examined the meaning and conditions of the problem, translate the several conditions into as many distinct algebraic equations. and, finally, by the resolution of these equations according to the rules laid down in Chapter IV, the quantities sought will be determined. It is proper to observe that, in certain cases, other methods of proceeding may be used, which practice and observation alone can suggest.

PROBLEM I.

There are two numbers, such, that three times the greater added to one-third the lesser is equal 36; and if twice the greater be subtracted from 6 times the lesser,and the remainder divided by 8, the quotient will be 4. What are the numbers?

Let x designate the greater number, and y the lesser number,

Then 3x+1=36,

and 6y¬2x=4;

8

S9x+y=108(A), 6y-2x= 32(B);

but 6y-2x=

32;

Multiplying equation (A) by 6, 6y+54x=648;

.. by subtraction, 56x=616, and by division, x=11.

From equation (A), y=108-9x ;

... by substitution, y=108-99, or y=9.

Prob. 2. After A had won four shillings of B, he had only half as many shillings as B had left. But had B won six shillings of A, then he would have three times as many as A would have had left. How many had each?

Let x

and y

designate the number of shillings A had, the number B had;

then y-4=2x+8,

and y+6=3x-18;

.. by subtraction, 10-x-26,

and by transposition, 36=x, or x=36; by substitution, y-4-3X36-18; and by transposition, y=84;

.. A had 36, and B 84.

Prob. 3. What fraction is that, to the numerator of which if 4 be added, the value is one-half, but if 7 be added to the denominator, its value is one-fifth?

Add 7 to the denom", then =},.. 5x=y+7;

4+7

by subtraction, 3x-8=7;

by transposition, 3x=15; .. x=5;

and y=2x+8; .. by substitution, y=10+8=18,

Prob. 4. A and B have certain sums of money, says A to B, give me 151 of your money, and I shall have 5 times as much as you have left says B to A, give me 51 of your money, and I shall have exactly as much as you will have left. What sum of money had each?

B.

Let x A's money, then x+15= what A would y= B's, Shave,after receiving 18/ from

Again, y+5

51 from A.

y-15 what B would have left. what B would have after receiving

x-5 what A would have left.

Hence, by the problem, x+15=5×(y−15)=5y

-75,

and y+5=x-5.

By transposition, 5y-x=90, and y--10;

... by subtraction, 4y=100, and by division, y=25 B's money,

From the second equation, x=y+10;

.. by substitution, x=25+10=35 A's money.

Prob. 6. A person was desirous of relieving a certain number of beggars by giving them 2s. 6d. each, but found that he had not meney enough in his pocket by 3 shillings; he then gave them 2 shillings each, and had four shillings to spare. What money had he in his pocket; and how many beggars did he relieve?

money in his pocket (in shillings); y= the number of beggars.

Let x

and 2xy, or 2y=

number of shillings which

;

would have been given at 2s. 6d. each

at 2s each.

5y

Hence, by the problem, x+3(A),

2

or y 14, the number of beggars.

From equation (B), x=2y+4=2×14+4, by substitution, .. x=32, the shillings in his pocket.

Prob. 6. There is a certain number, consisting of two digits. The sum of those digits is 5; and if 9 be added to the number itself, the digits will be inverted. What is the number?

Here it may be observed, that every number consisting of two digits is equal to 10 times the digit in the tens place, plus that in the units; thus, 24=2× 10+4=20+4.

Let x digit in the units place;

y= that in the tens.

Then 10x+y= the number itself,

and 10y+x the number with its digits inverted. Hence, by the problem, x+y=5(A),

and 10x+y+9=10y+x, or by transposition, 9x-9y =-9;.. by division, x-y=−1(B).

Subtracting equation (B) from (A), 2y=6;

..y-3, and x=5-y=5-3=2; ... the number is (10x+y)23. Add 9 to this number, and it becomes 32, which is the number with the digits inverted.

Prob. 7. A sum of money was divided equally amongst a certain number of persons; had there been four more, each would have received one shilling less, and had there been four fewer, each would have received two shillings more than he did required the number of persons, and what each received.

Let a designate the number of persons, y the sum each received in shillings;

then xy is the sum divided:

by the question;

.. (x+4)×(y-1)=xy,

and (x-4)X(y+2)=xy, S

.. xy+4y-x-4=xy, or

4y-x=4,

and xy-4y+2x-8=xy, or -4y+2x=8;

... by addition, x=12;

and 4y=4+x=4+12; .. y=4.

Prob. 8. A man, his wife, and son's years make 96, of which the father and son's equal the wife's and 15 years over, and the wife and son's equal the man's and two years over. What was the age of each?

Suppose x, y, and z = their respective ages. 1st condition x+y+z=96,

2nd

3d

x+2y+15,

• y+z=x+2,

by the problem.

Subtracting the 2nd from the 1st, y=96-y-15; 2y=81, and y=401 by division. Subtracting the 3d from the 1st, x=96-x-2; .. by transposition and division, x=47. And from the 1st, z=96-y-x; ··· z=81. And their ages are 47, 40, and 8 respectively.

Prob. 9. A labourer working for a gentleman during 12 days, and having had with him, the first seven days, his wife and son, received 74 shillings; he wrought afterwards 8 other days, during 5 of which he had with him his wife and son, and he received 50 shillings. Required the gain of the labourer per day, and also, that of his wife and son.

Let x the daily gain of the husband, y= that of the wife and son;

12 days work of the husband would produce 12x, 7 of the wife and son would be 7y;

.. by the first condition, 12x + 7y= 74; and by the second, Sx+5y= 50;

Multiplying the 1st equation by 2, 54x+14y=148;

2nd

by 3, 24x+15y=150;

.. by subtraction, y=2.

And from the 2nd, 8x=50-5y=50—10;

.. by division, x=5. Consequently the husband would have gained alone 5s. per day, and the wife and son 2 shillings in the same time.

267. Let us now suppose, that the first sum received by the workman was 46s, and the second 30s, the other circumstances remaining the same as before; The equations of the question would be

12x+7y=46, and 8x+5y=30.

From whence we find, by proceeding as above, x=5, and y=-2.

By putting in the place of its value 5, in the above equations, they become

60+7y=46, and 40+5y=30.

The inspection alone of these equations show an absurdity. In fact, it is impossible to form 46 by adding an absolute number to 60, which is already greater than it; and in like manner it is impossible to form 30 by adding an absolute number to 40.

Consequently what we attributed as a gain to the labour of the wife and son, must be an expense to the husband, which is also verified by the result y=—2.