The two lines 3x - 8y = 16 and 2 x + 4y = 6 intersect at (a, b) . Find the value of `(a^(2) - 4 b^(2))`

Options

A. 5
B. 10
C. 15
D. 20

Answer

To solve the problem of finding the intersection point of the lines given by the equations \(3x - 8y = 16\) and \(2x + 4y = 6\), and then calculating \(a^2 - 4b^2\), we will follow these steps:

### Step 1: Rewrite the equations
We have the two equations:
1. \(3x - 8y = 16\) (Equation 1)
2. \(2x + 4y = 6\) (Equation 2)

First, we can simplify Equation 2 by dividing every term by 2:
\[ x + 2y = 3 \quad \text{(Equation 2 simplified)} \]

### Step 2: Solve for one variable
From the simplified Equation 2, we can express \(x\) in terms of \(y\):
\[ x = 3 - 2y \]

### Step 3: Substitute into the first equation
Now we substitute \(x\) in Equation 1:
\[ 3(3 - 2y) - 8y = 16 \]
Expanding this gives:
\[ 9 - 6y - 8y = 16 \]
Combining like terms:
\[ 9 - 14y = 16 \]

### Step 4: Solve for \(y\)
Now, we isolate \(y\):
\[ -14y = 16 - 9 \]
\[ -14y = 7 \]
\[ y = -\frac{1}{2} \]

### Step 5: Substitute back to find \(x\)
Now that we have \(y\), we substitute it back into the expression for \(x\):
\[ x = 3 - 2\left(-\frac{1}{2}\right) \]
\[ x = 3 + 1 = 4 \]

### Step 6: Identify the intersection point
The intersection point \((a, b)\) is:
\[ (a, b) = (4, -\frac{1}{2}) \]

### Step 7: Calculate \(a^2 - 4b^2\)
Now we can calculate \(a^2 - 4b^2\):
\[ a^2 = 4^2 = 16 \]
\[ b^2 = \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \]
Thus,
\[ 4b^2 = 4 \times \frac{1}{4} = 1 \]
Now, substituting these values gives:
\[ a^2 - 4b^2 = 16 - 1 = 15 \]

### Final Answer
The value of \(a^2 - 4b^2\) is:
\[ \boxed{15} \]