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

Options

A. 1
B. 3
C. 6
D. `1//2`

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

### Step 2: Rearranging Equation 2
From Equation 2, we can rearrange it to:
\[ y - x = 5 \]
Let’s label this as Equation 2a.

### Step 3: Multiply Equation 2a by 4
To eliminate \( x \) when we add it to Equation 1, we multiply Equation 2a by 4:
\[ 4y - 4x = 20 \]
Let’s label this as Equation 3.

### Step 4: Add Equation 1 and the modified Equation 3
Now, we will add Equation 1 and Equation 3:
\[ (4x + 3y) + (4y - 4x) = 1 + 20 \]
This simplifies to:
\[ 7y = 21 \]
Thus, we find:
\[ y = \frac{21}{7} = 3 \]

### Step 5: Substitute \( y \) back into Equation 2 to find \( x \)
Now, substitute \( y = 3 \) back into Equation 2:
\[ 3 = x + 5 \]
Rearranging gives us:
\[ x = 3 - 5 = -2 \]

### Step 6: Substitute \( x \) and \( y \) into Equation 3
Now we have \( x = -2 \) and \( y = 3 \). We substitute these values into Equation 3:
\[ 5(3) + b(-2) = 3 \]
This simplifies to:
\[ 15 - 2b = 3 \]

### Step 7: Solve for \( b \)
Now, we solve for \( b \):
\[ -2b = 3 - 15 \]
\[ -2b = -12 \]
Dividing both sides by -2 gives:
\[ b = \frac{-12}{-2} = 6 \]

### Final Answer
Thus, the value of \( b \) is:
\[ \boxed{6} \]