Can you solve this balance problem in two ways?

I like it!
I wrote the expressions in the form of an equation.
1D +3S =2C
and so on.

I added the first two equations together and the answer was obvious.
I’ll PM you so I don’t spoil if for others.

So far I do not see another way.

Nice

Same methodology as @LuckyGuy Second equation 2s+3d=3c
Solve, Substitute, add weights.
Yes, I have the answer and will PM it so it lives on.

Interesting that I am using this type of Applied Math in my current Acoustics Calss at Berklee.

Here is the solution, which both @LuckyGuy and @Forever_Free found.

Consider what would happen if we combine the two weighings so that the two clubs from the first weighng are combined with the 3 clubs from the second weighing, and the diamond and 3 spades are combined with the 2 spades and 3 diamonds. The two sides balance, giving 5 clubs on one side and 5 spades and 4 diamonds on the other side.

For the third weighing we are given 5 clubs on one side and 5 spades on the other side. From the preivous paragraph we know that 5 clubs balance with 5 spades and 4 diamonds, so we can add 4 diamonds to get baoance.

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To do this algebraically, we assign variables c, s, and t to clubs spades and triangles. The two weighings form the equations:

d+3s= 2c
3d + 2s=3c

Multiply and add the two equations to eliminate the club term, giving s in terms of d.
Then multiply and add the two equations to eliminate the spade term, giving c in terms of d.

The difference between the two sides is 5c – 5s, which can be expressed in terms of d, tellibg how many diamonds must be added.

I doubt if anyone is paying attention to this anymore, but there is a clarification needed for the algebraic approach. What c, s and t represent are the weights of the clubs, spades and triangles. When clubs and spade weights are converted to triangle weights, it is found that the side with 4 clubs excedes the weight of 5 spades by the weight of 4 triangles. Balance can be achieved by adding 4 triangles to the 5 spades.