Answer :
Certainly! Let's go through the problem step-by-step to see if the point (6, 1850) is on the line that represents the instructor's savings over time.
### Step-by-Step Solution:
#### 1. Identify the Known Points
We are given:
- The initial savings amount: $3,500
- The monthly rent: $275
- The point to check: (6, 1850)
We will use the slope-intercept form of a line: \( y = mx + b \).
#### 2. Define the Points
- Initial point (when \( x = 0 \)): \((0, 3500)\)
- Given point: \((6, 1850)\)
#### 3. Find the Slope of the Line
To find the slope \( m \), use the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Here,
- \((x_1, y_1)\) = \((0, 3500)\)
- \((x_2, y_2)\) = \((6, 1850)\)
Substitute the values into the slope formula:
[tex]\[ m = \frac{1850 - 3500}{6 - 0} \][/tex]
[tex]\[ m = \frac{-1650}{6} \][/tex]
[tex]\[ m = -275 \][/tex]
Therefore, the slope \( m \) is \(-275\).
#### 4. Determine the Y-Intercept
From the initial point (0, 3500), we can see that the y-intercept \( b \) is the y-value when \( x = 0 \):
[tex]\[ b = 3500 \][/tex]
#### 5. Write the Equation of the Line
Using the slope \( m \) and the y-intercept \( b \), the equation of the line is:
[tex]\[ y = -275x + 3500 \][/tex]
#### 6. Check If the Point Is on the Line
To check if the point \((6, 1850)\) lies on this line, substitute \( x = 6 \) into the line equation and see if \( y = 1850 \).
Substitute \( x = 6 \) into the equation:
[tex]\[ y = -275(6) + 3500 \][/tex]
[tex]\[ y = -1650 + 3500 \][/tex]
[tex]\[ y = 1850 \][/tex]
Since this matches the y-coordinate of our point (6, 1850), the point lies on our line.
### Conclusion
The point (6, 1850) does indeed lie on the line represented by the equation [tex]\( y = -275x + 3500 \)[/tex]. Hence, the instructor is correct. The point (6, 1850) is on the line that models the instructor's savings over time, taking into account the monthly rent payment.
### Step-by-Step Solution:
#### 1. Identify the Known Points
We are given:
- The initial savings amount: $3,500
- The monthly rent: $275
- The point to check: (6, 1850)
We will use the slope-intercept form of a line: \( y = mx + b \).
#### 2. Define the Points
- Initial point (when \( x = 0 \)): \((0, 3500)\)
- Given point: \((6, 1850)\)
#### 3. Find the Slope of the Line
To find the slope \( m \), use the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Here,
- \((x_1, y_1)\) = \((0, 3500)\)
- \((x_2, y_2)\) = \((6, 1850)\)
Substitute the values into the slope formula:
[tex]\[ m = \frac{1850 - 3500}{6 - 0} \][/tex]
[tex]\[ m = \frac{-1650}{6} \][/tex]
[tex]\[ m = -275 \][/tex]
Therefore, the slope \( m \) is \(-275\).
#### 4. Determine the Y-Intercept
From the initial point (0, 3500), we can see that the y-intercept \( b \) is the y-value when \( x = 0 \):
[tex]\[ b = 3500 \][/tex]
#### 5. Write the Equation of the Line
Using the slope \( m \) and the y-intercept \( b \), the equation of the line is:
[tex]\[ y = -275x + 3500 \][/tex]
#### 6. Check If the Point Is on the Line
To check if the point \((6, 1850)\) lies on this line, substitute \( x = 6 \) into the line equation and see if \( y = 1850 \).
Substitute \( x = 6 \) into the equation:
[tex]\[ y = -275(6) + 3500 \][/tex]
[tex]\[ y = -1650 + 3500 \][/tex]
[tex]\[ y = 1850 \][/tex]
Since this matches the y-coordinate of our point (6, 1850), the point lies on our line.
### Conclusion
The point (6, 1850) does indeed lie on the line represented by the equation [tex]\( y = -275x + 3500 \)[/tex]. Hence, the instructor is correct. The point (6, 1850) is on the line that models the instructor's savings over time, taking into account the monthly rent payment.
