Which values of k would cause this system of equations to have no solution? Check all that apply.

Question

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

–2
5
7
10
21

Answer 1

To determine the values of \( k \) that would cause the system of equations:

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

to have no solution, we need to analyze the relationship between the two equations.

We can simplify the first equation by dividing all terms by 2:

\[ 3x + 2y = 7 \]

Now, we have the two equations:

1. \( 3x + 2y = 7 \)
2. \( 3x + 2y = k \)

For a system of linear equations to have no solution, the two equations must represent parallel lines. This occurs when their coefficients are proportional but their constants are different.

For the two equations we have:

• Coefficients of \( 3x \) and \( 2y \) are identical.
• The constants must be unequal.

Thus, for the system to have no solution:

\[ k \neq 7 \]

Now let's evaluate the provided values of \( k \):

• \( k = -2 \) (not equal to 7)
• \( k = 5 \) (not equal to 7)
• \( k = 7 \) (equal to 7)
• \( k = 10 \) (not equal to 7)
• \( k = 21 \) (not equal to 7)

The values of \( k \) that cause the system to have no solution are:

• \( -2 \)
• \( 5 \)
• \( 10 \)
• \( 21 \)

In summary, the values of \( k \) that will cause the system to have no solution are:

–2, 5, 10, 21. (Check all that apply.)