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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The product of a number and a number increased by three is 28. Find the two numbers.
![The text appears to be a series of algebraic problems with solutions demonstrated step-by-step. Below is a transcription that could appear on an educational website:
---
### Solving Algebraic Equations by Factoring
**Problem 1:**
"The product of twice a number and three less than the number is 0. Find the two numbers."
**Solution:**
1. Let the number be \( n \).
2. Set up the equation: \( 2n(n-3) = 0 \).
3. Use the zero-product property:
\[
2n = 0 \quad \text{or} \quad n-3 = 0
\]
4. Solve each equation:
\( 2n = 0 \rightarrow n = 0 \)
\( n-3 = 0 \rightarrow n = 3 \)
5. Therefore, the two numbers are \( \{0, 3\} \).
---
**Problem 2:**
"The product of a number and a number increased by three is 28. Find the two numbers."
**Solution:**
1. Let the number be \( n \).
2. Set up the equation: \( n(n + 3) = 28 \).
3. This solution process is not shown in the image.
---
**Problem 3:**
"The product of two consecutive integers is 110. Find the two pairs of numbers."
**Solution:**
1. Let the integers be \( n \) and \( n + 1 \).
2. Set up the equation: \( n(n+1) = 110 \).
3. Expand and rearrange:
\[
n^2 + n - 110 = 0
\]
4. Factor the quadratic equation:
\[
(n-10)(n+11) = 0
\]
5. Solve each equation:
\( n-10 = 0 \rightarrow n = 10 \)
\( n+11 = 0 \rightarrow n = -11 \)
6. The two pairs of consecutive integers are \( \{10, 11\} \) and \( \{-11, -10\} \).
---
This presentation helps students understand how to manipulate algebraic expressions and solve them by factoring.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b9a38c0-77e3-442f-bc6a-7392be38ce1a%2Faf5e5d92-02df-4cb0-83ed-76e56a27d7c0%2Fvp5yase_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The text appears to be a series of algebraic problems with solutions demonstrated step-by-step. Below is a transcription that could appear on an educational website:
---
### Solving Algebraic Equations by Factoring
**Problem 1:**
"The product of twice a number and three less than the number is 0. Find the two numbers."
**Solution:**
1. Let the number be \( n \).
2. Set up the equation: \( 2n(n-3) = 0 \).
3. Use the zero-product property:
\[
2n = 0 \quad \text{or} \quad n-3 = 0
\]
4. Solve each equation:
\( 2n = 0 \rightarrow n = 0 \)
\( n-3 = 0 \rightarrow n = 3 \)
5. Therefore, the two numbers are \( \{0, 3\} \).
---
**Problem 2:**
"The product of a number and a number increased by three is 28. Find the two numbers."
**Solution:**
1. Let the number be \( n \).
2. Set up the equation: \( n(n + 3) = 28 \).
3. This solution process is not shown in the image.
---
**Problem 3:**
"The product of two consecutive integers is 110. Find the two pairs of numbers."
**Solution:**
1. Let the integers be \( n \) and \( n + 1 \).
2. Set up the equation: \( n(n+1) = 110 \).
3. Expand and rearrange:
\[
n^2 + n - 110 = 0
\]
4. Factor the quadratic equation:
\[
(n-10)(n+11) = 0
\]
5. Solve each equation:
\( n-10 = 0 \rightarrow n = 10 \)
\( n+11 = 0 \rightarrow n = -11 \)
6. The two pairs of consecutive integers are \( \{10, 11\} \) and \( \{-11, -10\} \).
---
This presentation helps students understand how to manipulate algebraic expressions and solve them by factoring.
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