Help plz,
Set up an equation to solve the following problem, and then solve it.
On a sales trip, Irwin drove 330 miles averaging a certain speed. The return trip was at an average speed that is 11mph faster. Total time for the round trip was 11 hours. Find Irwin’s average speed on each part of the trip.

- anonymous

- chestercat

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- anonymous

well, its not hard , what dont we know , what we do we need to find , assign them to variables

- radar

Use distance equal rate times time.

- anonymous

let x equal the "certain speed"

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## More answers

- anonymous

so assume 330 miles for one way , that means in total he drove 660miles

- anonymous

we know the total time was 11
but time = distance / speed
time for first section = 330 / x
times for second section = 330/ ( x+11)
so ( 330/x) + [330 / (x+11) = 11

- anonymous

\[\frac{330}{x} + \frac{330}{x+11} =11\]

- anonymous

solve for x

- anonymous

i c. Thanks

- anonymous

it has two solutions x=-6, x=55
but because we are dealling with real world ( and yes, it says "speed", not velocity ) so a negative value is not allowed

- anonymous

so x=55
means for first half average was 55mph, and for second half average was 66mph ( ie 55 +11 )

- radar

Actually the average speed for the entire trip was 660/11 or 60 mi.
So wouldn't that made the avg speed for the first part 54.5 mph and the second part 65.5?

- anonymous

dnt think so

- radar

The avg of 54.5 + 65.5 is 120/2=60

- anonymous

firstly where did u magically get 54.5 and 65.5 from

- radar

x + (x+11) 660
---------- = ----- = 60
2 11

- radar

Nothing magic about it.

- anonymous

yes....
and.....?

- anonymous

yes there is , there are alot of different ways to get two numbers to sum to 120 , why did you pick those two numbers? , theres no calculations to suggest why you chose thoise two

- anonymous

ohh I think i see what you are doing now

- anonymous

bit of a dangerous way to think

- radar

x= first avg speed (certain speed)
x+11 is the second avg speed (return trip
Their sum divided by 2 equals the total avg speed
660/11 is also the total avg speed. Take it from there

- radar

Just sayin your way seems correct just wondering why they dont come out the same. Did you round off or take a root of or whatever/

- anonymous

no, the reason why they dont come out correct is because your way of thinking is wrong :P

- radar

Well they came out close.lol
gotta run

- anonymous

:\, the average of a sum is not the sum of the avergaes

- radar

Gotcha, I see the error of my logic.

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