Question

Percy said that any real number for k would cause the system of equations to have no solution. Explain the error in Percy’s statement.

6x + 4y = 14,
3x + 2y = k

Answer 1

To determine whether Percy's statement is correct, we need to analyze the system of equations:

1. \( 6x + 4y = 14 \)
2. \( 3x + 2y = k \)

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 \( k = 7 \).