If `c<1` and the system of equations `x+y-1=0,2x-y-c=0,and -bx+3b y-c=0` is consistent, then the possible real values of `b` are

Options

A. `b in (-3(3)/(4))`
B. ` b in (-(3)/(2) ,4)`
C. `b in (-(3)/(4),3)`
D. none of these

Explanation

To solve the problem, we need to determine the possible real values of \( b \) for which the given system of equations is consistent. The system of equations is:

1. \( x + y - 1 = 0 \)
2. \( 2x - y - c = 0 \)
3. \( -bx + 3by - c = 0 \)

### Step 1: Write the system in matrix form

We can express the system of equations in the form of a determinant:

\[ \begin{vmatrix} 1 & 1 & -1 \\ 2 & -1 & -c \\ -b & 3b & -c \end{vmatrix} = 0 \]

### Step 2: Expand the determinant

We will expand this determinant using the first row:

\[ = 1 \cdot \begin{vmatrix} -1 & -c \\ 3b & -c \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & -c \\ -b & -c \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & -1 \\ -b & 3b \end{vmatrix} \]

Calculating these 2x2 determinants:

1. \( \begin{vmatrix} -1 & -c \\ 3b & -c \end{vmatrix} = (-1)(-c) - (-c)(3b) = c + 3bc = c(1 + 3b) \)

2. \( \begin{vmatrix} 2 & -c \\ -b & -c \end{vmatrix} = (2)(-c) - (-c)(-b) = -2c - bc = -c(2 + b) \)

3. \( \begin{vmatrix} 2 & -1 \\ -b & 3b \end{vmatrix} = (2)(3b) - (-1)(-b) = 6b - b = 5b \)

Putting it all together, we have:

\[ c(1 + 3b) + c(2 + b) - 5b = 0 \]

### Step 3: Simplify the equation

Combining the terms gives:

\[ c(1 + 3b + 2 + b) - 5b = 0 \]

This simplifies to:

\[ c(3 + 4b) - 5b = 0 \]

### Step 4: Solve for \( c \)

Rearranging gives us:

\[ c(3 + 4b) = 5b \]

Thus,

\[ c = \frac{5b}{3 + 4b} \]

### Step 5: Apply the condition \( c < 1 \)

Since we know \( c < 1 \), we set up the inequality:

\[ \frac{5b}{3 + 4b} < 1 \]

### Step 6: Solve the inequality

Cross-multiplying gives:

\[ 5b < 3 + 4b \]

This simplifies to:

\[ 5b - 4b < 3 \implies b < 3 \]

### Step 7: Find the other condition

Now, we also need to ensure that the denominator \( 3 + 4b \) is not zero, so:

\[ 3 + 4b > 0 \implies 4b > -3 \implies b > -\frac{3}{4} \]

### Conclusion

Combining both conditions, we find:

\[ -\frac{3}{4} < b < 3 \]

Thus, the possible real values of \( b \) are:

\[ b \in \left(-\frac{3}{4}, 3\right) \]