If `4x+3y=k," "2x+3y=12` and `x+y=5` are concurrent lines then find k.

Answer

To find the value of \( k \) such that the lines \( 4x + 3y = k \), \( 2x + 3y = 12 \), and \( x + y = 5 \) are concurrent, we will follow these steps:

### Step 1: Solve the equations \( 2x + 3y = 12 \) and \( x + y = 5 \)

We start with the second equation:
\[ x + y = 5 \]
From this, we can express \( y \) in terms of \( x \):
\[ y = 5 - x \]

### Step 2: Substitute \( y \) in the first equation

Now, substitute \( y \) into the first equation \( 2x + 3y = 12 \):
\[ 2x + 3(5 - x) = 12 \]

### Step 3: Simplify and solve for \( x \)

Expanding the equation:
\[ 2x + 15 - 3x = 12 \]
Combine like terms:
\[ -1x + 15 = 12 \]
Subtract 15 from both sides:
\[ -1x = 12 - 15 \]
\[ -1x = -3 \]
Thus, we find:
\[ x = 3 \]

### Step 4: Find \( y \)

Now, substitute \( x = 3 \) back into the equation for \( y \):
\[ y = 5 - x = 5 - 3 = 2 \]

### Step 5: Substitute \( x \) and \( y \) into \( 4x + 3y = k \)

Now that we have \( x = 3 \) and \( y = 2 \), we can find \( k \) by substituting these values into the equation \( 4x + 3y = k \):
\[ k = 4(3) + 3(2) \]
Calculating this gives:
\[ k = 12 + 6 = 18 \]

### Final Answer

Thus, the value of \( k \) is:
\[ \boxed{18} \]