14. A man worked 10 days and his son 6, and they received $31; at another time he worked 9 days and his son 7, and they received $29.50. What were the wages of each?

15. A said to B, "Lend me one fourth of your money, and I can pay my debts." B replied, "Lend me $100 less than one half of yours, and I can pay mine." Now A owed $1200 and B $1900. each have in his possession ?

How much money did

Ans. A, $800; B, $1600.

16. If a is added to the difference of two quantities, the sum is b; and if the greater is divided by the less, the quotient will be c. What are the quantities?

17. A man owns two pieces of land. Three fourths of the area of the first piece minus two fifths of the area of the second is 12 acres; and five eighths of the area of the first is equal to four ninths of the area of the second. How many acres are there in each?

Ans. 1st, 64 acres; 2d, 90 acres.

18. A and B begin business with different sums of money; A gains the first year $350, and B loses $500, and then A's stock is to B's as 9: 10. If A had lost $500 and B gained $350, A's stock would have been to B's as 1:3. With what sum did each begin?

Ans. A, $1450; B, $2500.

19. If a certain rectangular field were 4 feet longer and 6 feet broader, it would contain 168 square feet more ; but if it were 6 feet longer and 4 feet broader, it would contain 160 square feet more. Required its length and breadth.

20. A market-man bought eggs, some at 3 for 7 cents and some at 2 for 5 cents, and paid for the whole $2.62; he afterward sold them at 36 cents a dozen, clearing $0.62. How many of each kind did he buy?

21. A and B can perform a piece of work together in 12 days. They work together 7 days, and then A finishes the work alone in 15 days. How long would it take each to do the work? Ans. A 36 and B 18 days.

22. "I was ten times as old as you 12 years ago,' said a father to his son; "but 3 years hence I shall be only two and one half times as old as you." What was the age of each?

23. If 3 is added to the numerator of a certain fraction, its value will be ; and if 4 is subtracted from the denominator, its value will be. What is the fraction?

24. A farmer sold to one man 7 bushels of oats and 5 bushels of corn for $12.76, and to another, at the same rate, 5 bushels of oats and 7 bushels of corn for $13.40. What was the price of each?

25. Find two quantities such that one third of the first minus one half the second shall equal one sixth of a; and one fourth of the first plus one fifth of the second shall equal one half of a.

Ans.

34 a

23

and

15 a

23

26. A person had a certain quantity of wine in two casks. In order to obtain an equal quantity in each, he poured from the first into the second as much as the second already contained; then he poured from the second into the first as much as the first then contained; and, lastly, he poured from the first into the second as much as the second still contained; and then he had 16 gallons in each cask. How many gallons did each originally contain? Ans. 1st, 22; 2d, 10 gallons.

10

SECTION XV.

EQUATIONS

OF THE FIRST DEGREE CONTAINING MORE THAN TWO UNKNOWN QUANTITIES.

117. The methods of elimination given for solving equations containing two unknown quantities apply equally well to those containing more than two unknown quantities.

x+y−z=4 (1) 2x+3y+4z=17 (2) 3x−2y+ 5x= 5 (3)

Multiplying equation (1) by 2 gives equation (4), which we subtract from (2), and obtain (6); multiplying (1) by 3 gives (5), and subtracting (5) from (3) gives (7). We have now obtained two equations, (6) and (7), containing but two unknown quantities. Multiplying (6) by 5, we obtain (8), and subtracting (7) from (8), we obtain (9), which reduced gives z=1. Substituting this value of z in (6), and reducing, we obtain y 3. Substituting these values

=

of y and z in (1), and reducing, we obtain x — 2.

Here we subtract (1) from (2), and obtain (6); then (6) from (3), and obtain (7); then (7) from (4), and obtain (8); then (8) from (5), and obtain (9), which reduced gives (10), or x = 9. Substituting this value of x in (1), (6), (7), and (8), and reducing, we obtain (11), (12), (13), and (14), or y = 17, z 12, w =

u = 37.

= 44, and

Hence, for solving equations containing any number of unknown quantities,

RULE.

From the given equations dedure equations one less in number, containing one less unknown quantity; and continue thus to eliminate one unknown quantity after another, until one equation is obtained containing but onc unknown quantity. Reduce this last equation so as to find the value of this unknown quantity; then substitute this value in an equation containing this and but one other unknown quantity, and reducing the resulting equation, find the value of this second unknown quantity; substitute again these values in an equation containing no more than these two and one other unknown quantity, and reduce as before; and so continue, till the value of each unknown quantity is found.

NOTE. The process can often be very much abridged by the exercise of judgment in selecting the quantity to be eliminated, the equations from which the other equations are to be deduced, the method of elimination which shall be used, and the simplest equations in which to substitute the values of the quantities which have been found.

Find the values of the unknown quantities in the following equations :

NOTE.

x + y + z +

If these equations are added together and the sum divided by 4, we shall have x + y + z + w + u = 20; and if from this the given equations are successively subtracted, the values of the unknown quantities become known.

( x = 2.