Question 31391
tara {{{x/y}}}
joe {{{(x - y) / y}}}
lynn {{{(2/3)*(x/y)}}}
they pool their money
{{{x/y + (x - y) / y + (2/3)*(x/y)}}}
If I put the 2/3 in the numerator of the 
last term, my common denominator is y
{{{(x + x - y + (2/3)*x) / y}}}
{{{(2*x + (2/3)*x - y) / y}}}
{{{((8/3)*x - y) / y}}}
divide through by y
{{{(8/3)*x / y - 1}}}
One way to check this is to plug in numbers
let x = 6
let y = 3
x has to be bigger than y so joe doesn't lose money
and y can't be 0. division by 0 not allowed
tara = 2
joe = 1
lynn = 4/3
add them up
2 + 1 + 4/3 = 13/3 
now use  {{{(8/3)*x / y - 1}}}
{{{(8/3)*6 / 3 - 1}}}
{{{2*8/3 - 1}}}
{{{16/s - 3/3}}}
{{{13/3}}}
checks OK