y=3, 6, 9. (7) x=11, 34, 57, &c.) y= 3, 11, 19, &c.f (8) x=29, 85, 141, &c.) y=20, 59, 98, &c.f (9) x=5, 36, 67, &c.) y=3, 23, 43, &c. f (10) 23, or 23+30t, for all positive integral values of t. (11) 173, or 173+385t, .... (12) 43. (13) x=15, 50, y= 1, 13 calves. y=82, 40, 2= 2. z=15, 30. (14) x= 5, 10, 15,) (16) 72, 70. y=42, 24, 6, (17) 269. 11 women, 16 children. 2=53, 66, 79. (18) 57. (21) 161. 566 (1) x=24-5t, y=11t-2.J (2) x=13-71) y=3t. y=10, 2, 1.j (22) x=2, 6, 8, 36.) y=36, 8, 6, 2.) (23) x=12, 2.1 y= 1, 3.) (5) x=2+9t, (24) x=3.) (14) 9. y=1+13t. y=4.S (6) x=19t-16,| y=271-25. (16) 4. (17) 4. (7) x=131-12, y=4t-1, (18) 3. y=6.5 (26) _x= 4, 5.1 y=21, 7.) (27) x=1,3,5,9,13.) y=10,6,4,2,1.J NOTES. NOTE 1. (p. 49.) In Ex. 7 it will not fail to be observed that the remainder is the same expression as the dividend, with a written in the place of x: this we shall shew to be true in all such cases. Let it be required to divide the expression Pox”+P1x”¬1+P¿x”¬2+.. ....+Pm-1x+Pn by x-a, and let Q be the quotient and R the remainder: then, writing for the sake of convenience f(x) for the dividend, we have, f(x)=Q.(x−a)+ R. Now R cannot contain a, for otherwise the division would not have been completely performed, and therefore R remains unaltered whatever value a assumes: also Q being evidently of the form i.e. R=f(a), the same expression as the dividend, with a written in the place of x. In order to find the quotient, we have f(x)=Q(x-a)+R, i.e. P ̧x”+P1x2¬1+P2x”~2+...+Pn-1x+P1= (¶。x2¬1+q1x"¬3+...+¶„−1)(x− a)+R, identically; .. performing the multiplication, and equating the coefficients of the several powers of x in each member of this equality, we get |