If the lines `3y + 4x = 1`, `y = x + 5` and `5y + bx = 3` are concurrent, then what is the value of b?

Options

A. 1
B. 3
C. 6
D. 0

Answer

To find the value of \( b \) such that the lines \( 3y + 4x = 1 \), \( y = x + 5 \), and \( 5y + bx = 3 \) are concurrent, we will follow these steps:

### Step 1: Find the intersection point of the first two lines.

We have the equations:
1. \( 3y + 4x = 1 \) (Equation 1)
2. \( y = x + 5 \) (Equation 2)

Substituting Equation 2 into Equation 1:

\[ 3(x + 5) + 4x = 1 \]

### Step 2: Simplify the equation.

Expanding the equation:

\[ 3x + 15 + 4x = 1 \]

Combining like terms:

\[ 7x + 15 = 1 \]

### Step 3: Solve for \( x \).

Subtracting 15 from both sides:

\[ 7x = 1 - 15 \]
\[ 7x = -14 \]

Dividing by 7:

\[ x = -2 \]

### Step 4: Find the corresponding \( y \) value.

Using Equation 2 to find \( y \):

\[ y = x + 5 = -2 + 5 = 3 \]

Thus, the intersection point of the first two lines is:

\[ (-2, 3) \]

### Step 5: Substitute the intersection point into the third line.

Now we will substitute \( x = -2 \) and \( y = 3 \) into the third line \( 5y + bx = 3 \):

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

### Step 6: Simplify and solve for \( b \).

Calculating the left side:

\[ 15 - 2b = 3 \]

Subtracting 15 from both sides:

\[ -2b = 3 - 15 \]
\[ -2b = -12 \]

Dividing by -2:

\[ b = 6 \]

### Conclusion

The value of \( b \) is:

\[ \boxed{6} \]