Simultaneous Indeterminate Equations Of the First Degree

In algebra, an indeterminate equation is an equation for which there is more than one solution.

Bhaskara II's Rule:

" Remove the first unknown from the second side of an equation and the others as well as the absolute number from the first side. Then on dividing the second side by the coefficient of the first unknown, its value will be obtained. If there be found in this way several values of the same unknown, from them, after reduction to a common denominator and then dropping it, values of another unknown should be determined. In the final stage of this process, the multiplier and quotient obtained by the method of the pulveriser will be the values of the unknowns associated with the dividend and the divisor (respectively). If there be several unknowns in the dividend, their values should be determined after assuming values of all but one arbitrarily. Substituting these values and proceeding reversely, the values of the other unknowns can be obtained. If on so doing there results a fractional value (at any stage),the method of the pulveriser should be employed again. Then determining the (integral) values of the latter unknowns accordingly and substituting them, the values of the former unknowns should be found proceeding reversely again.

Example from Bhaskara Il: " (Four merchants), woo have horses 5, 3, 6 and 8 respectively; camels 2, 7, 4 and 1 ; whose mules are 8, 2, 1 and 3 ; and oxen 7, 1, 2. and 1 in number; are all owners of equal wealth. Tell me instantly the price of a horse, etc."

If x, y, z, w denote respectively the prices of a horse, a camel, a mule and an ox, and W be the total wealth of each merchant, we have

${\begin{cases}5x+2y+8z+7w=W----(1)\\3x+7y+2z+w=W----(2)\\6x+4y+z+2w=W----(3)\\8x+y+3z+w=W----(4)\end{cases}}$

Then

${\begin{aligned}x&={\frac {1}{2}}(5y-6z-6w),\ from\ (1)\ and\ (2)\\&={\frac {1}{3}}(3y+z-w),\ from\ (2)\ and\ (3)\\&={\frac {1}{2}}(3y-2z+w),\ from\ (3)\ and\ (4)\\\end{aligned}}$

From the first and second values of x

$y={\frac {1}{9}}(20z+16w)$

From the second and third values

$y={\frac {1}{3}}(8z-5w)$

Equating these two values of y and on simplifying

$20z+16w=24z-15w$

Hence $z={\frac {31w}{4}}$

Consider $w=4t$ then $z=31t,\quad y=76t,\quad x=85t$