Consider the system

[−2314+8a−11−4a16+2b−18−b]
 
 
Given that  b≠−3
 , for which value of the parameter  a∈R
  does the system have no solutions?

a=
   
Number
 

We are given the system:

\[
-23a + 8a - 11 - 4a = 16 + 2b - 18 - b
\]

This simplifies to:

### Step 1: Combine like terms

On the left-hand side:

\[
-23a + 8a - 4a = -19a - 11
\]

On the right-hand side:

\[
16 - 18 + 2b - b = -2 + b
\]

Thus, the equation becomes:

\[
-19a - 11 = -2 + b
\]

### Step 2: Solve for \(b\)

We can rewrite this equation as:

\[
-19a = b + 9
\]

### Step 3: Analyze when there is no solution

The system will have no solutions if there is a contradiction or inconsistency. Since the parameter \( b \) is defined and constrained by \( b \neq -3 \), and the equation doesn't introduce any additional contradictions, there will be no specific value of \( a \) that leads to no solution. The system has solutions for all real values of \(a\), as long as \( b \neq -3 \). 

Thus, **there is no value of \( a \) that leads to no solutions**.

Would you like any further details or explanations? Here are five related questions you might consider:

1. How would the solution change if the restriction \( b \neq -3 \) were removed?
2. Can we solve for \( a \) in terms of \( b \)?
3. What happens if \( b = -3 \) after all?
4. How does changing the coefficients in the equation affect the solution set?
5. Can we explore a scenario where both \( a \) and \( b \) are restricted?

**Tip:** When solving systems with parameters, always check if there are implicit or explicit restrictions on variables!

Consider the system [-23a + 8a - 11 - 4a = 16 + 2b - 18 - b]. Given that b ≠ -3, for which value of the parameter a ∈ R does the system have no solutions? a= Number

Explore how to determine the value of parameter 'a' in the system of linear equations [-23a + 8a - 11 - 4a = 16 + 2b - 18 - b] under the condition b ≠ -3. This problem covers concepts in algebra and solving systems of linear equations. Suitable for students in Grades 9-12.

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