Vb - avb no> vb-V6.: V ve>1, and .. 1 Te>a. This second value gives a point P,, situated in the production of A B, and to *the right of the two lights. In fact, we suppose that the two lights give forth rays in all directions; there may, therefore, be a point in the production of A B equally illuminated by each, but this point must be situated in the production of A B to the right, in order that it may be nearer to the less powerful of the two lights. It is easy to perceive why the two values thus obtained are connected by the same equation. If, instead of assuming A P, for the unknown quantity x, b c we take A P3, then B P,=r-a, thus we have the equation =r ajz; but since (r—a)? is identical with (a—x)?, the new equation is the same as that already established, and which, consequently, ought to give A P, as well as A P. -ave The second value of a-X, Wo is negative, as it ought to be, being estimated in a contrary direction from the first, on the general principle already established, that quantities estimated in a contrary sense should be represented -avc with contrary signs; but changing the signs of the equation a-r=1 -=-vc avc we find x-a= noave and th -, and this value of ċ-a represents the absolute length of B P 2. II. Let b<c. The first value of x, vbi Vic" To is positive, and less than 5, for votvo Vb 1 >Vb+ vb .. Vb+ vc>2Vb, .: 7 *** Votvc 2• Votvc ave The corresponding value of a-X, Jhi to, is positive, and greater than 5 Hence the point P, is situated between the points A and B, and is nearer to A than to B. This is manifestly the true result, for the present hypothesis supposes that the intensity of B is greater than the intensity of A. - avb -avb. The second value of x, Jh to or De n is essentially negative. In order to interpret the signification of this result, let us resume the original equation, and substitute —x for +x, it thus becomes = 111z But since (a-x) expresses in the first instance the distance of B from the point required, atx ought still to express the same distance, and, therefore, the point required must be situated to the left of A, in P3, for example. In fact, since the intensity of the light B is, under the present hypothesis, greater than the intensity of A, the point required must be nearer to A than to B. -avc avc The corresponding value of a-X, vb-vior Jo-1615 the reason of this is, that r being negative, a-X expresses, in reality, an arithmetical sum. c But since III. Let b=c. The first two values of x and of a—x are reduced to 5, which gives the bisection of A B for the point equally illuminated by each light, a result which, is manifestly true, upon the supposition that the intensity of the two lights is the same. The other two values are reduced to , that is, they become infinite , that is to say, the second point equally illuminated is situated at a distance from the points A and B greater than any which can be assigned. This result perfectly corresponds with the present hypothesis; for if we suppose the difference b-c, without vanishing altogether, to be exceedingly small, the second point equally illuminated, exists, but at a great distance from the two lights ; this is indicated by the expression pression 26 J the denominator of which is exceedingly small in comparison with the numerator if we suppose b very nearly equal to c. In the extreme case, when b=c, or vb-vc=0, the point required no longer exists, or is situated at an infinite distance. IV. Let b=c and a=0. The first system of values of x and a-x in this case become 0, and the second system . This last result is here the symbol of indetermination; for if we recur to the equation of the problem b c the denominator of which (6—c).x? — 2abr=-a2b, it becomes, under the present hypothesis, 0.2-0.2=0, an equation which can be satisfied by the substitution of any number whatever for r. In fact, since the two lights are supposed to be equal in intensity, and to be placed at the same point, they must illuminate every point in the line A B equally. The solution 0, given by the first system, is one of those solutions, infinite in number, of which the problem in this case is susceptible. V. Let a=0, b not being =c. Each of the two systems in this case is reduced to 0, which proves that in this case there is only one point equally illuminated, viz., the point in which the two lights are placed. The above discussion affords an example of the precision with which algebra answers to all the circumstances included in the enunciation of a problem. We shall conclude this subject by solving one or two problems which require the introduction of more than one unknown quantity. PROBLEM 6. To find two numbers such that, when multiplied by the numbers a and b respectively, the sum of the products may be equal to 2s, and the product of the two numbers equal to p. Let x and y be the two numbers sought, the equations of the problem will be artby=2s ..................... (1) zy= p..................... (2) From (1) Substituting this value in (2) and reducing, we have ax? — 2s.c+bp=0. Whence The problem is, we perceive, susceptible of two direct solutions, for s is manifestly > V52—a’bp; but in order that these solutions may be real we must have's?>, or =a’bp. Let a=b=1; in this case the values of x and y are reduced to r=s+ V s2-p, y=sF V8-p. Here we perceive that the two values of y are equal to those of x taken in an inverse order; that is to say, if s-t v s2-p represent the value of x, then S-V -p will represent the corresponding value of y, and reciprocally. We explain this circumstance by observing that, in this particular case, the equations of the problem are reduced to rty=2s, xy=p, and the question then becomes, Required two numbers whose sum is 2s, and whose product is p, or, in other words, To divide a number 2s into two parts, such that their product may be equal to p. ........ (1) x+y=2s' .... xy=az ..... a+2+72+=4c? ................ (4) Squaring (1) and (2) and adding the results, ao +.x2 + y2 +z2+2az + 2xy=4(so+s?). But by (4), a +x2 + y2 +22 =4c. Subtracting, 2az +2ry=4(s2+'2 — (2). ... by (3), 4az=4(s+$"? —c)=4xy . . (5) Squaring (1), a’ + 2az+z?=4s. But by (5), 4az =4(s2 +8"? —c2). Subtracting, a’—2az+z:=4(co—s2). Extracting the root, a-z=+2 V6-s"? But by (1), a+z=2s. ECE (8) What two numbers are those whose sum is 20, and their product 36! Ans. 2 and 18. (9) To divide the number 60 into two such parts that their product may be to the sum of their squares in the ratio of 2 to 5. Ans. 20 and 40. (10) The difference of two numbers is 3, and the difference of their cubes is 117. What are those numbers ? Ans. 2 and 5. (11) A company at a tavern had £8 15s. to pay for their reckoning; but, before the bill was settled, two of them left the room, and then those who remained had 10s. apiece more to pay than before. How many were there in company ? Ans. 7. (12) A grazier bought as many sheep as cost him £60, and after reserving 15 out of the number, he sold the remainder for £54, and gained 2s. a head by them. How many sheep did he buy? Ans. 75. (13) There are two numbers whose difference is 15, and half their product is equal to the cube of the lesser number. What are those numbers ? Ans. 3 and 18. (14) A person bought cloth for £33 15s., which he sold again at £2 es. per piece, and gained by the bargain as much as one piece cost him. Required the number of pieces. Ans. 15. (15) What number is that, which when divided by the product of its two digits, the quotient is 3; and if 18 more be added to it, the digits will be transposed ? Ans. 24. (16) What two numbers are those whose sum, multiplied by the greater, is equal to 77, and whose difference, multiplied by the lesser, is equal to 12? Ans. 4 and 7. (17) To find a number such that, if you subtract it from 10, and multiply the remainder by the number itself, the product shall be 21. Ans. 7, or 3. (18) To divide 100 into two such parts that the sum of their square roots may be 14. Ans. 64 and 36. (19) It is required to divide the number 24 into two such parts that their product may be equal to 35 times their difference. Ans. 10 and 14. (20) The sum of two numbers is 8, and the sum of their cubes is 152. What are the numbers ? Ans. 3 and 5. (21) The sum of two numbers is 7, and the sum of their 4th powers is 641. What are the numbers ? Ans. 2 and 5. (22) The sum of two numbers is 6, and the sum of their 5th powers is 1056. What are the numbers ? Ans. 2 and 4. (23) Two partners, A and B, gained £140 by trade; A's money was 3 months in trade, and his gain was £60 less than his stock; and B's money, which was £50 more than A's, was in trade 5 months. What was A's stock ? Ans. £100. (24) To find two numbers such that the difference of their squares ma, be equal to a given number, 9%; and when the two numbers are multiplied by the numbers a and b respectively, the difference of the products may be equal to a given number, s. aso+b V 84-(a'—62)92 Ans. bs? Fa vst-la? — b4)q? a’ — 62 (25) There are two square buildings that are paved with stones a foot square each. The side of one building exceeds that of the other by 12 feet, and both their pavements taken together contain 2120 stones. What are the lengths of them separately? Ans. 26 and 38 feet. (26) A and B set out from two towns, which were at the distance of 247 miles, and traveled the direct road till they met. A went 9 miles a day, and the number of days at the end of which they met was greater by 3 than the number of miles which B went in a day. How many miles did each go? Ans. A went 117 and B 130 miles. (27) The joint stock of two partners was $2080; A's money was in trade 9 months, and B's 6 months ; when they shared stock and gain, A received $1140 and B $1260. What was each man's stock? Ans. $960 and $1120. (28) A square court-yard has a rectangular gravel walk round it. The side of the court wants 2 yards of being 6 times the breadth of the gravel walk, and the number of square yards in the walk exceeds the number of yards in the periphery of the court by 164. Required the area of the court. Ans. 256. (29) During the time that the shadow on a sun-dial, which shows true time, moves from 1 o'clock to 5, a clock, which is too fast a certain number of |