Elimination by the method of the Greatest Common Divisor. 15. Five friends, A, B, C, D, E, jointly spend $879 at an inn. This sum is to be paid by one of them; but, on consultation, they find that none of them had, alone, enough for this purpose. If, then, one of them is to pay it, the others must give him a part of their money. A can pay, if he receives one fourth; B, if he receives one fifth; C, if he receives one sixth; D, if he receives one seventh; and E, if he receives one eighth of the others' money. much has each? How Ans. A $319, B $ 459, C $543, D $ 599, E $ 639. 155. Third Method of solving the Problem of art. 142, called that of Elimination by the method of the greatest Common Divisor. Solution. This method is generally inapplicable to transcendental equations, but can be successfully applied in all other cases to eliminate one unknown quantity after another, until the given equations are reduced to one. In order to eliminate an unknown quantity from two equations which contain it, reduce them as in arts. 105 and 118, and arrange their terms according to the powers of the quantity to be eliminated, taking out each power as a factor from the terms which contain it. It being now recollected that the second member of each of these equations is zero, it will appear evident that, if the first members are divided one by the other, the remainder arising from this division must likewise be equal to zero; for this remainder is the difference between the dividend and a certain multiple of the divisor, that is, between zero and a certain multiple of zero. Elimination by the method of the Greatest Common Divisor. Hence, divide one of these first members by the other, and proceed, as in arts. 60, &c., to find their greatest common divisor; each successive remainder may be placed equal to zero. But a remainder will at last be obtained, which does not contain the quantity to be eliminated; and the equation, formed from placing this remainder equal to zero, is the equation from which this quantity is eliminated. By eliminating, in this way, the unknown quantity from either of the equations which contain it, taken with each of the others, a number of equations is formed one less than that of the given equations, and containing one less unknown quantity; and to which this process of elimination may be again applied until one equation is finally obtained. 156. Scholium. It sometimes happens, that the first members have a common divisor which contain the given unknown quantity; and, in this case, the process cannot be continued beyond this divisor. But as the given first members are multiples of their common divisor, they must be rendered equal to zero by those values of the unknown quantities which render the common divisor equal to zero; that is, the two given equations are satisfied by such values of the unknown quantities; so that, though they are in appearance distinct equations, they are, in reality, equivalent to but one equation, that is, to the equation formed by placing their common divisor equal to zero. 157. Scholium. Care must be taken that no factor be suppressed which may be equal to zero. Examples of Elimination by the method of the Greatest Common Divisor. 158. EXAMPLES. 1. Obtain one equation with one unknown quantity from the two equations x3y x2 - y3+5=0, x3 + y2 x − 5 = 0, by the elimination of x. Solution. Divide the first members as follows: x3 + y x2 - y3 +5|x3 + y2 x-5 x3 + y2 x 1st Rem. y x2. y2 x Divide the preceding divisor by this remainder after multiplying by y to render the first term divisible. Divide the preceding divisor by this remainder after multiplying by (3 y3 — 10) to render the first term divisible. (3 y3—10) (—4 y3+25 y2) x — 9 yo+150 y6—700 y3+1000 (3 y3—10) (—4 y5+25 y2) x — 4 y9+ 85 y6—375 y3 -5y965 yo - 325 y3+1000, whence the required equation is -5 y965 y6-325 y3 1000 = 0. Examples of Elimination by the method of the Greatest Common Divisor. 2. Obtain one equation with one unknown quantity from the two equations x3+ y3 = a, x5+ y5 = b, by the elimination of x. Ans. (y3a) 5 — (y5—b)3 = 0. 3. Obtain one equation with one unknown quantity from the two equations x2 + y2 = 2, x2 + x3 y + x2 y2+x y3 + y2 = 1, by the elimination of x. Ans. y8 — 4 y + 14 y1 — 20 y2 +9 = 0. 4. Obtain one equation with one unknown quantity from the two equations x2 + xy + y2 = 1, x3+ y3 = 0, by the elimination of x. Ans. 4 y66 y1+3y2-1=0. 5. Obtain one equation with one unknown quantity from the two equations x3 + y x2 + x + y = 4, x3 + x2 + y x = 3, by the elimination of x. Ans. Either y-10, or y2-3y+21=0. 6. Obtain one equation with one unknown quantity from the three equations x + y + z = a, x z + xy + y z = b, x y z = c, by the elimination of x and y. Ans. z3-a z2+ bz-c=0. Examples of Elimination by the method of the Greatest Common Divisor. 7. Obtain one equation with one unknown quantity from the three equations x + y + z = a, x2 + y2+z2=b, x y + x z + y z = c, by the elimination of x and y. Ans. These three equations involve an impossibility unless and in case this equation is satisfied by the given values of a, b, and c, the three given equations are equivalent to but two, one of them being superfluous, and, by the elimination of x, they give the indeterminate equation with two unknown quantities y2+ y z +x2-ay-az+c=0. 8. Obtain one equation with one unknown quantity from the three equations Ans. 28-8z6+16 z4+x-10=0. 9. Obtain one equation with one unknown quantity from the four equations x+y+z+u = a, x y + x z + xu+yz+yu+zu=b, x y z + xyu+xzu+yzu= c, xyzu = =e, by the elimination of x, y, and z. Ans. u4-a u3+bu2—cu+e=0. |