AlgebraHardValue of a constant for infinitely many solutions
Algebra practice question
In the equation 4(x + 2) = 4x + k, k is a constant. For which value of k does the equation have infinitely many solutions?
- A. 8Correct
- B. 2
- C. 4
- D. 0
Answer: A. 8
Expand the left side: 4x + 8 = 4x + k. The x terms are identical on both sides, so the equation is true for EVERY x exactly when the constants match too: k = 8. For any other k the equation reduces to 8 = k, which is false, so there would be no solution.
Why the other answers are wrong
- B. 2
- 2 is the number inside the bracket, before it's multiplied by 4. The constant on the left after expanding is 4 × 2 = 8.
- C. 4
- 4 is the coefficient of x on both sides; those already match. It's the constants that have to be made equal, and the left-hand constant is 8.
- D. 0
- With k = 0 the equation becomes 4x + 8 = 4x, which simplifies to 8 = 0 — never true. That's zero solutions, the opposite of what's asked.
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