On the first supposition the speed per hour is x + 6 miles, and the
time takeen is y — 4 hours. In this case the distance traversed will be
represented by (x + 6) (y — 4) miles.
On the second supposition the distance traversed will be represented by (x — 6) (y + 6) miles.
All these expressions for the distance must be equal ; xy= (x + 6) (y — 4) = (x — 6) (y + 6).
From these equations we have xy = xy + 6y — 4x — 24, or 6y-4x = 24 (1) ; and xy = xy — 6y + 6x — 36, or 6x-6y = 36 (2).
From (1) and (2) we obtain x = 30, y = 24.
Hence the distance is 720 miles.
Ex. 5. A person invests $3770, partly in 3 per cent Bonds at $102, and partly in Railway Stock at $ 84 which pays a dividend of 4{1}{2} per cent ; if his income from these investments is $136.25 per annum, what sum does he invest in each ?
Let x denote the number of dollars invested in Bonds, y the number of dollars invested in Railway Stock; then x + y = 3770 (1).
The income from Bonds is $ {3x}{102}, or ${x}{34} ; and that from Railway Stock is ${4{1}{2}y}{84} or ${3y}{56}.
Therefore {x}{34} + {3y}{56} = 136{1}{4}(2).
From (2) x + {51}{28}y = 4632 {1}{2}, and by subtracting (1) {23}{28}y = 862{1}{2}; whence y = 28 x 37{1}{2} = 1050 ; and from (1) x = 2720.
Therefore he invests $2720 in Bonds and $1050 in Railway Stock.
EXAMPLES XXV.
1. A sum of $100 is divided among a number of persons ; if the number had been increased by one-fourth each would have received a half-dollar less : find the number of persons.
