AlgebraVery hardValue of a constant giving no solution
Algebra practice question
In the equation 4(x + k) = 4x + 12 − 2k, k is a constant. For which value of k does the equation have NO solution?
- A. No such value exists; the equation has infinitely many solutions for k = 2 and no solution for every other k.Correct
- B. k = 2
- C. k = 3
- D. k = 0
Answer: A. No such value exists; the equation has infinitely many solutions for k = 2 and no solution for every other k.
Expand: 4x + 4k = 4x + 12 − 2k. The x-terms cancel, leaving 4k = 12 − 2k, i.e. 6k = 12. If k = 2 this is 0 = 0 (infinitely many solutions); for any other k it is a false statement (no solution). So no single k gives 'no solution' — every k except 2 does.
Why the other answers are wrong
- B. k = 2
- k = 2 gives 0 = 0: infinitely many solutions, not none.
- C. k = 3
- k = 3 gives no solution, but so does every k except 2; it is not THE value.
- D. k = 0
- k = 0 likewise gives no solution, as does any k ≠ 2.
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