To determine the value of \( Q \) for which the given system of equations
[tex]\[
\begin{cases}
x - 3y = 4 \\
Qx - 6y = 8
\end{cases}
\][/tex]
has the solution \(\{(x, y) : x - 3y = 4\}\), follow these steps:
1. Let's start with examining the first equation:
[tex]\[
x - 3y = 4
\][/tex]
Notice that this equation is already in a simplified form.
2. Now, consider the second equation:
[tex]\[
Qx - 6y = 8
\][/tex]
3. We need to rewrite the second equation in a form that is comparable to the first equation. Notice that \(Qx - 6y = 8\) can be factored out to show a relation comparable to the first equation. Let's divide the second equation by 2 to simplify it:
[tex]\[
\frac{Qx - 6y}{2} = \frac{8}{2}
\][/tex]
Simplifying both sides we get:
[tex]\[
\frac{Q}{2} x - 3y = 4
\][/tex]
4. Now observe that for this simplified version of the second equation to be consistent with the first equation \(x - 3y = 4\), the coefficients of \(x\) and \(y\) on the left side must match between the two equations.
5. By comparing the coefficients of \(x\), we have:
[tex]\[
\frac{Q}{2} x = x
\][/tex]
For this equality to hold true for all \(x\), it must be that:
[tex]\[
\frac{Q}{2} = 1
\][/tex]
6. Solving for \(Q\):
[tex]\[
Q = 2 \times 1
\][/tex]
[tex]\[
Q = 2
\][/tex]
Thus, the value of [tex]\(Q\)[/tex] that ensures the system of equations has the consistent solution [tex]\(\{(x, y) : x - 3y = 4\}\)[/tex] is [tex]\(\boxed{2}\)[/tex].
