A consultant traveled 10 hours to attend a meeting. The return trip took only 9 hours because the speed was 8 miles per hour faster. What was the consultants speed each way? No book. Thanks so much for your time! Make this chart DISTANCE RATE TIME Going Returning Fill in the times for going and returning, 10 hrs and 9 hrs: DISTANCE RATE TIME Going 10 Returning 9 Let x mph be the rate going. Fill that in DISTANCE RATE TIME Going x 10 Returning 9 The speed returning was 8 mph faster than x mph. So we add 8 mph to x mph and get x+8 mph. So fill that in as the rate returning DISTANCE RATE TIME Going x 10 Returning x+8 9 Now use D = RT to fill in the distances: DISTANCE RATE TIME Going 10x x 10 Returning 9(x+8) x+8 9 Now since the distance going was the same as the distance returning, we set the two distances equal to each other: 10x = 9(x+8) Solve that and get 72 mph going. Then the rate returning was 8 mph faster, or 80 mph! Lucky he didn't get a speeding ticket! :-) Edwin