Suppose three boba drinks and four burgers cost $28$ dollars, while two boba drinks and six burgers cost $\$ 37.70$. If you paid for one boba drink using only pennies, nickels, dimes, and quarters, determine the least number of coins you could use.
1. **Set up the system of linear equations:**
   We are given two equations based on the cost of boba drinks (denoted as \(d\)) and burgers (denoted as \(b\)):
   \[
   3d + 4b = 28 \quad \text{(1)}
   \]
   \[
   2d + 6b = 37.70 \quad \text{(2)}
   \]

2. **Solve the system of equations:**
   To eliminate one of the variables, we can multiply the first equation by 2 and the second equation by 3 to make the coefficients of \(d\) the same:
   \[
   2(3d + 4b) = 2 \cdot 28 \implies 6d + 8b = 56 \quad \text{(3)}
   \]
   \[
   3(2d + 6b) = 3 \cdot 37.70 \implies 6d + 18b = 113.10 \quad \text{(4)}
   \]

3. **Subtract equation (3) from equation (4):**
   \[
   (6d + 18b) - (6d + 8b) = 113.10 - 56
   \]
   \[
   10b = 57.10
   \]
   \[
   b = \frac{57.10}{10} = 5.71
   \]

4. **Substitute \(b = 5.71\) back into one of the original equations to find \(d\):**
   Using equation (1):
   \[
   3d + 4(5.71) = 28
   \]
   \[
   3d + 22.84 = 28
   \]
   \[
   3d = 28 - 22.84
   \]
   \[
   3d = 5.16
   \]
   \[
   d = \frac{5.16}{3} = 1.72
   \]

5. **Determine the least number of coins to make $1.72:**
   - Quarters: \(0.25\) dollars each
   - Dimes: \(0.10\) dollars each
   - Nickels: \(0.05\) dollars each
   - Pennies: \(0.01\) dollars each

   To minimize the number of coins:
   - Use as many quarters as possible: \(1.72 \div 0.25 = 6.88\), so use 6 quarters (\(6 \times 0.25 = 1.50\))
   - Remaining amount: \(1.72 - 1.50 = 0.22\)
   - Use as many dimes as possible: \(0.22 \div 0.10 = 2.2\), so use 2 dimes (\(2 \times 0.10 = 0.20\))
   - Remaining amount: \(0.22 - 0.20 = 0.02\)
   - Use pennies: \(0.02 \div 0.01 = 2\), so use 2 pennies

   Total number of coins:
   \[
   6 \text{ quarters} + 2 \text{ dimes} + 2 \text{ pennies} = 10 \text{ coins}
   \]

The final answer is \(\boxed{10}\).