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[answered] At x 1 we have those cars that were on x 3 at the time plus

I was told that the final inequality for x2 is incorrect and I'm stuck, What?is the final inequality for x2?

At x 1 we have those cars that were on x 3 at the time plus those that entered at x 1 subtracted x 1 . The equation then becomes: from the number of planes that left on x 1 +165= x 3 +200

In order to get x 1 by itself, I subtracted 165 from each side and then I get x 1 = x 3 + 200-165

I subtracted x 3 from each side and subtracted 165from 200 x 1?x3 =35 At x 2 ,we have the cars that were on from the cars that left on x 1 at the time plus those that entered at x 2 subtracted x 2 . The equation becomes: x 1+50=x 2+100

I subtracted 50 from each side, and I also subtracted x 2 from each side. x 1?x 2=50 For x 3 we have cars that were on x 2 plus those that entered at x3 . x 2+200=x 3+ 185

I subtracted 185 from each side and I did the same with x 3?x 2=15 With all three equations combined I have the following: x2 . x 3 minus the cars that left on x 1 - x 3 =35

x 1?x 2=50 ?x 2 + x 3=15 65 ? x 1 ? 81 This equation lets us know the maximum and minimum amount of cars that are at A,

with 65 being the minimum and 81 being the maximum.

Since x 1=65 I can use 65 to plug into my first equation 65?x 3=35 x 3=30

If we substitute 81 for x 1 our equation becomes 81?x 3=35 x 3=46

After solving the equation the solution is

Now that we have values for

The equation says that 30 ? x 3 ? 46 x 3 we can now solve for x 2 x 2?x 3=15 x 2?30=15

By adding 30 to each side the equation becomes : x 2=45 To solve for the next solution of the set I plugged in 46 for x 2?46=15

By adding 46 to each side the equation becomes x 2=61

The solution for

The solutions are: x 1=65 ? x 1 ? 81

x 2=45 ? x2 ? 61 x 3=30 ? x 3 ? 46 x 2 = 45 ? x2 ? 61 x3

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