQ36 divided by ℛQ12, gives for the quotient ℛQ3.

But if the roots be of different denomination, you muſt reduce them to the ſame denomination by the precedent Paragraph, and after multiply and divide as ſhall be ſhewed, for example, ℛQ3 multiplied by 2, the product is ℛQ12, and ℛQ12 divided by 2, the quotient is ℛQ3.

Sect. 3. How to know whether two ſurd roots be commenſurable or not.

You muſt divide the greateſt root by the leſſer, if the quotient be rational, the two roots are commenſurable: if otherwiſe, they are not. As becauſe ℛQ24 divided by ℛQ6, gives for the quotient ℛQ4, which is 2 a rational number, I conclude theſe two root: ℛQ24 and ℛQ6 to be commenſurable. In like manner, ſince root ſquare 24 divided by ℛQ8, the quotient will be 3, a ſurd number and irrational, you may conclude thoſe two roots, ℛQ24, and ℛQ8, to be incommenſurable.

Sect. 4. Addition of ſimple irrational roots.

If the roots be incommenſurable, you muſt adde them onelyonly by the ſigne+as ℛQ24 added unto ℛQ8 makes ℛQ24+ℛQ8. But if they p. 342be commenſurable, you muſt adde a unite to their quotient rational, and you ſhall have a ſumme, which being multiplied by the leſſer of the two roots to be added, will give a product which ſhall be the ſumme ſought. As ℛQ24 added with ℛQ6 makes ℛQ54, becauſe ℛQ24 divided by ℛQ6, gives 2 for the quotient rational, to which I adde a unite, and it is 3 by which (alwayes reducing them to the ſame denomination) I multiply ℛQ6, which is the leſſer of my two roots, and I finde for my ſumme ℛQ54.

Sect. 5. Subtraction of ſimple irrational roots.

If they be incommenſurable, you muſt ſubtract them by prefixing the ſigne−as ℛQ8 ſubtracted out of ℛQ24, the reſidue ſhall be ℛQ24−ℛQ8.

But if they be commenſurable, you muſt take away a unite out of the quotient rational, and you ſhall have the reſidue, the which being multiplied by the leſſer of the roots given, ſhall give a product which ſhall be the reſidue fought, as if I be to ſubtract ℛQ… from ℛQ4, dividing the greater by the leſſer, the quotient rational is 2, from which if you take 1, there remains 1, by which (reducing them firſt to one denomination) if you multiply the leſſer root, that is to ſay, ℛQ6, the reſidue will be ℛQ6.

p. 343Sect. 6. Addition and Subtraction of ſurd numbers, compoſed and diminiſhed.

I have here no new precepts, onelyonly advertiſe you, that you may remember what I have ſaid before of Coſſick numbers, touching the ſignes of + and − in the fifth and ſixth Paragraph of the firſt Chapter, and what is delivered in the fourth and fifth of this Chapter, touching the Addition and Subduction of ſimple ſurd numbers, and theſe will be no difficulty, as if you be to adde 5+ℛq. 24 with 3+ℛ q. 6, you will find 8+ℛ q. 54. In like manner, if you ſubduct 3−ℛ q. 6, from 5+ℛ q. 24, there reſts ℛ q. 54−42.

Sect. 7. Multiplication of numbers ſurd, compoſed and diminiſhed.

This multiplication hath no great difficulty, nor needs new precepts. Remember onelyonly that the ſame ſignes have in the Product+and different−with this, that your Multiplication is not good, if the particulars to be multiplied be not firſt reduced to the ſame denomination. For example. 5+ℛ q. 24 by …−ℛ q. 6, 'tis to be done after this manner:+ℛ q. …4 by−ℛ q. 6, maketh−ℛ q. 144, or−12 after+… by−ℛ q. 6, maketh−ℛ q. 150. Further, +ℛ q. 24 by 3 is +ℛ q. 216. Laſtly, +5 by 4… is+15, then the whole Product will be 15+ℛ q 216−ℛ q. 150−ℛ q. 144, or 3+ℛ q. 216−ℛ q. 150 becauſe p. 344ℛ q. 144 is a rational number, to wit 12, which being ſubducted out of 15, becauſe of the ſigne−leaveth 3.

Example: …

Sect. 8. Diviſion of ſurd numbers, compounded or diminiſhed.

If the Diviſor be ſimple, the diviſion is made by the interpoſition of a line between the Diviſor and the compound number to be divided, as if ℛ q. 2+ℛ q. 5 be divided by 8, the quotient will be ℛ q. 2+ℛ q. 5,/8 and ſo of others.

But becauſe it ſometimes may fall out (though very ſeldome) if the Diviſor alſo will be a Binome, p. 345or compound number, that is to ſay, a ſurd number compounded of two particulars with the ſigne +, or a Trinome that is compounded of three particulars, &c. See here the manner to divide in ſuch a caſe.

If the Diviſor be a Binome, you muſt multiply by his Apotome, as well the number to be divided, as the Diviſor (and if the Diviſor be an Apotome, you muſt divide by the Binome, as well the Dividend, as the Diviſor) by means of this Multiplication you ſhall have a new Dividend, and a new Diviſor. Now this new Diviſor will be alwayes rational, and therefore needs onelyonly to to be ſet under the Dividend with a line between. As for example: ℛ q. 6−2 by ℛ q. 5+ℛ q. 3, I take the Apotome of my Diviſor (that is) ℛ q. 5 −ℛ q. 3. by which I multiply both my Dividend and my Diviſor, by one of the multiplications is produced ℛ q. 30−ℛ q. 20−ℛ q. 18 +ℛ q. 18 +ℛ q. 12, for my new Dividend, and by the other is produced 2 for my new Diviſor. So that the quotient of my Diviſion will be ℛ q. 30−ℛ q. 20−ℛ q. 18+ℛ q. 12/4, becauſe 2 muſt be firſt ſquared, and then 4 put underneath the Dividend.

If your diviſor be a Trinome, you muſt obſerve the ſame method, multiplying the dividend and the Diviſor by the Apotome of the Diviſor, that is to ſay, by the ſame Diviſor onelyonly, changing the ſigne of the laſt particular. After this is done, you ſhall have a new Dividend and a new Diviſor, p. 346which ſhall be a Binome. Then you muſt again ſeek a new Dividend and Diviſor, which now will be ſimple and rational.

Laſt of all, if you will not take this pains, the Diviſion is good, if under the Dividend you ſubſcribe the Diviſor with a line between.

Sect. 9. Multiplieation of roots univerſal.

You muſt reduce the root to be multiplied, and the Multiplicator to their ſquares or cubes, according to the radical ſigne prefixed, and afterward perform you multiplication, as is taught in Sect. 7 of this Chapter. Afterward you muſt affix the ſigne radical, and incloſe all in a Parentheſis.

This is better underſtood by an example, as if you multiply ℛ q. (7+ℛ q. 3) by 2, the ſquares of the one and other numbers are 7+ℛ q. 3 and 4 then the firſt being multiplied by the laſt, maketh 28+ℛ q. 48, and therefore if you cloſe this number within a Parentheſis, and put before it the ſame radical ſigne, the product will be ℛ q. (28+ℛ q. 48.)

In like manner, if you would multiply this number 7/2−½R+ℛ q. (49/4−3/4 q.−7/2 R) by it ſelf, to get the ſquare of it, you muſt call to mind the fourth propoſition of the ſecond book of Euclide, which ſheweth that a line being divided into two parts, the ſquare to the whole is equal to the ſquare of the parts, and to double their Rectangles, you muſt therefore conceive this number, as divided into two parts, of which the firſt p. 347is 7/2−½ R, and the laſt ℛ q. (29/4−3/4 q.−7/2 R) take then the ſquares of the parts, which are 49/4+3/4 q.−7/2 R, and 49/4−3/4 q.−7/2 R, the double of the Rectangle of the parts is ℛ q. (9604/16+220/16 q.−2744/8R+7C+12/16 q. q.) then the ſquare of the number propoſed is the ſumme of theſe three numbers, that is to ſay 49/2−½ q.−7 R+ℛ q. (9604/16+220/16 q.−2744/8 R+7 C−11/16 q. q.) The ſquare of the Apotome is the ſame number, putting only the ſigne−before the univerſal root, & the ſum of the two ſquares is 49−1 q.−14 R.

Sect. 10. Diviſion of roots univerſal.

You muſt reduce the roots to be divided, and the Diviſor to their Squares, Cubes, &c. And after divide them as is taught in the 8 Sect. and when this is done, encloſe all in a Parentheſis, with the ſame radical ſigne which was before. As if you divide ℛ q. (13+ℛ q. 17) by the root ſquare of 5, their ſquares are 13+ℛ q. 17 and 5, then the firſt being divided by the laſt, the product will be 2 3/5+ℛ q. 17/25, then the quotient of the diviſion propoſed will be ℛ q. (2 3/5+ℛ q. 17/25.)

Sect. 11. Addition and Subduction of roots univerſal.

Many trouble themſelves to give precepts intricate enough, the ſhort and moſt certain is to p. 348adde them with the ſigne+, and ſubduct them with the ſigne−. As for example, ℛ q. (3+ℛ q. 2) added with ℛ q. (ℛ q. 5+6) will be ℛ q. (3+ℛ q. 2)+ℛ q. (ℛ q. 5+6) and the ſame firſt root ſubducted from the laſt, the reſidue is ℛ q. (ℛ q. 5+6)−ℛ q. (3+ℛ q. 2.)

Sect. 12. Extraction of the roots of Binomes and Apotomes.

1 Take the difference of the ſquares of the one and other part of the Binome. 2 Adde and ſubduct the ſquare root of this difference from the greateſt part of the Binome. 3 Conjoyn the ſquare root of the moity of the ſumme, with the ſquare root of the moity of the reſidue, by the ſigne+, if it be a Binome, and by the ſigne−, if it be an Apotome; and thus the extraction is finiſhed. As if you would extract the ſquare root of this Binome 3/2+ℛ q. 5/4, you ſhall firſt take the ſquare of the firſt part, which is 9/4, and the ſquare of the ſecond which is 5/4, the difference of theſe two is 4/4, that is to ſay 1. Secondly, you muſt extract the ſquare root of this difference, which is 1, you ſhall adde and take it away from the firſt part of the Binome, by the addition you ſhall have 5/2 for your ſum, and by ſubduction ½ for your reſidue. Thirdly, joyn the root ſquare of the ſum, with the root ſquare of the reſidue by the ſign+& you ſhall have ℛ q. 5/2+ℛ q. ½ for the ſquare root of your Binome propounded, and conſequently ℛ q. 5/2−ℛ q. ½ ſhall be the root ſquare of the Apotome 3/2−ℛ q. 5/4.

p. 349CHAP. V. The Ʋſe of Algebra.

'TIs much to have taken the pains to learn all that we have hitherto ſhewen or taught: but I dare boldly ſay, that thoſe that ſhall reſt here, do as yet know nothing to the purpoſe, although they may know all the precepts; it behoveth us then to make a ſtep further, to apply and bring thoſe precepts into uſe and exerciſe. 'Tis that which I deſire to demonſtrate in this Chapter, by ſome queſtions, the ſolution of which will give great light to the atteining of perfection in this Art. Wherefore I intreat thee (Reader) not to omit this Chapter, in which I pretend to yeeld thee ſome pleaſure and delight, as alſo an illuſtration of what hath been before treated of.

Sect. 1. Queſtions reſolved by one ſimple Equation.

Queſtion I.

A Lexander one day told Epheſtion, that he was elder than him by two years; thereupon Clitus tels them, that he was as old as both of them (their ages added together) and four years over and above. The Philoſopher Calliſthenes being preſent at this diſcourſe (ſaith he) I well remember that my father, who was 96 years old, had the p. 350age of you three. It is demanded here, how old Alexander was when he held this diſcourſe, as alſo how old Clitus and Epheſtion were. I put for the age of Epheſtion 1 R of years, whence it follows that Alexander had 1 R+2; therefore Clitus had, R+6, and thoſe three together, according to the condition of the queſtion, ought to be equal to 96: therefore there is an equality between 4 R+8 (which is the ſum of the three ages) and 96, take away 8 from both parts of the equation ſo there will remain on one ſide 4 R equal to 88, which divide by the number of the greateſt Coſſick Character, that is to ſay by 4, the quotient gives 22 for the value of one root, which was ſuppoſed for the age of Epheſtion. Therfore Epheſtion was at that time aged 22 years, Alexander 24, and Clitus 50, which all together make 96 years.

Queſtion II.

A Hare is 100 Geometrical paces diſtant from a Dog that ſwiftly purſues him, and the Dog runneth two times and an half faſter than the Hare: It is demanded how many Geometrical paces the Hare will have run when the Dog overtaketh her; I put for theſe Geometrical paces 1 R; therefore the Dog which runs 100 paces more than the Hare, will have run 100+1 R: and for that the Dog runs twice and an half ſwifter than the Hare, I take two numbers in like porportion to one another, that is to ſay 5 and 2, and conclude that there is the ſame proportion between 100+1 R to 1 R, as between 5 and 2; therefore p. 351the product of the firſt number 100+1 R multiplied by the laſt number 2 (which is 200+2 R) is equal to the product of the two means 1 R and 5 (which will be 5 R) therefore if you take from both parts 2 R, there will remain 200 equal to 3 R, and therefore I divide 200 by 3, which is the number of the greateſt Character, and find in the quotient 66⅔ which will be the value of the root. I ſay therefore that the Hare will have run 66 Geometrical paces, and ⅔ when the Dog ſhall have overtaken her, and the Dog will have run 166⅔ paces, which make twice and an half more than 66⅔.

Queſtion III.

The Architect Vitruvius in his ninth Book Chap. 3. tels us, that Archimedes found the quautity of ſilver that a Goldſmith had mixed in a golden Crown which he had made for the King Hiero (who was obliged by vow to preſent it to the gods) weighing 100 pounds. It is demanded by what means Archimedes could arrive to the knowledge of that ſecret. The common opinion is, that he took two maſſes, the one of gold, the other of ſilver, which weighed as much as the Crown; afterward he filled a veſſel up to the brim with water, which veſſel was placed in ſome grat baſon, that the water that ſhould be forced out of the firſt veſſel, might be preſerved and not loſt. Thirdly, he gently put in the two maſſes and the Crown, one after the other, into the prepared veſſel, taking exact notice of the quantity of water that iſſued out of the veſſel at p. 352each time, and concluding from thence, that the Goldſmith had mingled 6 pounds and ⅔ parts of ſilver. We will ſuppoſe then that the maſſe of gold weighing 100 pounds, did caſt out of the veſſel 60 pounds of water, & that the maſſe of ſilver alſo weighing 100 pounds, caſt forth of the veſſel 90 pounds of water, and that the Crown caſt forth 65 pounds. I put afterward for the ſilver mixed in the Crown 1R, and conſtitute twice the Rule of three after this manner.

If 100 pounds of gold give me 60 pounds of water, how much will 100−1.R? and I finde 6000−60R / 100 for my fourth number.

Secondly, if 100 pounds of ſilver give me 90 pounds of water, how much ſhall 1R? and I find 90R / 100. Now theſe pounds of water 6000−60R / 100 and 90R / 100 added together do make 6000+30R / 100 pounds of water caſt out, which ought to be equal to 65 pounds of water, caſt forth by the Crown, and therefore if we reduce them, we ſhall finde 6000+30R equal to 6500 (this reduction is made by multiplying the denominator 100 by 65, for ſeeing that this fraction 6000+30R / 100 is equal to 65 it ſhall be alſo equal to 65/1, Mark well this kinde of reduction once for all.and therfore there will be the ſame proportion of the numerator 6000+30R to the denominator 100, as of the ſecond numerator 65 to 1. Therefore the product under the extreams 6000+30R is equal to the product of the means 6500) take away therefore from both parts 6000, and there p. 352will remain an equation between 500 and 30 roots, and therefore divide 500 by 30, the number of the greateſt character, you ſhall have the value of the root 16⅔ for the pounds of ſilver mingled by the Goldſmith in the Crown:

Sect. II. Queſtions reſolved by an Equation compounded.

Queſtion I.

TO divide 8 into two ſuch numbers as their ſquares being added together, may make 34. I put for the firſt 1R, therefore the ſecond ſhall be 8−1R, their ſquares are 1Q, and 64+1Q−16R, which added together, do give for their ſum 64+2Q−16R, the queſtion imports that the ſum of the ſquares is 34. Therefore there is an Equation between 64+2Q−16R, and 34, which being reduced by addition and ſubtraction, there will remain alſo an equation between 2Q and 16R−30, and the whole divided by 2, which is the number of the greateſt Coſſick Character, there will yet remain an Equation between 1Q and 8R−15, from which I extract the root, as hath been ſhewn in the fifth Section of the ſecond Chapter. The half of the roots is 4, his ſquare 16, from which take the abſolute number 15, reſt 1, whoſe ſquare root 1 added to the half of the number of roots, gives for its ſum 5, which is the value of the root; therefore the two numbers ſought ſhall be 5 and 3.

p. 354Queſtion II.

To find two numbers whoſe product may be 12, and the difference of their ſquares 32, I put for the one of them 1R; therfore ſeeing that the product is 12, the other number ſhall be 12/1R (for if the product of two numbers be divided by one of thoſe two numbers, the quotient ſhall be the other number) their ſquares are 1Q and 144/1q whoſe difference is 144/19−1Q equal to 32, as appears by the queſtion; therefore there will be an equality between 144/1q and 32+1Q, & therefore if we make the reduction, as in the third Queſtion of the firſt Section, we ſhall alſo finde an Equation between 144 and 32Q+1QQ, alſo between 144−3Q and 1QQ; it behoveth then to extract the ſquare root of the number 144−32Q. The half of 32 is 16, whoſe ſquare is 256, to which adde 144, makes 400, whoſe ſquare root is 20, from which take the half of 32, to wit 16, there remains 4. See here the ſquare root but ſeeing that the ſquared ſquare root ought to be taken, I take again the root of 4, and I finde 2 for the value of the root. Therefore ſeeing that the ſecond number hath been put 12/2R the ſame ſecond number ſhall be 12/2, that is to ſay 6.

Queſtion III.

Two Merchants joyn in company, and together bring in the ſum of 165 Crowns; but the firſt mans money hath been expoſed twelve moneths intire, and the ſecond mans money onelyonly eight moneths: it happens that they gain but 28 p. 355Crowns, which added to 165 make 193, which they diſtribute to one another in ſuch ſort, as the firſt takes 67 Crowns, as well for his principal money as for his profit, and the ſecond takes 126 Crowns, the queſtion is what each of thoſe Merchants brought into ſtock. I put for the money of the firſt man 1R, therefore ſeeing that the ſum of both was 165 Crowns, the ſecond mans money is 165−1R. Now if you take away 1R, which is the ſum the firſt man brought in, from the ſum he received, which was compounded of the principal and profit, you will find that the firſt mans profit will be 67−1R, and by the ſame argument you ſhall finde, that the ſecond mans profit will be 1R−39. Now you muſt finde what one root gaineth in eight moneths, which will be done by the Rule of Three, thus, If in 12 moneths there be gained 67−1R, how much will there be gained in 8 moneths, and the fourth number ſhall be 134/3−2R / 3 for the firſt mans profit in 8 moneths; after that I ſeek what the ſecond man hath gained by another operation of the rule of three, ſaying; If 1R gain 135/3−3R / 2, what will 165−1R gain, and I finde for my fourth number 7370+2/3Q−464R / 3,/1R which is equal to the ſecond mans profit, which we have already found to be 1R−39, and therefore by reduction there will be an equation between 1Q−39R and 7370+2/3Q−464/3R, all which 39R being added, and ⅔Q taken away, there will be an p. 356equation between 7370 and 1/3Q+347/3R and conſequently between 7370−347/3R and ⅓Q, and therefore multiplying all by ⅓, which is the number of the greateſt character, there will be yet an equation between 1Q and 22110−347R out of which the ſquare root muſt be extracted. The half of the number of roots is 347/2, whoſe ſquare is 170409/4, which added to 22110, make 208849/4, whoſe ſquare root is 457/2, from which if you take half the number of roots, there will remain 110/2, that is to ſay, 55 for the value of one root, and was the money the firſt man put in bank; and for that we have found in the purſuit of theſe operations, that the firſt mans profit was 67−1R, that ſame profit will be 67−55, that is to ſay 12, by the ſame reaſon the ſecond mans ſtock ſhall be 110, and his profit 16.

Sect. III. Queſtions reſolved by ſurd numbers.

Queſtion I.

TO divide any given number (as for example 4) according to mean and extream reaſon, that is to ſay, to divide 4 into two numbers, in ſuch manner as that the whole 4 may bear the ſame proportion to its greateſt part, as the greateſt part bears to the leaſt. I put for the greateſt part 1R, therefore the leaſt ſhall be 4−1R. Therefore there is the ſame proportion of 4 to 1R, as of 1R to 4−1R, and therefore the ſquare of the p. 357middle part 1Q is equal to the product of the extreams 16−4R, from which I extract the root according to the rule before preſcribed. The half of the number of roots is 3, whoſe ſquare 4, added to 16, makes 20; out of which the ſquare root ought to be extracted according to the precept: but ſeeing it is no ſquare number, I muſt content my ſelf by putting the radical ſigne before it thus, ℛQ 20, from which I take the half of the number of roots, and I have for Reſidue ℛQ 20−2, which is the value of the root, by which I ſhall find with facility, that the other part wil be 6−ℛ Q 30. For the proof of this operation, it behoveth that theſe two parts added together make 4, and that the leſſer 6−ℛQ 20, being multiplied by 4, make the product equal to the ſquare of the greater part ℛ Q 20−2.

Queſtion II.

To divide 8 into two parts, between which, 2 may be a mean proportional. I put for the firſt part 1R, therefore the leſſer ſhall be 8−1R, and ſeeing that 1R and 2, and 8−1R ought to be proportional, it behoveth that the ſquare of 2, which is 4, be equal to the product of the extreams, which is 8 R−1Q, and therefore after the reduction it will be found, that 1Q is equal to 8R−4, whoſe ſquare root is ℛQ 12+4. For the one part of 8, and for the other part 4−ℛQ12, both the one and the other root do reſolve the queſtion, as you will find, if you take the pains to examine it.

From this practice may be framed an univerſal p. 358Canon, which may ſerve for the reſolution of an infinite number of Algebraical Problemes, which may be conceived after this manner. The ſum given, which conteineth the two extreams, ought to be diſtributed into two equal parts, that the ſquare of the half may be taken, from which the ſquare of the mean proportional given, muſt be taken, and the ſquare root of the Reſidue added, and taken from the half of the given ſum, will ſhew the two parts ſought. As for example, I take the half of 8, which is 4, whoſe ſquare is 16, from which I take 4, the ſquare of 2, which is the given mean, and there remains 12, whoſe ſquare root added to the ſame half, makes 4+ℛQ12▪ and taken from the ſame half, makes 4−ℛQ 12.

Queſtion III.

To divide any given number (as for example 4) into three numbers continually proportional, in ſuch ſort as that the ſquares of the extreams joyned together, may be triple the ſquare of the mean.

I put 1R for the number of the middle part, then ſeeing all three ought to make the ſame ſum of 4, we ſhall have for the ſum of the extreams 4−1R. Now ſeeing that of three numbers continually proportional, the ſquare of the ſum of the extreams is equal to the ſquare of the extreams, and to the double of the ſquare of the middle part. I take the ſquare of this ſum 4−1R, which is 16+1Q−8R, from which I take 2Q, which is the double of the ſquare of the p. 359middle part, and there will remain 16−1Q−8R for the ſum of the ſquares of the extreams: Therefore ſeeing that the condition of the queſtion requireth a triple proportion, there will be an equation between 16−1Q−8R and 3Q, adde therefore 1Q on both ſides of the equation, and you ſhall have 4Q equal to 6−8R, and dividing the whole by 4, which is the number of the greateſt character, the equation will be 1Q equal to 4−2R, whoſe ſquare root is ℛQ5−1 for the middle number ſought, and the ſum of the extreams ſhall be 5−ℛQ5, which being divided into two parts, by the Canon of the precedent queſtion, in ſuch ſort as ℛQ5−1, be the middle proportional, you will finde that the extreams are 2 and 2−ℛQ5, therefore the three numbers are 2 and ℛQ5−1, and 3−ℛQ5, which all together make 4, and are in continual proportion, and the ſquares of the extreams are triple the ſquare of the mean.

Sect. IV. Geometrical queſtions reſolved by Algebra.

Queſtion I.

THere is a piece of ground of a greater length than breadth, whoſe angles are right angles, and in a triple proportion, Before you attempt the reſolution of ſuch questions, you muſt draw their figures.and their ſquares taken together, are quintuple their ſum. The ſides, the diameter, and the capacity or ſuperficies of that piece of ground is demanded, I put for the leaſt ſide 1R, therefore p. 360ſeeing they are in a triple proportion, the other ſide ſhall be 3R, their ſquares ſhall be 1Q and 9Q, which added together make 10Q, which ought to be quintuple, the ſum of the numbers. Now the ſum of the numbers is 4R, and its quintuple 20 R, and by conſequence, ſee here an equation between 10Q and 20R, which are two collateral characters, and therefore dividing 20 by 10, which is the number of the greateſt character, you ſhall finde 2 for the leaſt ſide, therefore the greater ſide ſhall be 6. Therefore the ſuperficies ſhall be 12, and the diameter ℛQ40.

Queſtion II.

There is an equilateral triangle, whoſe ſuperficies is ℛQ243. The ſide and perpendicular is demanded, ſuppoſing that the perpendicular of an equilateral triangle doth alwayes divide the ſide into two equal parts, I put for the half of the ſide divided, 1R; therefore the ſide ſhall be 2R. Now ſeeing that in every equilateral triangle; the ſquare of the ſide is equal to the ſquare of the perpendicular joyned to the ſquare of the half of the ſide, which is 1Q of the ſquare of the whole ſide, which is 4. I ſhall have 3Q for the ſquare of the perpendicular, and ſo ℛQ3Q ſhall be perpendicular, which multiplied by the half of the ſide, which is 1R (reducing it firſt to its ſquare, becauſe of the radical ſigne which is in the number multiplied) you ſhal have ℛQ3QQ for the ſuperficies of the triangle: therefore there will be an equation between ℛQ243 and ℛQ3QQ, and therefore there will be alſo an equation p. 361between their ſquares, which are 243 and 3QQ, and the whole being divided by 3, the number of the greateſt character, there will be yet an equation between 81 and 1QQ. I extract therefore the ſquared ſquare root of 81, and have 3 for the half of the ſide, 6 for the ſide, ℛQ27 for the perpendicular, and ℛQ243 for the ſuperficies of the triangle.

Queſtion III.

There is a Semicircle, whoſe Diameter is divided according to mean & extream reaſon, on w^ch there is raiſed a perpendicular produced to the circumference, and the leſſer line which is drawn from the extremity of the diameter, to this point of the circumference, is ℛQ20−2, the quantity of the diameter, and of its parts, and of this perpendicular is demanded. To reſolve this queſtion, it is preſuppoſed that the greater part of the diameter ſhall be equal to the given line, as may with facility be Geometrically demonſtrated. That being done, I put for the leſſer part of the diameter 1R, therefore ſeeing that the other part is given ℛQ20−2, the whole diameter ſhall be ℛQ20−2+1R, which multiplied by 1R, giveth for the product ℛQ20−2R+1Q, equal to the ſquare of the given quantity, which is 24−ℛQ320, and by due tranſpoſition you ſhall finde 1Q equal to 24−ℛQ320+2R−ℛQ20, from which the ſquare root ought to be extracted, taking exact notice that the particles which have the Coſſick Characters may hold place with the number of roots. I conſider p. 362therefore in this term of the equation, the number of roots which is 2−ℛ Q 20, whereof I take the half, which is 1−ℛ Q 5, to whoſe ſquare 6−ℛ Q 20, the abſolute number 24−ℛ Q 320, ought to be added. and the ſum will be 30−ℛ Q 500, whoſe ſquare root ought to be extracted as from Apotomes, as hath been ſhewn in the laſt Section of the fourth Chapter, that root is 5−ℛ Q 5, which added to the half of the number of roots 1−ℛ Q 5 gives for the ſum 6−ℛ Q 20, which is the value of 1 root, that is to ſay, of the leſſer part of the diameter; and therefore if you adde it to the greater part, you ſhall have 4 for the quantity of the whole diameter, from whence the perpendicular is eaſily known, provided the rules of Geometry be in any reaſonable manner underſtood.

Sect. 5. Queſtions reſolved by the ſecond Roots.

Queſtion I.

THree men have amongſt them a ſum of money: The firſt ſaith to the ſecond, If you deliver me the half of your money, I ſhall have 100 Crowns: The ſecond ſaith to the third, If you deliver me ⅓ of your money, I ſhall have 100 Crowns: The third ſaith to the firſt, If you deliver me ¼ of your money, I ſhal have 100 Crowns. I demand how much money each one hath.

I put for the firſt mans money 1 R of Crowns, and for the ſecond mans money 1 A, and for the third mans money 1 B: therefore the firſt which p. 363hath 1 R with ½ of the ſecond mans money ſhall have 1 R+½ A equal to 100, and by conſequence ½ A ſhall be equal to 100−1 R, and multiplying the whole by 2, 1 A, ſhall be equal to 200−2 R. I begin again therefore the operation, and in ſtead of 1 A, I put for my ſecond number 200−2 R. Now the queſtion requires that the ſecond man, (with ⅓ of the third mans) ſhall have 100; therefore there will be an equation between 200−2 R+⅓ B, and between 100, adde to both parts of the equation 2 R, and take away 200, there will yet remain an equation between ⅓ B and 2 R−100, and multiplying the whole by 3, you will have 1 B equal to 6 R−300: That being found, I begin again the work, and in ſtead of 1 B, I put for the third number 6 R−300▪ which added to ¼ of the firſt mans, makes 25/4 R−300, which ought to be equal to 100, therefore if you adde on both parts 300, there will be an equation between 25/4 R and 400, therefore if divide 400 by 25/4, you ſhall finde 64 for one root; therefore the ſecond, which had 200−2 R ſhall have 200−128, that is to ſay 72, and the third ſhall have 84. Thoſe three numbers do perfectly ſatisfie all the conditions of the queſtion.

Queſtion II.

Two men divide between themſelves three hundred Crowns in ſuch ſort, as that the ſecond mans money divided by that of the firſt mans, makes …/…. It is demanded, how much each of them hath. I put for the firſt mans money 1 R, and for that p. 364of the ſecond 1 A, there is therefore an equation between 1 R+1 A, and 300, and therefore 1 A is equal to 300−1 R, therefore 300−1 R / 1 R is equal to 3/2, and by conſequence in croſſe multiplying theſe two fractions, I ſhall have 600−2 R equal to 3 R, therefore 600 ſhall be alſo equal to 5 R, and dividing 600 by 5, I ſhall find 120 for the firſt mans money, ſo ſhall the other have 180.

Queſtion III.

To find two numbers whoſe product may be 10, and the ſum of their ſquares 29; I put for the firſt 1 R, and for the other 1 A, the product is 1 R A, equal to 10; therefore in dividing the whole by 1 R, there will be an equation between 1 A and 10/1 R, and therefore I begin again the operation, and put for the firſt 1 R, and for the ſecond 10/1 R, their ſquares are 1 Q+100/19 equal to 29, and after the reductions and extractions of the roots, I find 5 and 2 for my ſought numbers.

Sect. VI. Queſtions reſolved indefinitely.

THat queſtion is ſaid to be reſolved indefinitely, in which the numbers are demonſtrated in ſuch Algebraical terms, as do ſatisfie all the conditions of the queſtion propoſed.

Queſtion I.

To divide 12 into four numbers Arithmetically and continually proportional, I preſuppoſe that when there is four numbers in Arithmetical proportion, the ſum of the extreams is alwayes p. 365equal to the ſum of the means: whence it follows, that in our queſtion the ſum of the extreams ſhall be 6, and the ſum of the means alſo 6. I put for the ſecond 1 R, therefore the third ſhall be 6−1 R, their difference is 2 R−6, preſuppoſing 1 R to be the greater number of the two; if therefore I adde this difference to 1 R, I ſhall have 3 R−6, and if I take it from 6−1 R, I ſhall have 12−3 R, and therefore the four numbers in continued Arithmetical proportion, ſhall be 3 R−6 | 1 R | 6−1 R | 12−3 R | and the queſtion is indefinitely reſolved, in purſuit of which you may take ſuch a number as you pleaſe for the value of 1 R, provided nevertheleſſe that you do admit of fained numbers leſſe than nothing: However, if you will have no other numbers than what are real, you ought to take the value of 1 R, beneath 4 and above 2, which will be eaſily underſtood by a little experience.

Take for example 5/2 for the value of 1 R, therefore the firſt number, which is 3 R−6, ſhall be 15/2−6, that is to ſay 3/2, the ſecond ſhall be 5/2, the third 7/2, and the fourth 9/2, which added together make 12, and are in continued Arithmetical proportion. And ſo you may take an infinite number of others.

Queſtion II.

A Vintner hath three ſorts of wine, the firſt is worth 4, the ſecond 6, and the third 10 pence the pinte; of theſe three ſorts of wine he deſires to fill a veſſel which conteins 80 pintes, which may be worth 8 pence the pint; I demand how many p. 366pintes he ought to take of each ſort. You ought here to conſider, that the number 80 muſt be divided into three ſuch numbers, as the firſt multiplied by 4, the ſecond by 6, and the third by 10, the ſums of the three products added together may make 640 (becauſe that all the wine which ſhal be in the veſſel to be filled wil coſt 640 pence, ſeeing that if one pint be worth 8 pence, 80 pintes will be worth 640 pence) I put therefore for the third number 1 R, which multiplied by 10, makes 10 R, which being taken from 640, doth leave for the reſidue 640−10 R, which is a number conteining the firſt 4 times, and the ſecond 6 times. On the contrary, ſeeing that the third hath been put 1 R, therefore 80−1 R ſhall make the ſum of the firſt and the ſecond, which multiplied by 4, will give 320−4 R, which being ſubſtracted from 640−10 R, will leave 320−6 R double to the ſecond number, and therefore the ſecond ſhall be 160−3 R. In like manner the ſame ſum 80−1 R multiplied by 6, will produce 480−6 R; therefore if you take away 640−10 R, there will remain 4 R−1…0, double to the firſt, and therefore the firſt ſhall be 2 R−80. See here the queſtion reſolved indefinitely, the firſt number is 2 R−80, the ſecond 160−3 R, and the third 1 R. The terms between which you ought to take the value of 1 R are 53⅓, and 40; if therefore you take 46 for the value of 1 R, you ſhall have 46 pints of wine of 10 pence the pinte, 22 of 6, and 12 of that of 4 pence the pinte.

p. 367Here I intreat you to conſider, A ſpecial Note.that it is impoſſible perfectly to underſtand the rule of Aligation, without the knowledge of Algebra: For if you propoſe this queſtion to one skilful onelyonly in Arithmetick, he will give you for the 3 ſought numbers, 40, 20, and 20. And if you tell him, that of the ſort of wine of 4 pence the pinte, you have but 16 pints, he will remain aſtoniſhed; whereas by your queſtion reſolved indefinitely by Algebra, you will be able to give ſatisfaction to this condition in infinite kindes and manners.

Queſtion III.

Two numbers are ſought, which have 56 for the difference of their Cubes, and which added together make 6. I put for the difference of thoſe numbers 1 R, and if from the difference of the two Cubes you take the Cube of the difference of the ſides, dividing this reſidue by the triple of the difference of the ſides, you have for the quotient the product of the ſides, it followes that if from 56 you take 1 C, and that the reſidual 56−1 C be divided by the difference of the ſides, which is 1 R, the quotient 56−1C / 1R ſhall be triple, the product of the ſides; and therefore if you divide this quotient by 3, you will have for the product of the ſides 56−1C / 3R, which is equivalent to 56/3R−19/3, therefore if you divide 6, (which is the ſum of your two ſought numbers) into two parts, whoſe product may be 36/3R−59/3, you will have the queſtion reſolved indefinitely. Now to attein this, I p. 368have given you a Canon in the ſecond queſtion of the third Section, the half of the ſum 6 is 3, whoſe ſquare is 9, from which if you take the product found, there will remain 9+19/3−56/3R, whoſe ſquare root added, and taken from the half of the ſum, gives for the reſolution 3+ℛ (9+19/3R−56/3R) which ſhall be the greateſt ſought number, and 3−ℛ Q (9+19/3−56/3R) which ſhall be the leaſt. Now theſe two numbers reſolves the queſtion indefinitely, in ſuch ſort, that if you take two for the value of the root, you will finde that your two ſought numbers ſhall be 4 and 2, and every other number (taken for the value of 1 R) above 2, will reſolve the queſtion.

p. 369Appendix. Queſtions in Algebra, moſt of which require the Rule of Three in their Operation.

Queſtion. I.

A Merchant receives in exchange for 568 Crowns four kindes of money; of the firſt 7 makes one Crown, of the ſecond 18, of the third 21; and of the fourth 28 make one Crown. Moreover, he received of each ſort of money a like number. I demand, How much he received of each kinde of money?

Put 1 R for the quantity of each ſorts of money, and then conſtitute the Rule of Three after this manner.

If 7 Pieces of mony be worth Crown 1 what is money 1 R I anſwer 1/7 R Crowns.

If 18 Pieces of mony be worth Crown 1 what is money 1 R I anſwer 1/12 R Crowns.

If 21 Pieces of mony be worth Crown 1 what is money 1 R I anſwer 1/21 R Crowns.

If 8 Pieces of mony be worth Crown 1 what is money 1 R I anſwer 1/28 R Crowns.

For if 7 pieces of money make 1 Crown, 1 R of money will make 1/7 R of Crowns, and ſo of the reſt. Now thoſe Fractions deſcribed in the fourth place, do make together 7…/2…2 R of Crowns, equal to the number 568 Crowns. Divide therefore p. 370568 by 71/252, and you have for the value of 1 R 2016, and ſo many pieces of each kind of money he received: which thus I prove.

If 7 Pieces are worth 1 Crown what Pieces 2016 Anſwer 288 Crowns.

If 18 Pieces are worth 1 Crown what Pieces 2016 Anſwer 112 Crowns.

If 21 Pieces are worth 1 Crown what Pieces 2016 Anſwer 96 Crowns.

If 28 Pieces are worth 1 Crown what Pieces 2016 Anſwer 72 Crowns.

For 2016 pieces of the firſt kind of money do make 288 Crowns, and as many of the ſecond kind make 112, as many of third kind make 96, and of the fourth kind 72, which added together do make 568 Crowns.

Queſtion II.

A Certain man hath two meaſures of wine, the one worth 12 Crowns, the other 15. Now he deſires of both theſe wines to fill another equal meaſure, whoſe worth may be 13 Crowns. I demand what part of each of thoſe wines he muſt take, to fill the other to be worth that price?

Put 1 R for the part of the meaſure of the worſt wine, and for the part of the meaſure of the beſt wine 1−1 R, then work by the rule of 3 thus.

For if one meaſure of the worſt wine be worth 12 Crowns, 1 R of one meaſure of the ſame wine will be worth 12 R of Crowns, & if one meaſure p. 371of the beſt wine be worth 15 Crowns 1=1 R of a meaſure of the ſame wine will be worth 15−15 R. Therefore 1 R meaſure of the worſt wine and 1−1 R meaſure of the beſt will be worth 15−3 R Crowns, which ought to be equal to to 13 Crowns. Adde therefore 3 R to each part of the equation, and the equation will be between 3 R+13 and 15; take away therefore 13 from both ſides, and the equation will be between 3 R and 2. Divide therefore 2 by 3, and you ſhall have 2/3 for the value of 1 R, and ſo much ought to be taken of the meaſure of the worſt wine, and ½ part of the meaſure of the beſt wine, which thus I prove by the Rule of Three.

For ⅔ of a meaſure of the worſt wine is worth 8 Crowns, and ⅓ of the meaſure of the beſt wine is worth 5 Crowns, which added together make three.

Queſtion III.

I Have a meaſure of wine worth ten Crowns; How much water muſt I mix with one meaſure, that a mixed (like) meaſure may be worth ſeven Crowns.

Put for the meaſure of water 1 R, then frame the queſtion by the rule of Three thus.

For if a meaſure of wine, together with 1 R of a meaſure of water, be worth 10 Crowns, one meaſure of wine and water mingled together will be worth 10/1+1 R & ſo the equation will be between 10/1+1 R and 7, which by croſſe multiplication is reduced to 10 and 7+7 R, take away 7 from each part of the equation, and it will be between 3 and 7 R; divide 3 by 7, and you have for 1 R, 3/7, and ſo much water ought to be mingled with a meaſure of wine, that a meaſure of the mixture may be worth ſeven Crowns, and is thus proved.

For if 1 of wine and 3/7 of water be worth 10 Crowns, 1 of that mixture is worth 7 Crowns.

Queſtion IV.

THere are in a certain veſſel 20 meaſures of wine, of which each of them are worth 12 Crowns the meaſure: Now this veſſel is filled up with water, and then one meaſure of this mixture is worth 10 Crowns, I demand the content of the veſſel.

p. 373Put for the meaſures 20+1 R, then work by the rule of Three thus.

For if one meaſure of wine be worth twelve Crowns, a meaſure of wine 20+1 R meaſure of water together will be worth 240 Crowns: therefore one meaſure mixt will be worth 240/20+1 R Crowns, ſo the equation is found between 240/20+1 R and 10, which by croſſe multiplication is reduced to 240 and 200+10 R; take away 200 from both parts, and there will remain an equation between 40 and 10 R: Divide 40 by 10, and you have 4 for the price of the root, and ſo many meaſures of water were put into the veſſel, and therefore the whole veſſel conteins 24 meaſures, thus proved: 24 meaſures wine and water worth 240 Crowns, 1 meaſure? worth 10 Crowns.

Queſtion V.

TWo Letter-carriers belonging to two cities diſtant each from other 140 leagues ſet forth towards one another, at one and the ſame time; the one travels eight leagues a day, the other ſix. I demand on what day they ſhall meet together.

p. 374Put 1 R for the day, then work by the rule of Three thus.

The firſt therefore in 1 R of dayes ſhall travel 8 R of leagues, and the latter 6 R, and both of them together will have meaſured 14 R of leagues, that is 140 leagues. There is therefore an equation between 14 R and 140. Divide then 140 by 14, and you ſhall have 10 for the value of 1 R. So the tenth day finiſhed, they met together, which thus I prove.

For the firſt in 10 dayes went 80 leagues, and the latter went 60 leagues, which both together make 140, the diſtance of the two cities from one another.

Queſtion VI.

A Certain Merchant bought a quantity of wooll, and another quantity of wax, which coſt him together 124 Crowns. Now 100 pound weight of wooll coſt him 7 Crowns, and 100 pounds of wax coſt him 14 Crowns; but the quantity of wooll that he bought was double to the quantity of wax I demand how many pounds of each ſort the Merchant bought?

p. 375Put for the wax 1 R, and for the wooll 2 R of pounds, then work by the rule of Three, thus.

Therefore there will be an equation between 28/100 R of Crowns, and 124 Crowns. Divide therefore 124 by 28/100, and you ſhall have for the value of 1 R 442 6/7, and ſo many pounds of wax he bought, and of wooll 885 5/7, which is thus proved.

And ſo for wooll and wax together he expended 124 Crowns.

Queſtion VII.

A Certain man buyes a number of Ells of Velvet, which he ſelleth again; he buyes 5 Ells for 7 crowns, and ſels 7 Ells for 11 crowns, and gained on the bargain 100 crowns. I demand how many Ells of Velvet he bought and ſold in all.

Put for the quantity of Ells 1 R, then work by the rule of Three thus.

p. 376You ſee if 7/5R of crowns which he laid out be ſubſtracted from 1 1/7R of crowns which he receiceived, there will remain 6/35R of crowns for the profit. Therefore the equation will be between 6/35R and 100 R, divide 100 by 6/35, and you have 583⅓ for the value of 1 R, and ſo many Ells he bought and ſold. Thus proved.

By which you may perceive there is 100 crowns gotten.

Queſtion VIII.

Certain man buyes a number of Ells of Velvet, paying 11 crowns for 7 Ells: Now he ſels the whole again after the rate of 5 Ells for 7 crowns, and loſt 100 crowns by the bargain. I demand how many Ells he bought and ſold in all.

Put for the number of Ells 1 R, and then work by the rule of Three thus.

Then if 7/5R of crowns which he received, be ſubſtracted from 1 1/7R of crowns which he expended, the loſſe will happen to be 6/35R of crowns. So the equation will be between 6/35R of crowns, p. 377and 100 crowns. Divide therefore 100 by 6/35, and you ſhall have for the value of 1 R, 583⅓ Ells, and ſo many Ells were bought and ſold, which thus I prove.

Where you ſee he loſt 100 crowns by the bargain.

Queſtion IX.

A Man buyes 100 pounds of wax for 17 crowns. I demand, how many pounds he muſt ſell for one crown, that ſo on 102 crowns he may gain 18 crowns.

Put 1R for the number of pounds, then work by the rule of Three thus framed.

For 102 crowns do give 600 pounds, and if for one crown there be given 1R of pounds there will be 1…0R of pounds given for 102+18, which ſaid 120R are equal to 600 pounds, the quantity of the wax ſold. Therefore the equation ſhall be between 120R of pounds and 600 pounds. Divide therefore 600 by 120, and ſo you have for the value of 1 R, 5 pounds, and ſo many pounds are ſold for one crown; ſo as that in 600 pounds 18 crowns may be gained on 102, thus proved.

Where in the firſt example 600 pound made 102, in this it makes 120, that is 102+18.

Queſtion X.

A Man buyes 100 pounds of wax for 17 crowns, in diſpoſing of which he loſeth 18 crowns, on 102 crowns. I demand how many pounds he ſold for one crown?

Put for the number of pounds 1 R, then work by the rule of Three, thus conſtituted.

So there will be equation between 84R of pounds and 600 pounds. Divide therefore 600 by 84, and the value of 1R will be 7 pounds and 1/7 and ſo many pounds, he ſold for one crown, and loſt 18 crowns on 102 by the bargain, which thus I prove.

Where you ſee he laid out 102 crowns for 600 pounds, and in ſtead thereof received but 84 crowns, that is 102−18 crowns.

p. 379Queſtion XI.

A Certain man agrees with a ſervant for 12 moneths ſervice, for ten crowns and a coat; but at the end of ſeven moneths, he gives him the coat and two crowns. I demand then at what rate he eſteemed the coat.

Put for the price of the coat 1R of crowns, and ſay by the rule of three. If 12 moneths require 1R+10 crowns, how much will one monethmonth require? and you ſhall find that it will require 1R+10/12 of crowns: again ſay, If 7 moneths require 1R+2 crowns, how much will one monethmonth require, and you ſhall finde it to be 1R+2/7 as here under appears.

Therefore there will be an equation between 1R+10/12 and 1R+2/7 ſeeing both are the reward of one monethmonth, which equation by croſſe multiplication is reduced to 7R+70 and 12R+24, take away 24 from both parts, and the equation will be between 7R+46 and 12R. p. 380Again, take away 7 from both parts, then it will be 46, equal to 5R; divide 46 by 5 and the price of a root, and ſo of the coat is 9⅕ crowns, as appears hereunder. The reward of 12 moneths is 19⅕, and of 7, 11⅕.

Queſtion XII.

A Certain Citizen agrees with a ſlothful ſervant for 30 dayes, that every day he wrought he would give him 7 Groats; but for every day that he idled, and wrought not, he was to allow his maſter five Groats: when the 30 dayes were paſt, it happens that the ſervant was to receive nothing from his maſter, nor the maſter from the ſervant. I demand then, how many dayes he laboured, and how many he idled?

Put 1R for the dayes of labour, and 30−1R for the dayes of idleneſſe, and then frame the rule of Three thus.

Now ſeeing that his work and play came to one reckoning, there will be an equation between 7R and 150−5R, adde 5R to each part, and the equation will be between 12R and 150. Divide therefore 150 by 12, and you have for the value of 1R, 12½, and ſo many dayes he laboured, and 17 dayes ½ he idled, which is thus proved.

Where you ſee the reward is the ſame with the mulct.

Queſtion XIII.

ONe ſels 20 pound weight, part Saffron and part Ginger, for 45 crowns; but he ſold 1 pound of Saffron for 3 crowns, and 1 pound of Ginger for ½ a crown. The queſtion is, how many pounds of each ſort he ſold?

Put for the Saffron 1R of pounds, and for the Ginger 20−1R of pounds, then by the rule of Three work thus.

Therefore the equation is between the ſum of 3R of crowns, and 20−1R / 2 of crowns added together, and 45 crowns, now that ſum is 10+5/2R of crowns (for 20−1R / 2 is equal to 10−½R, to which if you adde 3R of crowns, it makes the ſum 10+5/2R) therefore the equation ſhall be between 10+5/2R of crowns, and 45 crowns. Take away 10 from both parts, and it will be between 5/2R and 35. Divide therefore 35 by 5/2, and p. 382you ſhall have 14 for the value of 1 R. And ſo many pounds of Saffron were ſold, and 6 pound Ginger, which thus I prove.

Where you ſee that the price of 14 pound Saffron, and 6 pound Ginger added make 45 crowns.

Queſtion XIV.

A Certain Tradeſ-man hath 2 ſorts of coyn, in number 560 pieces, worth 160 crowns; a certain part thereof is worth each piece ⅓ of a crown, and each piece of the reſt ¼ of a crown. I demand the number of the firſt and latter ſort of money?

Put 1R for the firſt, and 560−1R for the latter, and then conſtitute the rule of Three after this manner.

The fourth number found is equal to 160 crowns, and the ſum of their numbers makes 140+1/12R (for 560−1R / 4 is equal to 140+¼R, to which if you adde ⅓R, makes the ſumme 140 +1/12R.) There is therefore an equation between 140+1/12R, and 160 crowns. Take away 140 p. 383from both parts, and then the equation is between 1/12 and 20. Divide therefore 20 by 1/12, and you have 240 for the value of 1 R, and ſo much money there was of the firſt ſort, of which each piece was worth ⅓ of a crown of the latter kinde 320, each worth ¼ of a crown. Thus proved.

Where you ſee the numbers in the fourth place make 160 crowns.

Queſtion XV.

IN the Army of the Emperour, the number of the Infantery were octuple to the number of the Cavallery, among them there is diſtributed 392000 crowns, ſo as that every Foot Souldier had 5 crowns, and every Horſeman 16. The queſtion is, Of how many Horſemen the Army conſiſted, and of how many Footmen.

Put 1R for the Horſmen, and 8R for the Foot, according to the condition of the queſtion, and then conſtitute the rule of Three thus:

Therefore 56R of crowns ſhall be equal to 392000 crowns: wherefore divide 392000 by 56, and you ſhall have for 1R 7000 for the number p. 384of Horſmen; therefore the Foot ſhall be 56000, eight times as many, and ſo there will be diſtributed to the Horſmen 112000 crowns, and to the Foot 280000, which together make 392000 crowns.

Queſtion XVI.

A Man hath a certain ſum of money in a purſe, which a ſtander by judgeth to be 600 crowns, whoſe errour he thus corrects. If to what I have in this purſe, there be added ½ ⅓ and ¼, and from the ſum there be ſubſtracted 1/12 part of my money, then I ſhould have 600 crowns. The queſtion is, How many crowns he had in the purſe!

Put 1R for the number of crowns. If the parts ½R, ⅓R, and ¼R, which together make 1 1/12R, be added to 1R, the whole makes 2 1/12R; take away 1/12R, and there will remain 2R equal to 600. Divide therefore 600 by 2, and you have 300 for the value of 1R, and ſo much money was in the purſe. For if you adde the ½, ⅓, and ¼, to wit 150, 100, and 75, it will make the ſum of 625, and taking away 1/12, to wit 25, the reſt will be the number 600, which reſolves the queſtion.

Queſtion XVII.

Certain Traveller goes 9 miles a day, another Traveller after the tenth day paſt, begins his journey from the ſame place, and goes every day 14 miles. I demand in how many dayes the latter will overtake the firſt.

p. 385Put 1R for the number of dayes, therefore the firſt over and above 90 miles, which he hath gone in ten dayes, goeth in 1R of dayes, beſides 9R of miles, ſeeing that every day he goeth nine miles. But the latter going 14 miles each day, goeth in 1R of dayes 14R of miles, and becauſe the firſt of neceſſity in 1R of dayes goes as many miles together with 90, which he went in ten dayes as the latter went in 1R of dayes; ſith that then they are to meet together, the equation ſhall be between 9R+90 and 14R. Take away 90 from both parts, and it will be between 90 and 5R. Divide therefore 90 by 5, and 1R makes 18. Therefore in 18 dayes they ſhall come together. For the firſt in 18 dayes went 162 miles, which added to 90, he made the firſt 10 dayes, makes 252 miles, which the latter went in 18 dayes.

Queſtion XVIII.

A Traveller goes nine miles a day, another Traveller after the end of ten dayes begins the ſame journey. I demand, how many miles a day the latter ought to travel, that ſo in 18 dayes he may overtake the firſt.

Put 1R for the miles. Therefore in 18 dayes he will have travelled 18R of miles. Seeing therefore that the firſt travelling every day nine miles, went in 18 dayes 162 miles, and adding thereto 90, which he went the firſt ten dayes before the ſecond ſet forth, it is manifeſt that he travelled 252 miles. Therefore the equation is between p. 38618R and 252. Divide 252 by 18, and you have 14 for the value of 1R, and ſo many miles the latter ought to travel, to overtake the firſt in 18 dayes.

Queſtion XIX.

A Certain man dying, made his will and teſtament, leaving 3000 crowns to be diſtributed between his Wife, Son, and two Daughters, on this condition, that the portion of the Son might be double to that of the Mother, and the portion of the Mother double alſo to the portion of each of the Daughters. The queſtion is, how much each ones portion was?

Put for the portion of one of the Daughters 1R, for the Mothers portion 2R, and for the Sons portion 4R. So there will be an equation between 8R and 3000 crowns.

The portion of one Daughter 1R that is 375 Cro.

The portion of th'other Daughter 1R that is 375 Cro.

The portion of the Mother 2R that is 750 Cro.

The portion of the Son 4R that is 1500 Cro.

Divide 3000 by 8, ſo the value of 1R will be 375, the portion of one of the Daughters, and therefore the Mothers portion will be 750, & the Sons 1500.

Queſtion XX.

A Certain man receiveth of a Merchant, a quantity of Saffron for 10 crowns; and again he receives of the ſame man 24 pounds more of p. 387Saffron, at length he returns to him 30 pounds thereof again, and the merchant computing the price of the Saffron, reſtoreth to him 14 crowns. I demand the price of a pound of that Saffron.

Here you ſee 10 crowns +24 pounds to be the whole debt which the buyer owed to the merchant, & in like manner 30 pounds −14 crowns. Therefore there will be an equation between 10 crowns +24 pounds, and 30 pounds −14 crowns. Adde 14 crowns on both parts, and the equation will be between 24 crowns +24 pounds and 30 pounds, take away 24 pounds from both parts, and it will be between 24 and 6 pounds. Divide 24 by 6, and you have 4 for the root, and ſo many crowns one pound of Saffron is worth, which I thus prove. 2½ pounds are bought for 10 crowns, and ſo the buyer received of the merchant 26½, which were worth 106 crowns. If therefore to the merchant there be reſtored 30 pounds, the merchant oweth to the buyer 3½, ſeeing he received onelyonly 26½, but 3½ pounds are worth 14 crowns, which the merchant rendered to the buyer.

Queſtion XXI.

TWo men enter into fellowſhip in trade, now the ſecond brings with him double the money that the firſt brings, and 5 crowns over and above. They gain by their traffick 960 crowns, of which the firſt takes to himſelf 300 crowns, and the ſecond 660. I demand how much each put in bank.

p. 388Put for the firſt 1R, and for the ſecond 2R +5, the ſum of both together is 3R +5, which have gained 960 crowns. Then work by the rule of three thus:

You ſhall finde that the firſt which brought in 1R, hath gained 960R / 3R+5 which number is equal to 300 crowns which he received. This equation by croſſe multiplying will be reduced to 960R, equal to 900R+1500. Take away therefore 900R from both parts, and there will Remain 60R equal to 1500, divide 1500 by 60, and you have 25 for one R, and ſo much the firſt put in bank, therefore the ſecond put in 55 crowns, and is thus proved. For both put in 80 crowns.

80. 960 25? 300

80. 960 55? 660

Queſtion XXII.

THree Merchants gain together 700 crowns, which thus they diſtribute amongſt themſelves (having regard to the ſumme each one brought into bank) ſo as that the portion of the ſecond ſurmounted the portion of the firſt by 12 crowns, and the portion of the third ſurpaſſed that of the ſecond by 16 crowns. I demand how much each mans portion was.

Put for the portion of the firſt man 1R, and then ſhall the portion of the ſecond be 1R+12, and the portion of the third 1R+28. Theſe 3 p. 389portions together make 3R+40, equal to 700. Take away 40 from both parts of the equation, and the equation will be between 3R and 660. Divide therefore 660 by 3, and you ſhall have 2…0 for the value of 1R, and ſo much was the firſt mans portion, ſo the portion of the ſecond ſhall be 232, and of the third 248, which all together make 700.

Queſtion XXIII.

A Caterer buyes a number of Fowls, ſo as that if he had bought the ⅓ ¼, and ⅕ of the number, and over and above 6, he would then have 100 juſt. I demand the number of Fowls he bought.

Put 1R for the number, whoſe ⅓, ¼, and ⅕ make 47/60R, which with 6, make 47/60R+6 equal to 100. Take away 6 from both parts of the equation, and the remaining equation will be between 47/60R and 94. Divide therefore 94 by 47/60, and the value of 1R will be 120, to wit, the number of Fowls that were bought.

For its ⅓ is 40, and its ¼ is 30, and its ⅓, 24; which all together with 6, make the ſum 100.

p. 390Some Examples in Algebra concerning Squares.

To finde two numbers in a given Exceſſe, ſo as their Squares may have alſo a given Exceſſe.

LEt there be ſought two numbers, whoſe Exceſſe is 4, and the Exceſſe of their Squares 144.

Put for the leſſer number 1R, and therefore the greater number muſt be 1R+4, whoſe ſquares are 1Q and 1Q+8R+16, their exceſſe is 8R+16, which ought to be equal to 144. Take away 16 from both parts of the equation, and the equation will be between 8R and 128. Divide therefore 128 by 8, and you ſhall have for the value of 1R, 16 for the leſſer number, the greater therefore will be 20, that the Exceſſe may be 4. The ſquares of thoſe 2 numbers are 256, and 400, whoſe difference or exceſſe is 144.

Two numbers being given, to finde another, with which multiplying both the given numbers, makes the greater number a ſquare, and its leſſer the ſide of that ſquare.

LEt the two given numbers be 200 & 5, and let the ſought number be put 1R. Now 1R multiplied in 200, produceth 200R, and 1R multiplied p. 391in 5, makes 5R, which ought to be the ſide, and ſo multiplied in it ſelf, ought to make a number equal to 200. But 5R multiplied in it ſelf, makes 25Q, the equation therefore is between 200R and 25Q. Divide therefore 200 by 25, and the value of 1R is 8, which multiplied in 200, make the ſquare 1600, whoſe ſide is the number 40, which is alſo produced by the multiplication of 8 in 5.

Some Examples relating to Cubes.

To finde a number, which multiplied in it ſelf, and the product multiplied by ſome given number, may produce a number in a given proportion to the Cube of the found number.

LEt the given number be 20, find another number, which being firſt multiplied in it ſelf, and then the product multiplied by the given number 20, may produce a number in a quintuple proportion to the Cube made of the found number.

Put for the ſought number 1R, which multiplied in its ſelf makes 1Q, which alſo multiplied by 20, makes 20Q, the Cube of 1R is 1C, to which 20Q ought to have a quintuple proportion; ſo the equation is between 20Q and 5C: Divide 20 by 5, and you have for the value of 1R, 4, the number ſonght. This 4 multiplied in it ſelf makes 16, and 16 multiplied by 20, makes 320, which is quintuple to 64, the Cube of 4.

p. 392To divide a given number in two parts, ſo as that their Cubes may make a given ſum, which ſhall be greater than the quarter part of the Cube deſcribed of the given number.

LEt the given number be 10, to be divided into two parts, whoſe Cubes may make 370, which number is greater than the ¼ part of the Cube of 10. Put for the firſt number compounded of 1R, and half of the given number 1R+5, and for the ſecond 5−1R. So theſe two numbers do make the given number. Their Cubes are 1C+15Q+75R+125, and 125−75R+15Q−1C; their ſum is 30Q+250: For +1C, and −1C, as alſo +75R, and −75R, do mutually deſtroy one another: and of 15Q, and 15Q, are made 30Q; alſo 125, & 125, make 250. Therfore the equation is between 30Q+250, and 370. Take away 250 from both parts, and the equation will be between 30Q and 120. Divide 120 by 30, and you have 4 for the value of 1Q, and the value of 1R, 2. The firſt part put 1R+5 ſhall be 7; and the ſecond put 5−1R ſhall be 3. The Cubes of thoſe parts are 343 and 27, which together make 370.

FINIS.

End of the text · William Leybourn, Arithmetic (1660)