To determine the equation for Lot A, we will start by observing the pattern in the given data.
The costs to park for different days for Lot A are as follows:
- On day 1, the cost is \[tex]$16.
- On day 2, the cost is \$[/tex]20.
- On day 3, the cost is \[tex]$24.
- On day 4, the cost is \$[/tex]28.
We can see that the cost increases by \[tex]$4 each additional day.
To find the linear equation representing the cost \(y\) versus the number of days \(x\) for Lot A, we use the form of a linear equation: \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
### Step-by-Step Process
1. Determine the slope \(m\):
- The slope \(m\) represents the rate of change in cost per day.
- From the data, we notice the cost increases by \$[/tex]4 each day. Therefore, [tex]\(m = 4\)[/tex].
2. Determine the y-intercept [tex]\(b\)[/tex]:
- To find the y-intercept [tex]\(b\)[/tex], we use one of the points given in the data. Let's use the point (1, 16) (i.e., on day 1, the cost is \$16).
- Plugging this point into the equation [tex]\(y = mx + b\)[/tex]:
[tex]\[
16 = 4(1) + b \\
16 = 4 + b \\
b = 16 - 4 \\
b = 12
\][/tex]
3. Write the equation:
- Using the slope [tex]\(m = 4\)[/tex] and the y-intercept [tex]\(b = 12\)[/tex], we can write the equation for Lot A:
[tex]\[
y = 4x + 12
\][/tex]
Therefore, the other equation in the system, representing the cost to park in Lot A, is:
[tex]\[
y = 4x + 12
\][/tex]
For Lot B, we are already provided with the equation [tex]\(y = 6x\)[/tex].
Thus, the system of linear equations used to determine on which day the cost to park is the same for both lots is:
[tex]\[
\begin{cases}
y = 6x \quad \text{(Lot B)} \\
y = 4x + 12 \quad \text{(Lot A)}
\end{cases}
\][/tex]
