Basic Algebra disguised in a tricky manner?

Hey guys,

Having trouble with this one - set up a little grid with distance, rate, and time - cannot figure out how to represent time and advance to the answer.

"Esther drove to work in the morning at an average speed of 45 miles per hour. She returned home in the evening along the same route and averaged 30 miles per hour. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning?

I feel silly because I've been able to figure out countless other problems like this before but this time I know that distance is what we're trying to find out in the morning, we know the rates, and the total time must equal 1 hour, i figured we could just do 45t=30t but i don't know how to represent t in this case - i feel like i need to use minutes but don't know how to set it up that they both must somehow equal 60.

The answer I believe is 18 miles she drove in the morning. Can anyone help point me in the right direction?

Let t represent the time going to work. Then, 1-t represents the time returning from work.
Distance = rate X time.
Distance to is the same as the distance from. That is the basis for an equation.

gusterguy said:

i figured we could just do 45t=30t

Click to expand...

Since the rates were different but the distances were the same, you cannot use the same time "t" for each. (Also, your equation solves as t = 0, which does not progress matters much....)

Try using a grid, like the instruction say:

Then plug in what you know:

Since d = rt, then naturally t = d/r. Use this fact to find expressions for the two times "t" in terms of the variable "d", and fill in the rest of the grid. Then note that you have:

. . . . .the sum of (time going) and (time coming) equalled (one hour)

Translate the above into an algebraic equation, using the expressions from your grid. Solve for the value of d, and you're done!

Eliz.