Use substitution to solve the system. 3x + 6y = -30 4x-y = 32 Select the correct choice below and, if necessary, fill in the answer box to complete your ch O A. The solution is (Type an ordered pair.) O B. There are infinitely many solutions. C. There is no solution.
Use substitution to solve the system. 3x + 6y = -30 4x-y = 32 Select the correct choice below and, if necessary, fill in the answer box to complete your ch O A. The solution is (Type an ordered pair.) O B. There are infinitely many solutions. C. There is no solution.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Solving Systems of Equations by Substitution
**Problem Statement:**
Use substitution to solve the system of equations:
\[ 3x + 6y = -30 \]
\[ 4x - y = 32 \]
**Instructions:**
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
**Options:**
- A. The solution is \(\_\_\_\_ \). (Type an ordered pair.)
- B. There are infinitely many solutions.
- C. There is no solution.
**Explanation:**
To solve this system using substitution, follow these steps:
1. Solve one of the equations for one variable.
2. Substitute this expression into the other equation.
3. Solve the resulting equation.
4. Substitute back to find the value of the other variable.
5. Check the solution in both original equations.
Given the equations:
\[ 4x - y = 32 \quad \text{(1)} \]
\[ 3x + 6y = -30 \quad \text{(2)} \]
First, isolate \( y \) in equation (1):
\[ 4x - 32 = y \]
\[ y = 4x - 32 \]
Next, substitute \( y = 4x - 32 \) into equation (2):
\[ 3x + 6(4x - 32) = -30 \]
\[ 3x + 24x - 192 = -30 \]
\[ 27x - 192 = -30 \]
\[ 27x = 162 \]
\[ x = \frac{162}{27} \]
\[ x = 6 \]
Now, substitute \( x = 6 \) back into \( y = 4x - 32 \):
\[ y = 4(6) - 32 \]
\[ y = 24 - 32 \]
\[ y = -8 \]
Thus, the solution is \( (6, -8) \).
**Correct Choice:**
- A. The solution is \( (6, -8) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57708f6a-cbd4-4295-9f28-330268fec57a%2Fd6c5aeb8-1759-40fa-9440-f1eb9cb610c5%2F788mjjm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Systems of Equations by Substitution
**Problem Statement:**
Use substitution to solve the system of equations:
\[ 3x + 6y = -30 \]
\[ 4x - y = 32 \]
**Instructions:**
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
**Options:**
- A. The solution is \(\_\_\_\_ \). (Type an ordered pair.)
- B. There are infinitely many solutions.
- C. There is no solution.
**Explanation:**
To solve this system using substitution, follow these steps:
1. Solve one of the equations for one variable.
2. Substitute this expression into the other equation.
3. Solve the resulting equation.
4. Substitute back to find the value of the other variable.
5. Check the solution in both original equations.
Given the equations:
\[ 4x - y = 32 \quad \text{(1)} \]
\[ 3x + 6y = -30 \quad \text{(2)} \]
First, isolate \( y \) in equation (1):
\[ 4x - 32 = y \]
\[ y = 4x - 32 \]
Next, substitute \( y = 4x - 32 \) into equation (2):
\[ 3x + 6(4x - 32) = -30 \]
\[ 3x + 24x - 192 = -30 \]
\[ 27x - 192 = -30 \]
\[ 27x = 162 \]
\[ x = \frac{162}{27} \]
\[ x = 6 \]
Now, substitute \( x = 6 \) back into \( y = 4x - 32 \):
\[ y = 4(6) - 32 \]
\[ y = 24 - 32 \]
\[ y = -8 \]
Thus, the solution is \( (6, -8) \).
**Correct Choice:**
- A. The solution is \( (6, -8) \).
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