Insert these values in (1) and reduce; then we obtain = - - 2x. The first equation is convertible into one having both sides squares, viz. x2 (2 + y)2 — 4 xy2 (2 + y) + 4y1 = 16y1, and extracting x (2 + y) 2y2 =±4y2; or taking them separately, and reducing, 6y2 — xy = 2x, and 2y2 + xy = Combining each of these with the second given equation, we shall readily obtain a = 2 and y from the first combination; and x =— and y from the second. 50 3 = 5 Substitute this in the first, and reduce; then 3y2+ 2y = 5. 1. Given æ” + y" = a”, and xy = b2, to find x and y. 2n Ans. x = {} a′′ ± } √/a2′′ – 4b2"}}, and y = {} a" = √ a2" — 4b2" } } . 2. Given x + y = a, and x1 + y2 = d1, to find x and y. Ans. æ= √3a2 + d +a*,andy= α 27 4 2 α 3. In the following equations find the values of ≈ and y = 2x (x2 + y) + 506. + 1, and 1/x3y + √xy3 = 78, find x and y. = a2, and (x + y + b)2 + (x y + b)2= c2, to find x, y. 7. Find the values of x, y, z in the three equations, x (y + z) = a2, y (z + x) = b2, and z (x + y) = c2 ; And likewise also from the three, x + y + z = 10, x2 + y2 + z2 = 38, and y2 + 1 = xz. 8. Find x, y, z n the three following equations, x2 + xy + y2 = c2, x2 + xz + z2 = b2, and y2 + yz + z2 = a2. (5). v + u3 − √/ uv = 2va + ‡ (21 — u), and u3 + v3 = 6. 13. Given x + y + z = a, x2 + y2 + z2 = b2, and y = √√xz. 14. Find u, x, y, z, from the equations u + x + y + z = 15, u2 +x2 + y2 + z2 = 85, x2 = uy, and y2 = x2. 15. uv + xz = 444, ux + vz = 180, uz + vx = 156, uvxz = 5184. 16. Given x + y = a, and x5 + y5 = b5 } to find x and y in both pairs. QUESTIONS PRODUCING QUADRATIC EQUATIONS. As none of these questions are difficult of solution after the equations are obtained, it has been considered unnecessary in general to do more than form the equations and put down the answers. 1. To find two numbers whose difference is 2, and product 80. Let x and y denote the numbers: then the conditions are a xy=80. y = 2, and 8 and y = - 10. Resolving, we have x = 10 and y = 8, or x = 2. To divide the number 14 into two such parts that their product may be 48. Let x and y be the numbers: then the conditions are x + y = 14 and xy=48. The answers are 6 and 8. 3. What two numbers are those whose sum, product, and difference of their squares are all equal. Let x and y be the numbers: then x + y = which are easy to resolve, and give x = x2 — y2, xy, and x + y = 1 + √5 2 VOL. I. 4. There are four numbers in arithmetical progression the product of whose extremes is 22, and that of whose means is 40. What are those numbers? Let x be the less extreme, and y the common difference: then the numbers are x, x + y, x+2y, x + 3y. Whence the conditions are x2 + 3 xy = 22, and x2 + 3 xy + 2y2 = 40. - The answers are 2, 5, 8, 11, or — 11, -8, — 5, 2. 5. To find four numbers in geometrical progression whose sum shall be 80, and the sum of whose squares shall be 3280. Here, taking u, x, y, z for the numbers, we shall have, by geometrical progression, uy = x, z= y2; and by the given conditions u + x + y + z = 80, and u2 + x2 + y2 + z2 =3280. These equations are precisely cases of Ex. 14, p. 192, and may be resolved as those were. But the problem admits of being expressed by only two equations. For let be the first term, and u the ratio of the proportion; then x, ux, u2x, and u3x are the four numbers, and the conditions are expressed by x (1 + u + u2 + u3) 80, and 2 (1 + u2 + u1 + uo) = 3280. By geometrical progression these are convertible into or, u) +u) (1 · u1) (1 = u2 (1 u )2 (1 — u1)2 = 1 + u1 3280 6400' (1 + u) (1 + u + u2 + u3) By multiplication and transposition this becomes 39 +39u* = 0, and dividing by u2 it becomes 39 (u2 + or 39 (u + 2)2 — 82 (u + = 160. 41 ter pair of which are imaginary, and therefore imply that all the four numbers are imaginary also. The two former values of u give the four numbers 2, 6, 18, 54, and 54, 18, 6, 2. In very nearly the same way may the problem be solved when there are five numbers instead of four. 6. There is a number composed of four digits which are in arithmetical progression. The sum of the digits is 20. If 6174 be subtracted from the number, the remainder will have the same digits in an inverse order; and the product of the extreme digits is two-thirds of the product of the intermediate ones. What is the number? Let u, x, y, z be the digits in order from right to left: then 1000u + 100x + 10y + z denotes the number, and 1000z + 100y + 10x + u denotes the number when the digits are inverted. But by the question we have 1000u + 100x + 10y + z − 6174 = 1000z + 100y + 10x + u; or, 999u + 90x · 999z6174; whence The other conditions are readily formed, and are u + z = x + y, u + x + y + z = 20, and uz = xy, which admits of solution as before. 90y 7. To find a number such that if 7 be subtracted from its square, and this root be added to twice the number, the sum shall be 5. Let x be the number; then the conditions are expressed by the equation 2x + √x2-7= 5. 2x, and squaring 3x2 By transp. √2 — 7 = 5 resolved by the Indian method, gives x = 4 and x = 23. 20x320, which, By substituting these values in the equation, they are found not to fulfil the condition, though they do fulfil the condition 2x √x2-75. No numbers, indeed, can be found to fulfil the given condition; and it must be carefully borne in mind that except the expressed condition be given free from radicals, a solution, even a symbolical one, cannot be depended on without subsequent verification. This circumstance has been a source of much perplexity to mathematicians, and was never cleared up till Mr. Horner addressed a letter to the present Editor of the Course, and which was published in the Philosophical Magazine for Jan. 1836. To this letter the reader is referred, as any intelligible account of the principles of his exposition would require more space than can be allotted to it here. After the student has solved the following questions for the particular data, he should be required to solve them when, instead of the given numbers, literal symbols are substituted. He should also, as he proceeds through the numerical solutions, put down all the results, even though they be imaginary; he should be accustomed to seek the interpretation of those results, and especially to point out those which are true solutions in the form proposed, and which are solutions of some collateral problems that are involved in the same algebraical expression, as well as to assign those which involve contradictory data, and which give therefore merely symbolical results. Sometimes questions are proposed, in which, though the equation has real answers, yet the question has not, from the answers, though real, being inconsistent with some condition either implied or expressed, which is not taken into account in the equation; as when a fractional number of men, or of terms of a progression, &c. results from the question; or when the number of things to be added turns out to be subtractive, or —; and so on. In the literal solution, the student must point out the conditions that render a problem impossible; that is, which give rise to imaginary roots, or are incongruous with the ideas implied in the subject of the problem. QUESTIONS FOR PRACTICE. 1. What number being added to its square will make 42? Ans. 6, or — 7. 2. Find two numbers such, that the less may be to the greater as the greater is to 12, and that the sum of their squares may be 45. Ans. 3 and 6. 3. What two numbers are those, whose difference is 2, and the difference of their cubes 98? Ans. 3 and 5. 4. What two numbers are those, whose sum is 6, and the sum of their cubes Ans. 2 and 4. 72? 5. What two numbers are those, whose product is 20, and the difference of their cubes 61? Ans. 4 and 5. 6. Divide the number 11 into two such parts, that the product of their squares may be 784. Ans. 4 and 7*. 7. Divide the number 5 into two such parts, that the sum of their alternate quotients may be 44, that is, of the two quotients of each part divided by the other. Ans. 1 and 4. 8. Divide 12 into two such parts, that their product may be equal to 8 times their difference. Ans. 4 and 8. 9. Divide the number 10 into two such parts, that the square of 4 times the less part may be 112 more than the square of twice the greater. Ans. 4 and 6. 10. Find two numbers such, that the sum of their squares may be 89, and their sum multiplied by the greater may produce 104. Ans. 5 and 8. 11. What number is that, which being divided by the product of its two digits, the quotient is 5; but when 9 is subtracted from it, there remains a number having the same digits inverted? Ans. 32, or 23. 12. Divide 20 into three parts, such that the continual product of all three may be 270, and that the difference of the first and second may be 2 less than the difference of the second and third. Ans. 5, 6, 9. 13. Find three numbers in arithmetical progression, such that the sum of their squares may be 56, and the sum arising by adding together 3 times the first, and twice the second, and 3 times the third, may be 32. Ans. 2, 4, 6. 14. Divide the number 13 into three such parts, that their squares shall have equal differences, and that the sum of those squares shall be 75. Ans. 1, 5, 7. 15. Find three numbers having equal differences, so that their sum shall be 12, and the sum of their fourth powers shall be 962. Ans. 3, 4, 5. 16. Find three numbers having equal differences, and such that the square of the least added to the product of the two greater shall make 28, but the square of the greatest added to the product of the two less shall make 44. Ans. 2, 4, 6. 17. Three merchants, A, B, C, on comparing their gains, find that among them all they have gained 14447; and that B's gain added to the square root of A's made 9207; but if added to the square root of C's it made 9127. What were their several gains? Ans. A 400, B 900, C 144. 18. Find three numbers in arithmetical progression, so that the sum of their squares shall be 93; also if the first be multiplied by 3, the second by 4, and the third by 5, the sum of the products may be 66. Ans. 2, 5, 8. 19. Find two numbers, such that their product added to their sum may make 47, and their sum taken from the sum of their squares may leave 62. Ans. 5 and 7. 20. (1) The sum of two numbers is 2, and their product is also 2; what are they? (2) Also find two numbers whose sum is a and whose product is b2. a+√a2 — 4b2 a = √ a2 — 4b2 2 y= 2 Ans. (1) impossible; (2) x = 21. The greatest term of an arithmetical series, the common difference, and * In a great number of the following questions the answers will appear in two forms or in four forms, and it will often happen in even more numerous ones still. The present question 11 ± 3 11 3 and x= 11 ± √233 11 √233 2 has for its solutions x = Though for the most part only the positive answers are set down to enable the student to verify his actual work, he should be required to give all the solutions that the equation admits of, in symbols at least, and reduced where they admit of reduction. |