Since ¾ cannot be taken from ½ or ²/₄, add 1_d._ to both quantities, which will not alter their difference; or, which is the same thing, add 4 farthings to the first, and 1_d._ to the second. The pence and farthings in the two lines then stand thus: 7⁶/₄_d._ and 11¾_d._ Now subtract ¾ from ⁶/₄, and the difference is ¾ which must be written under the farthings. Again, since 11_d._ cannot be subtracted from 7_d._, add 1_s._ to both quantities by adding 12_d._ to the first, and 1_s._ to the second. The pence in the first line are then 19, and in the second 11, and the difference is 8, which write under the pence. Since the shillings in the lower line were increased by 1, there are now 14_s._ in the lower, and 5_s._ in the upper one. Add 20_s._ to the upper and £1 to the lower line, and the subtraction of the shillings in the second from those in the first leaves 11_s._ Again, there are now £20 in the lower, and £24 in the upper line, the difference of which is £4; therefore the whole difference of the two sums is £4. 11. 8¾. If we write down the two sums with all the additions which have been made, the process will stand thus: £24 . 25 . 19⁶/₄ 20 . 14 . 11¾ ------------------ Difference £4 . 11 . 8¾ 225. The same method may be applied to any of the quantities in the tables. The following is another example: From 7 cwt. 2 qrs. 21 lbs. 14 oz. Subtract 2 cwt. 3 qrs. 27 lbs. 12 oz. After alterations have been made similar to those in the last article, the question becomes: From 7 cwt. 6 qrs. 49 lbs. 14 oz. Subtract 3 cwt. 4 qrs. 27 lbs. 12 oz. ---------------------------- The difference is 4 cwt. 2 qrs. 22 lbs. 2 oz. In this example, and almost every other, the process may be a little shortened in the following way. Here we do not subtract 27 lbs. from 21 lbs., which is impossible, but we increase 21 lbs. by 1 qr. or 28 lbs. and then subtract 27 lbs. from the sum. It would be shorter, and lead to the same result, first to subtract 27 lbs. from 1 qr. or 28 lbs. and add the difference to 21 lbs. 226. EXERCISES. A man has the following sums to receive: £193. 14. 11¼, £22. 0. 6¾, £6473. 0. 0, and £49. 14. 4½; and the following debts to pay: £200 . 19. 6¼, £305. 16. 11, £22, and £19. 6. 0½. How much will remain after paying the debts? _Answer_, £6190. 7. 4¾. There are four towns, in the order A, B, C, and D. If a man can go from A to B in 5ʰ 20ᵐ 33ˢ, from B to C in 6ʰ 49ᵐ 2ˢ and from A to D in 19ʰ 0ᵐ 17ˢ, how long will he be in going from B to D, and from C to D? _Answer_, 13ʰ 39ᵐ 44ˢ, and 6ʰ 50ᵐ 42ˢ. 227. In order to perform the process of MULTIPLICATION, it must be recollected that, as in (52), if a quantity be divided into several parts, and each of these parts be multiplied by a number, and the products be added, the result is the same as would arise from multiplying the whole quantity by that number.