30. A person's entire indebtedness to A, B, and C, was $270. His indebtedness to B was twice as much as to A, and his indebtedness to C was twice as much as to A and B.

did he owe each?

How much Ans A, $30; B, $60; C, $180. 31. A company of 4 laborers received $315. B received 11 times as much as A, C received 13 times as much as A and B, and D received 14 times as much as A, B, and C. What did each laborer receive? Ans. A, $24; B, $36; C, $80; D, $175.

NOTE.-Let 6x represent A's share, and 9x B's share.

32. A gamester, after losing of his money, won 4 shillings; he then lost of what he had, and afterward won 3 shillings; he then lost of what he had, and found that he had only 20 shillings remaining. How much had he at first? Ans. 30 shillings. 33. A gentleman spends of his yearly income for the support of his family, and of the remainder in improvements on his premises, and lays by $70 a year. What is his income?

Ans. $630.

34. Divide the number 60 into two such parts that the product of the two parts may be equal to 3 times the square of the less part. Ans. 15 and 45.

35. My horse and saddle are together worth horse is worth 8 times as much as my saddle. of each ?

90 dollars, and my What is the value

Ans. Saddle, $10; horse, 80.

36. Divide $462 between two persons, so that for every dime which one receives, the other may receive a dollar,

Ans. $42 and $420. 37. The rent of an estate is 8 per cent. greater this year than fast. This year it is 1890 dollars; what was it last year ?

Ans. $1750.

38. The sum of two numbers is 840, and their difference is equal to of the greater. What are the numbers?

Ans. 504 and 336.

39. A person, after spending 100 dollars more than of his income, had remaining 35 dollars more than of it. Required his income. Ars. $450.

40. Divide $1520 among A, B, and C, so that B shall have $100 more than A, and C $270 more than B.

Ans. A, $350; B, $450; C, $720.

A contracts an annual

41. A and B have the same income. debt amounting to of it; B lives upon of it; at the end of two years B lends to A enough to pay off his debts, and has 32 dollars to spare. What is the income of each ? Ans. $280.

42. A sets out from a certain place, and travels at the rate of 7 miles in 5 hours; and 8 hours afterward B sets out from the same place in pursuit, at the rate of 5 miles in 3 hours. How long and how far must B travel before he overtakes A? Ans. 42 hours; 70 miles.

5 x 43. A can perform a certain piece of work in 8 days, and B can do the same in 12 days; in how many days can both, working together, do it? Ans. 4.

44. A person has just 6 hours at his disposal; how far may he ride in a coach which travels 8 miles an hour, that he may return home in time, walking back at the rate of 4 miles an hour?

Ans. 16 miles.

45. A can dig a trench in one half the time required by B, B can dig it in two thirds of the time required by C, and all together can dig it in 6 days; find the time that each alone would require. Ans. A, 11 days; B, 22 days; C, 33 days.

46. A and B start from opposite points and travel toward each other, A at the rate of 3 miles an hour, and B at the rate of 4 miles an hour. At the same time, C sets out with A and travels at the rate of 5 miles an hour. After meeting B he turns back and travels until he meets A; he then finds that the whole time elapsed since starting is 10 hours. How far apart were A and B at the beginning? Ans. 72 miles.

47. Two farmers owning a flock of sheep, agree to divide them. A takes 72 sheep; B takes 92 sheep, and pays A $35. Required the value of the flock. Ans. $574.

48. A crew which can row at the rate of 12 miles an hour in still water, finds that it takes 7 hours to come up a river a certain distance, and 5 hours to go down again. river flow?

At what rate does the Ans. 2 miles per hour.

SIMPLE EQUATIONS

CONTAINING TWO UNKNOWN QUANTITIES.

161. We have seen that every simple equation containing only one unknown quantity can be satisfied with one value, and only one value, of the unknown quantity (155). But if we consider a single equation containing two unknown quantities, we shall find that for every value which we please to give to one of the unknown quantities, we can determine a corresponding value of the other unknown quantity, such that the set of values will satisfy the equation.

Put x=1, and substitute this value in the given equation; we have 2 + 3y = 17,

y= 5.

Now the set of values, x = 1, y = 5, will satisfy the equation; for, by substitution, we have

2+1517.

In the same manner, we may obtain the following sets of values, each one of which will satisfy equation (1):

It is evident that there is no limit to the number of sets of values that may be obtained. The equation, and also the quantities, in such cases, are said to be indeterminate. Hence,

162. An Indeterminate Equation is one which is satisfied by an infinite number of values of the unknown quantities. A single equation containing two unknown quantities is indeterminate.

163. If we take two equations with two unknown quantities, as 2x+5y31... (1),

3x+2y=19... (2),

it is evident that we may obtain as many sets of values as we please,

which will satisfy each equation, considered separately. Thus, proceeding as before, we find that the set,

will satisfy the first equation; and a different set,

X= 4,

will satisfy the second equation.

y = 31,

Now suppose we are required to satisfy both equations with the same set of values for x and y.

Multiplying (1) by 3, and (2) by 2,

which will satisfy both equations. For, let these values be substituted in the given equations; we shall have

62531,

9+10=19.

Equations thus related are said to be simultaneous. Hence, 164. Simultaneous Equations are those which must be satisfied by the same values of the unknown quantities which enter them.

When two or more simultaneous equations are given, the values of the unknown quantities are determined by a process called

ELIMINATION.

165. Elimination is the process of combining equations in such a manner as to cause one or more of the unknown quantities contained in them to disappear.

There are four principal methods of elimination:

1st, By addition and subtraction; 2d, By comparison; 3d, By substitution; 4th, By indeterminate multipliers

CASE I.

166. Elimination by addition and subtraction.

1. Given 3x+2y= 23, and 4x-3y= 8, to find the values

of x and y.

OPERATION.

3x+2y= 23... (1),

4x-3y 8... (2).

=

Multiplying (1) by 4, and (2) by 3, 12x+8y = 92... (3),

12x9y24... (4);

subtracting (4) from (3),

whence,

17y68... (5);

y= 4... (6).

Thus, we have eliminated x, and found the value of y.

Again, multiplying (1) by 3, and (2) by 2,

adding (7) to (8), whence,

9x+6y=69... (7),
8x6y 16... (8);
17x=85;

x = 5.

We have thus eliminated y, and found the value of x. Hence,

RULE.-I. Multiply or divide the equations by such numbers or quantities that the coefficients of the quantity to be eliminated shall be made equal in the two equations.

II. If these coefficients have like signs, subtract one of the prepared equations from the other, member from member; if they have unlike signs, add the equations, member to member.

NOTES.-1. In preparing the given equations by multiplication, find the least common multiple of the coefficients of the letter to be eliminated, and divide this multiple by each coefficient; the quotients will be the least multipliers that can be used. If the coefficients are prime to each other, it is evident that each equation must be multiplied by the coefficient in the other equation.

2. It is generally convenient to clear the equations of fractions, before applying the rule. This is not necessary, however. For if any letter has fractional coefficients in the two equations, the fractions may be reduced to a common denominator; it will then be necessary to render the numerators equal by multiplication or division, according to the rule.