Two cyclists left simultaneously from cities A and B heading towards each other at constant rates and met in 5 hours. The rate of the cyclist from A was 3 mph less than the rate of the other cyclist. If the cyclist from B had started moving 30 minutes later than the other cyclist, then the two cyclists would have met 31.8 miles away from A. What is the distance between A and B, in miles?

Answer:75 milesStep-by-step explanation:Let x mph be the cyclist A rate, then x+3 mph is the cyclist B rate.1. In 1 hour they both traveled x+x+3=2x+3 miles. In 5 hours they traveled[tex]5(2x+3)=10x+15\ miles.[/tex]2. Cyclist A spent [tex]\frac{31.8}{x}[/tex] hours to travel 31.8 miles. If the cyclist from B had started moving 30 minutes (1/2 hour) later than the cyclist A, then he spent [tex]\frac{31.8}{x}-\frac{1}{2}[/tex] hours to travel the rest of the distance. In total they both traveled the whole distance 10x+15 miles, thus[tex]31.8+\left(\dfrac{31.8}{x}-\dfrac{1}{2}\right)\cdot (x+3)=10x+15[/tex]Solve this equation. Multiply it by 2x:[tex]63.6x+(63.6-x)(x+3)=2x(10x+15)\\ \\63.6x+63.6x+190.8-x^2-3x=20x^2+30x\\ \\-x^2+124.2x+190.8-20x^2-30x=0\\ \\-21x^2+94.2x+190.8=0\\ \\210x^2-942x-1908=0\\ \\35x^2-157x-318=0\\ \\D=(-157)^2-4\cdot 35\cdot (-318)=69169\\ \\x_{1,2}=\dfrac{-(-157)\pm\sqrt{69169}}{2\cdot 35}=\dfrac{157\pm263}{70}=-\dfrac{106}{70},\ 6[/tex]The rate cannot be negative, thus, x=6 mph.Hence, the distance between cities A and B is[tex]10\cdot 6+15=60+15=75\ miles.[/tex]