Now to clear this of fractions, multiply by 6, and 6x=3x+2x+72; by 'transposition, 6x-5x=72; ... x= =72. It can be readily proved that 72 is the number re quired; thus, 7272 +=+12=36+24+12=72. 2 3 All other problems in this Section may be proved in like manner. Prob. 18 To find a number whose half minus 6, shall be equal fo its third part plus 10. Let x= the number required; x x then by the problem, 6= +10, 2 3 .. clearing of fractions, 3x-36=2x+60, by transposition, 3x-2x=60+36, ... x=96. Prob. 19. Two persons, A and B, set out from one place, and both go the same road, but A goes a hours before B, and travels n miles an hour, B follows and travels m miles an hour. In how many hours, and in how many miles travel will B overtake A. Let the hours that B travelled; then x+a=the hours that A travelled. Also mx the number of miles travelled by B; and n(x+a)=nx+na= the miles travelled by A. .. by the problem, mx=nx+na; by transposition, mx-nx=na, 261. This is a general or literal solution, because m, n, a, may be any numbers or quantities taken atpleasure: for example, Let a 9, n=5, and m=7; Then, A travels 9 hours at the rate of 5 miles an hour, before B sets out; and B follows after at the rate of 7 miles an hour. Now, by putting these values of a, n, and m, in the formula, found above; we have, by A. And mx=7×221-1571, the miles travelled by each. Again, suppose a=10, m=4, and n=4; then x= na 4 X 10 40 = = m-n 4-4 0 =∞, (Art. 165); hence, we con clude that the time is infinite, or that A will never overtake B, except at an infinite distance; because mx=4x 40 160 0 0 ∞; which would also appear evi dent without the aid of analysis. Now, if a=0, m=4, and n=4, then x= 4X0 0 0 0 na mn ; which is the mark of indetermination, (Art. 201), as it should be; since, in this case, A and B, setting out together and both travelling uniformly the same number of miles in every hour, must be together at any distance whatever from the place of depar ture. Prob. 20. Four merchants entered into a speculation, for which they subscribed 4755 dollars; of which B paid three times as much as A; C paid as much as A and B ; and D paid as much as C and B. What did each pay? Here, if we knew how much A paid, the sum paid by each of the rest could be easily ascertained. Let, therefore, x= number of dollars A paid; 3x number B paid; = 4x number C paid; and 7x number D paid; •• (x+3x+4x+7x=)15x=4755, and x=317. .. they contributed 317, 951, 1268, and 2219 dollars respectively. Prob. 21. Let it be required to divide 890 dollars between three persons, in such a manner, that the first may have 180 more than the second, and the second 115 more than the third. Here, it is manifest that if the least or third part were known, the remaining parts could be casily ascertained; therefore, Let the least or third part ... the greatest or first part. and part. .. 3x+115+115+180=890, or 3x+410-890; ... by transposition, 3a=890-410, or 3x=480, ..x= =160= least part. .. +115=160+115=275 second part. +115+180=160+115+180-455= greatest Prob. 22. A prize of 2329 dollars was divided between two persons A and B, whose shares therein were in proportion of 5 to 12. What was the share of each ? Let 5x-A's share; then 12x-B's share; ..5x+12x=3329, or 17x=2329; and x=137. ... their shares were 685 and 1644 dollars respectively. Prob. 23. A Fish was caught, whose tail weighed 9lbs.; his head weighed as much as his tail, and half his body; and his body weighed as much as his head and tail. What did the fish weigh Let 2x= the number of lbs. the body weighed ; then 9+x=the weight of the tail; .. 9+9+x=2x; by transposition, x=18; .. the fish weighed 36+27+9=72lbs. Prob. 24. A hare, 50 of her leaps before a greyhound, takes 4 leaps to the greyhound's three; but two of the greyhound's leaps are as much as three of the hare's. How many leaps must the greyhound take to catch the hare? Let 3x the number of leaps the greyhound must take; . 4x the number the hare takes in the same time, the whole number she takes, .. 4x+50 and 23: 3x: 4x+50; ..9x=8x+100; by transposition, x= =100, and the greyhound must take 300 leaps. Prob. 25. The number of soldiers of an army is such, that its triple diminished by 1000, is equal to its quadruple augmented by 2000. What is this number? Let x designate the number required; then, we are conducted to this equation, 3x-1000-4x+2000, whence x= =—3000, which gives an absurd answer with respect to the terms of the question, since that a number of soldiers cannot be negative. 262. We shall render this impossibility very plain, by observing that the triple of a number being less. than the quadruple of the same number, the triple diminished by 1000 is much less than the quadruple augmented by 2000. But by writing -x in the place of +x in the equation of the problem, then changing the signs of both sides, we find 3x+1000-4x-2000; .' x=3000. We can form the equation 3x+1000=4x-2000, re-establish the enunciation of the problem in such a manner that there results from the solution an absolute number, that is, x=3000. If in place of taking x for the representation of the unknown number, we had taken x'-6000, or x=x'-6000, we should find for the equation 3x-19000-4x-22000; by transposition, 22000-19000=4x-3x', A and ... x'=3000 as before. Thus the value x=-3000 being represented, on a line, by the length A'M, counted from A' towards M, or to the left of A', we pass by the substitution x= x-6000 from the origin A' to the origin A, to the left of A', and distant from A' by 6000-2A'M; then the length AM=x' is positive. Prob. 26. A Courier sets out from Trenton to Washington, and travels at the rate of 8 miles an hour; two hours after his departure another Courier sets out after him from New-York, supposed to be 68 miles distant from Trenton, and travels at the rate of 12 miles an hour. How far must the second Courier travel before he overtakes the first? Let x represent the number of miles which the second courier travels before he overtakes the first: then, by a little attention, we discover that this distance should be equal to the distance from New-York |