SOLUTION: I never seem to be able to set up equations from story problems. I think I have an idea about how to do it, then am usually disappointed. If I see an example, I can usually apply

Question 69814: I never seem to be able to set up equations from story problems. I think I have an idea about how to do it, then am usually disappointed. If I see an example, I can usually apply it to similar problems, but I rarely remember how they work for long.
Example:
The launch site for Trigon Balloon Co. is 250ft. above sea level. A hot-air balloon is launched from the site and begins to rise at a rate of 110ft/min. At the same time, another balloon, 2220 ft. above sea level begins to descend at a rate of 150 ft/min.
How long will it be until the balloons are at the same elevation?
What will their elevation be then?
I set it up this way, but it doesn't work out
250 + 110x = 2220 - 150x
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The launch site for Trigon Balloon Co. is 250ft. above sea level. A hot-air balloon is launched from the site and begins to rise at a rate of 110ft/min. At the same time, another balloon, 2220 ft. above sea level begins to descend at a rate of 150 ft/min
:.
How long will it be until the balloons are at the same elevation?
What will their elevation be then?
:
I set it up this way, but it doesn't work out
:
It looks OK to me, it has to work:
:
250 + 110x = 2220 - 150x
110x + 150x = 2220 - 250
260x = 1970
x = 1970/260
x = 7.6 min or 7 min 36 sec
:
Check by finding how many feet it will travel at a combined speed:
7.6(110+150) = 1976 ~ 1970 (we rounded off 7.5796 to 7.6)

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