To determine whether Percy's statement is correct, we need to analyze the system of equations:
- \( 6x + 4y = 14 \)
- \( 3x + 2y = k \)
Let's express both equations in standard form and see how they relate to each other.
The first equation can be rewritten as: \[ 6x + 4y - 14 = 0 \]
The second equation can be rewritten similarly: \[ 3x + 2y - k = 0 \]
Next, we can manipulate the first equation. Notice that the first equation is simply a multiple of the second equation:
If we multiply the second equation by 2, we get: \[ 2(3x + 2y) = 2k \implies 6x + 4y = 2k \]
Now, we can compare this with the first equation \( 6x + 4y = 14 \).
For the two equations to have the same line (i.e., an infinite number of solutions), we would need: \[ 2k = 14 \implies k = 7 \]
If \( k = 7 \), the system has an infinite number of solutions because the second equation then represents the same line as the first.
If \( k \neq 7 \), the two equations will represent two parallel lines, which means they will have no solution.
Therefore, Percy's error is in saying that "any real number for \( k \) would cause the system of equations to have no solution." This is not true because there exists a specific value (namely \( k = 7 \)) for which the system has infinitely many solutions. Thus, the system has no solution for all \( k \) except \( k = 7 \).
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