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Four balances are shown where the two sides of each contain exactly the same amount of weight. The first balance has 4 squares, 1 triangle and 1 ball on the left side, right side has 2 diamonds. The second balance has 1 triangle and 1 square on the left side, 1 ball on the right side. The third balance has 1 triangle and 1 diamond on the left side, 1 ball and 2 squares on the right side. The fourth triangle has 1 diamond, 1 triangle and 1 square on the left side. How many balls are on the right side? Explain/show how this was determined.

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Heaven

  1. balance 1: 4S+1T+1B=2D
    balance 2: 1S+1T=1B
    banace 3: 1T+1D=1B+2S
    balance 4: 1S+1T+1D= XXXX B

    solve for XXXX

    now, rearranging the four equations in order of BDST

    a) 1B-2D+4S+T=0
    b) -1B+0D+1S+T=0
    c) -1B+D-2S+1T=0
    d) XXXB+D+S+T=0

    subtract d from b
    (1-xx)B+D=0
    add a + C
    -D+2S+2T=0
    now add these two equations
    (1-xx)B+2S+2T=0 , now subtract 2*b from this
    (xx-1+2)B =0

    which means xxx=1

    check my work on this, it is easy to make an error.

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    bobpursley

  2. I got 3B on the last scale

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    Emily

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