Can some me on this problem? ? i think i’m setting it up

  60 miles per hour

Step-by-step explanation:

Let s represent the speed in miles per hour during the first part of the trip. Then s-10 will be the speed during the last part of the trip.

For speed/time/distance problems, the appropriate relation is the one that is posted on every speed limit sign:

  speed = miles/hour = distance/time

Rearranging this relation gives you the expression for time:

  time = distance / speed

You are given time and distance and you need to find speed. The time you're given is the total for the two parts of the trip, so ...

  4 = 84/s + 130/(s-10) . . . . . . total time = time1 + time2

Multiplying by the product of the denominators, this becomes ...

  4s(s-10) = 84(s-10) +130s

  4s^2 -254s +840 = 0 . . . . . . subtract the right side to put in standard form

This can be solved using any of several methods for solving quadratic equations. Solutions are ...

  s = 60, s = 3.5 . . . . . only the first solution makes any sense in the problem

The speed during the first part of the trip is 60 miles per hour.