Calculate Number of Apartments with 5 Tenants: Solving y-2 Average Equation

To solve this problem, we first note the setup: 5 apartments with 4 tenants and 6 apartments with 3 tenants are explicitly mentioned. That accounts for:

• $5 \times 4 = 20$ tenants from the 4-tenant apartments.
• $6 \times 3 = 18$ tenants from the 3-tenant apartments.

Now, the remaining $9$ apartments (since $20 - (5 + 6) = 9$) can house either 5 or 7 tenants. Let $x_1$ be the number of 5-tenant apartments and $x_2$ be the number of 7-tenant apartments:

• $x_1 + x_2 = 9$
• Total number of tenants = $20 + 18 + 5x_1 + 7x_2$

The average number of tenants per apartment is given by $y - 2$. We express the total tenant number equation:

$\frac{20 + 18 + 5x_1 + 7x_2}{20} = y - 2$

Substitute $x_2 = 9 - x_1$ into the tenant equation:

Total number of tenants = $38 + 5x_1 + 7(9 - x_1)$

= $38 + 5x_1 + 63 - 7x_1$

= $101 - 2x_1$

Average tenants = $\frac{101 - 2x_1}{20} = y - 2$

Multiplying throughout by 20:

101 - 2x_1 = 20(y - 2)

101 - 2x_1 = 20y - 40

Solving for $x_1$:

$-2x_1 = 20y - 40 - 101$

$-2x_1 = 20y - 141$

$x_1 = \frac{141 - 20y}{2}$

Therefore, $x_1 = 70.5 - 10y$.

Thus, the correct answer is $\boxed{70.5 - 10y}$.