Show all work to verify if the given point is a solution to the system of equations. [3y² + 3x²=6 2 4y² - 16x² + 12 = 0 Point: (-1,-1)

College Algebra (MindTap Course List)
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Chapter6: Linear Systems
Section6.1: Systems Of Linear Equations
Problem 85E
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**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.
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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