Elimination by Addition and Subtraction. 10. Solve the two equations y x3 13 +x= = 3, yx (y x2+1) — x3 + x = 6. Solution. The elimination of x gives 3 y 30, or y = 1; which, being substituted in the first of the given equations, produces x= = 3. 11. Solve the two equations x2 y1—8 y2x2+16x2 = 90 x y +60 (x—y2)—720 (y—1), (y2-4y+4) x 5 =3 12 159. Problem. To solve two equations of the first degree by Elimination by Addition and Subtraction. Solution. The given equations may, as in art. 146, be reduced to the forms Ax+By+ M = 0, A' x + B' y + M' = 0. The process of the preceding article, being applied to these equations in order to eliminate x, will be found to be the same as to Multiply the first equation by A' the coefficient of x in the second, multiply the second by A the coefficient of x in the first, and subtract the first of these products from the second. Examples of Elimination by Addition and Subtraction. Thus, these products are A A' xAB y + A' M = 0, and the difference is whence (A B'A' B) y + A MA' M=0; In the same way y might have been eliminated by multiplying the first equation by B', and the second by B, and the difference of these products is whence (A B' — A' B) x + B' M − B M' — 0'; the values of x and y thus obtained being the same as those given in art. 146. 160. Corollary. This process may be applied with the same facility to any equations of the first degree. 161. EXAMPLES. 1. Solve, by the preceding process, the two equations 13x + 7y341 = 71⁄2 y + 434 x, 2x+y=1. Ans. x - 12, y = 50. 2. Solve, by the preceding process, the two equations Examples of Elimination by Addition and Subtraction. 3. Solve, by the preceding process, the three equations x + y + z = 30, 8x+4y+2z = 50, 27x+9y+3x= 64. Ans. x = 3, y=-7, z= 36. 4. Solve, by the preceding process, the three equations Ans. x= = 360, y = 124, z= 100. 5. Solve, by the preceding process, the four equations = 21, = 2x + 7y-%- u = 683, Ans. x= =100, y = 60, z—— · 13, u—— - 50. 6. Solve, by the preceding process, the four equations 7. Solve, by the preceding process, the six equations Examples of Elimination by Addition and Subtraction. 8. A person has two large pieces of iron whose weight is required. It is known that ths of the first piece weigh 96 lbs. less than ths of the other piece; and that ths of the other piece weigh exactly as much as ths of the first. How much did each of these pieces weigh? Ans. The first weighed 720, and the second 512 lbs. 9. $2652 are to be divided amongst three regiments, in such a way, that each man of that regiment which fought best, shall receive $1, and the remainder is to be divided equally among the men of the other two regiments. Were the dollar adjudged to each man in the first regiment, then each man of the two remaining regiments would receive $; if the dollar were adjudged to the second regiment, then each man of the other two regiments would receive $; finally, if the dollar were adjudged to the third regiment, each man of the other two regiments would receive $4. What is the number of men in each regiment? Ans, 780 men in the first, 1716 in the second, and 2028 in the third regiment. 10. To find three numbers such that if 6 be added to the first and second, the sums are to one another as 2:3; if 5 be added to the first and third, the sums are as 7: 11; but if 36 be subtracted from the second and third, the remainders are as 6: 7. Ans. 30, 48, 50. 10 Indeterminate Coefficients. CHAPTER IV. NUMERICAL EQUATIONS. SECTION I. Indeterminate Coefficients. 162. Theorem. If a polynomial A + B x + C x2 + D x3 + Ex1+&c. is such, as to be equal to zero independently of x, that is, if it is equal to zero whatever values are given to x, it must always be the case that A=0, B=0, C=0, D=0, E = 0, &c. ; that is, that the aggregate of all the coefficients of each power of x is equal to zero, and also the aggregate of all the terms which do not contain x is equal to zero. Proof. Since the equation A+B+C x2 + D x3 + &c. = 0 is true for every value which can be given to x, it must be true when we make x = 0; in which case all the terms of the first member vanish except the first, and we have A = 0. |