then, by substituting these values in the first of the two proposed equations, it becomes or, rather, 6x+35t+60-18x-9t+2u=100; -12x+261+2u=40; ... u=20+6x-13t. The three unknowns, y, z, u, are thus found expressed in functions of 1, and of the indeterminate auxiliary t. In order to resolve the two proposed equations in positive numbers, it is evidently necessary to take r and t positive, since x is one of the primitive unknowns, and since y=5t. But it is necessary to satisfy also the inequalities 20-6x-3t>0, 20+6x-13t>0. In adding them, x disappears, and there remains 40-16t>0...t<21; then the values which we ought to give to t are t=0, 1, 2. With the value t=0 we should have Finally, with the value t=2 we should have y=10, z=14—6x, u=—6+6x. The only admissible values of x are x=1, 2; and from thence result the two further solutions In all, nine solutions. There would be but three if those were excluded in which one of the unknowns is zero. EXAMPLES. 1o. Two countrymen have together 100 eggs. The one says to the other, If I count my eggs by eights, there is a surplus of 7. If I count mine by tens, I find the same surplus of 7. each? The second answers, Ans. Number of eggs of the first, 63 or 23; of the second, =37 or 77. 2o. To find three whole numbers such that, if we multiply the first by 3, the second by 5, and the third by 7, the sum of the products shall be 560; and such, moreover, that if the first be multiplied by 9, the second by 25, and the third by 49, the sum of the products shall be 2920. 3o. A person purchased 100 animals at 100 dollars; sheep at 31 dollars a piece; calves at 13 dollars; and pigs at a dollar. of each kind? How many animals had he Ans. Sheep, 5, 10, 15. Pigs, 53, 66, 79. 4°. In a foundry two kinds of cannon are cast; each cannon of the first sort weighs 1600 lbs., and each of the second 2500 lbs. ; and yet for the second there are used 100 lbs. of metal less than for the first. How many cannons are there of each kind? Ans. Of the first, 11, 36...; of the second, 7, 23.... 5o. A farmer purchased 100 head of cattle for 4000 francs, to wit: oxen at 400 francs apiece, cows at 200, calves at 80, and sheep at 20. How many had he of each? Ans. In excluding the solutions which contain a zero the problem admits of the ten following: Oxen, 1, 1, 1, 1, 1, 1, 1, 1, 4, 4. Sheep, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92. QUADRATIC EQUATIONS. 178. Quadratic equations, or equations of the second degree, are divided into two classes. I. Equations which involve the square only of the unknown quantity. These are termed incomplete or pure quadratics. Of this description are the equations they are sometimes called quadratic equations of two terms, because, by transposition and reduction, they can always be exhibited under the general form II. Equations which involve both the square and the simple power of the unknown quantity. These are termed adfected or complete quadratics. Of this description are the equations 2x ax2+bx=c; x2-10x=7; 5x2 x 3 3 273 =8- -x2+· ; 12 they are sometimes called quadratic equations of three terms, because, by transposition and reduction, they can always be exhibited under the general form SOLUTION OF PURE QUADRATICS CONTAINING ONE UNKNOWN QUANTITY. 179. The solution of the equation ax2-b presents no difficulty. Dividing each member by a, it becomes whence x=+ b b a If be a particular number, either integral or fractional, we can extract its square root, either exactly or approximately, by the rules of arithmetic. If b be an algebraic expression, we must apply to it the rules established for the extraction of the square root of algebraic quantities. a It is to be remarked, that since the square both of +m and ―m is +m3, so, in like manner, 2 both (+√)R and b 2 is + Hence the above a equation is susceptible of two solutions, or has two roots; that is, there are two quantities which, when substituted for x in the original equation, will render the two members identical; these are for, substitute each of these values in the original equation ar2=b, it becomes Hence it appears that in pure quadratics the two values of the unknown quantity are equal with contrary signs.* hence the two values of r are +4 and 4, and either of these, if substituted for r in the original equation, will render the two members identical. Since 15 is not a perfect square, we can only approximate to the two values of x. We find the approximate values to be One might suppose that in extracting the square root of both members of such an equation as 2=b, the double sign should be prefixed to x, the root of 22, also. But it is to be observed, that it is the value of +x that is required. Besides, suppose we were to write ±±√b; combining these signs in all possible ways, there result the four equations, +x=+√b, +x=−√b,−x=+√b, —x=—√b, the last two of which may be deduced from the first two by changing the signs of the two members; the equation +x=±√b expresses nothing more, therefore, than the equation *=vb. We might always omit, since it is implied before v Render the denominator rational by multiplying both terms of the fraction by the numerator, the equation then becomes Transposing, Squaring, m+x+√2mx+x2= ±m√n. √2mx+x2= ±m√n—(m+x). 2mx+x=m2n2m √n(m+x)+(m+x)2. Transposing and reducing, 180. In the same manner we may solve all equations whatsoever, of any degree, which involve only one power of the unknown quantity; that is, all equations which are included under the general form For, dividing each member of the equation by a, it becomes If n be an even number, then the radical must be affected with the double |