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.
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.
2 answers
I got 3B on the last scale