it by the half of 4, or 2; the result is 6, according to the conditions of the problem.

Prob. 3. The sum of two numbers is 8, and their difference 2. What are those numbers?

Then x+2 will be the greater number.

The sum of these is 2x+2, which is required to equal 8.

The following is a generalization of the preceding Problem. Prob. 4. The sum of two numbers is a, and their difference b. What are those numbers?

Let x represent the least number.

Then x+b will represent the greater number.

The sum of these is 2x+b, which is required to equal a. Hence we have

By transposition,

2x+b=a.

2x=a-b,

or

a-b a b

x=

the less number.

2 2 2'

a b

α

b

Hence x+b=22+b=2+2, the greater number.

As these results are independent of any particular value attributed to the letters a and b, it follows that

Half the difference of two quantities, added to half their sum, is equal to the greater; and

Half the difference subtracted from half the sum is equal to the less.

b

a b 2 2

The expressions + and - are called formulas, because

they may be regarded as comprehending the solution of all questions of the same kind; that is, of all problems in which we have given the sum and difference of two quantities.

Given the sum of two

numbers,

12; 23;

100; their difference 50; required the numbers.

Prob. 5. From two towns which are 54 miles distant, two travelers set out at the same time with an intention of meeting. One of them goes 4 miles and the other 5 miles per hour. In how many hours will they meet?

Let x represent the required number of hours.

Then 4x will represent the number of miles one traveled, and 5x the number the other traveled; and since they meet, they must together have traveled the whole distance.

Proof. In 6 hours, at 4 miles an hour, one would travel 24 miles; the other, at 5 miles an hour, would travel 30 miles. The sum of 24 and 30 is 54 miles, which is the whole distance. This Problem may be generalized as follows:

Prob. 6. From two points which are a miles apart, two bodies move toward each other, the one at the rate of m miles per hour, the other at the rate of n miles per hour. In how many hours will they meet?

Let x represent the required number of hours.

Then mx will represent the number of miles one body moves, and nx the miles the other body moves, and we shall obviously have mx+nx=a.

This is a general formula, comprehending the solution of all problems of this kind. Thus,

We see that an infinite number of problems may be proposed, all similar to Prob. 5; but they are all solved by the formula of Prob. 6. We also see what is necessary in order that the answers may be obtained in whole numbers. The given distance (a) must be exactly divisible by m+n.

Prob. 7. A gentleman, meeting three poor persons, divided 60 cents among them; to the second he gave twice, and to the third three times as much as to the first. What did he give to each?

Let x=the sum given to the first; then 2x=the sum given to the second, and 3x=the sum given to the third.

Therefore he gave 10, 20, and 30 cents to them respectively. The learner should verify this, and all the subsequent results. The same problem generalized:

Prob. 8. Divide the number a into three such parts that the second may be m times, and the third n times as great as the first.

α

ma

na

Ans. 1+m+n' 1+m+n' 1+m+n'

What is necessary in order that the preceding values may be expressed in whole numbers?

Prob. 9. A bookseller sold 10 books at a certain price, and afterward 15 more at the same rate. Now at the last sale he

received 25 dollars more than at the first. What did he receive for each book? Ans. Five dollars.

The same Problem generalized:

Prob. 10. Find a number such that when multiplied successively by m and by n, the difference of the products shall be a.

Prob. 11. A gentleman, dying, bequeathed 1000 dollars to three servants. A was to have twice as much as B, and B three times as much as C. What were their respective shares? Ans. A received $600, B $300, and C $100.

Prob. 12. Divide the number a into three such parts that the second may be m times as great as the first, and the third n times as great as the second.

Ans.

1+m+mn 1+m+mn 1+m+mn Prob. 13. A hogshead which held 120 gallons was filled with a mixture of brandy, wine, and water. There were 10 gallons of wine more than there were of brandy, and as much water as both wine and brandy. What quantity was there of each?

Ans. Brandy 25 gallons, wine 35, and water 60 gallons. Prob. 14. Divide the number a into three such parts that the second shall exceed the first by m, and the third shall be equal to the sum of the first and second.

a 2m a+2m α

Ans.

;

4

4 2*

;

Prob. 15. A person employed four workmen, to the first of whom he gave 2 shillings more than to the second; to the second 3 shillings more than to the third; and to the third 4 shillings more than to the fourth. Their wages amount to 32 shillings. What did each receive?

Ans. They received 12, 10, 7, and 3 shillings respectively. Prob. 16. Divide the number a into four such parts that the second shall exceed the first by m, the third shall exceed the second by n, and the fourth shall exceed the third by p.

Ans. The first,

a-3m-2n-p
4

; the second,

a+m−2n−P;

4

a+m+2n-p

the third, a+m+

a+m+2n+3p 4

; the fourth, 4

Problems which involve several unknown quantities may often be solved by the use of a single unknown letter. Most of the preceding examples are of this kind. In general, when we have given the sum or difference of two quantities, both of them may be expressed by means of the same letter. For the difference of two quantities added to the less must be equal to the greater; and if one of two quantities be subtracted from their sum, the remainder will be equal to the other.

Prob. 17. At a certain election 36,000 votes were polled, and the candidate chosen wanted but 3000 of having twice as many votes as his opponent. How many voted for each?

Let x=the number of votes for the unsuccessful candidate; then 36,000-x=the number the successful one had, and 36,000-x+3000=2x. Ans. 13,000 and 23,000. Prob. 18. Divide the number a into two such parts that one part increased by b shall be equal to m times the other part.

Ans.

ma-b a+b
;
m+1 m+1'

Prob. 19. A train of cars, moving at the rate of 20 miles per hour, had been gone 3 hours, when a second train followed at the rate of 25 miles per hour. In what time will the second train overtake the first?

Let x=the number of hours the second train is in motion, and x+3=the time of the first train.

=

Then 25x the number of miles traveled by the second train, and 20(x+3)=the miles traveled by the first train.

But at the time of meeting they must both have traveled the same distance.

Proof. In 12 hours, at 25 miles per hour, the second train goes 300 miles; and in 15 hours, at 20 miles per hour, the first train also goes 300 miles; that is, it is overtaken by the second train.