10. A sum of money is to be divided equally among a certain number of persons. Now if there were three claimants less, each would receive 150 dollars more; and if there were 6 more, each would receive 120 dollars less. How many persons are there, and how much is each to receive ? 11. What fraction is that, to the numerator of which if i be added, its value will be , but if i be added to its denominator its value will be . 12. What fraction is that, to the numerator of which if a be added, its value will be m, but if a be added to its denomina Solve the 11th example by these formulas. 13. What fraction is that, from the numerator of which if a be subtracted, its value will be m, but if a be subtracted from n its denominator, its value will be L ? 9 N. B. The answers to the 12th and 13th differ only in the signs of the denominators. The learner will find by endeavouring to solve particular examples from these formulas, that he will not always succeed. If in making examples for the 12th, he selects his numbers, so that n p is greater than mq, the formula will fail; but if he takes the same numbers, and applies them according to the conditions of the 13th, they will answer those conditions. When m 9 is greater than n р the numbers will not suit the conditions of the 13th, but they will answer to those of the 12th. The numbers in example 11th will not form an example according to the 13th. The following numbers will form an example for the 13th but not for the 12th. 14. What fraction is that, from the numerator of which if 3 be subtracted, its value will be , but if 3 be subtracted from its denominator its value will be ? The reason why numbers chosen indiscriminately will not satisfy the conditions of the above formulas will be explained hereafter. Equations with several Unknown Quantities. XXIII. Questions involving more than two unknown Quan tities. Sometimes it is necessary to employ, in the solution of a question, more than two unknown quantities. In this case, the question must furnish conditions enough to form as many distinct equations as there are unknown quantities. 61. A market woman sold to one man, 7 apples, 10 pears, and 12 peaches, for 63 cents; and to another, 13 apples, 6 pears, and 2 peaches, for 31 cents ; and to a third, 11 apples, 14 pears, and 8 peaches for 63 cents. She sold 'them each time at the same rate. What was the price of each ? Let x = the price of an apple, y = a pear, a peach. Then we shall have 1. 7 x + 10 y + 12 z = 63 2. 13 x + 6y + 2z = 31 3. 11 x + 14 y + 8z = 63. The second being multiplied by 6, the z will have the same coefficient as in the first. 4. 78 x + 36 y + 12 z = 186 1. 7x + 10 y + 12 = 63 5. 71 x + 26 y = 123. If the second be multiplied by 4, the z will have the same coefficient as the 3d. 6. 3. 52 x + 24 y + 8z = 124 = 61 41 x + 10 y We have now the two equations 71 x + 26 y= 123 and 41 x + 10 y= 61 which contain only two unknown quantities. These may now be reduced in the same manner as others with two unknown quantities. Multiplying the 5th by 5, and the 7th by 13, the coefficient of y will be the same in both. 8. 355 x + 130 y = 615 9. 533 x + 130 y=793 10. 178 x = 178 We have now found an equation containing only one unknown quantity 178 x = 178 1. Putting the value of x into the 7th, it becomes 41 + 10 y = 61 10 y = 20 y = 2. Putting the values of x and y into the 2d, it becomes 13 + 12+2 z= 31 22= 6 z = 3. Ans. The apples 1, the pears 2. and the peaches 3 cents each. In the same manner, questions, involving four unknown quantities, may be solved. First combine them two by two till one of the unknown quantities is eliminated from the whole, and there will be three equations with three unknown quantities. Then combine these three two by two, until one of the un known quantities is elin inated, and then there will be two equations with two unknown quantities, and so on. Either of the methods of elimination may be used as is most convenient. It is not necessary that all the unknown quantities should enter into every equation. 2. A market woman sold at one time 7 eggs, 12 apples, and a pie for 26 cents ; at another time 12 eggs, 18 pears, and 3 pies, for 69 cents; at a third time 20 pears, 10 apples, and 17 eggs for 69 cents; and at a fourth time, 7 pies, 18 apples, and 10 pears for 66 cents. Each article was sold, at every sale, at the same price as at first. What was the price of each article ! y = 26 Let u = the price of an egg, an apple, y a pie, a pear. 1. 70 + 12x + 2. 12 u + 182 + 3y=69 3. 17 u + 20 = + 10x = 69 4. 10 z + 18 x + 7y= 66 5. In the 1st, y= 26 7U-12 x. Putting this value of y into the 2nd and 4th, they become 6. 12 u + 182 + 78 - 214-36 x=69 7. 10 2 + 18x + 182-49 u 84 x = 66. Uniting and transposing terms 8. 182 9U-36 x= 9 9. 102-49 u 66 x = 116 3. 20 z + 17u + 10x= 69 If the 9th be multiplied by 2, the coefficient of z will be the same as in the 3d; 10. 20% - 98 u 132 x = 232. Subtracting 10th from 3d 3. 202 + 17u+ 10x = 69 10. 132 x = -232 20 z - 98 u 11.* 90 z 45 U 90 z - 441 u X = 115 u + 142 x = 301 If the 8th be multiplied by 5, and the 9th by 9, the coefficients of z will be alike. 12. 180 x = - 45 13. 594 x = 1044. Subtracting 13th from 12th 14. 396 u + 414 x = 999. Deducing the value of x from 11th, and also from 14th. 301 -115 u 15. 142 999 - 396 u 16. 414 Making these values of x equal, we have an equation containing only one unknown quantity. 999 - 396 u 301 - 115 U 142 = 2 Putting this value of u into the 15th or 16th, we shall find 1 2* Putting these values of x and u into the 1st, 2nd, or 4th, and we shall find y = 6. Putting the values of x and u into the 3d, and we shall find z = 1; Ans. Eggs, 2 cents each, apples, cent, pears, 1f cent, and pies, 6 cents. * If the learner is at a loss how to subtract -233 from 69. let him transpose both into the first member, or some terms from the first to the second. |