Equations

Forming Equations
Before proceeding to the actual solution of an equation of any type, certain preliminary operations have necessarily to be carried out in order to prepare it for solution.

Still more preliminary work is that of forming the equation ( samī-karaṇa, samī-kāra or samī-kriyā; from sama, equal and kṛ , to do; hence literally, making equal) from the conditions of the proposed problem. Such preliminary work may require the application of one or more fundamental operations of algebra or arithmetic.

Bhāskara II says: "Let yāvat-tāvat be assumed as the value of the unknown quantity. Then doing precisely as has been specifically told-by subtracting, adding, multiplying or dividing the two equal sides of an equation should be very carefully built.

Algebraic Expressions and Algebraic Equations
What is an algebraic expression ? Let us try to understand this with an example.

Geeta says that she has 10 marbles more than Mala. We do not know exactly how many marbles Mala has. She may have any number of marbles. But we know that

Number of marbles of Geeta = Number of marbles of Mala = 10

We shall denote the ‘number of marbles of Mala' by the letter x. Here x is unknown which could be 1, 2, 3, 4, etc.

Using x, we write,

Number of marbles of Geeta = x+10.

Thus 'x + 10' is an algebraic expression.

Algebra abounds in the usage of symbols. These symbols represent the unknown quantities and operations performed with them. The following table gives the symbols which were used for some basic operations by the ancient Indian Mathematicians. The letter yā (an abbreviation of yāvat-tāvat) was the most popular representation of the unknown quantity. Its square was termed yāva, the abbreviation of yāvat-tāvat-varga (varga means square). While writing equations, the constant term was denoted by the letter rū, an abbreviation of rūpa as seen in the table above. Any negative sign in the equation is denoted by a dot above the term.

If there are three unknown quantities in an expression, the symbols used are yā, kā and nī. These are the abbreviations for yāvat tāvat, kālaka and nīlaka. The product of first two unknown quantities is represented as yākābha where yā and kā are the two unknowns and bha stands for their product.

The following table gives a representation of some of the algebraic expressions used by ancient Indian mathematicians. We will see how algebraic expressions are written by ancient Indian mathematicians.

Consider the equation 10 x - 8 = x2 +1

This can be written as,

0x2 + 10 x - 8 = 1x2 + 0x + 1

Observe the positions of x2, x1, x0 (constant term). Can you notice any pattern ? The standard way of writing an equation starts from the highest power of x. Then the powers of x were written in the descending order up to its lowest power. This format of writing equation was followed by mathematicians from ancient times.

Brahmagupta called an equation as samakaraṇa or samīkaraṇa. It means 'making equal. The two sides of an equation (LHS and RHS) were written one below the other. The symbol '=' was not used. Both the sides of an equation were made same by finding the appropriate value(s) for the unknown(s).

We shall see how this equation was written by Pṛthūdakasvāmin (864 CE) in his commentary on the Brāhma-sphuṭa-siddhānta. He writes the equation

10x - 8 = x2 + 1 as follows: Another example of an equation from Bījagaṇita of Bhāskara II is:

X4 - 2x2 - 400x = 9999

This is represented as,

यावव १  याव २.    या  ४.००    रू   ०

यावव ०  याव ०     या  ०        रू  ९९९९

Operations with Algebraic Expressions
Bhāskara II gives the operations using algebraic terms as follows:

स्याद्रूपवर्णाभिहतौ तु वर्णो द्वित्र्यादिकानां समजातिकानाम् ॥

वधे तु तद्वर्गघनादयः स्युस्तद्भावितं चासमजातिघाते।

भागादिकं रूपवदेव शेषं व्यक्ते यदुक्तं गणिते तदत्र ॥

“The product of a numerical constant and an unknown quantity is an unknown quantity. Products of two or three like terms are their squares or cubes (respectively). Product of unlike terms is bhāvita. Fractions etc. are as in the case of knowns. The other (processes) are same as explained in arithmetic."

Addition and Subtraction of Algebraic Expressions
Bhāskara II gives the rule for addition and subtraction of unknown quantities as follows:

योगोऽन्तरं तेषु समानजात्योर्विभिन्नजात्योश्च पृथक् स्थितिश्च।

“Addition and subtraction are performed amongst like terms. The unlike terms are to be kept separately."

Explanation:

It is well known that the addition and subtraction can be performed only amongst like terms and unlike terms are to be kept separately. Like terms are the terms that contain the same letter variables raised to the same powers. E.g., या ३, या ४, या ५ are like terms. याव २, याव ५, याव ७ are also like terms. का ३, का ७, का १५ are also like terms. Nowadays we say 3x, 4x, 5x are like terms. Similarly 2x2, 5x2, 7x2 are like terms. and 3y, 7y, 15y are also like terms. When we have like terms, the sum and difference can be simplified. E.g. 3x + 5x can be simplified as 8x. 10x2 - 4x2 can be simplified as 6x2.

Unlike terms are those terms having different variables or variables with different powers. E.g. या ३, याव ३, याघ ४, का ५, काव, याकाभा. In modern notation, these are represented as 3x, 3x2, 4x3, 5y, y2, xy.

Multiplication of Algebraic Expressions
Bījagaṇita gives a rule for multiplication -

गुण्यः पृथग्गुणकखण्डसमो निवेश्यस्तैः खण्डकैः क्रमहतः सहितो यथोक्त्या।

अव्यक्तवर्गकरणीगणनास चिन्त्यो व्यक्तोक्तखण्डगुणनाविधिरेवमत्र॥

“Place the multiplicand at as many places as the terms of the multiplier. Multiply with the terms of the multiplier in order separately and add the results as directed in the problem. This is applicable in the case of squares of unknown numbers and surds also. The method of partial products stated in the case of arithmetic numbers is applicable here also.”

Explanation If ax + b and cx + d are multiplicand and multiplier respectively, their product can be obtained as follows:

The multiplier has two terms, i.e., cx and d. Place the multiplicand at two places. Multiply them with the terms of the multiplier separately as shown.

(ax +b) x cx = acx2 + bcx

(ax + b) xd = adx + bd

Add the results.

The multiplication result is: acx2 + (bc + ad)x + bd

Algebraic Notations

 * The symbols used for unknown numbers are the initial syllables yа̄ of yа̄vat-tа̄vat (as much as), kа̄ of kа̄laka (black), nī of nīlaka (blue), pī of pīta (yellow) etc.


 * The product of two unknowns is denoted by the initial syllable bhā of bhāvita (product) placed after them. The powers are denoted by the initial letters va of varga (square), gha of ghana (cube); vava stands for vargavarga, the fourth power. Sometimes the initial syllable ghā of ghāta (product) stands for the sum of powers.


 * A coefficient is placed next to the symbol. The constant term is denoted by the initial symbol rū of rūpa (form).


 * A dot is placed above the negative integer


 * The two sides of an equation are placed one below the other. Thus the equation X4 - 2X2 - 400x = 9999; is written as

यावव​ १ याव २● या ४००● रू ०

यावव​ ० याव ० या ० रू ९९९९

which means writing x for या

x4 -2x2 -400x+0 = 0x4 +0x2+0x+9999

If there be several unknowns, those of the same kind are written in the same column with zero coefficients, if necessary. Thus the equation

197x - 1644y - z = 6302 is represented by

yа̄ 197 kа̄ 1644● ni 1● ru 0

yа̄ 0 kа̄ 0 ni 0 ru 6302

which means, putting y for kа̄ and z for ni

197x - 1644Y - z + 0 = 0x + 0y + 0z + 6302.

Bhāskara II says:

"Then the unknown on one side of it (the equation) should be subtracted from the unknown on the other side; so also the square and other powers of the unknown;

the known quantities on the other side should be subtracted from the known quantities of another side."

The following illustration is from the Bījagaṇita of Bhāskara II:

"Thus the two sides are

yā va 4  yā 34●   rū 72

yā va 0  yā 0      rū 90

On complete clearance (samaśodhana), the residues of the two sides are

yā va 4  yā 34●   rū 0

yā va 0  yā 0      rū 18

i.e., 4x2 - 34x = 18

Classification of Equations
The earliest Hindu classification of equations seems to have been according to their degrees, such as simple (technically called yāvat tāvat ), quadratic (varga), cubic (ghana) and biquadratic (varga-varga). Reference to it is found in a canonical work of circa 300 B.C. But in the absence of further corroborative evidence, we cannot be sure of it. Brahmagupta (628) has classified equations as: (I) equations in one unknown (eka-varna-samīkaraṇa), (2) equations in several unknowns (aneka-varna-samīkaraṇa), and (3) equations involving products of unknowns (bhaivita).

The first class is again divided into two sub classes, viz.,(i) linear equations, and (ii) quadratic equations (avyakta-varga-samīkaraṇa). Here then we have the beginning of our present method of classifying equations according to their degrees. The method of classification adopted by Pṛthūdakasvāmī (860) is slightly different. His four classes are: (1) linear equations with one unknown, (2) linear equations with more unknowns, (3) equations with one, two or more unknowns in their second and higher powers, and (4) equations involving products of unknowns. As the method of solution of an equation of the third class is based upon the principle of the elimination of the middle term, that class is called by the name madhyamāharaṇa (from madhyama, "middle", aharana "elimination", hence meaning "elimination of the middle term"). For the other classes, the old names given by Brahmagupta have been retained. This method of classification has been followed by subsequent writers.

Bhāskara II distinguishes two types in the third class, viZ" (i) equations in one unknown in its second and higher powers and (ii) equations having two or more unknowns in their second and higher powers.' According to Krsna (1580) equations are primarily of two classes: (1) equations in one unknown and (z) equations in two or more unknowns. The class (1), again, comprises two subclasses: (i) simple equations and (ii) quadratic and higher equations. The class (2) has three subclasses: (i) simultaneous linear equations, (ii) equations involving the second and higher powers of unknowns, and (iii) equations involving products of unknowns. He then observes that these five classes can be reduced to four by including the second subclasses of classes (1) and (2) into one class as madhyamāharaṇa.

Linear Equations in One Unknown
A Linear equation is an equation containing only the first power of the variables, coefficients and constants. For example, the equation 2x + 4 = 5 is a linear equation in one variable. This is called first-order equation since the power of the variable (x) is one. If the equation has highest power of x as two, i.e. x2, then it will be a quadratic (second order) equation.

Early Solutions:
As already stated, the geometrical solution of a linear equation in one unknown is found in the śulba, the earliest of which is not later than 800 B.C. There is a reference in the Sthānāṅga-Sūtra (c. 300 B.C.) to a linear equation by its name (yāvat -tāvat ) which is suggestive of the method of solution! followed at that time.We have, however, no further evidence about it. The earliest Hindu record of doubtless value of problems involving simple algebraic equations and of a method for their solution occurs in the Bakhshālī treatise, which was written very probably about the beginning of the Christian Era.

One problem is "The amount given to the first is not known. The second is given twice as much as the first; the third thrice as much as the second; and the fourth four times as much as the third. The total amount distributed is 132. What is the amount of the first?"

If x be the amount given to the first, then according to the problem,

x + 2X + 6x + 24X = 132.

Rule of False Position:
The solution of this equation is given as follows:

" 'Putting any desired quantity in the vacant place' ; any desired quantity is 1 ; 'then construct the series. 'multiplied' added' 33. "Divide the visible quantity' (which) on reduction becomes (This is) the amount given (to the first)."

The solution of another set of problems in the Bakhshālī treatise, leads ultimately to an equation of the type ax+ b=p. The method given for its solution is to put any arbitrary value g for x, so that

ag+ b =p' say.

Then the correct value will be

$${\displaystyle x = {\frac{(p - p')}{a}} + g}$$

Solution of Linear Equations
āryabhaṭaI(499) says:

"The difference of the known "amounts" relating to the two persons should be divided by the difference of the coefficients of the unknown. The quotient will be the value of the unknown, if their possessions be equal."

This rule contemplates a problem of this kind: Two persons, who are equally rich, possess respectively a, b times a certain unknown amount together with c, d

units of money in cash. What is that amount?

If x be the unknown amount, then by the problem

ax+ c= bx+ d.

Therefore     $${\displaystyle x = {\frac{(d - c)}{(a- b)}}}$$

Hence the rule.

Rule for solving the linear equation of the form bx + c = dx + e where b, c, d and e are given numbers is given by Brahmagupta as follows.

अव्यक्तान्तरभक्तं व्यस्तं रूपान्तरं समेऽव्यक्तः।

वर्गाव्यक्ताः शोध्या यस्माद्रूपाणि तदधस्तात् ॥

"The difference of absolute numbers, inverted and divided by the difference of unknown, is the [value of the) unknown in an equation.”

Explanation: Consider the equation, bx + c = dx + e

Here x is the unknown quantity whose value is to be found. The letters b and d are its coefficients. The remaining letters c and e are numerical constants.

Difference of absolute numbers = c-e

Difference of absolute numbers inverted = e-c

Difference of coefficients of unknown = b - d

x is found as

$${\displaystyle x = {\frac {(e-c)}{(b-d)} }}$$

Bhāskara II explains how the above formula is obtained.

यावत्तावत् कल्प्यमव्यक्तराशेर्मानं तस्मिन् कुर्वतोद्दिष्टमेव ।

तुल्यौ पक्षौ साधनीयौ प्रयत्नात्त्यक्त्वा क्षिप्त्वा वाऽपि संगुण्य भक्त्वा ॥

एकाव्यक्तं शोधयेदन्यपक्षाद्रूपाण्यन्यस्येतरस्माच्च पक्षात्

शेषाव्यक्तेनोद्धरेद्रूपशेषं व्यक्तं मानं जायतेऽव्यक्तराशेः॥

“Assume the unknown quantity (x). Perform the desired process by transposing the factors involving unknown terms to one side and constant terms to the other side after cancelling or reducing or multiplying or dividing. Divide the terms by the coefficient of the unknown and calculate the value of the unknown factor."

Explanation: For instance, let us consider the following equation:

6x - 5 = 2x + 3

(i) Transposing the factors involving unknown terms to one side and the constants to the other side, we get,

6x - 2x = 3 + 5

Hence, 4x = 8

ii) Dividing the terms by the coefficient of the unknown, we get

x = 2

Sripati writes :

"First remove the unknown from anyone of the sides (of the equation) leaving the known term; the reverse (should be done) on the other side. The difference of the absolute terms taken in the reverse order divided by the difference of the coefficients of the unknown will be the value of the unknown.

Naraya writes:

"From one side clear off the' unknown and from the other the known quantities; then divide the residual known by the residual coefficient of the unknown. Thus will certainly become known the value of the unknown. "

For illustration we take a problem proposed by Brahmagupta :

"Tell the number of elapsed days for the time when four times the twelfth part of the residual degrees increased by one, plus eight will be equal to the residual

degrees plus one."

It has been solved by Pṛthūdakasvāmī as follows:

"Here the residual degrees are (put as) yāvat -tāvat ,

ya increased by one, ya 1 ru 1; twelfth part of it, (ya 1 ru 1) / 12

four times this, (ya 1 ru 1) / 3 ; plus the absolute quantity eight, (ya 1 ru 25) / 3. This is equal to the residual degrees plus unity. The statement of both sides

tripled is

ya 1 ru 25

ya 3 ru 3

The difference between the coefficients of the unknown is 2. By this the difference of the absolute terms, namely 22, being divided, is produced the residual of the degrees of the sun 11. These residual degrees should be known to be irreducible. The elapsed days can be deduced then, (proceeding) as before."

In other words, we have to solve the equation

$${\displaystyle {\frac{4}{12}(x+1)+8 = x+1}}$$

which gives x + 25 = 3x + 3

2x = 22

Therefore x= 11

The following problem and its solution are from the Bijaganita of Bhāskara II :

"One person has three hundred coins and six horses. Another has ten horses (each) of similar value and he has further a debt of hundred coins. But they

are of equal worth. What is the price of a horse?

"Here the statement for equi-clearance is :

6x + 300 = 10x - 100.

Now, by the rule, 'Subtract the unknown on one side from that on the other etc.,' unknown on the first side being subtracted from the unknown on the other side,

the remainder is 4x. The absolute term on the second side being subtracted from the absolute term on the first side, the remainder is 400. The residual known

number 400 being divided by the coefficient of the residual unknown 4x, the quotient is recognized to be the value of x, (namely) 100."

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

In other words sankramana is the solution of the simultaneous equations

x+ y= a, x-y= b.

Brahmagupta's rule for solution is: "The sum is increased and diminished by the difference and divided by two; (the result will be the two unknown quantities): (this is) concurrence. The same rule is restated by him on a different occasion in the form of a problem and its solution.

"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.

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

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."

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

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.

Thus $${\displaystyle x = {\frac{(am - bn)}{(a^2 -b^2)}} }$$, $${\displaystyle y = {\frac{(an - bm)}{(a^2 -b^2)}} }$$

The following example with its solution is taken from the BfjagatJita of Bhāskara II :

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 ?"

The equations are

x + 100 = 2(y - 100), (I)

y + 10 = 6(x - 10). (2)

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

First Method: Assume x = 2z.- 100, y = z + 100,

so that equation (I) is identically satisfied. Substituting

these values in the other equation, we get

z + 110 = 12z- 660;

Hence z =.70 Therefore, x = 40, y = 170.

Second Method: From equation (I), we get

x =2y - 300,

and from equation (2)

$${\displaystyle x = {\frac{1}{6}}(y+70) }$$

Equating these two values of x, we have

$${\displaystyle 2y -300 = {\frac{1}{6}}(y+70) }$$

$${\displaystyle 12y -1800 = y+70 }$$

Hence y= 170. Substituting this value of y in any of the two expressions for x, we get x = 40.

A Type of Linear Equations
The earliest Hindu treatment of systems of linear equations involving several unknowns is found in the Bakhshālī treatise. One problem in it runs as follows:

"[Three persons possess a certain amount of riches each.] The riches of the first and the second taken together amount to 13; the riches of the second and

the third taken together are 14; and the riches of the first and the third mixed are known to be 15. Tell me the riches of each."

If x1, x2, x3 be the wealths of the three merchants respectively, then x1 + x2 = 13, x2 + x3 = 14, x3 + x1 = 15.

Another problem is "Five persons possess a certain amount of riches each. The riches of the first and the second mixed together amount to 16; the riches of the second and the third taken together are known to be 17; the riches of the third and the fourth taken together are known to be 18; the riches of the fourth and the fifth mixed together are 19; and the riches of the first and the fifth together amount to 20. Tell me what is the amount of each. x₁ x₂ x₃ x₄ x₅

x₁ + x₂ = 16, x₂ + x₃ = 17, x₃+ x₄ = 18, x₄ + x₅ = 19, x₅ + x₁= 20.

There are in the work a few other similar problems. Every one of them belongs to a system of linear equations of the type

x₁ + x₂ = a1, x₂ + x₃ = a2 ..., xn + x₁ = an n being odd.

Solution by False Position
A system of linear equations of this type is solved in the Bakhshālī treatise substantially as follows:

Assume an arbitrary value p for x₁ and then calculate the values of x₂, x₃, ... corresponding to it. Finally let the calculated value of xn + x₁ be equal to b

(say). Then the true value of x₁ is obtained by the formula

$${\displaystyle x_1 = p +{\frac{1}{2}}(a_n- b)}$$

In the particular case (1) the author assumes the arbitrary value 5 for x ; then are successively calculated the values x₂ = 8, x₃ = 6 and x₃ + x₁ = 11. The

correct values are, therefore,

x₁= 5 + (15 - 11)/2 = 7, x₂ = 6, x₃= 8

Rationale. By the process of elimination we get from

equations (I)

(a2-a1)+(a4-a3)+· ... +(an-1 - an-2) + 2x1 = an

Assume x1 = p; so that

(a2-a1)+(a4-a3)+· ... +(an-1 - an-2) + 2p = b say.

Subtracting 2(x1 - p) = an - b.

Therefore $${\displaystyle x_1 =p + {\frac{1}{2}}(a_n-b) }$$

Second Type
A particular case of the type of equations (I) for which n = 3, may also be looked upon as belonging to a different type of systems of linear equations.

Σx - x1 = a1, Σx - x2 = a2, Σx - xn = an

Where Σx stands for x1 + x2 +....+xn

But it will not be proper to say that equations of this. type have been treated in the Bakhshili treatise. l They have however, been solved by āryabhaṭa(499) and Mahivira (850). The former says: "The (given) sums of certain (unknown) numbers, leaving out one number in succession, are added together separately and divided by the number of terms less one; that (quotient) will be the value of the whole.

$${\displaystyle \sum x = \sum_{r=1}^n a_r/(n-1) }$$

Mahāvīra states the solution thus: "The stated amounts of the commodities added together should be divided by the number of men less one. The quotient will be the total value (of all the commodities). Each of the stated amounts being subtracted from that, (the value) in the hands (of each will be found).

In formulating his rule Mahāvīra had in view the following example:

"Four merchants were each asked separately by the customs officer about the total value of their commodities.

The first merchant, leaving out his own investment, stated it to be 22; the second stated it to be 23, the third 24 and the fourth 27; each of them deducted

his own amount in the investment. O friend, tell me separately the value of (the share of) the commodity owned by each."

Here $${\displaystyle x_1 + x_2+x_3+x_4 = {\frac{22+23+24+27}{4-1}}=32}$$

Therefore x1 = 10, x2 = 9, x3 = 8, x4 = 5.

Nārāyanasays: "The sum of the depleted amounts divided by the number of persons less one, is the total amount. On subtracting from it the stated amounts severally will be found the different amounts."

Third Type
A more generalized system of linear eguations will be

$${\displaystyle b_1\sum x - c_1x_1=a_1 }$$, $$b_2\sum x - c_2x_2=a_2$$..........,

$$b_n\sum x - c_nx_n=a_n$$ (III)

Therefore $${\displaystyle \sum x  =  {\frac { \sum (a/c)}{ \sum(b/c) -1}}}$$

Hence $${\displaystyle x_r ={\frac{b_r}{c_r}}. {\frac { \sum (a/c)}{ \sum(b/c) -1}} - {\frac{a_r}{c_r}}}$$ --(1)

r = I, 2, 3,..... ,n.

A particular case of this type is furnished by the following example of Mahāvīra:

"Three merchants begged money mutually from one another. The first on begging 4 from the second and 5 from the third became twice as rich as the others.The second on having 4 from the first and 6 from the third became thrice as rich. The third man on begging 5 from the first and 6.from the second became five times as rich as the others. O mathematician, if you know the citra-kuttaka-misra,1 tell me quickly what was the amount in the hand of each."

That is, we get the equations

x + 4 + 5 = 2(y + z - 4 - 5),

y + 4 + 6 = 3(z + x - 4 - 6),

z + 5 + 6 = 5 (x + y - 5 - 6);

or 2(x + y + z) - 3x = 27,

3(x + y + z) - 4y = 40 ;

5 (x + y +z) - 6z = 66;

a particular case of the system (III). Substituting In

(I), we get

x = 7,. Y = 8, Z = 9·

Brahmagupta's Rule: Brahmagupta (628) states the following rule for solving linear equations involving several unknowns:

"Removing the other unknowns from the side of the first unknown and dividing by the coefficient of the first unknown, the value of the first unknown is obtained. In the case of more values of the first unknown, two and two (of them) should be considered after reducing them to common denominators. And so on repeatedly. If more unknowns remain in the final equation, the method of the pulveriser should be employed. Then proceeding reversely the values of other unknowns can be found."

Pṛthūdakasvāmī (860) has explained it thus: "In an example in which there are two or more unknown quantities, colours such as yāvat -tāvat, ,etc.should be assumed for their values. Upon them should be performed all operations conformably to the statement of the example and thus should be carefully framed two or more sides and also equations. Equi-clearance should be made first between two and two of them and so on to the last: from one side one unknown should be cleared, other unknowns reduced to a common denominator and also the absolute numbers should be cleared from the side opposite. The residue of other unknowns being divided by the residual coefficient of the first unknown will give the value of the first unknown. If there be obtained several such values, then with two and two of them, equations should be formed after reduction to common denominators. Proceeding in this way to the end find out the value of one unknown. If that value be in terms of another unknown then the coefficients of those two will be reciprocally the values of the two unknowns. If, however, there be present more unknowns in that value, the method of the pulveriser should be employed. Arbitrary values may then be assumed for some of the unknowns." It will be noted that the above rule embraces the indeterminate as well as the determinate equations. In fact, all the examples given by Brahmagupta in illustration of the rule are of indeterminate character. We shall mention some of them subsequently at their proper places. So far as the determinate simultaneous equations are concerned, Brahmagupta's method for solving them will be easily recognized to be the same as our present one.

Bhāskara's Rule. Bhāskara II has given practically the same rule as that of Brahmagupta for the solution of simultaneous linear equations involving several unknowns.

We take the following illustrations from his works.

Example 1. "Eight rubies, ten emeralds and a hundred pearls which are in thy ear-ring were purchased by me for thee at an equal amount; the sum of the-price rates of the three sorts of gems is three less than the half of a hundred. Tell me, 0 dear auspicious lady, if thou be skilled in mathematics, the price of each."

If x, y, z be the prices of a ruby, emerald and pearl respectively, then 8x = 10y = 100z

x+y+z = 47·

Assuming the equal amount to be w, says Bhāskara II, we shall get •

x = w/8, y = w/10, z = w/100.

Substituting in the remaining equation, we easily get w = 200. Therefore

x = 25, y = 20, z = 2.

Quadratic Equations
The geometrical solution of the simple quadratic equation

$$4h^2 -4dh =- c^2$$ is found in the early canonical works of the Jainas (500- 300 B. C.) and also in the Tattvathadhigama-Sūtra of Umasvati (c. 150 B. C.) as.

$${\displaystyle h = {\frac {1}{2}} (d-\sqrt{d^2-c^2})}$$

Sridhara's Rule. Sddhara (c. 750) expressly indicates his method of solving the quadratic equation.

His treatise on algebra is now lost. But the relevant portion of it is preserved in quotations by Bhāskara II and others. Sridhara's method is:.

"Multiply both the sides (of an equation) by a known quantity equal to four times the coefficient of the square of the unknown; add to both sides a known

quantity equal to the square of the (original) coefficient of the unknown: then extract the root."

That is, to solve

$$ax^2 + bx = c$$

we have $$4a^2x^2 + 4abx = 4ac$$

or

$$(2ax+b)^2 = 4ac + b^2$$

$$2ax+b = \sqrt{4ac + b^2}$$

$$x= \frac{\sqrt{4ac+b^2} -b}{2a}$$

Sripati's Rules. Sripati (1039) indicates two methods of solving the quadratic. There is a lacuna in our manuscript in the rule describing the first method, but it can be easily recognized to be the same as that of Sridhara. "Multiply by four times the coefficient of the square of the unknown and add the square of the coefficient of the unknown; then extract the square-root divided by twice the coefficient of the square of the unknown, is said to be the value of the unknown." "Or multiplying by the coefficient of the square of the unknown and adding the square of half the coefficient of the unknown, extract the square-root. Then proceeding as before, it is diminished by half the coefficient of the unknown and divided by the coefficient of the square of the unknown. This quotient is said to be the value of the unknown."

$$ax^2 + bx = c$$

or $${\displaystyle a^2x^2+ abx +\left ( \frac{b}{2} \right )^2 =  ac +\left ( \frac{b}{2} \right )^2 } $$

Therefore

$${\displaystyle ax+ \left ( \frac{b}{2} \right ) = \sqrt{ac + \left ( \frac{b}{2} \right ) ^2} } $$

$${\displaystyle x= \frac{\sqrt{ac + \left ( \frac{b}{2} \right ) ^2} - \frac{b}{2}}{a}  } $$

Bhāskara II's Rules. Bhāskara- II (1150) says: "When the square of the unknown, etc., remain, then, multiplying the two sides (of the equation) by some suitable quantities, other suitable quantities should be added to them so that the side containing the unknown becomes capable of yielding a root (pada-prada). The equation should then be formed again with the root of this side and the root of the known side. Thus the value of the unknown is obtained from that equation.

This rule has been further elucidated by the author as follows:.

"When after perfect clearance of the two sides, there remain on one side the square, etc., of the unknown and on the other side the absolute term only, then, both the sides should be multiplied or divided by some suitable optional quantity; some equal quantities should further be added to or subtracted from both the sides so that the unknown side will become capable of yielding a root. The root of that side must be equal to the root of the absolute terms on the other side. For, by simultaneous equal additions, etc., to the two equal sides the equality remains. So forming an equation again with these roots the value of the unknown is found."

It may be noted that in his treatise on arithmetic Bhāskara II has always followed the modern method of dividing by the coefficient of the square of the unknown.

Jnanaraja (1503) and Ganesa (1545) describe the same general methods for solving the quadratic as Bhāskara II.

Elimination of the Middle Term. The method of solving the quadratic was known amongst the Hindu algebraists by the technical designation madhyamāharaṇa or "The Elimination of the Middle" (from madhyama = middle and aharana = removal, or destroying, that is, elimination). The origin of the name will be easily found in the principle underlying the method. By it a quadratic equation which, in its general form, contains three terms and so has a middle term, is reduced to a pure quadratic equation or a simple equation involving only two terms and so having no middle term. Thus the middle term of the original quadratic is eliminated by the method generally adopted for its solution. And hence the name. Bhāskara II has observed, "It is also specially designated by the learned teachers as the madhyamāharaṇa. For by it, the removal of one of the two terms of the quadratic, the middle one, takes place." The name is, however, employed also in an extended sense so as to embrace the methods for solving the cubic and the biquadratic, where also certain terms are eliminated. It occurs as early as the works of Brahmagupta (628).

Two Roots of the Quadratic. The Hindus recognized early that the quadratic has generally two root. In this connection Bhāskara II has quoted the following rule from an ancient writer of the name of Padmanabha whose treatise on algebra is not available now. "if after extracting roots the square-root of the absolute side of the quadratic be less than the negative absolute term on the other side, then taking it negative as well as positive, two values of the unknown are found."

Bhāskara points out with the help of a few specific illustrations that though these double roots of the quadratic are theoretically correct, they sometimes lead to incongruity and hence should not always be accepted. So he modifies the rule as follows: "If the square-root of the known side of the quadratic be less than the negative absolute term occurring in the square-root of the unknown side, then making it negative as well as positive, two values of the unknown should be determined. This is to be done occasionally."

Example 1. "The eighth part of a troop of monkeys, squared, was skipping inside the forest, being delightfully attached to it. Twelve were seen on the hill delighting in screaming and rescreaming. How many were they ?"

Solution. "Here the troop of monkeys is x. The square of the eighth part of this together with 12, is equal to the troop. So the two sides are

x²/64 + 0x + 12 = 0x² + x + o.

Reducing these to a common denominator and then deleting the denominator, and also making clearance the two sides become

x² - 64x + 0 = 0x2 + 0x - 768.

Adding the square of 32 to both sides and extracting square-roots, we get

x- 32. = ± (0x + 16).

In this instance the absolute term on the known side is smaller than the negative absolute term on the side of the unknown; hence it is taken positive as well as negative; the two values of x are found to be 48, 16.

Equations Of Higher Degrees
Cubic and Biquadratic. The Hindus did not achieve much in the solution of the cubic and biquadratic equations. Bhāskara II (1150) attempted the application of the method of the madhyamāharaṇa (elimination of the middle) to those equations also so as to reduce them by means of advantageous transformations and introduction of auxiliary quantities to simple and quadratic equations respectively. He thus anticipated one of the modern methods of solving the biquadratic. "If, however," observes Bhāskara II, "due to the presence of the cube, biquadrate, etc., the work of reduction cannot proceed any further, after the performance of such operations, for want of a root of the unknown side (of an equation), then the value of the unknown must be obtained by the ingenuity (of the mathematician)." He has given two examples, one of the cubic and the other of the biquadratic, in which such reduction is possible.

Example 1. "What is that number, which being multiplied by twelve and increased by the cube of the number, is equal to six times the square of the number added with thirty-five.

Solution. "Here the number is x. This multiplied by twelve and increased by the cube of the number becomes x³ + 12x. It is equal to 6x² + 35. On making clearance, there appears on the first side x³ - 6x²+ 12x; on .the other side 35 . Adding negative eight to both the sides and extracting cube-roots, we get x - 2. = 0x + 3. And from this equation the number is found to be 5.

Example 2. "What is that number which being multiplied by 200 and added to the square of the number, and then multiplied by 2 and subtracted from the fourth power of the number will become one myriad less unity? Tell that number.

Solution. "Here the number is x; multiplied by 200 it becomes 200x; added to the square of the number, becomes x² + 200x; this being multiplied by two, 2x² + 400x; by this being diminished the fourth power of the number, namely, this x4 becomes x4- 2x² - 400x. This is equal to a myriad less unity. Equiclearance having been made, the two sides will, be

x4- 2x² - 400x. = 0x4 + 0x² + 0x + 9999.

Here on adding four hundred x plus unity to the first side, the root can be extracted, but on adding the same to the other side, there will be no root of it. Thus the work (of reduction) does not proceed. Hence here, ingenuity (is called for). Here adding to both the sides four times the square of x, four hundred x and unity and then extracting roots, we get

x² + 0x+ 1 = 0x² + 2x + 100.

Again, forming equation with these and proceeding as before, the value of x is obtained as 11 In similar instances the value of the unknown must be determined

by the ingenuity of the mathematician."

Simultaneous Quadratic Equations
Common Forms Various problems involving simultaneous quadratic equations of the following forms have been treated by Hindu writers:

x - y = d ; xy = b ......(1)

x + y = a ; xy = b ......(2

x² + y² = c ; xy = b ......(3)

x² + y² = c ; x + y = a ......(4)

For the solution of (1) āryabhaṭaI (499) states the following rule:

"The square-root of four times the product (of two quantities) added with the square of their difference, being added and diminished by their difference and halved gives the two multiplicands."

i.e., x =½(√(d² + 4b + d)) y=½(√(d² + 4b - d))

Brahmagupta (628) says: "The square-root of the sum of the square of the difference of the residues and two squared times the product of the residues, being added and subtracted by the difference of the residues, and halved (gives) the desired residues severally ."

Nārāyana(1357) writes: "The square-root of the square of the difference of two quantities plus four times their product is their sum."

"The square of the difference of the quantities together with twice their product is equal to the sum of their squares. The square-root of this result plus twice the product is the sum."

For the solution of (2) the following rule is given by Mahāvīra(850) : "Subtract four times the area (of a rectangle) from the square of the semi-perimeter; then by sankramana between the square-root of that (remainder) and the semi-perimeter, the base and the upright are obtained."

x =½(a +√(a2 - 4b ))

y=½(a -√(a2 - 4b ))

Nārāyana says: "The square-root of the square of the sum minus four times the product is the difference."

For (3) Mahāvīra gives the rule: "Add to and subtract twice the area (of a rectangle) from the square of the diagonal and extract the square roots. By sankramana between the greater and lesser of these (roots), the side and upright are found.

x =½(√(c + 2b) + √(c - 2b))

y =½(√(c + 2b) - √(c - 2b))

For equations (4) āryabhaṭa I writes: "From the square of the sum (of two quantities) subtract the sum of their squares. Half of the remainder is their product."

The remaining operations will be similar to those for the equations (2); so that

x =½(a + √(2c –a2))

y =½(a - √(2c –a2))

Brahmagupta says: "Subtract the square of the sum from twice the sum of the squares; the square-root of the remainder being added to and subtracted from the sum and halved, gives the desired residues."

Nārāyanahas given two other forms of simultaneous quadratic equations, namely,

x - y = d......(5)

x² - y² =m; xy = b ......(6)

For the solution of (5) he gives the rule: "The square-root of twice the sum of the squares decreased by the square of the difference is equal to the

sum."

x + y = √(2c- d²)

Therefore

x = ½(√(2c- d²) + d)

y = ½(√(2c- d²) - d)

x = ~(V 2c - d2 + d), _'Y = -Hv 2.C - d2 - d).

For (6) Nārāyanawrites: "-

"Suppose the square of the product as the product (of two quantities) and the difference of the squares as their difference. From them by sankrama will be obtained the (square) quantities. Their square-roots severally will give the quantities (required)."

We have

x² - y² =m

x² y² = b2

These are of the form (1). Therefore

x² = ½(√(m2 + 4b2) + m )

y² = ½(√(m2 + 4b2) - m )

hence we get the values of x and y.