https://file.aiquickdraw.com/mathbot/file/1727986937158vv4gpa45.jpg

The problem asks to find all values of \( a \) for which the system has no solution. Let's focus on exercise 36 in the image:

The system of equations given is:

\[
\begin{aligned}
1. \quad x_1 + 2x_2 &= -3 \\
2. \quad ax_1 - 2x_2 &= 5
\end{aligned}
\]

To determine the values of \( a \) that make the system inconsistent (i.e., it has no solution), we can approach this by examining the system in terms of linear dependence between the two equations.

### Step 1: Elimination Method

We aim to eliminate one variable. Multiply the first equation by \( a \) to align the \( x_1 \) terms:

\[
a(x_1 + 2x_2) = a(-3)
\]
\[
ax_1 + 2ax_2 = -3a
\]

Now, subtract the second equation from this new equation:

\[
(ax_1 + 2ax_2) - (ax_1 - 2x_2) = -3a - 5
\]

Simplifying:

\[
2ax_2 + 2x_2 = -3a - 5
\]
\[
(2a + 2)x_2 = -3a - 5
\]

### Step 2: Determine when there is no solution

For the system to have no solution, the value of \( x_2 \) must be inconsistent. This happens if the left side is zero but the right side is not. Therefore, set the coefficient of \( x_2 \) to zero:

\[
2a + 2 = 0
\]
\[
a = -1
\]

### Step 3: Substitute \( a = -1 \)

Substitute \( a = -1 \) into the second equation to check if the system becomes inconsistent:

\[
-3x_2 = -3(-1) - 5 = 3 - 5 = -2
\]

This simplifies to:

\[
-4x_2 = -2
\]

which gives \( x_2 = \frac{1}{2} \).

Thus, the system is consistent when \( a = -1 \). So, the only value of \( a \) that leads to no solution is when the right-hand side and the left-hand side contradict. Therefore, there are no specific conditions leading to contradictions for \( a = -1 \).

Would you like more details or have any questions?

Here are five related questions:
1. How would you solve the system if \( a = 0 \)?
2. What does it mean graphically if a system has no solution?
3. How do you determine if a system of linear equations is inconsistent?
4. How can you solve the system using matrix methods?
5. What changes in the system when \( a \) takes positive or negative values?

**Tip:** When solving systems of equations, always check for cases where variables cancel out, leading to contradictions or infinitely many solutions.

Find all values of a for which the system has no solution:
1. x1 + 2x2 = -3
2. ax1 - 2x2 = 5

Explore how to find values of 'a' that make a system of linear equations inconsistent. Learn about the elimination method and criteria for systems with no solutions. Suitable for students in Grades 10-12.

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