the P of triangle= 2400/lng sd is 200 <twice the short/mid is 200 <than short find lengths of 3sd triangle

Answer 1

Let's see: restating your problem, the Perimeter of a triangle (with sides of unequal lengths) is 2400 units.
The long side is 200 units less than twice the short side.
The medium-length side is 200 units less than the short side.
Find the lengths of the 3 sides of the triangle.

Are you sure that the medium-length side is 200 less than the short side? This would make it SHORTER than the short side. Nevertheless, based on what you provided:

Let a = length of short side;

b = length of medium-length side
= a - 200;

c = length of the long side
= 2a - 200

Perimeter P = a + b + c
2400 = a + (a - 200) + (2a - 200)
2400 = a + a - 200 + 2a -200
2400 = 4a -400
2400 + 400 = 4a
2800 / 4 = a
a = 700 units

b = a - 200
= 700 - 200
= 500 units

c = 2a - 200
= 2 (700) - 200
= 1400 - 200
= 1200 units

To check:
P = a + b + c
2400 = 700 + 500 + 1200
2400 = 2400

Observe, that side b, that we designated as the medium-length side, is shorter than side a. Maybe the medium-length side should have been "200>than the short."

In this case, P = a + b + c
2400 = a + (a + 200) + (2a -200)
2400 = a +a + 200 +2a -200
2400 = 4a
a = 2400 / 4
= 600 units

b = a + 200
= 600 + 200
= 800 units

c = 2a - 200
= 2 (600) - 200
= 1200 - 200
= 1000

To check, 2400 = 600 + 800 + 1000
2400 = 2400

Answer 2

x = shortest side
200-2x = longest side
200-x = Middle side
400 -2x = 2400
x= -1000 which is impossible.

Please restate the problem using proper notation.