A triple beam balance, like the scales we use every week in lab, works using rotational equilibrium. If a 253 g mass is placed on the pan, 1 cm from the pivot, and there are three masses, 1 g, 10 g, and 50 g masses on the other side can be slid back and forth between 0 cm and 10 cm from the pivot, where would the masses need to be placed? Full credit only if you provide values at whole-integers of centimeters for the three masses. So far I've got: .253-.05r-.01(r+x)-.001(r+x+y)=0 but I can't figure out how to solve for all the variables. any help is appreciated!