Problem 32 - Algebra 1: Concepts and Skills

Question

In Exercises 32 and 33, use the following information. The graph below represents the need for low-income rental housing in the United States and the number of affordable rental units available. Use the points (0, 6200) and (25, 10,500) to write an equation for the number of housing units needed. Then use the points (0, 6500) and (25, 6100) to write an equation for the number of affordable units available.

Step-by-Step Solution

Answer

The equation for the number of housing units needed is \( y = 172x + 6200 \), and for the number of affordable units available is \( y = -160x + 6500 \).

Step 1: Calculate the slope of the housing units needed

The slope can be found using the formula \( m = \frac{{y2 - y1}}{{x2 - x1}} \), where \( (x1, y1) = (0, 6200) \) and \( (x2, y2) = (25, 10500) \). Thus, the slope \( m = \frac{{10500 - 6200}}{{25 - 0}} = 172 \).

Step 2: Form the equation of the housing units needed

To find the equation \( y = mx + b \), we know m is 172 from the previous step. By substituting the point (0, 6200) into the equation, we find that b = 6200. Therefore, the equation is \( y = 172x + 6200 \).

Step 3: Calculate the slope of the affordable housing units available

Using the same formula for the slope, and the points \( (x1, y1) = (0, 6500) \) and \( (x2, y2) = (25, 6100) \), we calculate \( m = \frac{{6100 - 6500}}{{25 - 0}} = -160 \).

Step 4: Form the equation of the affordable housing units available

With the slope -160, by substituting the point (0, 6500) into the equation \( y = mx + b \), we find that b = 6500. Therefore, the equation is \( y = -160x + 6500 \).

Key Concepts

Slope Calculation

Slope is a measure of how steep a line is. It's calculated using the formula \( m = \frac{{y_2 - y_1}}{{x_2 - x_1}} \), where \((x_1, y_1)\) and \((x_2, y_2)\) are two different points on the line.
To easily find the slope:

• Subtract the y-coordinates: \( y_2 - y_1 \).
• Subtract the x-coordinates: \( x_2 - x_1 \).
• Divide the difference in y by the difference in x: \( \frac{{y_2 - y_1}}{{x_2 - x_1}} \).

In the exercise, for the housing units needed, using the points \((0, 6200)\) and \((25, 10500)\), the slope was calculated as 172. Similarly, for the affordable housing units available, the slope was found to be -160 using points \((0, 6500)\) and \((25, 6100)\). This implies that the first situation is increasing, whereas the second is decreasing over time.

Graph Interpretation

Graphs are visual representations of data, important for understanding relationships between variables.
When interpreting graphs, look for:

• The slope, which indicates the rate of change.
• The y-intercept, where the line crosses the y-axis, telling us the starting value.
• Comparisons between different lines to understand how they vary.

In our specific graph, the line representing the need for housing units shows a positive slope, indicating an increase in need over time. Conversely, the slope of the line for affordable units available is negative, indicating a decrease in available units over time.
This contrast can be crucial in making predictions or supporting further analysis. It confirms that while the need is increasing, the available units are decreasing, which may highlight a potential issue or challenge to address.

Equation Formation

Forming a linear equation involves using the slope-intercept form, \(y = mx + b\), where:

• \(m\) is the slope.
• \(b\) is the y-intercept.

With the slope already calculated, substitute a known point from the line into the equation to find \(b\):

• Insert the slope into the equation.
• Use a point \((x, y)\) from the line and solve for \(b\) by substituting these values in.

For example, in the housing units needed case, the slope \(m\) was 172 and the point \((0, 6200)\) was used, resulting in the equation \(y = 172x + 6200\).
For the affordable housing units, the slope \(m\) was -160 and the point \((0, 6500)\) was chosen, which formed the equation \(y = -160x + 6500\).
This formula helps determine the line's behavior over time and assists in locating new points on the graph by relating \(x\) values to predicted \(y\) outputs.