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than the numerator, and consequently the fraction itself is greater than 1, that is, the second value of x is greater than c; which shows that the second point of equal illumination is in the prolongation of A B, and on the side of the feebler light.
Second Supposition. c>0, and a < b. In this case the fractional factor of the first value of x is positive and less than $; hence, the first value of x is positive and less than $c; which shows that the first point of equal illumination is between the two lights, and nearer the feebler one.
The fractional factor of the second value of x is negative, and consequently the second value of x is negative; which shows that the second point of equal illumination is to the left of A, that is, it is on the prolongation of BA, and on the side of the feebler light.
The results of the first and second suppositions agree with each other. In both cases the first point is between the two lights, and nearer the feebler one, and the second point is in the prolongation of the line joining the lights and on the side of the feebler one.
Third Supposition. c>0, and a = b.
In this case the fractional factor of the first value of x is equal to 1, and consequently the first value of x is equal to fc; which shows that the first point of equal illumination is midway between the two lights. The fractional factor of the second value of x is
Na equal to or to ; which shows that the second
point of equal illumination is at a distance from A greater than any assignable distance, that is, there is but one point on the line of the lights that is equally illuminated by them.
The preceding results have been interpreted in accordance with the principles already explained. It is obvious that these interpretations are in strict agreement with the conditions implied in the several suppositions.
II. EQUATIONS CONTAINING TWO OR MORE
Explanation. 148. Two equations, each of which contains two unknown quantities, cannot be solved by preceding methods. Every attempt at their solution leads to an equation of the fourth degree in terms of one unknown quantity, which we have not yet learned how to solve. There are, however, special cases in which the solution can be effected.
First Special Case.
149. Having given one equation of the second degree and one of the first degree, each containing two unknown quantities, we may proceed as follows:
1. Take the two equations,
202 + 12xy + y = 85
Find the value of x in terms of y from (2), and substituting it in (1), we have,
121 – 66y +9y2 + 132y - 3642 + y2 = 85;
Second Special Case. 150. Having given two equations of the second degree which are homogeneous with respect to the unknown quantities, we may proceed as follows:
Take the equations,
Assume y=nx, n being an auxiliary unknown quantity. Substituting in (1) and (2), we have,
10 2? + no2 = 10.
.:. 22 =
n = n(n + 1):
$ ; substituting this value of n in (3), and the resulting value of a in (1), we have,
x = 2, and y = 3.
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.
Solve the following groups of equations :
151. 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 principles can be laid down for making these transformations, each case requiring to be treated in a manner peculiar to itself. A few examples are given, to illustrate some of the methods of solution.
Squaring (2), and multiplying by 2, 2xy = 800 ;
(8) adding (1) and (3),
(x2 + 2xy + y) + (x + y) = 1722 ; regarding (C + y) as a single quantity, we have, by the rule, 2 + y
- 1 V1722 + 1 = 1 + $; x + y = 41, and x + y taking the first value of x + y, and combining with (3), we have,
2C = 25,
- 42 ;
y = 16.
x + y + V x + y =
From (1) taking only the first value, we have,
- 4 + }
Væ + y = 1 + V12 + 1
x + y = 9;