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Solve the problem provided in the screenshot and provide a step-by-step solution

Let's solve the systems of linear equations provided step-by-step.

---

### (i) General System of Linear Equations:
\[
\begin{aligned}
ax + by &= c, \\
dx + ey &= f, \quad \text{where } a, b, c, d, e, f \in \mathbb{R}.
\end{aligned}
\]
This system of equations can be solved using either **substitution** or **elimination method**. We'll use the elimination method.

1. Multiply the first equation by \(e\) and the second equation by \(b\), to make the coefficients of \(y\) the same:
   \[
   \begin{aligned}
   e(ax + by) &= e(c), \\
   b(dx + ey) &= b(f),
   \end{aligned}
   \]
   which gives:
   \[
   \begin{aligned}
   eax + eby &= ec, \\
   bdx + bey &= bf.
   \end{aligned}
   \]

2. Subtract the second equation from the first to eliminate \(y\):
   \[
   eax - bdx = ec - bf.
   \]
   Factor out \(x\):
   \[
   x(ea - bd) = ec - bf,
   \]
   so:
   \[
   x = \frac{ec - bf}{ea - bd}.
   \]

3. Substitute \(x\) into one of the original equations to solve for \(y\). Using \(ax + by = c\), we have:
   \[
   a\left(\frac{ec - bf}{ea - bd}\right) + by = c.
   \]
   Solve for \(y\):
   \[
   by = c - a\left(\frac{ec - bf}{ea - bd}\right),
   \]
   \[
   by = \frac{c(ea - bd) - a(ec - bf)}{ea - bd}.
   \]
   Simplify:
   \[
   y = \frac{c(ea - bd) - a(ec - bf)}{b(ea - bd)}.
   \]

So, the solutions for \(x\) and \(y\) are:
\[
x = \frac{ec - bf}{ea - bd}, \quad y = \frac{c(ea - bd) - a(ec - bf)}{b(ea - bd)}.
\]

---

### (ii) Specific System of Linear Equations:
\[
\begin{aligned}
2x - y &= 1, \\
4x + 4y &= 20.
\end{aligned}
\]
1. Simplify the second equation by dividing it by 4:
   \[
   x + y = 5.
   \]
   
2. Now, we have the system:
   \[
   \begin{aligned}
   2x - y &= 1, \\
   x + y &= 5.
   \end{aligned}
   \]
   
3. Add both equations to eliminate \(y\):
   \[
   (2x - y) + (x + y) = 1 + 5,
   \]
   \[
   3x = 6,
   \]
   so:
   \[
   x = 2.
   \]

4. Substitute \(x = 2\) into \(x + y = 5\):
   \[
   2 + y = 5,
   \]
   \[
   y = 3.
   \]

Thus, the solution is \(x = 2\), \(y = 3\).

---

### (iii) System of Three Equations:
\[
\begin{aligned}
x_1 + x_2 &= 1, \\
2x_1 + 2x_2 + x_3 &= 1, \\
2x_1 + 2x_2 &= 2.
\end{aligned}
\]
1. From the third equation, simplify:
   \[
   2x_1 + 2x_2 = 2 \quad \Rightarrow \quad x_1 + x_2 = 1.
   \]
   This is the same as the first equation, so the third equation does not add new information.

2. Now we are left with:
   \[
   \begin{aligned}
   x_1 + x_2 &= 1, \\
   2x_1 + 2x_2 + x_3 &= 1.
   \end{aligned}
   \]

3. From the first equation, \(x_1 + x_2 = 1\), substitute this into the second equation:
   \[
   2(1) + x_3 = 1,
   \]
   \[
   2 + x_3 = 1,
   \]
   \[
   x_3 = -1.
   \]

Thus, the solutions are:
\[
x_1 + x_2 = 1, \quad x_3 = -1.
\]
This means \(x_1\) and \(x_2\) can take any values as long as they satisfy \(x_1 + x_2 = 1\).

---

Would you like further details or explanations on any step? Here are some related questions you might want to explore:

1. What happens if the determinant in system (i) is zero?
2. Can the elimination method be applied to larger systems?
3. How can we use matrix methods to solve systems of linear equations?
4. What is the difference between consistent and inconsistent systems?
5. How can the substitution method be applied to system (iii)?

**Tip:** Always check the determinant of the coefficient matrix when solving a system of linear equations to ensure it has a unique solution.

Solve the systems of linear equations provided in the image.

Explore how to solve systems of linear equations using substitution, elimination, and determinants. The example includes both two-variable and three-variable systems, covering step-by-step methods suitable for high school algebra students.

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