To solve the given system of equations using Gaussian elimination, we need to convert the system into an augmented matrix and then perform row operations to reduce it to row-echelon form. The given system is:
[tex]\[
\begin{aligned}
5x + 4y &= -38 \quad \text{(Equation 1)} \\
-5x + 2y &= 11 \quad \text{(Equation 2)}
\end{aligned}
\][/tex]
Step-by-Step Solution:
### Step 1: Write the Augmented Matrix
First, write the augmented matrix for the system of equations:
[tex]\[
\left[\begin{array}{ccc}
5 & 4 & -38 \\
-5 & 2 & 11
\end{array}\right]
\][/tex]
### Step 2: Make the Leading Coefficient of Row 1 Equal to 1
To simplify Row 1, we divide the entire row by 5:
[tex]\[
\left[\begin{array}{ccc}
1 & \frac{4}{5} & -\frac{38}{5} \\
-5 & 2 & 11
\end{array}\right]
\][/tex]
### Step 3: Eliminate the [tex]\(x\)[/tex]-term in Row 2
Add 5 times Row 1 to Row 2 to make the [tex]\(x\)[/tex]-coefficient in Row 2 zero:
[tex]\[
\left[\begin{array}{ccc}
1 & \frac{4}{5} & -\frac{38}{5} \\
0 & 6 & -27
\end{array}\right]
\][/tex]
### Step 4: Make the Leading Coefficient of Row 2 Equal to 1
Divide Row 2 by 6:
[tex]\[
\left[\begin{array}{ccc}
1 & \frac{4}{5} & -\frac{38}{5} \\
0 & 1 & -4.5
\end{array}\right]
\][/tex]
### Step 5: Eliminate the [tex]\(y\)[/tex]-term in Row 1
Subtract [tex]\(\frac{4}{5}\)[/tex] times Row 2 from Row 1 to make the [tex]\(y\)[/tex]-coefficient in Row 1 zero:
[tex]\[
\left[\begin{array}{ccc}
1 & 0 & -4 \\
0 & 1 & -4.5
\end{array}\right]
\][/tex]
### Step 6: Extract the Solutions
From the final matrix, we can directly read the solutions for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
\begin{aligned}
x &= -4 \\
y &= -4.5
\end{aligned}
\][/tex]
Therefore, the unique solution to the system of equations is:
[tex]\[
(-4, -4.5)
\][/tex]
So, the correct choice is:
A. There is one solution. The solution is [tex]\((-4, -4.5)\)[/tex].
