Question 770163
You have all the right equations.  At least try a substitution.  Eliminate z through substitution into the first two equations, and then solve for x and y.


Tickets: x+y+z=1,000        
Money:  3x+4y+5z=3,800
Adults versus students:  z=y+100


{{{x+y+y+100=1000}}} and {{{3x+4y+5(y+100)=3800}}}
{{{x+2y=900}}} and {{{3x+9y=3300}}}
{{{x+2y=900}}} and {{{x+3y=1100}}}


Add the opposite of the 900 equation to the 1100 equation.
{{{x+3y-x-2y=1100-900}}}
{{{highlight(y=200)}}},
So, what is x?
and then what is z?