Cocker's decimal arithmetick wherein is shewed the nature and use of decimal fractions ... together with tables of interest and rebate ... : whereunto is added, his Artificial arithmetick, shewing the genesis ... of the logarithmes ... : also, his Algebraical arithmetick, containing the doctrine of composing and resolving an equation, with all other rules requisite for the understanding of that mysterious art according to the method used by Mr. John Kerley in his incomparable treatise of algebra / composed by Edward Cocker ... ; perused, corrected, and published by John Hawkins ...

CHAP. XIII. The Resolution of Arithmeti∣cal Questions (Algebraically) which produce Simple Equa∣tions (Book 13)

I. AN Equation is two-fold, viz. First, Simple, and secondly, Adfected, or Compounded.

II. A Simple Equation is when the Quantity sought (solely possessing one part of the Equation) is either expressed by a Single or Simple Root, as a, or by a Single or Simple Power as aa, or aaa, &c. as in these Equations, viz. a=32, and aa=64, or 4aaa=256, and such like.

III. When a Question is propounded, and to be resolved Algebraically, Then for the Answer put a, and for each of the given Numbers put Consonants, then proceed according to the Te∣nure of the Question, by Addition, Subtracti∣on, Multiplication, or Division, until an Equa∣tion is Composed; and when the Equation is composed, then proceed to reduce it, (accord∣ing to the Rules contained in the 11th Chap∣ter) until the Quantity unknown (being a or some power of a) do solely possess one part of

the Equation, and the known or given Quanti∣ties the other part, and then will the Quanti∣ty sought be also known.

IV. I shall in the Resolution of every Que∣stion proceed (gradatim) step by step, according to the method used by Mr. Kersey, each step be∣ing numbred orderly in the Margent, from the beginning to the end by 1, 2, 3, 4, &c. And I shall only proceed in the operation literally, because otherwise this Treatise would swell to a bigger Volume than is at present intended; but I shall give the Learner a taste of Numeral Algebra, in the solution of two or three of the first Questions thereby.

Quest. 1. There are two numbers whose sum is 48 (or b) and the excess of the greater above the lesser is 14 (or c) I demand what are the Numbers.

The Solution literally.

1 For the greater number put	〈 math 〉〈 math 〉
2 From which if you sub∣tract the difference (c) you will have the lesser, which is	〈 math 〉〈 math 〉
3 The greater and lesser be∣ing added together, will be equal to (b) the sum whence this equation	〈 math 〉〈 math 〉
4 And by the Transposition of −c the equation is	〈 math 〉〈 math 〉
5 Then dividing each part of the equation by 2, it is	〈 math 〉〈 math 〉

And if from 〈 math 〉〈 math 〉 you sub∣tract (c) the excess of the greater above the lesser, the lesser will be

〈 math 〉〈 math 〉

So that the numbers sought are 31 and 17, for by the fifth step (a) the greater is found to be=to 〈 math 〉〈 math 〉 and b is given 48, and c is given 14, the sum of which is 62, which divided by 2, gives 3•…•…, for the greater, and by the sixth step, if from the greater you subtract the difference (c) the remainder will give the lesser, which is 17, for 〈 math 〉〈 math 〉.

Now if the fifth and sixth steps are duly con∣sidered, they will present you with this

Theorem.

The sum of the sum and difference of any two Numbers being divided by 2, will give the grea∣ter Number, and the difference of any two Numbers being subtracted from half the sum of the sum and difference, the remainder will give the lesser number.

The Solution Numerally.

1 For the greater number put	〈 math 〉〈 math 〉
2 From which if you subtract the difference (14) the lesser is	〈 math 〉〈 math 〉

3 Which added together, will be the sum, whence this Equation.	〈 math 〉〈 math 〉
4 And by transposition of of 14 it will be	〈 math 〉〈 math 〉
5 And both parts of the E∣quation being divided by 2, will give the value of (a) the greater.	〈 math 〉〈 math 〉
6 From which if you subtract (14) the Difference, the re∣mainder will give the lesser by the 2 step.	〈 math 〉〈 math 〉

So that the Numbers sought are 31 and 17, which will satisfie the conditions of the Question.

Quest. 2.

There are two Numbers whose Sum is 56 (or b) and the lesser hath such proportion to the greater, as 2 to 5, (or c to d) I demand what are the Numbers?

1 For the lesser number put	〈 math 〉〈 math 〉
2 Then by the Rule of Three find the greater, viz. 〈 math 〉〈 math 〉	〈 math 〉〈 math 〉
3 Wherefore the sum of the 2 numbers sought is	〈 math 〉〈 math 〉
4 Which sum must be equal to the given sum, whence this equation.	〈 math 〉〈 math 〉

5 Which equation being re∣duced by the fourth and fifth Rules of the eleventh Chap. the value of a will be found to be	〈 math 〉〈 math 〉
6 And by the first, second, and fifth steps the greater number will be discovered to be	〈 math 〉〈 math 〉

So that the numbers sought are 40 and 16 for (a) the lesser is found to be by the fifth step 〈 math 〉〈 math 〉 viz. the Product of (cb) 56 by 2 divided by (c+d) the sum of 2 and 5, viz. 7, which is 16, &c.

And if (according to the third Rule of the twelfth Chap.) the two last steps be turned into proportionals, it will give this

Theorem.

As the sum of the Terms which represent the ratio of two Numbers, is to the sum of the num∣bers themselves, so is the lesser term to the lesser number; and so is the greater Term to the grea∣ter Number.

Therefore if the sum of two Numbers is gi∣ven, and also their Ratio, the Numbers them∣selves are also given by this Theorem.

The same Question solved Numerally.

1 For the lesser number put	〈 math 〉〈 math 〉
2 Then by the Rule of Three the greater number is found 〈 math 〉〈 math 〉	〈 math 〉〈 math 〉
3 Then will their sum be	〈 math 〉〈 math 〉
4 And according to the tenure of the Question, their sum must be equal to the given sum whence this Equation	〈 math 〉〈 math 〉
5 And that Equation being reduced by the fifth and sixth Rules of the ele∣venth Chap. the value of a will be found to be	〈 math 〉〈 math 〉
6 Which being subtracted from the given sum, the greater number is	〈 math 〉〈 math 〉

Quest. 3.

A Gentleman asked his Friend (that had 4 purses in his hand) what money he had in each Purse? To whom he answered that he knew not but (quoth he) this I know, that in the second purse there are 8 (or b) Crowns more than in the first or least purse, and in the third 8 Crowns more than the second, and in the fourth or big∣gest purse there are 8 Crowns more than in the third, and twice as many as in the first or least, I demand what number of Crowns he had in each purse?

1 For the number of Crowns in the first purse put	〈 math 〉〈 math 〉
2 Then in the second there is	〈 math 〉〈 math 〉
3 And in the third there is	〈 math 〉〈 math 〉
4 And in the fourth	〈 math 〉〈 math 〉
5 Which according to the Tenure of the Question is double to that in the first, whence this Equation	〈 math 〉〈 math 〉
6 Then by the Transposition of a from the first side of the equation, it is	〈 math 〉〈 math 〉

which discovereth the value of a to be 3b, or 3 times 8, which is 24, &c. which is the num∣ber of Crowns in the first purse, and conse∣quently the number of Crowns in each purse is 24, 32, 40, and 48, which will satisfie the con∣ditions of the Question.

The same Question solved Numerally.

1 For the Crowns in the first purse put	〈 math 〉〈 math 〉
2 Then in the second there is	〈 math 〉〈 math 〉
3 And in the third	〈 math 〉〈 math 〉
4 And in the fourth	〈 math 〉〈 math 〉
5 Which is double to the number of Crowns in the first, whence this equation	〈 math 〉〈 math 〉
6 Which equation being re∣duced by the transposition of a discovers the value of a, viz.	〈 math 〉〈 math 〉

Quest. 4.

Three men build a Ship which cost them 2700l. (or b) Pounds, of which B must pay double to what A must pay, and C must pay three times as much as B, I demand the share that each must pay?

1 For the sum to be paid by A, put	〈 math 〉〈 math 〉
2 Then B must pay	〈 math 〉〈 math 〉
3 And C must pay	〈 math 〉〈 math 〉
4 The sum of these 3 quanti∣ties are equal to the total charge, whence this Equa∣tion	〈 math 〉〈 math 〉
5 Which being reduced, dis∣covers the value of a, viz.	〈 math 〉〈 math 〉

which is the sume that A must pay, viz. 300l. Therefore B must pay 〈 math 〉〈 math 〉. which is twice as much as A, and C must pay 〈 math 〉〈 math 〉. which is three times as much as B.

Quest. 5.

There is a Fish whose head is supposed to be 9 (or b) Inches, and his Tail is as long as his Head and half his Body, and his Body is as long as his Head and his Tail, I demand the length of such a Fish?

1 For the length of the bo∣dy put	〈 math 〉〈 math 〉
2 Then will the Tail be	〈 math 〉〈 math 〉

3 Then if to the Tail you add the length of the head, viz. b, the sum is	〈 math 〉〈 math 〉
4 Which according to the te∣nure of the Question is equal to the length of the body, whence this equation	〈 math 〉〈 math 〉
5 And the second part of the equation being cleared of the unknown quantity a by Reduction, gives the value of a the length of the body, viz.	〈 math 〉〈 math 〉
6 Then according to the Te∣nure of the question, if therefrom you subtract (b) the length of the head, the remainder will be the length of the Tail, which is	〈 math 〉〈 math 〉

By the fifth step the length of the Body is found to be 4b=36, And by the sixth step the length of the Tail is discovered to be 3b=3×9 =27. So that the length of the head is (given) 9 inches, the length of the Tail 27 Inches, and the length of the Body 36 Inches, which num∣bers will satisfie the conditions of the question, 〈 math 〉〈 math 〉 So that the whole length of the Fish is 9+27+36=72 Inches.

Quest. 6.

A Father lying at the point of Death, left to his 3 Sons A, B, and C all his Estate in Money, and divided it thus, viz. to A he gave ½, wanting 44 (or b) pounds, and to B he gave 1/3 and 14, (or c) pounds over, and to C he gave the rest, which was 82 (or d) pounds less than the share of B, Now I demand what was the Fathers Estate?

1 For the Fathers Estate put	〈 math 〉〈 math 〉
2 Then will the share left to A be	〈 math 〉〈 math 〉
3 And the share of B	〈 math 〉〈 math 〉
4 And by the third step the share of C is	〈 math 〉〈 math 〉
5 The Quantities in the 3 last steps being added together, give	〈 math 〉〈 math 〉
6 Which must be equal to the Fathers Estate, whence this Equation.	〈 math 〉〈 math 〉
Which equation after due reduction and Transpositi∣on of quantities the value of a is discovered to be	〈 math 〉〈 math 〉

And 6b=6×44=264, and 6d=6×82=492, and 12c=12×14=168, now 264+492−168=588, so that the Fathers estate was 588 pounds, of which A had 250 l. B 210 l, and C 128, which Numbers do answer the conditions of the Que∣stion.

Quest. 7.

Two persons thus discoursed together con∣cerning their Money, quoth A to B, give me 3 (or b) of your Crowns, and I shall have as many as you; nay quoth B to A, but if you will give me 3 of your Crowns, I shall have 5 times as many as you. Now I demand how many Crowns had each person?

1 For the number of Crowns which A had put	〈 math 〉〈 math 〉
2 Then forasmuch as adding 3 (or b) Crowns to A will be equal to the Crowns remain∣ing to B after he had given 3 Crowns to A, therefore B will then have left	〈 math 〉〈 math 〉
3 And consequently if you add thereto the 3 (or b) Crowns which he gave to A the sum will be the number of Crowns which B had at first, which is	〈 math 〉〈 math 〉
4 Then if from the number of Crowns A had at first (a) you subtract 3 (or b) crowns, there will remain to A a−b crowns, and giving the same to B, he will then have	〈 math 〉〈 math 〉
5 Which according to the te∣nure of the Question is five times as much as what A had left, whence there ari∣seth this Equation.	〈 math 〉〈 math 〉

6 Which equation being re∣duced by the second and seventh Rules of the ele∣venth Chapter, the value of a is discovered to be	〈 math 〉〈 math 〉
7 And by the sixth and third steps the number of Crowns which B had at first are found to be	〈 math 〉〈 math 〉

So that it is found that A had 6 Crowns, and B had 12 Crowns, which numbers will satisfie the conditions of the Question. For,

〈 math 〉〈 math 〉

Quest. 8.

A Labourer had 576 (or b) pence for thresh∣ing 60 (or c) Quarters of Corn, viz. Wheat and Barly; for the wheat he had 12 (or d) pence per Quarter, and for the Barly he had 6 (or f) pence per Quarter, I demand how many Quarters of each he threshed?

1 For the quarters of wheat which he threshed put	〈 math 〉〈 math 〉
2 Then the quarters of Bar∣ley will be	〈 math 〉〈 math 〉
3 The quantity of Wheat in the first step being multiply∣ed by its price produceth	〈 math 〉〈 math 〉

4 The quantity of Barley in the second step being mul∣tiplyed by its price, pro∣duceth	〈 math 〉〈 math 〉
5 The sum of the quantities in the two last steps must be equal to the given price of the 60 quarters, whence this equation	〈 math 〉〈 math 〉
6 Which being reduced by the second, third, and fifth Rules of the eleventh Chap. the quantity of Wheat will be discovered to be	〈 math 〉〈 math 〉
7 And by the second and fifth steps the quantity of Barly is discovered to be	〈 math 〉〈 math 〉

So that the quarters of Wheat which he threshed were 36, and the quarter of Barly 24.

The Proof.

〈 math 〉〈 math 〉

Quest. 9.

A Gentleman bought a Cloak of a Sales-man, which cost him 3 l.−10 s. or 70 (or b) shillings, and desiring the Sales-man to tell him what he

gained thereby, he said he gained ¼ (or c) of what it cost him, the question is what the Cloak cost the first penny?

1 Suppose the Cloak cost	〈 math 〉〈 math 〉
2 Then he gained	〈 math 〉〈 math 〉
3 The first and Second steps being added together, their sum will be equal to the sum which the Gentleman gave for it, whence this equation	〈 math 〉〈 math 〉
4 Which Equation being re∣duced by the ninth Rule of the eleventh Chap. the value of a will be discove∣red to be	〈 math 〉〈 math 〉

So that it cost 56 shillings, ¼ of which is 14 shillings, and 56+14=70.

And if the quantity in the fourth step be duly considered, you will find that if the gain had been any other part or parts of the first cost, if the price it was sold for had been divided by the Fraction representing the part of gain, increa∣sed by 1, the quote would have been the an∣swer.

Quest. 10.

A Gentleman hired a Labourer to work for him for 40 (or b) days, and made this agree∣ment with him that for every day he wrought he should have 20 (or c) pence, and for every day that he played he should forfeit 8 (or d) pence, and at the end of the said 40 dayes he received

184 (or f) pence, which was his full due. Now I demand how many dayes he wrought, and how many dayes he played?

1 For the number of days he wrought, put	〈 math 〉〈 math 〉
2 Then the number of days he played will be	〈 math 〉〈 math 〉
3 And if the time he wrought (in the first step) be multi∣plyed by 20 (c) it will pro∣duce the total he gained by work, viz.	〈 math 〉〈 math 〉
4 And if the time he played (in the 2d step) be drawn into 8 (d) the product will be what he lost by play	〈 math 〉〈 math 〉
5 And if the total loss (in the fourth step) be subtracted from the gain (in the third step) the remainder will be what he received, whence this Equation	〈 math 〉〈 math 〉
6 Which being reduced by the second and ninth Rules of the eleventh Chapter, it will discover the value of a to be eighteen which is the days that he wrought.	〈 math 〉〈 math 〉
7 And from the sixth and se∣cond steps the number of days he played are discove∣red to be 22 dayes, viz.	〈 math 〉〈 math 〉

So that by the sixth step it appears he wrought 18 dayes, and by the seventh step it appears that he played 22 dayes.

The proof.

〈 math 〉〈 math 〉

Quest. 11.

A person (in the Afternoon) being asked what a Clock it was, answered that 3/5 (or b) parts of the time from Noon was equal to 5/8 (or c) parts of the time remaining to midnight, now, (sup∣posing the time from Noon to Midnight to be divided in 12 (or d) equal parts or hours) I de∣mand what was the present hour of the day?

1 For the hour sought put	〈 math 〉〈 math 〉
2 Then the time to midnight will be	〈 math 〉〈 math 〉
3 Then will 3/5 (or b) parts of the hour from Noon be	〈 math 〉〈 math 〉
4 And 5/8 (or c) parts of the time remaining till mid∣night will be	〈 math 〉〈 math 〉
5 Therefore from the third and fourth steps there ari∣seth this equation.	〈 math 〉〈 math 〉
6 Which equation being re∣duced according to the se∣cond and ninth Rules of the eleventh Chap. gives the va∣lue of a (to be 〈 math 〉〈 math 〉 the hour sought) viz.	〈 math 〉〈 math 〉

So that the hour sought was 〈 math 〉〈 math 〉, and conse∣quently the time remaining till midnight was 〈 math 〉〈 math 〉 hours, which two numbers will answer the conditions of the question, for, 3/5 parts of 〈 math 〉〈 math 〉, which is 〈 math 〉〈 math 〉 is equal to 5/8 parts of 〈 math 〉〈 math 〉, as you may prove at your leisure.

Moreover, If the last step be converted into proportionals by the third Rule of the twelfth Chap. it will give this

Theorem.

As the sum of the parts of any two Numbers (wherein there is an equality) is to the sum of those Numbers, so is the given parts of any one of those Numbers, to the other Number.

As suppose it were required to find out two Numbers, whose sum is 27, and such, that ¾ of the one may be equal to 3/5 of the other, the same may be found out by the said Theorem. For, 〈 math 〉〈 math 〉 which number so found is the number sought, whereof 3/5 is to be taken; and the other is 27−15=12, or it may be found by the follow∣ing proportion, viz.

〈 math 〉〈 math 〉

Quest. 12.

One asked a Shepherd what was the price of his hundred Sheep, quoth he, I have not an hun∣dred, but if I had as many more, and half as ma∣ny

more, and 7½ (or b) sheep, then I should have just 100 (or c) I demand how many sheep he had?

1 For the number of sheep he had, put	〈 math 〉〈 math 〉
2 Which being doubled is	〈 math 〉〈 math 〉
3 And if to the second step you add half the first, it is	〈 math 〉〈 math 〉
4 And if to the third step there be added 7½ (or b) the sum is	〈 math 〉〈 math 〉
5 Which quantity in the fourth step is equal to 100 (or c) whence this equation	〈 math 〉〈 math 〉
6 Which equation being re∣duced by the 5th and 7th Rules of the 11th. Chap. the value of a will be disco∣vered to be 37, viz.	〈 math 〉〈 math 〉

So that the number of sheep he had were 37 for 37+37+37/2+7½=100