9. Pr. Two pocket-books contain together $100. If onehalf of the contents of one pocket-book and one-third of the contents of the other be removed, the amount of money left in both will be $70. How many dollars does each pocket-book contain?

Let x stand for the number of dollars contained in the first pocket-book; then the number of dollars contained in the second is 100 x. When one-half of the contents of the first and one-third of the contents of the second are removed, the 1 number of dollars remaining in the first is, and in the second 2

(100-x). By the conditions of the problem, we have x+(100-x)= 70, whence x=-20.

Substituting for x in the given equation, we obtain

x + (100+x) = 70, or (100 + x) − x = 70.

This equation corresponds to the following conditions:

If a stand for the number of dollars in one pocket-book, then 100+ stands for the number of dollars in the other; that is, one pocket-book contains $100 more than the other. The second condition of the problem, obtained from the equation, is two-thirds of the contents of one pocket-book exceeds one-half of the contents of the other by $70. Therefore the modified problem reads as follows:

Two pocket-books contain a certain amount of money, and one contains $100 more than the other. If one-third of the contents be removed from the first pocket-book, and one-half of the contents from the second, the first will then contain $70 more than the second. How much money is contained in each pocket-book?

10. These problems show that the required modification of an assumption, question, or condition of a problem which has led to a negative result, consists in making the assumption, question, or condition the opposite of what it originally was. Thus, if a positive result signify a distance toward the right from a certain point, a negative result will signify a distance toward the left from the same point; and vice versa; etc.

Zero Solutions.

11. A zero result gives in some cases the answer to the question; in other cases it proves its impossibility.

Pr. A merchant has two kinds of wine, one worth $7.25 a gallon, and the other $5.50 a gallon. How many gallons of each kind must be taken to make a mixture of 16 gallons worth $88?

Let x stand for the number of gallons of the first kind; then 16 - will stand for the number of gallons of the second kind. Therefore, by the condition of the problem, we have

7.25x+5.5(16x) = 88; whence x = 0.

That is, no mixture which contains the first kind of wine can be made to satisfy the condition. In fact, 16 gallons of the second kind are worth $88.

EXERCISE III.

Solve the following problems, and interpret the results. Modify those problems which have negative solutions so that they will be satisfied by positive solutions.

1. A and B together have $100. If A spend one-third of his share, and B spend one-fourth of his share, they will then have $80 left. What are their respective shares?

2. A father is 40 years old, and his son is 13 years old; after how many years will the father be four times as old as his son ? 3. The sum of the first and third of three consecutive numbers is equal to three times the second. What are the numbers?

4. In a number of two digits, the tens' digit is two-thirds of units' digit. If the digits be interchanged, the resulting number will exceed the original number by 36. What is the number?

5. A teacher proposes 30 problems to a pupil. The latter is to receive 8 marks in his favor for each problem solved, and 12 marks against him for each problem not solved. If the number of marks against him exceed those in his favor by 420, how many problems will he have solved?

6. In a number of two digits the tens' digit is twice the units' digit. If the digits be interchanged, the resulting number will exceed the original number by 18. What is the number?

7. A has $100, and B has $30. A spends twice as much money as B, and then has left three times as much as B. How much does each one spend?

or zero.

Discuss the solutions of the following general problems. State under what conditions each solution is positive, negative, Also, in each problem, assign a set of particular values to the general numbers which will give an admissible solution. 8. A father is a years old, and his son is b years old. After how many years will the father be n times as old as his son?

9. Having two kinds of wine worth a and b dollars a gallon, respectively, how many gallons of each kind must be taken to make a mixture of n gallons worth c dollars a gallon?

10. Two couriers, A and B, start at the same time from two stations, distant d miles from each other, and travel in the same direction. A travels n times as fast as B. Where will A overtake B?

We may substitute in this equation any particular numerical value for x, and obtain a corresponding value for y. Thus,

when x=1, y=4; when x=2, y = 3; when x =

3, y=2; etc.

In like manner the equation could have been solved for x in terms of y, and corresponding sets of values obtained.

Any set of corresponding values of x and y satisfies the given equation, and is therefore a solution.

Now, observe that equations (1) and (2) have the common solution, x = 2, y = 3. It seems evident, and it is proved in School Algebra, that these equations have only this solution. in common.

Equations (1) and (2) express different relations between the unknown numbers, and are called Independent Equations.

Also, since they are satisfied by a common set of values of the unknown numbers, they are called Consistent Equations.

3. A System of Simultaneous Equations is a group of equations which are to be satisfied by the same set, or sets, of values of the unknown numbers.

A Solution of a system of simultaneous equations is a set of values of the unknown numbers which satisfies all of the equations.

4. The examples of Arts. 1-2 are illustrations of the following general principles:

A system of linear equations has a definite number of solutions. (i.) When the number of equations is the same as the number of unknown numbers.

(ii.) When the equations are independent and consistent.

5. Two systems of equations are equivalent when every solution of either system is a solution of the other.

E.g., the systems (I.) and (II.):

3x+2y=8,

(I.)

x − y = 1,

3x+2y=8,
2x-2y=2,

(II.)

are equivalent. For they are both satisfied by the solution, x = 2, y = 1, and, as we shall see later, by no other solution.

in which the unknown number y does not appear. We say that y has been eliminated from the given equations.

7. Elimination is the process of deriving from two or more equations an equation which has one less unknown number.

Elimination by Addition and Subtraction.

8. Ex. 1. Solve the system 3x+4y= 24,

(1)

(2)

To eliminate y, we multiply the equations by such numbers

as will make the coefficients of y numerically equal.