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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Question
![**Verifying a Solution to a System of Equations**
To determine whether the point \((-1, -1)\) is a solution to the given system of equations, we will substitute the coordinates of the point into each equation and check if the equations are satisfied.
### Given System of Equations:
1. \(3y^2 + 3x^2 = 6\)
2. \(4y^2 - 16x^2 + 12 = 0\)
### Given Point:
\((-1, -1)\)
### Step-by-Step Verification:
**Equation 1: \(3y^2 + 3x^2 = 6\)**
1. Substitute \(x = -1\) and \(y = -1\) into the first equation:
\[
3(-1)^2 + 3(-1)^2 = 6
\]
2. Calculate the squares:
\[
3(1) + 3(1) = 6
\]
3. Simplify the equation:
\[
3 + 3 = 6
\]
4. Verify if the equation holds true:
\[
6 = 6 \quad \text{(True)}
\]
**Equation 2: \(4y^2 - 16x^2 + 12 = 0\)**
1. Substitute \(x = -1\) and \(y = -1\) into the second equation:
\[
4(-1)^2 - 16(-1)^2 + 12 = 0
\]
2. Calculate the squares:
\[
4(1) - 16(1) + 12 = 0
\]
3. Simplify the equation:
\[
4 - 16 + 12 = 0
\]
4. Combine the terms:
\[
0 = 0 \quad \text{(True)}
\]
Since both equations are satisfied, the point \((-1, -1)\) is a solution to the system of equations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ddf5237-61ec-402c-bcef-2feae18f38a9%2Fbe6f2177-a6a2-4f22-9ec7-56d3fd9c3b04%2F2lm8xf_processed.png&w=3840&q=75)
Transcribed Image Text:**Verifying a Solution to a System of Equations**
To determine whether the point \((-1, -1)\) is a solution to the given system of equations, we will substitute the coordinates of the point into each equation and check if the equations are satisfied.
### Given System of Equations:
1. \(3y^2 + 3x^2 = 6\)
2. \(4y^2 - 16x^2 + 12 = 0\)
### Given Point:
\((-1, -1)\)
### Step-by-Step Verification:
**Equation 1: \(3y^2 + 3x^2 = 6\)**
1. Substitute \(x = -1\) and \(y = -1\) into the first equation:
\[
3(-1)^2 + 3(-1)^2 = 6
\]
2. Calculate the squares:
\[
3(1) + 3(1) = 6
\]
3. Simplify the equation:
\[
3 + 3 = 6
\]
4. Verify if the equation holds true:
\[
6 = 6 \quad \text{(True)}
\]
**Equation 2: \(4y^2 - 16x^2 + 12 = 0\)**
1. Substitute \(x = -1\) and \(y = -1\) into the second equation:
\[
4(-1)^2 - 16(-1)^2 + 12 = 0
\]
2. Calculate the squares:
\[
4(1) - 16(1) + 12 = 0
\]
3. Simplify the equation:
\[
4 - 16 + 12 = 0
\]
4. Combine the terms:
\[
0 = 0 \quad \text{(True)}
\]
Since both equations are satisfied, the point \((-1, -1)\) is a solution to the system of equations.
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