Or thus, to divide 3 a c-3 by c2, is the same as to

1

multiply it by or c2, which gives the same

C2

re

3ac-1d-1x 4ec-3d-1 12 a c-4d-2 e=

=

12 a e

In this example the exponents to be subtracted had the sign, which in subtracting was changed to +.

When a question has been resolved generally, that is, by representing the known quantities by letters, we sometimes propose to determine what values the unknown quantities will take, for particular suppositions made upon the known quantities.

The two following questions offer nearly all the circumstances that can ever occur in equations of the first degree.

Two couriers set out at the same time from the points A and B, distant from each other a number m of miles, and travel towards each other until they meet. The courier who sets out from the point A, travels at the rate of a miles per hour; the other travels at the rate of b miles per hour. At what distance from the points A and B will they meet?

Suppose C to be the point, and

Let the distance A C

Since the first courier travels x miles, at the rate of

a miles per hour, he will be hours upon the road.

The second courier will be

α

hours upon the road.

Putting this value of x into the first equation, it be

Since neither of the quantities in these values of x and y has the sign, it is impossible for either value to become negative. Therefore whatever numbers may be put in place of a, b, and m, they will give an answer according to the conditions of the question. In fact, since they travel towards each other, whatever be the distance of the places, and at whatever rate they travel, they must necessarily meet.

Suppose now that the two couriers setting out from the points A and B situated as before, both travel in the same direction toward D, at the same rates as before. At what distances from the points A and B will the place of their meeting, C, be?

The second equation expressing only the equality of the time will not be altered.

Here the values of x and y will not be positive unless a is greater than b; that is, unless the courier, that sets out from A, travels faster than the other.

In this case the point C, where they come together, is distant from A twice the distance A B.

Here the values of x and y are both negative; hence there is some absurdity in the enunciation of the question for these numbers. In fact, it is impossible that the courier setting out from A, and travelling slower than the other should ever overtake him.