Examples of Elimination by Addition and Subtraction. Thus, these products are AA'x+A'By+A'M = 0, AA'x+AB'y+AM' = 0; and the difference is whence (AB'A' B)y+AM-A'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 (AB'A' B)x + B'M-BM' = 0; the values of x and y thus obtained being the same as those given in art. 110. 120. Corollary. This process may be applied with the same facility to any equations of the first degree. EXAMPLES. 1. Solve, by the preceding process, the two equations Ans. X=- 12, y = 50. 2. Solve, by the preceding process, the two equations fx + 4y = 6, Examples of Elimination by Addition and Subtraction. 2-53. Solve, by the preceding process, the three equations x + y + z= 30, 8x+4y+2x= 50, Ans. x, y = -7, a8-51 4. Solve, by the preceding process, the three equations 3x- 1005y + 360, 23x + 200 = 163z — Ans. x = 360, y = 610, 124, z = 100. 5. Solve, by the preceding process, the four equations x 9y+3z 10 u = 21, 2x+7y 2x + 7y u = 683, 3x + y +52 + 2 u = 195, 4x6y-2z+ 9u516, Ans. x= 100, y = 60, =- 13, u = 6. Solve, by the preceding process, the four equations 7. Solve, by the preceding process, the six equations Ans. y + z + u + t + w = 25. x = 2, y = 3, z = 4, u = 5, t = 6, y = 7. 8. A person has two large pieces of iron whose weight is required. It is known that ths of the first piece weigh Examples of Elimination by Addition and Subtraction. 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, 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. Power of a Monomial. CHAPTER IV. Powers and Roots. SECTION I. Powers and Roots of Monomials. 121. Problem. To find any power of a monomial. Solution. The rule of art. 25, applied to this case, in which the factors are all equal, gives for the coefficient of the required power the same power of the given coefficient, and for the exponent of each letter the given exponent added to itself as many times as there are units in the exponent of the required power. Hence Raise the coefficient of the given monomial to the required power; and multiply each exponent by the exponent of the required power. 122. Corollary. An even power of a negative quantity is, by art 27, positive, and an odd power is negative. EXAMPLES. 1. Find the third power of 2 a2 b5 c. Ans. 8a6 b15 c3. 2. Find the mth power of a". Ans. amn. 4. Find the mth power of a-". 5. Find the ―mth power of a—". 6. Find the 6th power of the 5th power of a3 b c2. Ans. a-mn Ans. amn. Root of a Monomial; imaginary quantity. 7. Find the qth power of the -pth power of the mth power of a n ̧ 8. Find the rth power of am b-n cr d. Ans. 9. Find the 3d power of a 2 b3 c Ans. Ans. amnpq. amr b―nr cp r dr. 5 ƒ3 x 1. a6b9c15f18 x3. Ans. 81 b-12. 13. Find the 4th power of -36-3. 14. Find the 5th power of the 4th power of the 3d power Ans. a60. of 123. To find any root of a monomial. Solution. Reversing the rule of art. 121, we obtain im mediately the following rule. Extract the required root of the coefficient; and divide each exponent by the exponent of the required root. |