Question 80947
each day, Pat travels 2 miles more than the previous day ... on the first day he traveled 1 mile


so the distance traveled on day x is 1+2(x-1) ... for Mike, 2+4(x-1)


let d=days traveled ... distance equals (average per day)*(days) ... ((first day+last day)/2)*(days)


so Pat's distance is (1+(1+2(d-1))/2)*d ... similarly, Mike's distance is (2+(2+4(d-1))/2)*d


the combined distance is 363 ... so (1+(1+2(d-1))/2)*d+(2+(2+4(d-1))/2)*d=363 ... (d)*d+(2d)*d=363


{{{3d^2=363}}} ... {{{d^2=121}}} ... d=11