Two cyclists 95 mi apart start riding towards each other at

Two cyclists, 95 mi apart, start riding towards each other at the same time. One cycles twice as fast as the other. If they meet 4 hr later, at what average speed is each cyclist traveling?

Solution

D=RT
95=(X+2X)4
95=3X*4
95=12X
X=95/12
X=7.91 MPH IS THE SLOWER SPEED RIDER.
2*7.91=15.83 MPH FOR THE FASTER RIDER.
PROOF
95=(7.91+15.83)4
95=23.74*4
95~95