$\text { A real estate office handles an apartment complex with } 50 \text { units. When the rent is }$ $\$ 800$ per month, all 50 units are occupied. However, when the rent is $\$ 845$, the average number of occupied units drops to 47 . Assume that the relationship between the monthly rent $p$ and the demand $x$ is linear. Write a linear equation giving the demand $x$ in terms of the rent $p$.
A) $x = \frac { 1 } { 15 } ( 1595 - p )$
B) $x = \frac { 1 } { 15 } ( 1505 + p )$
C) $x = \frac { 1 } { 45 } ( 1550 + p )$
D) $x = \frac { 1 } { 15 } ( 1550 - p )$
E) $x = \frac { 1 } { 45 } ( 1595 - p )$

Question

Quizplus · Accepted Answer

Let's use the point-slope form for a linear equation:

$$y - y_1 = m(x - x_1)$$

where $m$ is the slope of the line and $(x_1,y_1)$ is a point on the line.

We know that when the rent is \$800, 50 units are occupied. So we have the point $(800,50)$.

When the rent is \$845, the average number of occupied units drops to 47. This means that if we graph the number of occupied units against the rent, we will get a line that passes through the points $(800,50)$ and $(845,47)$.

We first need to find the slope:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{47 - 50}{845 - 800} = -\frac{1}{9}$$

Now we can use the point-slope form:

$$y - 50 = -\frac{1}{9}(x - 800)$$

Simplifying:

$$y = -\frac{1}{9}x + \frac{4450}{9}$$

This is the equation of the line that gives the number of occupied units ($x$) as a function of the monthly rent ($p$). To make the substitution, we need to find $x$ when $p$ is given. We know that $p$ and $y$ are related by the equation:

$$p = 800 + \frac{45}{47}(y - 50)$$

or

$$p = 800 - \frac{45}{47}(50 - y)$$

We can simplify this into the form $y = mx + b$:

$$y = -\frac{47}{45}p + \frac{1595}{9}$$

This is in the form $y = mx + b$ and shows the relationship between the monthly rent ($p$) and the number of occupied units ($x$). The answer is (D).

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