Solving Equations Containing Algebraic Fractions Worksheet for 8th ... - Free Printable

Problem Analysis:

The task involves solving a system of linear equations using the Gauss-Jordan elimination method. The goal is to transform the augmented matrix into its reduced row echelon form (RREF) to find the solution(s) for the variables.

#### Given System of Equations:
We are tasked with solving two systems of equations:

1. Example 1:
\[
\begin{cases}
x + y = 7 \\
2x - y = 8
\end{cases}
\]

2. Example 2:
\[
\begin{cases}
x + y + z = 6 \\
x - y + z = 2 \\
2x + y - z = 5
\end{cases}
\]

#### Steps to Solve:
1. Write the augmented matrix for each system.
2. Use elementary row operations to transform the matrix into RREF.
3. Interpret the RREF to find the solution(s).

---

Solution for Example 1:

#### Step 1: Write the Augmented Matrix
The given system is:
\[
\begin{cases}
x + y = 7 \\
2x - y = 8
\end{cases}
\]
The augmented matrix is:
\[
\begin{bmatrix}
1 & 1 & | & 7 \\
2 & -1 & | & 8
\end{bmatrix}
\]

#### Step 2: Transform to RREF
1. Eliminate \( x \) from the second row:
- Subtract \( 2 \times \text{(Row 1)} \) from Row 2:
\[
R_2 \leftarrow R_2 - 2 \cdot R_1
\]
\[
\begin{bmatrix}
1 & 1 & | & 7 \\
2 - 2(1) & -1 - 2(1) & | & 8 - 2(7)
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & | & 7 \\
0 & -3 & | & -6
\end{bmatrix}
\]

2. Normalize the second row:
- Divide Row 2 by \(-3\):
\[
R_2 \leftarrow \frac{R_2}{-3}
\]
\[
\begin{bmatrix}
1 & 1 & | & 7 \\
0 & 1 & | & 2
\end{bmatrix}
\]

3. Eliminate \( y \) from the first row:
- Subtract \( 1 \times \text{(Row 2)} \) from Row 1:
\[
R_1 \leftarrow R_1 - R_2
\]
\[
\begin{bmatrix}
1 & 1 - 1 & | & 7 - 2 \\
0 & 1 & | & 2
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & | & 5 \\
0 & 1 & | & 2
\end{bmatrix}
\]

#### Step 3: Interpret the Solution
The RREF corresponds to the system:
\[
\begin{cases}
x = 5 \\
y = 2
\end{cases}
\]

Thus, the solution is:
\[
\boxed{(5, 2)}
\]

---

Solution for Example 2:

#### Step 1: Write the Augmented Matrix
The given system is:
\[
\begin{cases}
x + y + z = 6 \\
x - y + z = 2 \\
2x + y - z = 5
\end{cases}
\]
The augmented matrix is:
\[
\begin{bmatrix}
1 & 1 & 1 & | & 6 \\
1 & -1 & 1 & | & 2 \\
2 & 1 & -1 & | & 5
\end{bmatrix}
\]

#### Step 2: Transform to RREF
1. Eliminate \( x \) from the second and third rows:
- Subtract \( 1 \times \text{(Row 1)} \) from Row 2:
\[
R_2 \leftarrow R_2 - R_1
\]
\[
\begin{bmatrix}
1 & 1 & 1 & | & 6 \\
1 - 1 & -1 - 1 & 1 - 1 & | & 2 - 6 \\
2 & 1 & -1 & | & 5
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 1 & | & 6 \\
0 & -2 & 0 & | & -4 \\
2 & 1 & -1 & | & 5
\end{bmatrix}
\]

- Subtract \( 2 \times \text{(Row 1)} \) from Row 3:
\[
R_3 \leftarrow R_3 - 2 \cdot R_1
\]
\[
\begin{bmatrix}
1 & 1 & 1 & | & 6 \\
0 & -2 & 0 & | & -4 \\
2 - 2(1) & 1 - 2(1) & -1 - 2(1) & | & 5 - 2(6)
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 1 & | & 6 \\
0 & -2 & 0 & | & -4 \\
0 & -1 & -3 & | & -7
\end{bmatrix}
\]

2. Normalize the second row:
- Divide Row 2 by \(-2\):
\[
R_2 \leftarrow \frac{R_2}{-2}
\]
\[
\begin{bmatrix}
1 & 1 & 1 & | & 6 \\
0 & 1 & 0 & | & 2 \\
0 & -1 & -3 & | & -7
\end{bmatrix}
\]

3. Eliminate \( y \) from the first and third rows:
- Subtract \( 1 \times \text{(Row 2)} \) from Row 1:
\[
R_1 \leftarrow R_1 - R_2
\]
\[
\begin{bmatrix}
1 & 1 - 1 & 1 & | & 6 - 2 \\
0 & 1 & 0 & | & 2 \\
0 & -1 & -3 & | & -7
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 1 & | & 4 \\
0 & 1 & 0 & | & 2 \\
0 & -1 & -3 & | & -7
\end{bmatrix}
\]

- Add \( 1 \times \text{(Row 2)} \) to Row 3:
\[
R_3 \leftarrow R_3 + R_2
\]
\[
\begin{bmatrix}
1 & 0 & 1 & | & 4 \\
0 & 1 & 0 & | & 2 \\
0 & -1 + 1 & -3 & | & -7 + 2
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 1 & | & 4 \\
0 & 1 & 0 & | & 2 \\
0 & 0 & -3 & | & -5
\end{bmatrix}
\]

4. Normalize the third row:
- Divide Row 3 by \(-3\):
\[
R_3 \leftarrow \frac{R_3}{-3}
\]
\[
\begin{bmatrix}
1 & 0 & 1 & | & 4 \\
0 & 1 & 0 & | & 2 \\
0 & 0 & 1 & | & \frac{5}{3}
\end{bmatrix}
\]

5. Eliminate \( z \) from the first row:
- Subtract \( 1 \times \text{(Row 3)} \) from Row 1:
\[
R_1 \leftarrow R_1 - R_3
\]
\[
\begin{bmatrix}
1 & 0 & 1 - 1 & | & 4 - \frac{5}{3} \\
0 & 1 & 0 & | & 2 \\
0 & 0 & 1 & | & \frac{5}{3}
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 & | & \frac{7}{3} \\
0 & 1 & 0 & | & 2 \\
0 & 0 & 1 & | & \frac{5}{3}
\end{bmatrix}
\]

#### Step 3: Interpret the Solution
The RREF corresponds to the system:
\[
\begin{cases}
x = \frac{7}{3} \\
y = 2 \\
z = \frac{5}{3}
\end{cases}
\]

Thus, the solution is:
\[
\boxed{\left( \frac{7}{3}, 2, \frac{5}{3} \right)}
\]

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Final Answers:

1. Example 1: \(\boxed{(5, 2)}\)
2. Example 2: \(\boxed{\left( \frac{7}{3}, 2, \frac{5}{3} \right)}\)