# Simultaneous equations substitution method

## Homework Statement:

A) Use the clue and the diagram to write a pair of simultaneous equations.
B) Solve the equations to find x and y
C) Check your solution to see if it makes the seesaw balance

Problem 16. The sum of the weights is 24.

## Relevant Equations:

Fulcrum equation Ax = By where A and B are the respective weights and x and y are the distant between them needed for balancing

I'm really stuck on this one, I was able to get the answer but not by the substitution method.

So its the weight as A and B so I get A + B = 24
A(3) = B(5) so in my head I calculate a few pairs, 3 x 5 = 15 but 3 + 5 only = 8 so the next pair would be 10 and 6 which is still to small so I move on to 15 and 9 which does = 24 and solves the equation but I failed to use the substitution method which to me means I haven't really solve the problem.

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DaveE
Gold Member
So, your unknown variables are the two weights, x and y (or A and B, as you said). You've already told us that x + y = 24, that is one of the two equations that you need to solve this. Can you think of another relevant equation? It would represent another requirement on the distribution of weight. For example x = 0 doesn't work, y = 0 doesn't work, why not?

archaic
There has to be some weight on both sides for it to balance so neither x or y can be 0.

I think I got it. So I try 3x = 5y which when gives me x = 1 and (2/3) y so
x+ y is (8/3) = 24 which gives me 9, then its x + 9 = 24 subtract 9 from both sides and get x = 15. Thanks for the help!

DaveE
DaveE
Gold Member
Yes! good work.

I'm not sure I followed your description of the process though. Here's one way I would do it:
1) x + y = 24 => y = 24 - x
2) 3x =5y => x = (5/3)y
substitute 2) into 1) => y = 24 - (5/3)y then solve for y, y + (5/3)y = 24, y = 24/(1 + 5/3), y = 9
then find x with one of the first two equations, x = 24 - 9 = 15.

BTW, there are a lot of different but equally good way of solving these equations.

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