Marcelo was watching Math Nation videos and he solved a Try It! exercise that asked him to solve a linear system in three variables. His work is shown below. Try It! 3. Solve the following system of equations: -3a + 4b + 2c =1 e+2b – c = 6) (2a - b+ 3c = 7) a +26-c =6 + 4a -26 +bc = 14 |-3& +46 +2c = 1 -)a -46 +2c :-1a 5a + -Sa +Vc = -11 -5a + 4c= -11 5a + Sc = 20 5a +5l-1) : 20 Sa -5= 20 Sa= 25 C = -1 frnal stop at a6-c =6 S+26 -(-1) =6 5+26+1 =6 26 =0 6=0 Check Marcelo's work. Identify and correct any mistakes, if needed. If his work is perfect (no mistakes), then check that the solutions work in every equation of the system.
Marcelo was watching Math Nation videos and he solved a Try It! exercise that asked him to solve a linear system in three variables. His work is shown below. Try It! 3. Solve the following system of equations: -3a + 4b + 2c =1 e+2b – c = 6) (2a - b+ 3c = 7) a +26-c =6 + 4a -26 +bc = 14 |-3& +46 +2c = 1 -)a -46 +2c :-1a 5a + -Sa +Vc = -11 -5a + 4c= -11 5a + Sc = 20 5a +5l-1) : 20 Sa -5= 20 Sa= 25 C = -1 frnal stop at a6-c =6 S+26 -(-1) =6 5+26+1 =6 26 =0 6=0 Check Marcelo's work. Identify and correct any mistakes, if needed. If his work is perfect (no mistakes), then check that the solutions work in every equation of the system.
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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Question
![**Marcelo's Solution to the Linear System**
Marcelo was watching Math Nation videos and solved a Try It! exercise involving a linear system with three variables. The task is to solve the system, and his work is shown below.
### Try It!
3. **Solve the following system of equations:**
\[
\begin{align*}
-3a + 4b + 2c &= 1 \quad \quad \quad \text{(Equation 1)} \\
a + 2b - c &= 6 \quad \quad \quad \text{(Equation 2)} \\
2a - b + 3c &= 7 \quad \quad \quad \text{(Equation 3)}
\end{align*}
\]
**Steps:**
1. **Substitute and Simplify:**
- From Equation 2:
\[
a + 2b - c = 6
\]
Solve to express \( a \) in terms of \( b \) and \( c \):
\[
a + 2b = c + 6
\]
2. **Combination of Equations:**
- Use Equation 2:
\[
a + 2b - c = 6
\]
- Use Equation 1 to eliminate variable:
\[
4a - 2b + 4c = 14
\]
Adding these, results in:
\[
5a + 5c = 20 \quad \Rightarrow \quad 5a = 20 \quad \Rightarrow \quad a = 5
\]
3. **Substitute Back to Find Other Variables:**
- Substitute \( a = 5 \) into Equation 3:
\[
2(5) - b + 3c = 7 \quad \Rightarrow \quad 10 - b + 3c = 7
\]
- Simplify:
\[
-b + 3c = -3
\]
- Solve for \( b \):
\[
b = 0
\]
4. **Solve for the Remaining Variable:**
- Substitute values of \( a \) and \( b \) into Equation 1:
\[
-3(5) +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe5ed07d5-7a8f-4097-9c81-82df457ca627%2F286c3669-36b2-4c80-b988-9e379647b5f6%2Fn0wmn5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Marcelo's Solution to the Linear System**
Marcelo was watching Math Nation videos and solved a Try It! exercise involving a linear system with three variables. The task is to solve the system, and his work is shown below.
### Try It!
3. **Solve the following system of equations:**
\[
\begin{align*}
-3a + 4b + 2c &= 1 \quad \quad \quad \text{(Equation 1)} \\
a + 2b - c &= 6 \quad \quad \quad \text{(Equation 2)} \\
2a - b + 3c &= 7 \quad \quad \quad \text{(Equation 3)}
\end{align*}
\]
**Steps:**
1. **Substitute and Simplify:**
- From Equation 2:
\[
a + 2b - c = 6
\]
Solve to express \( a \) in terms of \( b \) and \( c \):
\[
a + 2b = c + 6
\]
2. **Combination of Equations:**
- Use Equation 2:
\[
a + 2b - c = 6
\]
- Use Equation 1 to eliminate variable:
\[
4a - 2b + 4c = 14
\]
Adding these, results in:
\[
5a + 5c = 20 \quad \Rightarrow \quad 5a = 20 \quad \Rightarrow \quad a = 5
\]
3. **Substitute Back to Find Other Variables:**
- Substitute \( a = 5 \) into Equation 3:
\[
2(5) - b + 3c = 7 \quad \Rightarrow \quad 10 - b + 3c = 7
\]
- Simplify:
\[
-b + 3c = -3
\]
- Solve for \( b \):
\[
b = 0
\]
4. **Solve for the Remaining Variable:**
- Substitute values of \( a \) and \( b \) into Equation 1:
\[
-3(5) +
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