#### Answer

The number of tickets are: $200$ (8 dollars), $150$ (10 dollars), $50$ (12 dollars)

#### Work Step by Step

Step 1. Assume the number of tickets are: $x$ (8 dollars), $y$ (10 dollars), $z$ (12 dollars)
Step 2. Based on the given conditions, we have
$\begin{cases} x+y+z=400\\ 8x+10y+12z=3700 \\ x+y=7z \end{cases}$
Step 3. Combining the first and the third equations, we have $8z=400$; thus $z=50$
Step 4. Back substitute the known $z$ value; we have $\begin{cases} x+y=350\\ 8x+10y=3100 \end{cases}$
Step 5. Multiply 4 to the first and simply the second equation: $\begin{cases} 4x+4y=1400\\ 4x+5y=1550 \end{cases}$
Step 6. Eliminate $x$; we get $y=150$, thus $x=200$ and $z=50$
Step 7. We conclude the number of tickets are: $200$ (8 dollars), $150$ (10 dollars), $50$ (12 dollars)