system of linear equations help please

[babe20042004]

How would I go about solving these question?
This table represents 4 lead blocks balancing on a fulcrum in different positions. These are the three equilibrium positions found.
_____________ Left Side_________________Right Side

_________Weight 1 Weight 2__________Weight 3 Weight 4
Position 1___ 50_______20________________10______30
Position 2___ 25_______50_______________ 50______30
Position 3___20 _______30________________30______20

The first question asks to write out the systems of equations that correspond to the results.

The way I interpreted it to be was:
50+20-10=30
25-50-30=-50
20+30-20=30
by putting one each of the weights on the other side of the equal side once.

Then the second question asks to determine the ratio of the weights. ???

 

[wjm11]

This table represents 4 lead blocks balancing on a fulcrum in different positions. These are the three equilibrium positions found.
_____________ Left Side_________________Right Side

_________Weight 1 Weight 2__________Weight 3 Weight 4
Position 1___ 50_______20________________10______30
Position 2___ 25_______50_______________ 50______30
Position 3___20 _______30________________30______20

The first question asks to write out the systems of equations that correspond to the results.

The way I interpreted it to be was:
50+20-10=30
25-50-30=-50
20+30-20=30
by putting one each of the weights on the other side of the equal side once.

Then the second question asks to determine the ratio of the weights. ???

_________Weight 1 Weight 2__________Weight 3 Weight 4
Position 1___ 50_______20________________10______30
Position 2___ 25_______50_______________ 50______30
Position 3___20 _______30________________30______20

Equations:
It is necessary to multiply the weight times the distance in order to develop the equations.
50(W1) + 20(W2) = 10(W3) + 30(W4)
25(W1) + 50(W2) = 50(W3) + 30(W4)
20(W1) + 30(W2) = 30(W3) + 20(W4)

Determine ratios:
Let W1 be arbitrarily set equal to 1; the other weights will then be solved in terms of W1.

50(1) + 20(W2) = 10(W3) + 30(W4)
25(1) + 50(W2) = 50(W3) + 30(W4)
20(1) + 30(W2) = 30(W3) + 20(W4)

20(W2) - 10(W3) - 30(W4) = -50
50(W2) - 50(W3) - 30(W4) = -25
30(W2) - 30(W3) - 20(W4) = -20

2(W2) - 1(W3) - 3(W4) = -5
5(W2) - 5(W3) - 3(W4) = -2.5
3(W2) - 3(W3) - 2(W4) = -2

You now have 3 equations and 3 unknowns, which can be solved in a variety of ways: Gaussian elimination, matrices, etc.

Solutions:
W2 = 1.5 (meaning W2/W1 = 3/2)
W3 = .5 (meaning W3/W1 = 1/2)
W4 = 2.5 (meaning W2/W1 = 5/2)

or

W1:W2:W3:W4 = 2:3:1:5

 

[babe20042004]

thanks.