Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Graphing a System of Inequalities**
On the set of axes below, we will graph the following system of inequalities:
\[
2x + y \ge 8
\]
\[
y - 5 < 3x
\]
To graph these inequalities, follow these steps:
### Step-by-Step Instructions:
1. **Graphing the Boundary Line for \(2x + y \ge 8\):**
- Rewrite the inequality in slope-intercept form: \( y \ge -2x + 8 \).
- Plot the boundary line \( y = -2x + 8 \). This line has a slope of -2 and a y-intercept of 8.
- Since the inequality is \( \ge \), shade the region above the line.
2. **Graphing the Boundary Line for \(y - 5 < 3x\):**
- Rewrite the inequality in slope-intercept form: \( y < 3x + 5 \).
- Plot the boundary line \( y = 3x + 5 \). This line has a slope of 3 and a y-intercept of 5.
- Since the inequality is \( < \), shade the region below the line with a dashed line to indicate that points on the line are not included in the solution set.
3. **Determine the Solution Region:**
- The solution region will be the area where the shaded regions of both inequalities overlap.
### Determine if the Point (1, 8) is in the Solution Set:
Now, let's determine if the point \( (1, 8) \) is in the solution set:
1. Substitute \( x = 1 \) and \( y = 8 \) into \( 2x + y \ge 8 \):
\[
2(1) + 8 \ge 8 \implies 2 + 8 \ge 8 \implies 10 \ge 8
\]
This inequality is true.
2. Substitute \( x = 1 \) and \( y = 8 \) into \( y - 5 < 3x \):
\[
8 - 5 < 3(1) \implies 3 < 3
\]
This inequality is false since 3 is not less than 3.
Therefore, the point \( (1, 8](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdbc53fb1-a2da-4d15-b47e-51a8a2b53fbe%2F1e4d93d3-3efc-482f-bced-debef0dac9a9%2Fv1x40bq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Graphing a System of Inequalities**
On the set of axes below, we will graph the following system of inequalities:
\[
2x + y \ge 8
\]
\[
y - 5 < 3x
\]
To graph these inequalities, follow these steps:
### Step-by-Step Instructions:
1. **Graphing the Boundary Line for \(2x + y \ge 8\):**
- Rewrite the inequality in slope-intercept form: \( y \ge -2x + 8 \).
- Plot the boundary line \( y = -2x + 8 \). This line has a slope of -2 and a y-intercept of 8.
- Since the inequality is \( \ge \), shade the region above the line.
2. **Graphing the Boundary Line for \(y - 5 < 3x\):**
- Rewrite the inequality in slope-intercept form: \( y < 3x + 5 \).
- Plot the boundary line \( y = 3x + 5 \). This line has a slope of 3 and a y-intercept of 5.
- Since the inequality is \( < \), shade the region below the line with a dashed line to indicate that points on the line are not included in the solution set.
3. **Determine the Solution Region:**
- The solution region will be the area where the shaded regions of both inequalities overlap.
### Determine if the Point (1, 8) is in the Solution Set:
Now, let's determine if the point \( (1, 8) \) is in the solution set:
1. Substitute \( x = 1 \) and \( y = 8 \) into \( 2x + y \ge 8 \):
\[
2(1) + 8 \ge 8 \implies 2 + 8 \ge 8 \implies 10 \ge 8
\]
This inequality is true.
2. Substitute \( x = 1 \) and \( y = 8 \) into \( y - 5 < 3x \):
\[
8 - 5 < 3(1) \implies 3 < 3
\]
This inequality is false since 3 is not less than 3.
Therefore, the point \( (1, 8
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