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We are tasked with determining for which value of the parameter \( a \in \mathbb{R} \) the given system has no solutions. The system of linear equations is represented by the augmented matrix:

\[
\begin{pmatrix}
-8 & 4 - 6a & | & 2 + 4b \\
-8 & 6 & | & 4
\end{pmatrix}
\]

### Condition for No Solutions
A system of linear equations has no solutions if the equations are inconsistent. This typically happens if, after performing Gaussian elimination, we get a row where the left-hand side consists of all zeroes but the right-hand side is nonzero, i.e.,

\[
\begin{pmatrix} 0 & 0 & | & c \end{pmatrix} \quad \text{with} \quad c \neq 0
\]

### Step 1: Eliminate One Variable (Row Reduction)

We start by eliminating the first element in the second row (i.e., making the second row's first entry zero) using row operations.

To eliminate the \(-8\) in the second row, we subtract the first row from the second. The new second row becomes:
\[
R_2 \to R_2 - R_1
\]
This gives:
\[
-8 - (-8) = 0 \\
6 - (4 - 6a) = 6 - 4 + 6a = 2 + 6a \\
4 - (2 + 4b) = 4 - 2 - 4b = 2 - 4b
\]

Thus, the system becomes:
\[
\begin{pmatrix}
-8 & 4 - 6a & | & 2 + 4b \\
0 & 2 + 6a & | & 2 - 4b
\end{pmatrix}
\]

### Step 2: Check for Inconsistency
For the system to have no solutions, we need the second row to become of the form \( \begin{pmatrix} 0 & 0 & | & c \end{pmatrix} \), with \( c \neq 0 \). This requires both the coefficients of \( a \) and \( b \) in the second row to cancel out, i.e.,

\[
2 + 6a = 0 \quad \Rightarrow \quad a = -\frac{1}{3}
\]

Now, substitute \( a = -\frac{1}{3} \) into the second row:

\[
2 - 4b \quad \text{(we need this to be non-zero for inconsistency)}
\]

Thus, the system will have no solutions when \( a = -\frac{1}{3} \) and \( 2 - 4b \neq 0 \), or equivalently, \( b \neq \frac{1}{2} \), which is already given in the problem.

### Final Answer:
The system has no solutions when \( a = -\frac{1}{3} \).

---

Do you want any further details or clarifications? Here are some follow-up questions:

1. How would the solution process change if \( b = \frac{1}{2} \)?
2. What are the possible solution types (unique, infinite, or none) for a system of two equations in two variables?
3. How do Gaussian elimination and row reduction generally help in solving linear systems?
4. Could we solve this system graphically? What would it mean geometrically for the system to have no solutions?
5. What happens if we set \( a = 0 \)? Would the system still have a solution?

**Tip**: Inconsistency in linear systems often arises from parallel lines when working with two-variable systems or planes when dealing with three dimensions.

Given that b ≠ 1/2, for which value of the parameter a ∈ R does the system have no solutions?

Explore the conditions under which a system of linear equations has no solutions by determining the parameter a. This problem involves Gaussian elimination, row reduction, and the analysis of inconsistent systems, ideal for students in Grades 10-12.

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