Understanding System Consistency

A system of linear equations is considered consistent if it has at least one solution. This means the equations do not contradict each other. A system can have a unique solution (exactly one) or infinitely many solutions (many).

We are given the following system of linear equations:

1. x + 2y - 3z = 2
2. (k + 3)z = 3
3. (2k + 1)y + z = 2

We need to find the values of k for which this system is consistent.

Identifying Critical Values of K

Let's look at equations 2 and 3, as they involve k directly and fewer variables:

• From equation (2), if k + 3 is not zero, we can find a unique value for z: $z = \frac{3}{k+3}$. If k + 3 is zero, the equation changes significantly.
• From equation (3), if 2k + 1 is not zero, we can potentially solve for y in terms of z. If 2k + 1 is zero, the term with y vanishes.

The critical values of k to examine are those where the coefficients involving k become zero:

• k + 3 = 0 $\implies$ k = -3
• 2k + 1 = 0 $\implies$ k = -1/2

Let's analyze the system's consistency at these critical values.

Case 1: K = -3

Substitute k = -3 into the system of equations:

1. x + 2y - 3z = 2
2. (-3 + 3)z = 3 $\implies$ 0 * z = 3 $\implies$ 0 = 3
3. (2(-3) + 1)y + z = 2 $\implies$ (-6 + 1)y + z = 2 $\implies$ -5y + z = 2

Equation (2) simplifies to 0 = 3, which is a contradiction. A system of equations that includes a contradiction has no solution. Therefore, when k = -3, the system is inconsistent.

Case 2: K = -1/2

Substitute k = -1/2 into the system of equations:

1. x + 2y - 3z = 2
2. (-1/2 + 3)z = 3 $\implies$ (5/2)z = 3 $\implies$ z = $3 \times \frac{2}{5} = \frac{6}{5}$
3. (2(-1/2) + 1)y + z = 2 $\implies$ (-1 + 1)y + z = 2 $\implies$ 0 * y + z = 2 $\implies$ z = 2

From equation (2), we get z = 6/5. From equation (3), we get z = 2. Since 6/5 is not equal to 2, there is a contradiction in the value of z. The system has no solution. Therefore, when k = -1/2, the system is also inconsistent.

Concluding for System Consistency

Our analysis shows that the system of equations is inconsistent when k = -3 and when k = -1/2. For any other value of k (where k ≠ -3 and k ≠ -1/2), equations (2) and (3) can be solved for unique values of z and y, which can then be used in equation (1) to find a unique value for x. Thus, for k ≠ -3 and k ≠ -1/2, the system is consistent (specifically, it has a unique solution).

Based on the provided options for the values of K for which the system is consistent, the values are -3 and -1/2.