Question 287839
Set-Up
------------
Let A = # of seats in theater 1
Let B = # of seats in theater 2
Let C = # of seats in theater 3

Equation 1: {{{A  = 204}}}
Equation 2: {{{B = 108 + C}}}
Equation 3: {{{A + B + C = 1000}}}

Solution:
------------
We were given that A = 204, And equation 2 is solved for B.  You can substitute these values into equation 3.
Equation 3: {{{A + B + C = 1000}}}
{{{204 + (108 + C) + C = 1000}}} Simplify
{{{312 + 2C = 1000}}} Subtract 312 from both sides
{{{2C = 688}}} Divide both sides by 2
{{{highlight(C = 344)}}}

Now plug 344 into equation 2 for C.
Equation 2: {{{B = 108 + C}}}
{{{B = 108 + 344}}}
{{{highlight(B = 452)}}}