**Solving Systems of Equations by Substitution** Solve the following system of equations by substitution: \[ \begin{align*} x + y &= 0 \\ 3x - 2y &= 10 \end{align*} \] **Choices for the Solution:** - \(\left(\frac{3}{1/3}, 0 \right)\) - \((2, -2)\) - \(\emptyset \) (the null set) - \((2, 2)\) To solve this system using substitution, follow these steps: 1. From the first equation \( x + y = 0 \), solve for \( y \): \[ y = -x \] 2. Substitute \( y = -x \) into the second equation \( 3x - 2y = 10 \): \[ 3x - 2(-x) = 10 \] Simplify the equation: \[ 3x + 2x = 10 \] \[ 5x = 10 \] \[ x = 2 \] 3. Substitute \( x = 2 \) back into the equation \( y = -x \) to find \( y \): \[ y = -2 \] Therefore, the solution to the system of equations is \((2, -2)\). **Answer:** \((2, -2)\)
**Solving Systems of Equations by Substitution** Solve the following system of equations by substitution: \[ \begin{align*} x + y &= 0 \\ 3x - 2y &= 10 \end{align*} \] **Choices for the Solution:** - \(\left(\frac{3}{1/3}, 0 \right)\) - \((2, -2)\) - \(\emptyset \) (the null set) - \((2, 2)\) To solve this system using substitution, follow these steps: 1. From the first equation \( x + y = 0 \), solve for \( y \): \[ y = -x \] 2. Substitute \( y = -x \) into the second equation \( 3x - 2y = 10 \): \[ 3x - 2(-x) = 10 \] Simplify the equation: \[ 3x + 2x = 10 \] \[ 5x = 10 \] \[ x = 2 \] 3. Substitute \( x = 2 \) back into the equation \( y = -x \) to find \( y \): \[ y = -2 \] Therefore, the solution to the system of equations is \((2, -2)\). **Answer:** \((2, -2)\)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Solving Systems of Equations by Substitution**
Solve the following system of equations by substitution:
\[ \begin{align*}
x + y &= 0 \\
3x - 2y &= 10
\end{align*} \]
**Choices for the Solution:**
- \(\left(\frac{3}{1/3}, 0 \right)\)
- \((2, -2)\)
- \(\emptyset \) (the null set)
- \((2, 2)\)
To solve this system using substitution, follow these steps:
1. From the first equation \( x + y = 0 \), solve for \( y \):
\[
y = -x
\]
2. Substitute \( y = -x \) into the second equation \( 3x - 2y = 10 \):
\[
3x - 2(-x) = 10
\]
Simplify the equation:
\[
3x + 2x = 10
\]
\[
5x = 10
\]
\[
x = 2
\]
3. Substitute \( x = 2 \) back into the equation \( y = -x \) to find \( y \):
\[
y = -2
\]
Therefore, the solution to the system of equations is \((2, -2)\).
**Answer:** \((2, -2)\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb4c9257-c3a4-42ab-bf0e-0de9c4c5a001%2F5080fdf1-7dcb-42dd-aca7-21fa13153db9%2F4ikm9pq.jpeg&w=3840&q=75)
Transcribed Image Text:**Solving Systems of Equations by Substitution**
Solve the following system of equations by substitution:
\[ \begin{align*}
x + y &= 0 \\
3x - 2y &= 10
\end{align*} \]
**Choices for the Solution:**
- \(\left(\frac{3}{1/3}, 0 \right)\)
- \((2, -2)\)
- \(\emptyset \) (the null set)
- \((2, 2)\)
To solve this system using substitution, follow these steps:
1. From the first equation \( x + y = 0 \), solve for \( y \):
\[
y = -x
\]
2. Substitute \( y = -x \) into the second equation \( 3x - 2y = 10 \):
\[
3x - 2(-x) = 10
\]
Simplify the equation:
\[
3x + 2x = 10
\]
\[
5x = 10
\]
\[
x = 2
\]
3. Substitute \( x = 2 \) back into the equation \( y = -x \) to find \( y \):
\[
y = -2
\]
Therefore, the solution to the system of equations is \((2, -2)\).
**Answer:** \((2, -2)\)
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