Answer :
[tex]v=\frac{s}{t}\\\\v-velocity;\ s-distance;\ t-time\\\\v=\frac{s}{t}\to vt=s\to t=\frac{s}{v}\\---------------------\\v-Rock's\ veloocity\ (mph)\\v+8-Jack's\ velocity\ (mph)\\s_J=164\ mi-Jack's\ distance\\s_R=106\ mi-Rock's\ distance\\t_J=t_R-times\ are\ equal\\\\therefore:\frac{s_R}{v}=\frac{s_J}{v+8}\\\\subtitute\ values\ of\ distance:[/tex]
[tex]\frac{106}{v}=\frac{164}{v+8}\ \ \ \ |cross\ multiply\\\\106(v+8)=164v\\\\106v+848=164v\ \ \ \ |subtract\ 106v\ from\ both\ sides\\\\848=58v\to 58v=848\ \ \ |divide\ btoh\ sides\ by\ 58\\\\v\approx14.6\ (mph)\\\\Answer:\boxed{v+8=(14.6+8)mph=22.6\ mph}[/tex]
[tex]\frac{106}{v}=\frac{164}{v+8}\ \ \ \ |cross\ multiply\\\\106(v+8)=164v\\\\106v+848=164v\ \ \ \ |subtract\ 106v\ from\ both\ sides\\\\848=58v\to 58v=848\ \ \ |divide\ btoh\ sides\ by\ 58\\\\v\approx14.6\ (mph)\\\\Answer:\boxed{v+8=(14.6+8)mph=22.6\ mph}[/tex]
