The Compendious Book on Calculation by Completion and Balancing/Various questions

by Abū ʿAbdallāh Muḥammad ibn Mūsā al-Khwārizmī

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VARIOUS QUESTIONS.

If a person puts such a question to you as: “I have (30) divided ten into two parts, and multiplying one of these by the other, the result was twenty-one;”^[1] then you know that one of the two parts is thing, and the other ten minus thing. Multiply, therefore, thing by ten minus thing; then you have ten things minus a square, which is equal to twenty-one. Separate the square from the ten things, and add it to the twenty-one. Then you have ten things, which are equal to twenty-one dirhems and a square. Take away the moiety of the roots, and multiply the remaining five by itself; it is twenty-five. Subtract from this the twenty-one which are connected with the square; the remainder is four. Extract its root, it is two. Subtract this from the moiety of the roots, namely, five; there remain three, which is one of the two parts. Or, if you please, you may add the root of four to the moiety of the roots; the sum is seven, which is likewise one of the parts. This is one of the problems which may be resolved by addition and subtraction.

If the question be: “I have divided ten into two parts, and having multiplied each part by itself, I have subtracted the smaller from the greater, and the remainder was forty;”^[2] then the computation is—you multiply ten (31) minus thing by itself, it is a hundred plus one square minus twenty things; and you also multiply thing by thing, it is one square. Subtract this from a hundred and a square minus twenty things, and you have a hundred, minus twenty things, equal to forty dirhems. Separate now the twenty things from a hundred, and add them to the forty; then you have a hundred, equal to twenty things and forty dirhems. Subtract now forty from a hundred; there remains sixty dirhems, equal to twenty things: therefore one thing is equal to three, which is one of the two parts.

If the question be: “I have divided ten into two parts, and having multiplied each part by itself, I have put them together, and have added to them the difference of the two parts previously to their multiplication, and the amount of all this is fifty-four;”^[3] then the computation is this: You multiply ten minus thing by itself; it is a hundred and a square minus twenty things. Then multiply also the other thing of the ten by itself; it is one square. Add this together, it will be a hundred plus two squares minus twenty things. It was stated that the difference of the two parts before multiplication should be added to them. You say, therefore, the difference between them is ten minus two things. The result is a hundred and ten and two squares minus twenty-two things, which are equal to fifty-four dirhems. Having reduced and equalized this, you may say, a hundred and ten dirhems and two squares are equal to fifty-four dirhems and twenty-two things. Reduce now the two squares to one square, by taking the moiety of all you have. Thus it becomes fifty-five dirhems and a square, equal to twenty-seven dirhems and eleven things. Subtract twenty-seven from fifty-five, there remain (32) twenty-eight dirhems and a square, equal to eleven things. Halve now the things, it will be five and a half; multiply this by itself, it is thirty and a quarter. Subtract from it the twenty-eight which are combined with the square, the remainder is two and a fourth. Extract its root, it is one and a half. Subtract this from the moiety of the roots, there remain four, which is one of the two parts.

If one say, “I have divided ten into two parts; and have divided the first by the second, and the second by the first, and the sum of the quotient is two dirhems and one-sixth;”^[4] then the computation is this: If you multiply each part by itself, and add the products together, then their sum is equal to one of the parts multiplied by the other, and again by the quotient which is two and one-sixth. Multiply, therefore, ten less thing by itself; it is a hundred and a square less ten things. Multiply thing by thing; it is one square. Add this together; the sum is a hundred plus two squares less twenty things, which is equal to thing multiplied by ten less thing; that is, to ten things less a square, multiplied by the sum of the quotients arising from the division of the two parts, namely, two and one-sixth. We have, therefore, twenty-one things and two-thirds of thing less two squares and one-sixth, equal to a hundred plus two squares less twenty things. Reduce this by adding the two squares and one-sixth to a hundred plus two squares less twenty things, and add the twenty negative things from the hundred plus the two squares to the twenty-one things and two-thirds of thing. Then you have a hundred plus four squares (33) and one-sixth of a square, equal to forty-one things and two-thirds of thing. Now reduce this to one square. You know that one square is obtained from four squares and one-sixth, by taking a fifth and one-fifth of a fifth.^[5] Take, therefore, the fifth and one-fifth of a fifth of all that you have. Then it is twenty-four and a square, equal to ten roots; because ten is one-fifth and one-fifth of the fifth of the forty-one things and two-thirds of a thing. Now halve the roots; it gives five. Multiply this by itself; it is five-and-twenty. Subtract from this the twenty-four, which are connected with the square; the remainder is one. Extract its root; it is one. Subtract this from the moiety of the roots, which is five. There remains four, which is one of the two parts.

Observe that, in every case, where any two quantities whatsoever are divided, the first by the second and the second by the first, if you multiply the quotient of the one division by that of the other, the product is always one.^[6]

If some one say: “You divide ten into two parts; multiply one of the two parts by five, and divide it by the other then take the moiety of the quotient, and add this to the product of the one part, multiplied by five; the sum is fifty dirhems;”^[7] then the computation is this: Take thing, and multiply it by five. This is now to be divided by the remainder of the ten, that is, by ten less thing; and of the quotient the moiety is to be taken.

(34) You know that if you divide five things by ten less thing, and take the moiety of the quotient, the result is the same as if you divide the moiety of five things by ten less thing. Take, therefore, the moiety of five things; it is two things and a half: and this you require to divide by ten less thing. Now these two things and a half, divided by ten less thing, give a quotient which is equal to fifty less five things: for the question states: add this (the quotient) to the one part multiplied by five, the sum will be fifty. You have already observed, that if the quotient, or the result of the division, be multiplied by the divisor, the dividend, or capital to be divided, is restored. Now, your capital, in the present instance, is two things and a half. Multiply, therefore, ten less thing by fifty less five things. Then you have five hundred dirhems and five squares less a hundred things, which are equal to two things and a half. Reduce this to one square. Then it becomes a hundred dirhems and a square less twenty things, equal to the moiety of thing. Separate now the twenty things from the hundred dirhems and square, and add them to the half thing. Then you have a hundred dirhems and a square, equal to twenty things and a half. Now halve the things, multiply the moiety by itself, subtract from this the hundred, extract the root of the remainder, and subtract this from the moiety of the roots, which is ten and one-fourth the remainder is eight; and this is one of the portions.

If some one say: “You divide ten into two parts: multiply the one by itself; it will be equal to the other taken eighty-one times.”^[8] Computation: You say, ten less thing, multiplied by itself, is a hundred plus a (35) square less twenty things, and this is equal to eighty-one things. Separate the twenty things from a hundred and a square, and add them to eighty-one. It will then be a hundred plus a square, which is equal to a hundred and one roots. Halve the roots; the moiety is fifty and a half. Multiply this by itself, it is two thousand five hundred and fifty and a quarter. Subtract from this one hundred; the remainder is two thousand four hundred and fifty and a quarter. Extract the root from this; it is forty-nine and a half. Subtract this from the moiety of the roots, which is fifty and a half. There remains one, and this is one of the two parts.

If some one say: “I have purchased two measures of wheat or barley, each of them at a certain price. I afterwards added the expences, and the sum was equal to the difference of the two prices, added to the difference of the measures.”^[9]

Computation: Take what numbers you please, for it is indifferent; for instance, four and six. Then you say: I have bought each measure of the four for thing; and accordingly you multiply four by thing, which gives four things; and I have bought the six, each for the moiety of thing, for which I have bought the four; or, if you please, for one-third, or one-fourth, or for any other quota of that price, for it is indifferent. Suppose that you have bought the six measures for the moiety of thing, then you multiply the moiety of thing by six; this gives three things. Add them to the four things; the sum is seven things, which must be equal to the difference of the two quantities, which is two measures, plus the difference of the two prices, which is a moiety of thing. You have, therefore, seven things, equal to two and a moiety of thing. Remove, now, this moiety of thing, by subtracting it from the seven things. There remain six things and a half, equal to two dirhems: (35) consequently, one thing is equal to four-thirteenths of a dirhem. The six measures were bought, each at one-half of thing; that is, at two-thirteenths of a dirhem. Accordingly, the expenses amount to eight-and-twenty thirteenths of a dirhem, and this sum is equal to the difference of the two quantities; namely, the two measures, the arithmetical equivalent for which is six-and-twenty thirteenths, added to the difference of the two prices, which is two-thirteenths: both differences together being likewise equal to twenty-eight parts.

If he say: “There are two numbers,^[10] the difference of which is two dirhems. I have divided the smaller by the larger, and the quotient was just half a dirhem.”^[11] Suppose one of the two numbers^[10] to be thing, and the other to be thing plus two dirhems. By the division of thing by thing plus two dirhems, half a dirhem appears as quotient. You have already observed, that by multiplying the quotient by the divisor, the capital which you divided is restored. This capital, in the present case, is thing. Multiply, therefore, thing and two dirhems by half a dirhem, which is the quotient; the product is half one thing plus one dirhem; this is equal to thing. Remove, now, half a thing on account of the other half thing; there remains one dirhem, equal to half a thing. Double it, then you have one thing, equal to two dirhems. Consequently, the other number^[12] is four.

If some one say: “I have divided ten into two parts; I have multiplied the one by ten and the other by itself, and the products were the same;” ^[13] then the computation is this: You multiply thing by ten; it is ten things. Then multiply ten less thing by itself; it is a hundred (37) and a square less twenty things, which is equal to ten things. Reduce this according to the rules, which I have above explained to you.

In like manner, if he say: “I have divided ten into two parts; I have multiplied one of the two by the other, and have then divided the product by the difference of the two parts before their multiplication, and the result of this division is five and one-fourth;"^[14] the computation will be this: You subtract thing from ten; there remain ten less thing. Multiply the one by the other, it is ten things less a square. This is the product of the multiplication of one of the two parts by the other. At present you divide this by the difference between the two parts, which is ten less two things. The quotient of this division is, according to the statement, five and a fourth. If, therefore, you muliply five and one-fourth by ten less two things, the product must be equal to the above amount, obtained by multiplication, namely, ten things less one square. Multiply now five and one-fourth by ten less two squares. The result is fifty-two dirhems and a half less ten roots and a half, which is equal to ten roots less a square. Separate now the ten roots and a half from the fifty-two dirhems, and add them to the ten roots less a square; at the same time separate this square from them, and add it to the fifty-two dirhems and a half. Thus you find twenty roots and a half, equal to fifty-two dirhems and a half and one square. Now continue reducing it, conformably to the rules explained at the commencement of this book.

(38) If the question be: “There is a square, two-thirds of one-fifth of which are equal to one-seventh of its root;” then the square is equal to one root and half a seventh of a root; and the root consists of fourteen-fifteenths of the square^[15]. The computation is this: You multiply two-thirds of one-fifth of the square by seven and a half, in order that the square may be completed. Multiply that which you have already, namely, one-seventh of its root, by the same. The result will be, that the square is equal to one root and half a seventh of the root; and the root of the square is one and a half seventh; and the square is one and twenty-nine one hundred and ninety-sixths of a dirhem. Two-thirds of the fifth of this are thirty parts of the hundred and ninety-six parts. One-seventh of its root is likewise thirty parts of a hundred and ninety-six.

If the instance be: “Three-fourths of the fifth of a square are equal to four-fifths of its root,”^[16] then the computation is this: You add one-fifth to the four-fifths, in order to complete the root. This is then equal to three and three-fourths of twenty parts, that is, to fifteen eightieths of the square. Divide now eighty by fifteen; the quotient is five and one-third. This is the root of the square, and the square is twenty-eight and four-ninths.

If some one say: “What is the amount of a square-root,^[17] which, when multiplied by four times itself, amounts to twenty?^[18]” the answer is this: If you multiply it by itself it will be five: it is therefore the root of five.

If somebody ask you for the amount of a square-root,^[19] which when multiplied by its third amounts to ten,^[20] the solution is, that when multiplied by itself it will amount to thirty; and it is consequently the root of thirty.

(39) If the question be: “To find a quantity^[19], which when multiplied by four times itself, gives one-third of the first quantity as product,”^[21] the solution is, that if you multiply it by twelve times itself, the quantity itself must re-appear: it is the moiety of one moiety of one-third.

If the question be: “A square, which when multiplied by its root gives three times the original square as product,”^[22] then the solution is that if you multiply the root by one-third of the square, the original square is restored; its root must consequently be three, and the square itself nine.

If the instance be: “To find a square, four roots of which, multiplied by three roots, restore the square with a surplus of forty-four dirhems,”^[23] then the solution is: that you multiply four roots by three roots, which gives twelve squares, equal to a square and forty-four dirhems. Remove now one square of the twelve on account of the one square connected with the forty-four dirhems. There remain eleven squares, equal to forty-four dirhems. Make the division, the result will be four, and this is the square.

If the instance be: “A square, four of the roots of which multiplied by five of its roots produce twice the square, with a surplus of thirty-six dirhems;”^[24] then the solution is that you multiply four roots by five roots, which gives twenty squares, equal to two squares and thirty-six dirhems. Remove two squares from the twenty on account of the other two. The remainder is eighteen squares, equal to thirty-six dirhems. Divide now thirty-six dirhems by eighteen; the quotient is two, and this is the square.

(40) In the same manner, if the question be: “A square, multiply its root by four of its roots, and the product will be three times the square, with a surplus of fifty dirhems.”^[25] Computation: You multiply the root by four roots, it is four squares, which are equal to three squares and fifty dirhems. Remove three squares from the four; there remains one square, equal to fifty dirhems. One root of fifty, multiplied by four roots of the same, gives two hundred, which is equal to three times the square, and a residue of fifty dirhems.

If the instance be: “A square, which when added to twenty dirhems, is equal to twelve of its roots,”^[26] then the solution is this: You say, one square and twenty dirhems are equal to twelve roots. Halve the roots and multiply them by themselves; this gives thirty-six. Subtract from this the twenty dirhems, extract the root from the remainder, and subtract it from the moiety of the roots, which is six. The remainder is the root of the square: it is two dirhems, and the square is four.

If the instance be: “To find a square, of which if one-third be added to three dirhems, and the sum be subtracted from the square, the remainder multiplied by itself restores the square;”^[27] then the computation is this: If you subtract one-third and three dirhems from the square, there remain two-thirds of it less three dirhems. This is the root. Multiply therefore two-thirds of thing less three dirhems by itself. You say two-thirds by two-thirds is four ninths of a square; and less two-thirds by three dirhems is two roots: and again, two-thirds by three dirhems is two roots; and less three dirhems by less three dirhems is nine dirhems. You (41) have, therefore, four-ninths of a square and nine dirhems less four roots, which are equal to one root. Add the four roots to the one root, then you have five roots, which are equal to four-ninths of a square and nine dirhems. Complete now your square; that is, multiply the four-ninths of a square by two and a fourth, which gives one square; multiply likewise the nine dirhems by two and a quarter; this gives twenty and a quarter; finally, multiply the five roots by two and a quarter; this gives eleven roots and a quarter. You have, therefore, a square and twenty dirhems and a quarter, equal to eleven roots and a quarter. Reduce this according to what I taught you about halving the roots.

If the instance be: “To find a number,^[28] one-third of which, when multiplied by one-fourth of it, restores the ^[28]number,”^[29] then the computation is: You multiply one-third of thing by one-fourth of thing, this gives one-twelfth of a square, equal to thing, and the square is equal to twelve things, which is the root of one hundred and forty-four.

If the instance be: “A number,^[28] one-third of which and one dirhem multiplied by one-fourth of it and two dirhems restore the number,^[28] with a surplus of thirteen dirhems;”^[30] then the computation is this: You multiply one-third of thing by one-fourth of thing, this gives half one-sixth of a square; and you multiply two dirhems by one-third of thing, this gives two-thirds of a root; and one dirhem by one-fourth of thing gives one-fourth of a root; and one dirhem by two dirhems gives two dirhems. This altogether is one-twelfth of a square and two dirhems and (42) eleven-twelfths of a thing, equal to thing and thirteen dirhems. Remove now two dirhems from thirteen, on account of the other two dirhems, the remainder is eleven dirhems. Remove then the eleven-twelfths of a root from the one (root on the opposite side), there remains one-twelfth of a root and eleven dirhems, equal to one-twelfth of a square. Complete the square: that is, multiply it by twelve, and do the same with all you have. The product is a square, which is equal to a hundred and thirty-two dirhems and one root. Reduce this, according to what I have taught you, it will be right.

If the instance be: “A dirhem and a half to be divided among one person and certain persons, so that the share of the one person be twice as many dirhems as there are other persons;”^[31] then the Computation is this:^[32] You say, the one person and some persons are one and thing: it is the same as if the question had been one dirhem and a half to be divided by one and thing, and the share of one person to be equal to two things. Multiply, therefore, two things by one and thing; it is two squares and two things, equal to one dirhem and a half. Reduce them to one square: that is, take the moiety of all you have. You say, therefore, one square and one thing are equal to three-fourths of a dirhem. Reduce this, according to what I have taught you in the beginning of this work.

If the instance be: “A number,^[33] you remove one-third of it, and one-fourth of it, and four dirhems: then you multiply the remainder by itself, and the number,^[34] is restored, with a surplus of twelve dirhems:” then the computation is this: You take thing, and subtract from it one-third and one-fourth; there remain five- twelfths of thing. Subtract from this four dirhems: (43) the remainder is five-twelfths of thing less four dirhems. Multiply this by itself. Thus the five parts become five-and-twenty parts; and if you multiply twelve by itself, it is a hundred and forty-four. This makes, therefore, five and twenty hundred and forty-fourths of a square. Multiply then the four dirhems twice by the five-twelfths; this gives forty parts, every twelve of which make one root (forty-twelfths); finally, the four dirhems, multiplied by four dirhems, give sixteen dirhems to be added. The forty-twelfths are equal to three roots and one-third of a root, to be subtracted. The whole product is, therefore, twenty-five-hundred-and-forty-fourths of a square and sixteen dirhems less three roots and one-third of a root, equal to the original number,^[35] which is thing and twelve dirhems. Reduce this, by adding the three roots and one-third to the thing and twelve dirhems. Thus you have four roots and one-third of a root and twelve dirhems. Go on balancing, and subtract the twelve (dirhems) from six-teen; there remain four dirhems and five-and-twenty-hundred-and-forty-fourths of a square, equal to four roots and one-third. Now it is necessary to complete the square. This you can accomplish by multiplying all you have by five and nineteen twenty-fifths. Multiply, therefore, the twenty-five-one-hundred-and-forty-fourths of a square by five and nineteen twenty-fifths. This gives a square. Then multiply the four (44) dirhems by five and nineteen twenty-fifths; this gives twenty-three dirhems and one twenty-fifth. Then multiply four roots and one-third by five and nineteen twenty-fifths; this gives twenty-four roots and twenty-four twenty-fifths of a root. Now halve the number of the roots: the moiety is twelve roots and twelve twenty-fifths of a root. Multiply this by itself. It is one hundred-and-fifty-five dirhems and four hundred-andsixty-nine six-hundred-and-twenty-fifths. Subtract from this the twenty-three dirhems and the one twenty-fifth connected with the square. The remainder is one-hundred-and-thirty-two and four-hundred-and-forty six-hundred-and-twenty-fifths. Take the root of this it is eleven dirhems and thirteen twenty-fifths. Add this to the moiety of the roots, which was twelve dirhems and twelve twenty-fifths. The sum is twenty-four. It is the number^[36] which you sought. When you subtract its third and its fourth and four dirhems, and multiply the remainder by itself, the number^[36] is restored, with a surplus of twelve dirhems.

If the question be: “To find a square-root,^[36] which, when multiplied by two-thirds of itself, amounts to (45) five;”^[37] then the computation is this: You multiply one thing by two-thirds of thing; the product is two-thirds of square, equal to five. Complete it by adding its moiety to it, and add to five likewise its moiety. Thus you have a square, equal to seven and a half. Take its root; it is the thing which you required, and which, when multiplied by two-thirds of itself, is equal to five.

If the instance be: “Two numbers,^[38] the difference of which is two dirhems; you divide the small one by the great one, and the quotient is equal to half a dirhem;^[39] then the computation is this: Multiply thing and two dirhems by the quotient, that is a half. The then the computation is this: Multiply thing and two dirhems by the quotient, that is a half. The product is half a thing and one dirhem, equal to thing. Remove now half a dirhem on account of the half dirhem on the other side. The remainder is one dirhem, equal to half a thing. Double it: then you have thing, equal to two dirhems. This is one of the two numbers^[40], and the other is four.

Instance: “You divide one dirhem amongst a certain number of men, which number is thing. Now you add one man more to them, and divide again one dirhem amongst them; the quota of each is then one-sixth of a dirhem less than at the first time.”^[41] Computation: You multiply the first number of men, which is thing, by the difference of the share for each of the other number. Then multiply the product by the first and second number of men, and divide the product by the difference of these two numbers. Thus you obtain the sum which shall be divided. Multiply, therefore, the first number of men, which is thing, by the one-sixth, which is the difference of the shares; this gives one-sixth of root. Then multiply this by the original number of the men, and that of the additional one, that is to say, by thing plus one. The product is one-sixth of square and one-sixth of root divided by one (46) dirhem, and this is equal to one dirhem. Complete the square which you have through multiplying it by six. Then you have a square and a root equal to six dirhems. Halve the root and multiply the moiety by itself, it is one-fourth. Add this to the six; take the root of the sum and subtract from it the moiety of the root, which you have multiplied by itself, namely, a half. The remainder is the first number of men; which in this instance is two.

If the instance be: “To find a square-root,^[42] which when multiplied by two-thirds of itself amounts to five”^[43] then the computation is this: If you multiply it by itself, it gives seven and a half. Say, therefore, it is the root of seven and a half multiplied by two-thirds of the root of seven and a half. Multiply then two-thirds by two-thirds, it is four-ninths; and four-ninths multiplied by seven and a half is three and a third. The root of three and a third is two-thirds of the root of seven and a half. Multiply three and a third by seven and a half; the product is twenty-five, and its root is five.

If the instance be: “A square multiplied by three of its roots is equal to five times the original square;”^[44] then this is the same as if it had been said, a square, which when multiplied by its root, is equal to the first square and two-thirds of it. Then the root of the square is one and two-thirds, and the square is two dirhems and seven-ninths.

If the instance be: “Remove one-third from a square, then multiply the remainder by three roots of the first square, and the first square will be restored.”^[45] Computation: If you multiply the first square, before (47) removing two-thirds from it, by three roots of the same, then it is one square and a half; for according to the statement two-thirds of it multiplied by three roots give one square; and, consequently, the whole of it multiplied by three roots of it gives one square and a half. This entire square, when multiplied by one root, gives half a square; the root of the square must therefore be a half, the square one-fourth, two-thirds of the square one-sixth, and three roots of the square one and a half. If you multiply one-sixth by one and a half, the product is one-fourth, which is the square.

Instance: “A square; you subtract four roots of the same, then take one-third of the remainder; this is equal to the four roots.” The square is two hundred and fifty-six.^[46] Computation: You know that one-third of the remainder is equal to four roots; consequently, the whole remainder must be twelve roots; add to this the four roots; the sum is sixteen, which is the root of the square.

Instance: “A square; you remove one root from it; and if you add to this root a root of the remainder, the sum is two dirhems.”^[47] Then, this is the root of a square, which, when added to the root of the same square, less one root, is equal to two dirhems. Subtract from this one root of the square, and subtract also from the two dirhems one root of the square. Then two dirhems less one root multiplied by itself is four dirhems and one square less four roots, and this is equal to a square less one root. Reduce it, and you find a square and four dirhems, equal to a square and three roots. Remove square by square; there remain three roots, equal to four dirhems; consequently, one root is equal to one dirhem and one-third. This is the root of the square, and the square is one dirhem and seven-ninths of a dirhem. (48)

Instance: “Subtract three roots from a square, then multiply the remainder by itself, and the square is restored.”^[48] You know by this statement that the remainder must be a root likewise; and that the square consists of four such roots; consequently, it must be sixteen.

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1. ${\begin{aligned}(10-x)x&=21\\10x-x^{2}&=21\\{\text{which is to be reduced to}}\\x^{2}+21&=10x\\x=5\pm {\sqrt {25-21}}&=5\pm 2\end{aligned}}$
2. ${\begin{aligned}(10&-x)^{2}-x^{2}=40\\100&-20x=40\\100&=20x+40\\60&=20x\\3&=x\end{aligned}}$
3. ${\begin{aligned}(10-x)^{2}+x^{2}+(10-x)-x&=54\\100-20x+2x^{2}+10-2x&=54\\100-22x+2x^{2}&=54\\55-11x+x^{2}&=27\\x^{2}+28&=11x\\x={\tfrac {11}{2}}\pm {\sqrt {{\tfrac {121}{4}}-28}}={\tfrac {11\pm 3}{2}}&=7\;{\text{or}}\;4\end{aligned}}$
4. ${\begin{aligned}&{\tfrac {10-x}{x}}+{\tfrac {x}{10-x}}=2{\tfrac {1}{6}}\\&100+2x^{2}-20x=x(10-x)\times 2{\tfrac {1}{6}}=21{\tfrac {2}{3}}x-2{\tfrac {1}{6}}x^{2}\\&100+4{\tfrac {1}{6}}x^{2}=41{\tfrac {2}{3}}x\\&24+x^{2}=10x\\&x=5\pm {\sqrt {25-24}}=5\pm 1=4\;{\text{or}}\;6\end{aligned}}$
5. $4{\tfrac {1}{6}}={\tfrac {25}{6}},\;{\text{and}}\;{\tfrac {6}{25}}={\tfrac {1}{5}}+{\tfrac {1}{5}}\times {\tfrac {1}{5}}$
6. ${\tfrac {a}{b}}\times {\tfrac {b}{a}}=1$
7. ${\begin{aligned}{\tfrac {5x}{2(10-x)}}+5x&=50\\{\tfrac {x}{2(10-x)}}+x&=10\\x^{2}+100&=20{\tfrac {1}{2}}x\\x={\tfrac {41}{4}}-{\tfrac {9}{4}}&=8\end{aligned}}$
8. ${\begin{aligned}(10-x)^{2}&=81x\\100-20x+x^{2}&=81x\\x^{2}+100&=101x\\x={\tfrac {101}{2}}-{\sqrt {{\tfrac {101^{2}}{4}}-100}}&=50{\tfrac {1}{2}}-49{\tfrac {1}{2}}=1\end{aligned}}$
9. The purchaser does not make a clear enunciation of the terms of his bargain. He intends to say, “I bought $m$ bushels of wheat, and $n$ bushels of barley, and the wheat was $r$ times dearer than the barley. The sum I expended was equal to the difference in the quantities, added to the difference in the prices of the grain.” If $x$ is the price of the barley, $rx$ is the price of the wheat; whence, $mrx+nx=(m-n)+(rx-x);\;\therefore \;x={\tfrac {m-n}{mr+n+r-1}}$ and the sum expended is ${\tfrac {(mr+n)\times (m-n)}{mr+n+r-1}}$.
10. In the original, “squares.” The word square is used in the text to signify either, 1st, a square, properly so called, fractional or integral; 2d, a rational integer, not being a square number; 3d, a rational fraction, not being a square; 4th, a quadratic surd, fractional or integral.
11. ${\begin{aligned}&{\tfrac {x}{x+2}}={\tfrac {1}{2}}\\&x={\tfrac {x+2}{2}}={\tfrac {x}{2}}+1\\&{\tfrac {x}{2}}=1\;{\text{and}}\;x+2=4\end{aligned}}$
12. “Square” in the original.
13. ${\begin{aligned}10x=(10-x)^{2}=100-20x+x^{2}\\x=15-{\sqrt {225-100}}=15-{\sqrt {125}}\end{aligned}}$
14. ${\begin{aligned}{\tfrac {x(10-x)}{(10-2x)}}=5{\tfrac {1}{4}}\\10x-x^{2}=52{\tfrac {1}{2}}-10{\tfrac {1}{2}}x\\20{\tfrac {1}{2}}x=x^{2}+52{\tfrac {1}{2}}\\x=10{\tfrac {1}{4}}-7{\tfrac {1}{4}}=3\\\end{aligned}}$
15. ${\begin{aligned}{\tfrac {2}{3}}\times {\tfrac {1}{5}}x^{2}={\tfrac {x}{7}}\\x^{2}=7{\tfrac {1}{2}}\times {\tfrac {x}{7}}=1{\tfrac {1}{14}}x\\x=1{\tfrac {1}{14}}\\x^{2}=1{\tfrac {29}{196}}\\{\tfrac {2}{15}}x^{2}={\tfrac {30}{196}}\times {\tfrac {x}{7}}\end{aligned}}$
16. ${\begin{aligned}&{\tfrac {3}{4}}\times {\tfrac {1}{5}}x^{2}={\tfrac {4}{5}}x\\&{\tfrac {3{\tfrac {3}{4}}}{20}}x,\;{\text{or}}\;{\tfrac {15}{80}}x,\;{\text{or}}\;{\tfrac {3}{16}}x=1\\&\therefore x={\tfrac {16}{3}}=5{\tfrac {1}{3}}\end{aligned}}$
17. “Square” in the original.
18. ${\begin{aligned}4x^{2}=20\\x={\sqrt {5}}\end{aligned}}$
19. “Square” in the original.
20. ${\begin{aligned}&x\times {\tfrac {x}{3}}=10\\&x^{2}=30\\&x={\sqrt {30}}\end{aligned}}$
21. ${\begin{aligned}&x\times 4x={\tfrac {x}{3}}\\&x={\tfrac {1}{12}}\end{aligned}}$
22. ${\begin{aligned}&x^{2}\times x=3x^{2}\\&x=3\end{aligned}}$
23. ${\begin{aligned}4x\times 3x&=x^{2}+44\\11x^{2}&=44\\x^{2}&=4\\x&=2\end{aligned}}$
24. ${\begin{aligned}4x\times 5x&=2x^{2}+36\\18x^{2}&=36\\x^{2}&=2\end{aligned}}$
25. ${\begin{aligned}4x^{2}&=3x^{2}+50\\x^{2}&=50\end{aligned}}$
26. ${\begin{aligned}x^{2}+20&=12x\\x=6\pm {\sqrt {36-20}}&=6\pm 4=10\;{\text{or}}\;2\end{aligned}}$
27. ${\begin{aligned}\left[x-({\tfrac {x}{3}}+3)\right]^{2}&=x\\{\text{or}}\;[{\tfrac {2x}{3}}-3]^{2}&=x\\{\tfrac {4x^{2}}{9}}+9&=5\\x^{2}+20{\tfrac {1}{4}}&=11{\tfrac {1}{4}}x\\x=9,\;{\text{or}}\;2{\tfrac {1}{4}}\end{aligned}}$
28. “Square” in the original.
29. ${\begin{aligned}{\tfrac {x}{3}}\times {\tfrac {x}{4}}&=x\\x^{2}&=12x\\x&=12\end{aligned}}$
30. ${\begin{aligned}&({\tfrac {x}{3}}+1)-({\tfrac {x}{4}}+2)=x+13\\&{\tfrac {x^{2}}{12}}+{\tfrac {11}{12}}x+2=x+13\\&{\tfrac {x^{2}}{12}}={\tfrac {x}{12}}+11\\&x^{2}=x+132\\&x={\tfrac {1}{2}}+{\tfrac {23}{2}}=12\end{aligned}}$
31. The enunciation in the original is faulty, and I have altered it to correspond with the computation. But in the computation, $x$, the number of persons, is fractional! I am unable to correct the passage satisfactorily.
32. ${\begin{aligned}{\tfrac {1{\tfrac {1}{2}}}{1+x}}=2x\\x2+x={\tfrac {3}{4}}\\x=1-{\tfrac {1}{2}}\\x={\tfrac {1}{2}}\end{aligned}}$
33. “Square” in the original.
34. ${\begin{aligned}(x-{\tfrac {1}{3}}x-{\tfrac {1}{4}}x-4)^{2}=x+12\\({\tfrac {5}{12}}x-4)^{2}=x+12\\{\tfrac {25}{144}}x^{2}+16-3{\tfrac {1}{3}}x=x+12\\{\tfrac {25}{144}}x^{2}+4=4{\tfrac {1}{3}}x\\x^{2}+23{\tfrac {1}{25}}=24{\tfrac {24}{25}}x\\{\sqrt {\left[({\tfrac {24{\tfrac {24}{25}}}{2}})^{2}-23{\tfrac {1}{25}}\right]}}+{\tfrac {24{\tfrac {24}{25}}}{2}}-x\\11{\tfrac {13}{25}}+12{\tfrac {12}{25}}=24=x\end{aligned}}$
35. “Square” in the original.
36. “Square” in the original.
37. ${\begin{aligned}x\times {\tfrac {2}{3}}x=5\\{\tfrac {2}{3}}x^{2}=5\\x^{2}=7{\tfrac {1}{2}}\\x={\sqrt {7{\tfrac {1}{2}}}}\\\end{aligned}}$
38. “Squares” in the original.
39. ${\begin{aligned}{\tfrac {x}{x+2}}={\tfrac {1}{2}}\\{\tfrac {1}{2}}x+1=5\\{\tfrac {1}{2}}x=1\\x=2,x+2=4\end{aligned}}$
40. “Square” in the original.
41. ${\begin{aligned}{\tfrac {1}{x}}-{\tfrac {1}{x+1}}&={\tfrac {1}{6}}\\1&={\tfrac {x(x+1)}{6}}\\x^{2}+x&=6\\{\sqrt {({\tfrac {1}{2}})^{2}+6}}-{\tfrac {1}{2}}&=x=2\end{aligned}}$
42. “Square” in the original.
43. ${\begin{aligned}{\tfrac {2}{3}}x^{2}&=5\\x^{2}&=7{\tfrac {1}{2}}\\x&={\sqrt {7{\tfrac {1}{2}}}}\\{\sqrt {7{\tfrac {1}{2}}}}\times {\tfrac {2}{3}}{\sqrt {7{\tfrac {1}{2}}}}&=5\\{\sqrt {{\tfrac {4}{9}}\times 7{\tfrac {1}{2}}}}={\sqrt {3{\tfrac {1}{3}}}}&={\tfrac {2}{3}}{\sqrt {7{\tfrac {1}{2}}}}\\{\sqrt {3{\tfrac {1}{3}}\times 7{\tfrac {1}{2}}}}&={\sqrt {25}}=5\end{aligned}}$
44. ${\begin{aligned}x^{2}\times 3x&=5x^{2}\\x^{2}\times x&=1{\tfrac {2}{3}}x^{2}\\x&=1{\tfrac {2}{3}}\\x&=2{\tfrac {7}{9}}\end{aligned}}$
45. ${\begin{aligned}(x^{2}-{\tfrac {1}{3}}x^{2})\times 3x=x^{2}&\;\therefore \;{\tfrac {2}{3}}x^{2}\times 3x=x^{2}\\x^{2}\times 3x&=1{\tfrac {1}{2}}x^{2}\\x={\tfrac {1}{2}}&\;\therefore \;x^{2}={\tfrac {1}{4}}\end{aligned}}$
46. ${\begin{aligned}{\tfrac {x^{2}-4x}{3}}&=4x\\x^{2}-4x&=12x\\x^{2}&=16x\\x=16&\;\therefore \;x^{2}=256\end{aligned}}$
47. ${\begin{aligned}{\sqrt {x^{2}-x}}+x&=2\\{\sqrt {x^{2}-x}}&=2-x\\x^{2}-x&=4+x^{2}-4x\\x^{2}+3x&=4+x^{2}\\3x&=4\\x&=1{\tfrac {1}{3}}\end{aligned}}$
48. ${\begin{aligned}(x^{2}-3x)^{2}&=x^{2}\\x^{2}-3x&=x\\x^{2}&=4x\\x&=4\\\end{aligned}}$