19.Rose drove her car from her home to the province and back, a total distance of 120km. Her average speed returning was 3kph slower than her average speed going to the province. If her total driving time was 9 hours, what was her average speed in going to the province? thanks a lot tutorz!! Rose drove very slowly to have taken 9 hours to travel only 120km. I will assume the traffic was extremely heavy both ways. The words "a total distance of 120km" are ambiguous. Is it 60km going and 60km returning for a total of 120km? Or is the total distance between her home and the province 120km? Either way she is still driving very slowly. I will assume she is driving the slowest and the distance between her home and the province is 60km. Make this chart: DISTANCE RATE TIME Going Returning Fill in the two distances, which are both 60: DISTANCE RATE TIME Going 60 Returning 60 Since the question is >>...what was her average speed in going to the province...<< We will let x represent her speed going, so we fill in x for the rate going: DISTANCE RATE TIME Going 60 x Returning 60 >>...Her average speed returning was 3kph slower...<< 3kph slower than x kph is x-3 kph. So we fill that in for the rate returning: DISTANCE RATE TIME 60 x Returning 60 x-3 Now we use the fact that TIME = DISTANCE/RATE to fill in the times: DISTANCE RATE TIME Going 60 x 60/x Returning 60 x-3 60/(x-3) To find the equation we use this: >>...her total driving time was 9 hours...<< We add the two TIME's and put that equal to 9 60/x + 60/(x-3) = 9 Can you solve that by first multiplying through by LCD = x(x-3) ? If not post again. You will get two answers: x = 16 and x = 4/3 We must discard the 4/3, since that would cause her rate returning to be negative. So the answer is a very slow 16 kph. If we assume it was 120km from her home to the province the equation is not factorable and the answer is 28.251 kph, which is still very slow. Edwin