Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.7: Solving Systems Of Linear And Quadratic Equations
Problem 3GP
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Question
![### Solve the System of Equations by Graphing
Consider the following system of linear equations:
\[
\begin{cases}
y = -2x + 2 \\
y = \frac{1}{2}x - 3
\end{cases}
\]
To graphically solve this system, we plot each equation on the coordinate grid.
1. **Graph of \( y = -2x + 2 \):**
- **Y-Intercept:** When \( x = 0 \), \( y = 2 \). So, the point (0, 2) is on the line.
- **Slope:** The slope is -2, which means for every 1 unit increase in \( x \), \( y \) decreases by 2 units.
2. **Graph of \( y = \frac{1}{2}x - 3 \):**
- **Y-Intercept:** When \( x = 0 \), \( y = -3 \). So, the point (0, -3) is on the line.
- **Slope:** The slope is \(\frac{1}{2}\), which means for every 2 units increase in \( x \), \( y \) increases by 1 unit.
### Graph Interpretation
The provided coordinate grid ranges from -6 to 6 on both the x-axis and y-axis.
- **Steps to Plot \( y = -2x + 2 \):**
- Start at the y-intercept (0, 2).
- Use the slope to find another point. For example, moving one unit to the right (x = 1) gives you \( y = -2(1) + 2 = 0 \). Plot the point (1, 0).
- Connect these points with a straight line.
- **Steps to Plot \( y = \frac{1}{2}x - 3 \):**
- Start at the y-intercept (0, -3).
- Use the slope to find another point. For example, moving two units to the right (x = 2) gives you \( y = \frac{1}{2}(2) - 3 = -2 \). Plot the point (2, -2).
- Connect these points with a straight line.
### Intersection Point
The intersection point of the two lines represents the solution to the system of](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa369931a-fe5c-47f4-90de-c0f8ffd54c08%2Fe8c6c111-24dd-410b-b746-341a7b69878b%2Faej959j_processed.png&w=3840&q=75)
Transcribed Image Text:### Solve the System of Equations by Graphing
Consider the following system of linear equations:
\[
\begin{cases}
y = -2x + 2 \\
y = \frac{1}{2}x - 3
\end{cases}
\]
To graphically solve this system, we plot each equation on the coordinate grid.
1. **Graph of \( y = -2x + 2 \):**
- **Y-Intercept:** When \( x = 0 \), \( y = 2 \). So, the point (0, 2) is on the line.
- **Slope:** The slope is -2, which means for every 1 unit increase in \( x \), \( y \) decreases by 2 units.
2. **Graph of \( y = \frac{1}{2}x - 3 \):**
- **Y-Intercept:** When \( x = 0 \), \( y = -3 \). So, the point (0, -3) is on the line.
- **Slope:** The slope is \(\frac{1}{2}\), which means for every 2 units increase in \( x \), \( y \) increases by 1 unit.
### Graph Interpretation
The provided coordinate grid ranges from -6 to 6 on both the x-axis and y-axis.
- **Steps to Plot \( y = -2x + 2 \):**
- Start at the y-intercept (0, 2).
- Use the slope to find another point. For example, moving one unit to the right (x = 1) gives you \( y = -2(1) + 2 = 0 \). Plot the point (1, 0).
- Connect these points with a straight line.
- **Steps to Plot \( y = \frac{1}{2}x - 3 \):**
- Start at the y-intercept (0, -3).
- Use the slope to find another point. For example, moving two units to the right (x = 2) gives you \( y = \frac{1}{2}(2) - 3 = -2 \). Plot the point (2, -2).
- Connect these points with a straight line.
### Intersection Point
The intersection point of the two lines represents the solution to the system of
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