39. An undergraduate rowed down the river a distance of 11 miles in 1 hrs. with the stream, and on his return met the same stream, and with a uniform stroke throughout he rowed back again in 32 hrs. Find the rate of the stream per hour.

40. A can dig a trench in 5 hours, B in 9 hours, and C in 15 hours. How long would A and B take to dig the trench, C working for 1 hour?

41. Two ladies A and B were talking of their ages. A says to B, the sum of our years is 90: and your age is now th of my age; how long is it since my age was 3 times your age?

42. Two passengers have together 5 cwt. of luggage, and are charged for the excess above the weight allowed 5s. 2d. and 9s. 10d. respectively; had the luggage all belonged to one of them he would have been charged 19s. 2d.; how much luggage is each passenger allowed without charge?

43. Two persons walk at the rate of 5 and 6 miles an hour respectively, they set out to meet each other from two places 22 miles apart; having passed each other once, find the place of their second meeting, supposing them to continue their journey between the two places. Also find the time when the second meeting takes place.

44. A farmer has a certain number of hurdles; he finds that by arranging them so as to enclose a given space of ground, if he place them one foot distant from each other, he has not enough hurdles by 80; but if he place them a yard apart he has 50 hurdles to spare. How many hurdles has he?

45. A can dig a trench in the time that B can; B can dig it in rds the time that C can; altogether they can dig it in 6 days; find the time it would take each of them alone to dig it.

46. A person after paying a poor-rate, and also an income-tax of 7d. in the pound, has £486 remaining; the poor-rate amounts to £22. 10s. more than the income-tax: find the original income and the number of pence in the pound in the poor-rate.

47. Divide the number a into 4 such parts, that if to the first you add n, from the second subtract n, multiply the third by n, and divide the fourth by n, the results will be all equal. If a=90, n=2, what will the results be?

48. A greyhound spying a hare at the distance of 60 of his own leaps from him, pursues her, making 4 leaps for every 5 leaps of the hare, but he passes over as much ground in 3 leaps as the hare did in 4; how many leaps did each make during the whole course?

49. At the review of an army the troops were drawn up into a solid mass 40 deep; when there were just 1th as many men in front as there were spectators. Had the depth however been increased by 5, and the spectators drawn up in the mass with the army, the number of men in front would have been 100 fewer than before. Find the number of men in the army.

50. If A can do a piece of work in 2m days, and B and A in n days,

days, find the number of days in which A, B, and

C together would do the work.

51. Two couriers A and B leave two towns distant from each other m miles, and travel, in the same direction, at the rate of a and b miles an hour respectively. When will the second courier overtake the first? Interpret the result by taking

1. m 20 miles, a=12 miles, b= 8 miles.

52. If a men or b boys can dig m acres in n days, required the number of boys whose assistance will be required to enable (a-p) men to dig (m+p) acres in (n-p) days.

SIMULTANEOUS EQUATIONS.

106. In each of the examples which we have hitherto solved in equations, we have had only one equation to solve, and one unknown quantity x to determine by means of it. In each case we have reduced a to the form, some known quantity; and by this means we have found the value of x.

If, however, we have an equation of the form ax + by=c where T and y are both unknown quantities, and a, b, c known quantities, it is

clear that we shall obtain from it x=

-by

C

a

; and since y is an unknown

quantity, the value of x must still be unknown also: but if we have a second equation a'x+b'y=d, where x and y are supposed to have the

same values as in the first equation, we may reduce this second equation

And since the value of x is supposed to be the same in both equations, we have

an equation containing only one unknown quantity y, from which therefore the value of y may be found in known terms; and then a can be determined by substituting the value of y thus found in either of the equations ax + by=c, or a'x+b'y=d.

As these two equations coexist so as to be true at the same time, they are called 'SIMULTANEOUS EQUATIONS'; and generally, if we have two or more equations, which involve two or more unknown quantities, and which are to be satisfied by the same values of the unknown quantities, these equations are called SIMULTANEOUS EQUATIONS.

Hence it appears that, if we have two unknown quantities, we must have two independent equations; so that we may, by certain methods which are to be explained, eliminate x, or cause it to disappear, and thus obtain a third equation involving only one unknown quantity.

There are three methods by which such equations may be solved. First Method. Let the two equations be ax + by=c and a'x+b'y=c', where a, b, c, a', b', d' are known quantities and x and y unknown.

ax + by = c, (1)
ax+b'y=c', (2)

Multiply each term of equation (1) by a', the coefficient of x in equation (2): then multiply each term of equation (2) by a, the coefficient of x in equation (1), and place the results as underneath,

Note. If the second equation had been -a'x+b'y=d', we should have added (4) to (3), instead of subtracting (4) from (3).

Therefore y is found in terms of known quantities; and on substituting this value of (y) in either of the original equations, the value of a will be found in terms of known quantities.

Thus, for instance, from equation (1) we obtain by substitution,

Therefore the value of x is determined.

The first Rule for solving such equations may therefore be thus expressed.

RULE 1. "Multiply the first equation by the coefficient of x in the second equation, and the second equation by the coefficient of x in the first equation subtract the last of these resulting equations from the first, or add them together, according as the equation (2) is of the form ax+b′y=c', or-a'x+b'y=d; there will arise an equation which contains only y and known quantities, from which the value of y is determined. The value of a is found by substituting the value of y just found, in either of the given equations."

Second Method. Let the equations be, as before,

Then, putting these values of x equal to each other,

The value of x can be deduced as in the former case.

The second Rule for solving such equations may therefore be thus expressed.

RULE 2. "Put the value of x in the first equation equal to its value in the second, and there will arise a new equation involving only y; from which the same value of y is found, as in Case I. The value of x is obtained by substituting the value of y just found, in either of the given equations."

Third Method. Let the equation be, as before,

ax + by = c, (1)

a'x+y=d, (2))

c-by

From equation (1), as before, x= a

Now substituting this value of x in equation (2), we get

The value of x can be deduced as before.

The third Rule for solving such equations may therefore be thus expressed.

RULE 3. "Find the value of x from the first equation, and substitute this value for a in the second equation; there will arise an equation which gives the same value of y as in the two former instances. The value of a is found as before."