Answer :
To find the value of [tex]\( Q \)[/tex] in the given system of equations so that the solutions to the system are consistent, let's analyze the equations step-by-step.
We start with the system of equations:
[tex]\[ \begin{array}{l} x - 3y = 4 \quad \text{(1)} \\ 2x - 6y = Q \quad \text{(2)} \end{array} \][/tex]
We need to determine the value of [tex]\( Q \)[/tex] such that every solution [tex]\((x, y)\)[/tex] to equation (1) is also a solution to equation (2).
First, consider equation (1):
[tex]\[ x - 3y = 4 \][/tex]
We aim to see if this equation can help us determine what [tex]\( Q \)[/tex] must be.
Notice that equation (2) can be seen as a multiple of equation (1). To see this, let's multiply the entire equation (1) by 2:
[tex]\[ 2(x - 3y) = 2 \times 4 \][/tex]
That simplifies to:
[tex]\[ 2x - 6y = 8 \][/tex]
Now, compare this result with equation (2):
[tex]\[ 2x - 6y = Q \][/tex]
[tex]\[ 2x - 6y = 8 \][/tex]
From comparison, it is clear that for the two equations to be equivalent, [tex]\( Q \)[/tex] must be equal to 8.
Therefore, the value of [tex]\( Q \)[/tex] that makes the system consistent and ensures that the solution to the system is [tex]\(\{(x, y): x - 3y = 4\}\)[/tex] is:
[tex]\[ \boxed{8} \][/tex]
We start with the system of equations:
[tex]\[ \begin{array}{l} x - 3y = 4 \quad \text{(1)} \\ 2x - 6y = Q \quad \text{(2)} \end{array} \][/tex]
We need to determine the value of [tex]\( Q \)[/tex] such that every solution [tex]\((x, y)\)[/tex] to equation (1) is also a solution to equation (2).
First, consider equation (1):
[tex]\[ x - 3y = 4 \][/tex]
We aim to see if this equation can help us determine what [tex]\( Q \)[/tex] must be.
Notice that equation (2) can be seen as a multiple of equation (1). To see this, let's multiply the entire equation (1) by 2:
[tex]\[ 2(x - 3y) = 2 \times 4 \][/tex]
That simplifies to:
[tex]\[ 2x - 6y = 8 \][/tex]
Now, compare this result with equation (2):
[tex]\[ 2x - 6y = Q \][/tex]
[tex]\[ 2x - 6y = 8 \][/tex]
From comparison, it is clear that for the two equations to be equivalent, [tex]\( Q \)[/tex] must be equal to 8.
Therefore, the value of [tex]\( Q \)[/tex] that makes the system consistent and ensures that the solution to the system is [tex]\(\{(x, y): x - 3y = 4\}\)[/tex] is:
[tex]\[ \boxed{8} \][/tex]
