To determine with more facility the values of t", which will give entire values for x and y, we express x and y immediately in terms of t". In order to this we substitute for tits value in the equation for t, which gives t=-(3-5)+t"=6t"—3 substituting next for t and t' their values in the equation for x, and for x and their values in the equation for y, we obtain finally x = 6—11 ť" If then we make successively t"=0, 12, 3, or otherwise, t = 0, 1, — 2, — 3, in the above, we shall obtain all the entire values of x and y proper to satisfy the equation proposed. But if entire and positive values only are required for x and y, it will be necessary to give to t" such values only, as will render 6-11 t", 40 +17 t" positive. It is evident, that t" = 0, t": 2 are the only values of t", that will fulfil this condition; for, every positive value of t" will render x negative, and every negative value of t" numerically greater than 2 will render y negative. Putting therefore t" 0, 1, -2 successively, we have - 1, x= 6, 17, 28 The proposed therefore admits of three different solutions in entire and positive numbers, and of three only. 2. Let it be proposed, as a second example, to divide the number 159 into two such parts, that the first may be divisible by 8, and the second by 13. Designating by x and y the quotients, arising from the division of the parts sought by the numbers 8 and 13 respectively, we have by the question 8x + 13 y = 159. Pursuing with this equation the same process, as in the preceding example, we have the five equations Expressing x and y in terms of t", we have x= 15+ 13 t y= 3-8t!!! Here it is evident, that t"=0, and t"" =— - 1 are the only values of ""', which will give entire and positive values for x and y. Making successively t"=0, t"" = 1, we have x= 15, 2 y= 3, 11 Since then 8 x and 13 y represent the parts required, the proposed admits of two solutions, viz. 120 and 39 for the first solution, and 16 and 143 for the second. 3. It is required to divide 25 into two parts, one of which may be divisible by 2, and the other by 3. 4. A person has in his pocket pieces of 5 shillings and 3 shillings only, and wishes to pay a bill of 58 shillings. How many pieces must he give of each? 5. A sum of $81 was distributed among some poor persons, men and women; each woman received $5, and each man $7. How many men and women were there? 81. Let it be required next to solve in entire numbers the equation 49 x — · 35 y = 11. 11 Here it will be observed, that the coefficients of x and y have a common factor 7; dividing therefore both members by 7, we have 7 x—5y=1, an equation which is evidently impossible in entire numbers; the proposed therefore does not admit of entire and positive values for x and y. In general, the proposed equation being reduced to the form ax+by=c, if the coefficients of x and y have a common factor, which does not enter into the second member, the equation is impossible in entire numbers. If there be a factor, common to the coefficients of x and y, which does not enter into the second member, and this factor be not perceived at first, the course of the calculation will make known sooner or later the impossibility of solving the question in entire numbers. Applying the process explained above to the equation 49x35 y 11, we obtain finally the equation = an equation, which is evidently impossible in entire numbers for t and t, from which it is readily inferred, that the proposed will not admit of entire solutions. If the equation of the proposed question has therefore a factor common to both members, we suppress it; the coefficients of x and y will then be prime to each other, if the question admits of solution in entire and positive numbers. This being the case, the process explained above will always lead to a final equation, in which the coefficient of one of the indeterminates is equal to unity. Indeed, it will readily be perceived, that in the course pursued we apply to the coefficients of x and y in the proposed the process of the greatest common divisor; since then these coefficients are by hypothesis prime to each other, we arrive finally at a remainder equal to unity, which will be the coefficient of the last but one of the indeterminates introduced in the course of the calculation. 82. In certain cases the preceding process admits of simplifications, which it is important to introduce in practice. 1. Let it be required to solve in entire numbers the equation 5 x + 3y =49. Proceeding as before, we have 3y=49-5x 2x-1 y=16- -X 3 but the quotient on dividing 5 x by 3 being nearer 2 x, we put the equation under the form By means of the simplification, here introduced, the number of indeterminates, employed in the calculation, is one less, than would otherwise be necessary. 2. A person purchases wheat at 16s. and barley at 9s. a bushel, and pays in all 167s. How many bushels of each did he purchase? 3. To find two numbers such, that if the first be multiplied by 7 and the second by 13, the sum of the products will be 128. 4. Let it be proposed next to resolve in entire numbers the equation 13 x 57 y 101. In order that x and y in this equation may be entire num bers, 5 (y+2) must be equal to an entire number; but since 5 and 13 are prime to each other, it is necessary in order to this, that y+2 13 should be an entire number; putting t for this Here, every entire and positive value for t will give similar values for x and y; but if we suppose t 0, or to be nega The number of tive, the values of x and y will be negative. entire and positive solutions of the proposed is therefore infinite, and the smallest system of values for x and y is x= 56, y By means of the simplifications, here introduced, one indeterminate only is employed instead of 3, which would otherwise be necessary. 5. To divide 100 into two such parts, that if the first be divided by 5, the remainder will be 2; and if the second be divided by 7 the remainder will be 4. 6. To find two numbers such, that 11 times the first, diminished by 7 times the second, may be equal to 53. 7. A person purchases some horses and oxen; he pays $30 for each horse, and $23 for each ox; and he finds, that the oxen cost him $12 more than the horses. How many horses and oxen did he buy? 8. To find two numbers such, that if 8 be added to 17 times the first, the sum will be equal to 49 times the second. 9. Let it be proposed next to resolve the equation Deducing from this equation the value of x, we have x=y+ 17y+11 Here, in the expression 17y+11 it will be observed that the difference between 17, the coefficient of y, and the divisor 39, contains the other term 11 as a factor; on this account we take the quotient 56 y divided by 39 in excess, which gives x=2y 22 y 11 39 27 · 19 10. To find two numbers such, that if the first be multiplied by 11 and the second by 17, the first product is 5 greater than the second. 11. In how many ways can a debt of 546 livres be paid, by paying pieces of 15 livres, and receiving in exchange pieces of 11 livres. 12. The difference between two numbers is 309, and if the greater be divided by 37 the remainder will be 5, and if the less be divided by 54 the remainder will be 2; what are the numbers? 83. From what has been done, it will be perceived, that if the equation proposed be of the form ax+by=c, the number of solutions in entire and positive numbers will be limited; but if the equation be of the form a x-by=c, c being either positive or negative, the number of solutions will be infinite. If moreover we compare the formulas for x and y with the |