Question:

Step 1: The pair of lines $6x^2 - axy - 3y^2 - 24x + 3y + b = 0$ intersects on the x-axis. Let the point of intersection be $(h, 0)$. Since $(h, 0)$ lies on the lines, it must satisfy the equation: $$6h^2 - a(h)(0) - 3(0)^2 - 24h + 3(0) + b = 0$$ $$6h^2 - 24h + b = 0 \quad (*)$$ Step 2: For a general second-degree equation $F(x,y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ to represent a pair of straight lines, the point of intersection $(x_0, y_0)$ is given by the solution to the system of partial derivatives: $$\frac{\partial F}{\partial x} = 0 \quad \text{and} \quad \frac{\partial F}{\partial y} = 0$$ For the given equation, $F(x,y) = 6x^2 - axy - 3y^2 - 24x + 3y + b$. The partial derivatives are: $$\frac{\partial F}{\partial x} = 12x - ay - 24$$ $$\frac{\partial F}{\partial y} = -ax - 6y + 3$$ At the point of intersection $(h, 0)$: $$12h - a(0) - 24 = 0 \implies 12h - 24 = 0 \implies h = 2$$ Step 3: Substitute $h=2$ into the second partial derivative equation: $$-a(2) - 6(0) + 3 = 0 \implies -2a + 3 = 0 \implies a = \frac{3}{2}$$ Step 4: Substitute $h=2$ into equation $(*)$ from Step 1 to find $b$: $$6(2)^2 - 24(2) + b = 0$$ $$6(4) - 48 + b = 0$$ $$24 - 48 + b = 0$$ $$-24 + b = 0 \implies b = 24$$ Step 5: Calculate the value of $20a - b$ using $a = 3/2$ and $b = 24$: $$20a - b = 20\left(\frac{3}{2}\right) - 24$$ $$= 10(3) - 24$$ $$= 30 - 24$$ $$= 6$$