whence x= x y a y 2abc y= + = 2 с 1 1 1 1 1 + -= -+ с 2abc 2abc ac + bc - ab' ab+bc-ac' ac+ab-bc 8. Suppose A completes the work in x days, B in y days, and C in z days, 1 1 1 m+1 1 1 m 1 1 n 1 1 p Then + y 2 20 2 2 y x y 2 + -= • m+1 xy+xz+yz and +-+ = 1 = p+1 xy+xz+yz” 1 XXVIII. 1. This may most readily be effected by the method of indeterminate multipliers. 3. By eliminating ≈ from the first and second equations, and z between the second and third equations, c1+by_a. Next eliminate x and y from these two equations, and aa,b,c,_bb, ac, from this may be deduced; 7. First eliminate x between the first and second equations, and between the second and third; next eliminate y between the resulting equations, and the value of z is determined: then find the values of x and y. Having found the values o x, y, z, then a*x+by+c1z may be readily shewn equal to (a3x+b2y+c2z)2. b+1 c+1 d+1 Otherwise, since y+z+u=ax, add x to each side, then x+y+z+u=(1+a)x. ; similarly for the rest of the equations; then taking x+y+z+u +a the sum, there results 1 1 6. By successive substitutions of y, z, u, in terms of x, it will be found that x=t. 7. See pp. 1, 2. XXX. See the two examples solved in the note, page 13. The general values of x and y in terms of a third unknown quantity, p, are here given, from which the numerical values of x and y can be found. =3p-2, y=2p-1. Here p must not be less than unity. 1. x= 2. x=5p-2, y=7p-3. 4. x=117p+56, y=19p+9. 6. x=17p+12, y=13p+6. 8. x=28-13p, y=9p-4. 10. x= =228-13p, y=9p—4. 12. x=5p, y=20-11p. 14. x 21-17p, y=11p+1. 16. x=17p+5, y=5-23p. 3. x=19p+6, y=3p-19. 5. x=19p-2, y=17p-4. -7. x=7p-1, y=22-13p. 9. x=34-13p, y=11p+2. 11. x=92-7p, y=7p+1. 13. x=43p+4, y=5-30p. 15. y-34-11p, x-11p-6. 17. x-7000, y=6000 is one solution of the equation. Any number can readily be found when one is known. XXXI. 1. 3x+7y+172=100, then 3x+7y=100-17%. In order that 100 -17% may be positive, z must not be greater than 5. If z be taken equal to 1, 2, 3, 4, 5, then will arise the following five different equations: 3x+7y=83, 3x+7y=66, 3x+7y=49, 3x+7y=32, 3x+7y=15. And from these equations the values of x and y may be determined. 2. 6x-4y+7z=190, 6x-4y=190-7z. In order that 190 - 7z may be a positive integer, z must be an even number less than 27. And by substituting for z, 1, 2, 3, &c., there will be found 13 values of 190 - 7%, to each of which 6x-4y may be made equal. Each of these equations will give an unlimited number of solutions. and x 20-2y+3z= - 1, 0, 1, 5. 4, 1. 3. IIence the equations admit of only three solutions in positive integers 2. 6x+7y+4=122, 11x+8y-6z=145. Eliminate z, 40x+37y=656, from which y=8-40p, x=37p+9. Here p must not be greater nor less than 0, and x=9, y=8, z=3, or the equations adinit of only one solution in positive integers. 3 and 4 require no remarks. XXXIII. 1. Let x be the number, then -1, -1 x 5 If 3-p, then 2=3p; substitute this in 2-1, then 3p-1 6p-2 5 5 is an integer, also is an integer, but op then p=5q+2. Hence x=3(5q+2). 157+6-1 15q+5 or 4 is an integer, ... 39+1 is an integer, but 44 is an integer, 4 2-1 is an integer. Let q=1=r, .'.q=4r+1, .'. x=15q+6=15(4r+1)+6= 60r +21, and in order that x may be a positive integer, r may be 0, 1, 2, &c. And the least value of x=21. The number of solutions is unlimited. 2. The number is 175. 3. The least number is 104, and the number of solutions is unlimited. x-5x-4x-3 x-2x- 1 are 2 4. If x denote the number required, then integers; the least value of x may be obtained by the same process as above in the first example, and will be found to be 59. 6. Let x, y be the two numbers, then x+y=196, also 27 - 5 then x=7p+5 and y=5q+1, ... x+y=7p+5+5q+1=196, and 7p+5q=190, which admits of 5 solutions in positive integers. x= 40, 75, 110, 145, 180. 8. Let a=mp, b=mq, then the equation ax±by=±c becomes mpx±mqy=±c, and px±qy-±, but m is not a divisor of c, .'. 3. If x be the digit in the place of tens, y in that of units, then 10x+y=6x+38, and there are nine solutions. 5. If x, y denote the numerators of the fractions, then 7. If x, y be the multiples, 355x-452y=m any number assignable at pleasure. XXXV. 1. If there be a weights of 91b. each, and y of 141b. each, then 9x+14y=2240. 2. If a denote the number of half-guineas, and y the half-crowns, then 21x+5y=400, and x=5p, y=160-21p, from which it appears p must not 2 2 exceed 7. 3. In 1999 different ways. 4. The least number required is 2 one-pound notes and 4 seven-shilling pieces. 20 25 5 5. Here 25 francs are equal to 20 shillings, .. 1 shilling is equal to or of one franc, and ... 28 shillings is equivalent to 35 francs. Let x denote the sovereigns, y the Napoleons, then 25x-20y=35, and x= =4p+3, y=5p+2, p must not be less than 0, then x=3, y=2, are the least values of x and y. The number of solutions is unlimited. 5x 2 6. It will be found that £13 6s. is the double of £6 13s. 7. Let x half-crowns and y shillings be required to pay the debt, then +y=19, or 5x+2y=39, whence x=5p+2, p not less than 1, and y=7-2p, p not greater than 3. 64 16 8. Here £100 stock in 3 per cents. =£64 sterling, ... £1 stock= £1 sterling, and £100 stock in 4 per cents. =£84 sterling, . £1 stock= 84 100 25 100 25 denote the stock in the 3 per cents., and y that in the 4 per cents., then from which x 5448-21p, p not greater than 259, y=16p-8, p not less than 1. Hence there are 259 solutions, the least is when p=1. 9. The interest of £1 in 4 per cents. is £ 4 1 " 99 24 = the interest of £1 in the 5 per Let x pounds be invested in the 4 per cents., and y in the x y 5 per cents. Then + 4, and 7x+8y=756, which gives 13 different solu tions. 24' 21 XXXVI. 1. The prime factors of 30 are 2, 3, 5. Let x, y, z be the respective numerators, 2 ข 2 53 2. Let x, y, z be the respective numerators of the fractions whose denominators are x y 2 8 6, 9, 18. Then 6 9 18 3 3x-4y+z=0, .. y=8, and z=32-3x. In order that z be positive, x must not be greater than 10, nor less than 1. 6. Suppose x months of 28 days, y of 30, and ≈ of 31, then x+y+z=12, and 28x+30y+31x=-365, if the year be not leap year. 7. Let v, x, y, z, denote the value of a ruby, sapphire, pearl, and diamond respectively. The result is ≈=96y=6x=4v, and the problem is indeterminate. If the value of any one be assigned, the values of each of the rest can be determined. 8. Let v, x, y, z denote respectively the price of a horse, camel, mule, and ox. Then 5v+2x+8y+7z=3v+7z+2y+z=6v+4x+y+2x=8v+x+3y+z. Suppose the property of each denoted by p, the values of v, x, y, z can be found in terms of p, which being unassigned, the problem is indeterminate. The least values of v, x, y, z are 85, 76, 31, 4 respectively. |