Questions: Part (c) Substitute the value of C obtained in Part (a), distribute it to all terms in the first equation, and add the equations together: C(2 x+6 y =0) (3 x+15 y)/(0 x+α y) = 8/β Enter the value of β, the constant term on the right-hand side. β=

Part (c)
Substitute the value of C obtained in Part (a), distribute it to all terms in the first equation, and add the equations together:

C(2 x+6 y =0)
(3 x+15 y)/(0 x+α y) = 8/β

Enter the value of β, the constant term on the right-hand side.

β=

Transcript text: Part (c) Substitute the value of $\mathcal{C}$ obtained in Part (a), distribute it to all terms in the first equation, and add the equations together: \[ \begin{aligned} \mathcal{C}(2 x+6 y & =0) \\ \frac{3 x+15 y}{0 x+\alpha y} & =\frac{8}{\beta} \end{aligned} \] Enter the value of $\beta$, the constant term on the right-hand side. \[ \beta= \] $\square$

Solution

Solution Steps

Step 1: Substitute and Distribute

We start by substituting the value of $\mathcal{C}$ into the first equation. The first equation becomes: \[ \mathcal{C}(2x + 6y) = 0 \] After substituting, we have: \[ C(2x + 6y) \]

Step 2: Define the Second Equation

The second equation is given as: \[ \frac{3x + 15y}{0x + \alpha y} = \frac{8}{\beta} \] This can be rearranged to: \[ \frac{3x + 15y}{\alpha y} - \frac{8}{\beta} = 0 \]

Step 3: Combine the Equations

We combine the modified first equation and the second equation: \[ C(2x + 6y) - \frac{8}{\beta} + \frac{3x + 15y}{\alpha y} = 0 \]

Step 4: Solve for $\beta$

From the combined equation, we can isolate $\beta$: \[ \beta = \frac{8\alpha y}{2C\alpha xy + 6C\alpha y^2 + 3x + 15y} \]