SOLUTION: From towns 363 miles apart, Pat and Mike set out to meet each other. If Pat travels 1 mile the first day, 3 miles the second, 5 miles the third, etc. and Mike travels 2 miles the f

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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

... ... d=11