Question

Average rate of the train

Answer 1

Explanation:

First of all, I'm having a bit of trouble seeing what the exponent on the leading term in the distance is, but I think its an x-squared so that's what I used to solve this. Keep in mind that you do NOT get a number answer here; you get another expression in terms of x for an "answer".

Let's simplify down the rate first by multiplying. What I did was put the denominator under each of the terms in the numerator and then simplified by dividing each term:

`r=(2x^2)/(2x)-(5x)/(2x)-(3)/(2x)` which simplifies to

`r=x-2.5-(3)/(2x)` Now let's get rid of that denominator there by multiplying everything by 2x to get

`r=2x^2-5x-3` and we're ready to solve for time.

If d = rt, then

`t=(d)/(r)` so filling in:

`t=(6x^2+3x)/(2x^2-5x-3)` and we can divide this using the method of long division. It would be impossible to show you how to do that here in this forum, but when you divide, you get

`t=3+(18x+9)/(2x^2-5x-3)` ...as crazy as that seems.