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to Trenton, or NT=68 miles, plus the distance travelled by the first courier in two hours which his departure preceded that of the second, together with the number of miles which the first travels whilst the second courier is on rout; that is, NM, or x=NT+ TR+RM.
Let us translate the two last distances, that is, TR and RM; in the first place, 2x8=16=TR= the number of miles which the first courier travels before the second sets out; then, in order to find an expression for MR; we shall say, since the distances passed over in an hour are as 8 : 12, or 2:3; as, 2 : 3 : MR
2x : x; and consequently MR= So that we obtain
3 for a translation of the enunciation,
3 by multiplication, 3x=252+2r; ...=252, that is to say, the two couriers would meet when the second shall have travelled 252 miles. In fact, while the second travelled 252 miles, the first travelled 168
2x miles; since is the expression for the number of
3 miles which the first travelled while the second was
22 2 X 252 on rout; that is substituting 252 for x;
3 3 504
- 168 miles. 3
Now, the place from whence the first courier departed, being 68 miles distant from New York, besides, he has the advantage of having travelled 10 miles before the second set out. Consequently 68 +16+168 must be equal to the number of miles which the second courier travels before they meet; that is, 68+16+168=252.
We see here an example of verification of the value of the unknown; it is a proof which the student can, and should always make.
263. In order to have a general solution of this problem. Let us therefore represent in general, by a the distance between the two places of departure, which was 68 miles in the preceding question, by b the the number of hours which the departure of the first precedes that of the second, by c the number of miles that the first courier travels per hour, and by d the number which the second travels in the same time. Let x= the distance which the second courier must travel before they meet; then, we shall have the distance travelled by the first courier during the time that the second has been travelling, by calculating the fourth term of a proportion that commences thus ;
cxx dic:: 2 :
d d The first courier travelling c miles an hour, he will have travelled.c Xb. miles before the second sets out.
Therefore by the condition of the problem, we shall have.
x= d which gives the solution of all questions of the same kind.
In order to show the use of this formula, let us resume again the preceding enunciation, and by ollecting that we must replace a by 68, 6 by 2, с by 8, and d by 12. Then the value of x becomes 12(16+68)
= 252 miles as before. ,12-8
+ bc+a; whence r=d(cb+a)
264. Such is therefore the use of these general solutions, that by substituting in the place of the letters, the numbers which they are designed to represent, and making the operations indicated by the signs, we have the answer to a particular enunciation.
Let us now suppose, in the above formula, that da c, or, what is the same thing, that both couriers go
d(dbta) over equal spaces in equal times; it becomes = c ; which signifies that they will never meet, or that the two couriers will meet at a distance greater than any given quantity whatever; this distance cannot be constructed, and we learn by this infinity that the problem proposed to be resolved is impossible, (Art. 238). This impossibility is not relative to the position of the problem (Art. 199), as it happens when the value of x is negative, it is an absolute impossibility.
But if, at the same time that c=d, we suppose that a=0,b=0, that is to say, if the couriers set out together from the same place, and travel with the same
0 velocity, we shall find x=; since here the result is , =
0 given by three hypotheses, it announces an indetermination, as we have already seen, (Art. 201); that is, x admits of an infinite number of answers.
265. Hitherto we have supposed that both couriers go the same way; if we now suppose that the two couriers set out to meet one another; by changing the sign of c (Art. 262), in the formula, and preserving the above denominations, we shall have w= da-bc)
Let, in order to verify this formula, d=c, b=0; then both couriers set out at the same time, and travel with the same velocity : we find, in this case,
which indicates that that the two couriers ought
2 to meet at one half of the distance between the places of departure.
265. We should here demand how does the value ddb ta)
which answers to the case where do 0 satisfy the above equation ; for it is an essential pro.
perty of Algebra that the symbol expressing the value of the unknown quantity, whatever it may be, being submitted to the operations indicated upon this unknown quantity, should satisfy the equation of the problem. By substituting this value of x in the equation
d we have d(db +-a)_cd(db+a)
+ octa; 0
dxo therefore, clearing of fractions,
c(db +a)=c(db +.a)+(bc+a) xdx0; but, since dx0=0, (beta) xdx0=0; .. it becomes dota=db+a, an equation whose two members are identically equal, and therefore this value of a satisfies the equation of the problem.
Prob. 27. What two numbers are those whose difference is 10, and if 15 be added to their sum, the whole will be 43 ?
Aps. 9 and 19. Prob. 29. What two numbers are those, whose diference is 14, and if 9 times the lesser be subtracted from six times the greater, the remainder will be 33 ?
Aps. 17 and 31. Prob. 29. What number is that, which being divided by 6, and 2 subtracted from the quotient, the remainder will be 2 ?
Aps. 24. Prob. 30. What two numbers are those, whose difference is 14, and the quotient of the greater divided by the lesser 3?
Ans. 21 and 7. Prob. 31. What two numbers are those, whose sum 60, and the greater is to the lesser as 9 to 3 ?
Ans. 45 and 15. Prob. 32. What number is that, which being added
to 5, and also multiplied by 5, the product shall be-4 times the sum ?
Ans. 20. Prob. 33. What number is that, which being multiplied by 12, and 48 added to the product, the sum shall be 18 times the number required ?
Ans. 3. Prob. 34. What number is that, whose part exceeds its į part by 32?
Ans. 640. Prob. 35. A Captain sends out of his men, plus 10; and there remained è, plus 15; how many bad he ?
Ans. 150. Prob. 36. What number is that, from which if 8 be subtracted, three-fourths of the remainder will be 60 ?
Ans. 88. Prob. 37. What number is that, the treble of which is as much above 40, as its half is below 51 ?
Ans. 26. Prob. 38. What number is that, the double of which exceeds four-fifths of its half by 40 ?
Ans. 25. Prob. 39. At a certain election, 946 men voted, and the candidate chosen had a majority of -558. How many men voted for each.
Ans. 194 for one, and 720 for the other. Prob. 40. After paying away ; of my money, and and then } of the remainder, I had 140 dollars left: what had I at first?
*Ans. 180 dollars. Prob. 41. One being asked how old he was, answered, that the product of as of the years he had lived, being multiplied by of the same, would be his age. What was his
Ans. 30. Prob. 42. After A had lent 10 dollars to B, he wanted 8 dollars in order to have as much money as B; and together they had 60 dollars. What money had each at first?
Ans. A 36, and B 24.