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,...
Related questions
Question
![**Problem 8:**
Solve, graph, and write the solution in interval notation for the compound inequalities:
\[ 2x - 1 < 9 \ \text{and} \ -3x + 1 \leq 7 \]
**Solution Steps:**
1. **Solve the first inequality:**
\[
2x - 1 < 9
\]
- Add 1 to both sides:
\[
2x < 10
\]
- Divide by 2:
\[
x < 5
\]
2. **Solve the second inequality:**
\[
-3x + 1 \leq 7
\]
- Subtract 1 from both sides:
\[
-3x \leq 6
\]
- Divide by -3 (note the inequality sign flips):
\[
x \geq -2
\]
**Interval Notation:**
- Combine the solutions: \(-2 \leq x < 5\)
- In interval notation, the solution is \([-2, 5)\)
**Graph Explanation:**
- The graph is a number line with a closed dot at \(-2\) and an open dot at \(5\), shaded in between.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39f8eda8-4420-4bda-ab27-1b53a7c9ec36%2F01066845-e3db-4d48-b1d4-324261a5561c%2Fnjvwa0r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 8:**
Solve, graph, and write the solution in interval notation for the compound inequalities:
\[ 2x - 1 < 9 \ \text{and} \ -3x + 1 \leq 7 \]
**Solution Steps:**
1. **Solve the first inequality:**
\[
2x - 1 < 9
\]
- Add 1 to both sides:
\[
2x < 10
\]
- Divide by 2:
\[
x < 5
\]
2. **Solve the second inequality:**
\[
-3x + 1 \leq 7
\]
- Subtract 1 from both sides:
\[
-3x \leq 6
\]
- Divide by -3 (note the inequality sign flips):
\[
x \geq -2
\]
**Interval Notation:**
- Combine the solutions: \(-2 \leq x < 5\)
- In interval notation, the solution is \([-2, 5)\)
**Graph Explanation:**
- The graph is a number line with a closed dot at \(-2\) and an open dot at \(5\), shaded in between.
![**Problem 7:**
Find a line perpendicular to \(3x - 2y = -4\) and passes through the point \((4, -2)\).
**Discussion:**
To find the equation of the line perpendicular to a given line, we first need to determine the slope of the original line. The given line is in standard form: \(Ax + By = C\).
1. **Convert to Slope-Intercept Form:**
- The given equation is: \(3x - 2y = -4\).
- Solve for \(y\) to put it in slope-intercept form \(y = mx + b\).
\[
-2y = -3x - 4
\]
\[
y = \frac{3}{2}x + 2
\]
The slope (\(m\)) of the given line is \(\frac{3}{2}\).
2. **Find the Perpendicular Slope:**
- The slope of a line perpendicular to another is the negative reciprocal of the original slope.
- Perpendicular slope \(m_{\perp}\) is \(-\frac{2}{3}\).
3. **Equation of the Perpendicular Line:**
- Use the point-slope form \(y - y_1 = m(x - x_1)\).
- Point \((4, -2)\) and perpendicular slope \(-\frac{2}{3}\):
\[
y + 2 = -\frac{2}{3}(x - 4)
\]
Distribute and simplify:
\[
y + 2 = -\frac{2}{3}x + \frac{8}{3}
\]
\[
y = -\frac{2}{3}x + \frac{8}{3} - 2
\]
\[
y = -\frac{2}{3}x + \frac{2}{3}
\]
**Conclusion:**
The equation of the line perpendicular to \(3x - 2y = -4\) and passing through \((4, -2)\) is \(y = -\frac{2}{3}x + \frac{2}{3}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39f8eda8-4420-4bda-ab27-1b53a7c9ec36%2F01066845-e3db-4d48-b1d4-324261a5561c%2Fvz4m1cd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 7:**
Find a line perpendicular to \(3x - 2y = -4\) and passes through the point \((4, -2)\).
**Discussion:**
To find the equation of the line perpendicular to a given line, we first need to determine the slope of the original line. The given line is in standard form: \(Ax + By = C\).
1. **Convert to Slope-Intercept Form:**
- The given equation is: \(3x - 2y = -4\).
- Solve for \(y\) to put it in slope-intercept form \(y = mx + b\).
\[
-2y = -3x - 4
\]
\[
y = \frac{3}{2}x + 2
\]
The slope (\(m\)) of the given line is \(\frac{3}{2}\).
2. **Find the Perpendicular Slope:**
- The slope of a line perpendicular to another is the negative reciprocal of the original slope.
- Perpendicular slope \(m_{\perp}\) is \(-\frac{2}{3}\).
3. **Equation of the Perpendicular Line:**
- Use the point-slope form \(y - y_1 = m(x - x_1)\).
- Point \((4, -2)\) and perpendicular slope \(-\frac{2}{3}\):
\[
y + 2 = -\frac{2}{3}(x - 4)
\]
Distribute and simplify:
\[
y + 2 = -\frac{2}{3}x + \frac{8}{3}
\]
\[
y = -\frac{2}{3}x + \frac{8}{3} - 2
\]
\[
y = -\frac{2}{3}x + \frac{2}{3}
\]
**Conclusion:**
The equation of the line perpendicular to \(3x - 2y = -4\) and passing through \((4, -2)\) is \(y = -\frac{2}{3}x + \frac{2}{3}\).
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