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History of Hindu Mathematics: A Source Book v2.c17.s6: 8. Linear Equations with Two Unknowns

History of Hindu Mathematics: A Source Book v2.c17.s6

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8. Linear Equations with Two Unknowns

v2.c17.s6.1

**Rule of Concurrence.** One topic commonly discussed by almost all Hindu writers goes by the special name of *saṅkramaṇa* (concurrence). According to Nârâyaṇa (1350), it is also called *saṅkrama* and *saṅkrâma*.² Brahmagupta (628) includes it in algebra while others consider it as falling within the scope of arithmetic. As explained by the commentator Gaṅgâdhara (1420), the subject of discussion here is "the investigation of two quantities concurrent or grown together in the form of their sum and difference."

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¹ *BMs*, Folio 4, verso.

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² *GK*, i. 31.

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Or, in other words, *saṅkramaṇa* is the solution of the simultaneous equations $x+y=a,$ $x-y=b.$ So Brahmagupta and Śrîpati are perfectly right in thinking that concurrence is truly a topic for algebra.

v2.c17.s6.5

Brahmagupta's rule for solution is :

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"The sum is increased and diminished by the difference and divided by two; (the result will be the two unknown quantities): (this is) concurrence."¹

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¹ *BrSpSi*, xviii. 36.

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The same rule is restated by him on a different occasion in the form of a problem and its solution.

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"The sum and difference of the residues of two (heavenly bodies) are known in degrees and minutes. What are the residues? The difference is both added to and subtracted from the sum, and halved; (the results are) the residues."²

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² *BrSpSi*, xviii. 96.

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Similar rules are given also by other writers.³

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³ *GSS*, vi. 2; *MSi*, xv. 21; *ŚiŚe*, xiv. 13; *L*, p. 12; *GK*, i. 31.

v2.c17.s6.13

**Linear Equations.** Mahâvîra gives the following examples leading to simultaneous linear equations together with rules for the solution of each.

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*Example.* "The price of 9 citrons and 7 fragrant wood-apples taken together is 107; again the price of 7 citrons and 9 fragrant wood-apples taken together is 101. O mathematician, tell me quickly the price of a citron and of a fragrant wood-apple quite separately."⁴

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⁴ *GSS*, vi. 140½-142½.

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If x, y be the prices of a citron and of a fragrant wood-apple respectively, then $9x + 7y = 107,$ $7x + 9y = 101.$ Or, in general, $ax + by = m,$ $bx + ay = n.$

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*Solution.* “From the larger amount of price multiplied by the (corresponding) bigger number of things subtract the smaller amount of price multiplied by the (corresponding) smaller number of things. (The remainder) divided by the difference of the squares of the numbers of things will be the price of each of the bigger number of things. The price of the other will be obtained by reversing the multipliers.”¹

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¹ *GSS*, vi. 139½.

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$Thus x = (am - bn)/(a² - b²), y = (an - bm)/(a² - b²).$

v2.c17.s6.20

*Example.* “A wizard having powers of mystic incantations and magical medicines seeing a cock-fight going on, spoke privately to both the owners of the cocks. To one he said, ‘If your bird wins, then you give me your stake-money, but if you do not win, I shall give you two-thirds of that.’ Going to the other, he promised in the same way to give three-fourths. From both of them his gain would be only 12 gold pieces. Tell me, O ornament of the first-rate mathematicians, the stake-money of each of the cock-owners.”²

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² *GSS*, vi. 270-2½.

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$i.e., x - ¾y = 12, y - ⅔x = 12.$ Or, in general, $x - c/dy = p, y - a/bx = p.$

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*Solution:*¹ $x = (b(c + d))/((c + d)b - (a + b)c)p,$ $y = (d(a + b))/((a + b)d - (c + d)a)p.$

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¹ *GSS*, vi. 268½-9½.

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The following example with its solution is taken from the *Bîjagaṇita* of Bhâskara II :

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*Example.* "One says, 'Give me a hundred, friend, I shall then become twice as rich as you.' The other replies, 'If you give me ten, I shall be six times as rich as you.' Tell me what is the amount of their (respective) capitals ?"²

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² *BBi*, p. 41.

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The equations are $x + 100 = 2(y - 100), (1)$ $y + 10 = 6(x - 10). (2)$

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Bhâskara II indicates two methods of solving these equations. They are substantially as follows :

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*First Method.*³ Assume $x = 2z - 100, y = z + 100,$ so that equation (1) is identically satisfied. Substituting these values in the other equation, we get $z + 110 = 12z - 660;$ whence z = 70. Therefore, x = 40, y = 170.

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³ *BBi*, p. 46.

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*Second Method.*⁴ From equation (1), we get $x = 2y - 300,$ and from equation (2) $x = ⅙(y + 70).$

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⁴ *BBi*, pp. 78f.

v2.c17.s6.34

Equating these two values of *x*, we have $2y - 300 = ⅙(y + 70),$ or $12y - 1800 = y + 70;$ whence y = 170. Substituting this value of *y* in any of the two expressions for *x*, we get x = 40.

v2.c17.s6.35

It is noteworthy that the second method of solution of the problem under consideration is described by Bhâskara II in the section of his algebra dealing with "linear equations with several unknowns," while the first method in that dealing with "linear equations in one unknown." In this latter connection he has observed that the solution of a problem containing two unknowns can sometimes be made by ingenious artifices to depend upon the solution of a simple linear equation.

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