3a(n2-2n+1)+(2an+a) (n-1)+an2+3an-a

n2-2n+1

Or performing the operations indicated and reducing 6an2-4an+a a(6n2-4n+1)

= n2-2n+1

(n-1)2

which agrees with the preceding result.

This solution is more complete than the preceding, since we obtain from it the estate of the father, and the expressions for the three parts.

9. A father ordered in his will, that the eldest of his children should have a sum a, out of his estate, plus the nth part of the remainder; that the second should have a sum 2a, plus the nth part of what remained after having subtracted from it the first part and 2a; that the third should have a sum 3a, plus the nth part of the new remainder and so on. It is moreover supposed that the children share equally. Required, the value of the father's estate, the share of each child, and the number of children.

This problem is remarkable, because the number of conditions contained in the enunciation is greater than the number of unknown values required to be found.

Let the estate of the father be represented by x: then will x-a express what remains after having taken from it the sum a. Therefore the share of the eldest is

Subtracting the first part, and 2a, from x, we have

In like manner, the other parts might be formed, but as all the parts should be equal, it suffices to form the equation of the problem, to equate the two first parts, which gives

Substituting this value of x in the expression for the first part,

and as the parts are equal, by dividing the whole estate by the first part, we will obtain a quotient that will show the number of child

It yet remains to be shown, that the other conditions of the problem are satisfied; that is, that by giving to the second child, 2a plus the nth part of what remains; to the third, 3a plus the nth part of what remains, &c., the share of each child is in fact (n−1)a.

The difference between the estate of the father and the first part being a(n-1)-a(n-1), the share of the second child will be a(n−1)3—a(n—1) — 2a 2a(n-1)+a(n-1)2-a(n-1)

2a+

and reducing

n

a(n−1)+a(n−1)?

n

or

In like manner, the difference between a(n-1)2 and the two first parts being, a(n—1)3—2a(n−1), the third part will be

In the same way we would obtain for the fourth part

Hence all the conditions of the enunciation are satisfied.

10. What number is that from which, if 5 be subtracted, of the remainder will be 40?

183

Ans. 65.

11. A post is in the mud, in the water, and ten feet above the water what is the whole length of the post?

Ans. 24 feet.

12. After paying my purse: how many guineas were in it at first?

and of my money, I had 66 guineas left in

Ans. 120.

13. A person was desirous of giving 3 pence a piece to some beggars, but found he had not money enough in his pocket by 8 pence he therefore gave them each 2 pence and had 3 pence remaining required the number of beggars. Ans. 11.

14. A person in play lost of his money, and then won 3 shillings; after which he lost of what he then had; and this done, found that he had but 12 shillings remaining: what had he at first ? Ans. 208.

15. Two persons, A and B, lay out equal sums of money in trade; A gains $126, and B loses $87, and A's money is now double of B's: what did each lay out? Ans. $300.

16. A person goes to a tavern with a certain sum of money in his pocket, where he spends 2 shillings; he then borrows as much mo

ney as he had left, and going to another tavern, he there spends 2 shillings also; then borrowing again as much money as was left, he went to a third tavern, where likewise he spent two shillings and borrowed as much as he had left; and again spending 2 shillings at a fourth tavern, he then had nothing remaining. What had he at first? Ans. 3s. 9d.

Of Equations of the First Degree involving two or more unknown quantities.

95. Although several of the questions hitherto resolved, contained in their enunciation more than one unknown quantity, we have resolved them by employing but one symbol. The reason of this is, that we have been able, from the conditions of the enunciation, to express easily the other unknown quantities by means of this symbol; but this is not the case in all problems containing more than one unknown quantity.

To ascertain how problems of this kind are resolved: first, take some of those which have been resolved by means of one unknown quantity.

1. Given the sum a, of two numbers, and their difference b, it is required to find these numbers.

Let x= the greater, and y the less number.

Then by the conditions

x+y=a.

x-y=b.

2x=a+b.

2y=a-b.

Each of these equations contains but one unknown quantity.

For a second example, let us also take a problem that has been already solved.

2. A person engaged a workman for 48 days. For each day that he labored he was to receive 24 cents, and for each day that he was idle he was to pay 12 cents for his board. At the end of the 48 days, the account was settled, when the laborer received 504 cents. Required the number of working days and the number of days he was idle.

Let

x = the number of working days.

y = the number of idle days.

= 48.

n = the whole number of days
a = what he received per day for work

= 24 cts. what he paid per day for board = 12 cts.

what he received at the end of the time = 504.

what he earned,

And by = what he paid for his board.

We have by the question

{

x+y=n.

ax-by= c.

It has already been shown that the two members of an equation can be multiplied by the same number, without destroying the equality; therefore the two members of the first equation may be multiplied by b, the co-efficient of y in the second, and we have

In like manner, multiplying the two members of the first equa.

tion by a, the co-efficient of x in the second, it becomes