I'.RA. 



ft Tind tho cube root of * - &e* + l&t* - 2(Xi* -H&* - <te + 1. 

 U.1 the square root of 4* - 4r + 13e - 6* + 8. 

 1 tli.- 4!h root of 10a*- v '*- 21" 



.it of a 1 + 5x + lOjr* 4- 10x + &*+!. 



t of a 8 - Oa'O -f I5au 1 - 20 A + 15a6 - CoW + t*. 



INDETERMINATE EQUATIONS. 



\Vheiitheroaro more unknown quantities than independent. 



ho number of corresponding values which those 



<inantities admit is indefinite. This number may bo loSHi-ned l.y 



rejeeting all tho values which are not integers, and it may bo 



further lessened by rejecting all the negative values. 



An indeterminate equation of two unknown quantities of tho 



first degree is of tho form mx + ny = d ; and hero it may bo 



. d that this equation cannot be solved in whole numbers 



unless m and n are prime to each other, and that if one solution 



n or found, all tho others may bo derived from it. 

 In tho equation mx ny = + d, the solution in whole num- 

 always possible if m and n be prime to each other, and 

 then an indefinite number of integral values may be assigned to 



d y, which satisfy tho equation. 



In tiio equation mx + ny = d, tho solution in whole numbers 

 is always possible provided d bo greater than mn - m - n. 



EXAMPLES. 



1 . In how many ways can a guinea bo paid by using shillings 

 and crown pieces only ? 



Lot = the number of shillings, and y =-= the number of 

 crowns, then * + 5y =21 (taking all tho values in shillings); 

 /. x = 21 5;/, where wo see that to know the value of x we may 

 assume y at pleasure ; but since x must be a positive integer 

 from the nature of the problem, it is natural to assume y equal 

 to tho consecutive whole numbers 1, 2, 3, etc. 



If y = 1, 2, 3, 4, then x = 1G, 11, 6, 1 ; and if y be taken 

 = 5, this gives x a negative value ; therefore the problem 

 admits only of the above four solutions, and it is easy to test the 

 correctness of tho values thus found. 



1 crown and 16 shillings; 2 crowns and 11 shillings; 3 

 crowns and 6 shillings ; 4 crowns and 1 shilling, all equal 

 1 guinea, as required. A>is. 



2. Tho difference between the values of ^ of 5 Ibs. of salt 

 butter, and } of 2 Ibs. of fresh ditto, is J of a penny. Find the 

 price per Ib. of each in integers ? 



Let x = the price per Ib. of the salt butter, 

 And y = ,, fresh butter; 



Then, by the question, - ( - J = \. 

 12 7 



Now, multiplying by 84 (the least common multiple of 12, 7 



and 4), wo have 35a; - 24y = 21 ; from which y ; 



24 



and here it may be remarked that it is best to find the value of 

 that unknown which has tho Jeasf co-efficient (in this case y). 



Now -*" = x -\ (A). Now, since x is to be an 



integer, must be one also ; /. put - = m, whence 



- 1 24 



= 2m + 1 + 2n - " f - 10 (B). Now, for a similar 

 ?-_MO = n> ^ Q m m lln-10 = 5n _ 5 



+ - (C). Now assume _ = r, then n = 2r. Remembering that 



all tho indeterminates, y, x, m, n, have an integral value, we 

 retrace our steps by substitution, and have from (C), substi- 

 tuting 2r for n, m lOr - 5 + r = llr - 5 ; from (B), by 

 substituting llr-5 for m, x = 22r 10 + 1 + 2r = 24r 9; 

 from (A), by substituting 24r 9 for x, y = 24r 9 + llr - 5 

 = 35r 14. The two unknowns in terms of r are, therefore, 

 x = 24r 9, and y = 35r 14, from which we see that any 

 positive integer, from 1 to infinity, for the value of r, will give 

 positive integers for the values of x and y. Thus, 

 If r 1, 2, 3, 4, etc., increasing by 1 ; 

 x = 15, 39, 63, 87, etc., increasing by 24 ; 

 y = 21, 56, 91, 126, etc., increasing by 35. 



Observe that tho different values of x and y form arithmetical 

 progressions, whose common differencss are 24 and 35 respec- 



tively : that formed by tho value* of z has the co-efficient of y 

 (in the given equation) for iU common difference; and toe one 

 formed by the value* of y lout the oo-efflcict.: of t tor its 

 common difference. Wo ohJo nee from these two example*, that 

 when tho two unknown* have the tame sign, the number of 

 solutions in limited ; but that the number is unlimited whan 

 they have different sign*. 



3. At a country railway station the money taken WM 3. 

 Tho tickets iasuod were for fare* of 2*., SB., and 5*. each. How 

 many of each were issued t also find how many solutions the 

 question admits of. 



Let x, y, * represent the numbers of the different kinds; then 



, - 



by tho question, 2* + 3y -f 5* = 60 ; whence * = - 



= 30 - y - 2z - 1L+..' ; put ? = t ; then y = 2t - *, and 



* = 30-y-2z-( = 30-3-z; if * = 1, then x = 29 - 3, and 

 y = 2t - 1 ; to make x and y whole numbers, t cannot be mort 

 than 9 nor less than 1 (thus giving 9 solutions). 



If t = 1, 2, 3, 4, 5, 6, 7, 8, 9; 



x = 26, 23, 20, 17, 14, 11, 8, 5, 2; 

 y = 1, 3, 5, 7, 9, 11, 13, 15, 17 ; 

 =1, 1, 1, 1, 1, 1, 1, 1, 1. 



If z = 2, then ar= 28 - 3t, and y = 2t - 2 ; t cannot be greattr 

 than 9 nor less than 2. 



If t = 2, 3, 4, 5, 6, 7, 8, 9 ; 

 ar = 22, 19, 16, 13, 10, 7, 4; 

 y = 2, 4, 6, 8, 10, 12, 14, 10; 

 * = 2, 2, 2, 2, 2, 2, 2, 2; 



thus giving 8 more solutions. 



If a = 3, x = 27 - 3, and y = 2t - 3, from which we see t 

 cannot bo more than 8 nor less than 2 ; thus giving 7 more 

 solutions. 



If 2 4, x = 26 - 3t, and y = 2t - 4, where t cannot bo nor* 

 than 8 nor less than 3 ; thus giving 6 more solutions. 



If z = 5, x = 25 - 3t, and y = 2t - 5, where t cannot be more 

 than 8 nor less than 3 ; thus giving 6 more solutions. 



If 2 = 6, x = 24 3t, and y = 2t - 6, where t cannot bo tnor 

 than 7 nor less than 4; thus giving 4 more solution 



If z 7, x = 23 - 3t, and y = 2t 7, where t can: 

 than 7 nor Jess than 4 ; thus giving 4 more solutions. 



If z = 8, x = 22 - 3<, and y = 2t - 8, where t cannot bo more 

 than 7 nor Jess than 5 ; thus giving 3 more solutions. 



If z = 9, then x = 21 - 3t, and y = 24 - 9, where t cannot be 

 more than 6 nor Jess than 5 ; thus giving 2 more solutions. 



If z = 10, then x = 20 - 3t, and y = 2t - 10, where t can oni;/ 

 be 6 ; thus giving 1 more solution. 



If z = 11, then x = 19 - 3t, and y = 2t - 11, where t can only 

 be 6 ; thus giving 1 more solution. 



From the given equation, 2x + 3y -f 5z = 60, we eoe z cannot 

 be more than 11 ; hence there are 9 + 8 + 7 + 6 + 0-1-4 + 4 

 + 3 + 2 + 1 + 1 = 51 solutions to this question. 



From the foregoing examples we may deduce the following 



Rule. If a simple equation express the relation of two 

 unknown quantities, and their corresponding integral values be 

 required, divide the whole equation by the co-efficient which is 

 the Jess of the two, and suppose that part of the result which is 

 in a fractional form equal to some whole number ; thus a new 

 simple equation is found, with which we must proceed as before, 

 and so on till tho co-efficient of one of the unknown quantities 

 is 1 , and the co-efficient of the other a whole number ; then an 

 integral value of the former may be found by substituting 0, or 

 any whole number for the other; and from the preceding 

 equations integral values of the original unknown quantities 

 may be found. 



EXERCISE 74. 



1 . Find tho number of solutions, and the rcspectiTO values of t, v. and 

 z, in tho equation SJE + 8y + 7* = 50. 



2. Find the number of solutions, and tho values of r, v, and , in the 

 equations 



(5* + 7y - 3i = 10. 

 "(o.r - -2j + 4i = IS. 



3. Find tho number of solution*, and the values of x, y, and m, in 

 the equations 



+ y + * + m = 1C, 



6y. 



