greater than 1; hence, the value of x is greater than c which shows that the second point of equal illumina tion, is in the prolongation of AB, and on the side of the lesser light. These results correspond with the supposition. For, if A is the greater light, it is plain that the first point of equal illumination ought to be nearer the second light, and a little reflection will show that there ought to be a second point of equal illumination beyond the second light. Va or, co. Hence, the first value of x is equal to c, and the second is infinite which shows that the first point of equal illumination is midway between the two lights, and the second is at an infinite distance, that is, there is no second point at any finite distance. These results correspond with the supposition. For, if the lights are of equal intensity, the first point ought to be midway between them, and it is equally plain that there should be no other point on the line of the lights. 3. c > 0, and a < b. In this case, the fraction, is less than †, √a √a + √o' because the denominator is more than twice the numer. ator. The fraction, is negative, because, √a < √b. Hence, the first value of x is positive, and less than c, and the second value of x is negative: which shows that the first point is between the lights, and nearer the lesser light, and that the second point is on the prolongation of the lights to the left of A, that is, on the side of the feebler light. These results correspond with the supposition, the case being just the same as the first, the position of the lights being reversed. The coincidence of the mathematical results with those of ordinary reasoning, shows that the methods of interpretation of positive and negative symbols are correct. c = 0, a > b, α = b, and a < b. The hypothesis of c = 0, places the two lights at the same point, in which case they form one and the same light, whose intensity is equal to the sum of their intensities taken separately. The conditions of the problem involve the necessity of two lights, and the equation of the problem is found under this hypothesis. This equation ought not, therefore, to respond to the case of a single light. For, the interpretation of the results obtained from an equation, can only give the cases which fall under the hypothesis. The hypothesis of two lights, and the hypothesis of a single light, are not connected by any law which affords a common equation of condition. Hence, the results obtained on the supposition of = 0, do not belong to the problem. 10 EQUATIONS INVOLVING MORE THAN ONE UNKNOWN QUANTITY. 149. In the general case, two equations involving two unknown quantities, which are both of the second degree, cannot be solved by any of the preceding methods. Every attempt to solve two such equations, leads to an equation of the fourth degree, which as yet we have not learned how to solve. When, however, one of the equations is of the first degree, they can always be solved, and there are many particular eases in which they can be solved when both are of the second degree. Two equations containing two unknown quantities, one of which is of the first and the other of the second degree, can always be solved. 1. Take the two equations, x2 + 12xy + y2 = 85. Find the value substituting it in of x in terms of y from (2), and (1), we have, 121 — 66y+ 9y2+ 132y — 36y2 + y2 = 85; (1.) (2.) 2, and x" = 12. In like manner, other similar groups of equations may be solved. Solve the following groups of simultaneous equations: = 202 x' = 11, x" = 2. x2 - y2 } y' 10 26) = "} Ans. Ans. Ans. 15 n(n + 1); Ans. = x' = 15, x'' = −13. y' = 13, y" -15. Two equations of the second degree, can always be solved when they are homogeneous with respect to the unknown quantities. To illustrate, let us take the equations: 7. x2 + xy = 10 y2 + xy = 15 10 n+ 1 15 n(n + 1) Equating these values of x2, we find, = 9. 9, y' = 11. Assume ynx, n being an auxiliary unknown quan tity. Substituting in (1) and (2), 2. n = 10. ... 2.2 =====33 2 + n = 15. 312 = = 4. = 7. x'' = 1. y' = 5. x' 5. y' = 3. (1.) (2.) (3.) (4.) Substituting this value of n in (3), and the resulting value of x in (1), we have, 8. x2 + y2 Only a single pair of values of x and y are deduced. The complete solution would give four pairs of values. In the same way, similar groups of equations may be solved. x = = 61 xy = 6 x2 - xy + y2 = 13 y2+ x2 2xy = 5 = 29 4xy 2, and y = 3. 3x2 = 2xy + 24 y2 = xy 3 - 3y2 = 64 } 13. x2 + y2 + x + y 20 = 3. y = 4 x = 5. y = 2. 28 = 4. = 3. Many other equations of the second degree may be so transformed as to come within the rules for solution. Equations of a higher degree than the second may, also, in certain cases, be reduced, by transformation, to such forms as to come within the rules already given. No general rules can be laid down for making the transformations, each case, in general, requiring to be treated in a manner peculiar to itself. A few examples are given, to illustrate some of the methods of solution. Sx = 7. y = 4. (1.) } |