To determine which equation has exactly one solution, we can simplify each equation to see if it yields a unique value for \( x \).
Option A: \( \frac{8 + 3x}{2} = 13 + x \)
Multiply both sides by 2: \[ 8 + 3x = 26 + 2x \] Subtract \( 2x \) from both sides: \[ 8 + 3x - 2x = 26 \] So: \[ x + 8 = 26 \] Subtract 8 from both sides: \[ x = 18 \] This has exactly one solution.
Option B: \( \frac{3 + 12x}{2} = 6x + 15 \)
Multiply both sides by 2: \[ 3 + 12x = 12x + 30 \] Subtract \( 12x \) from both sides: \[ 3 = 30 \] This is a contradiction (false statement), meaning this equation has no solutions.
Option C: \( \frac{10 + 15x}{5} = 3x + 10 \)
Multiply both sides by 5: \[ 10 + 15x = 15x + 50 \] Subtract \( 15x \) from both sides: \[ 10 = 50 \] This is also a contradiction, meaning this equation has no solutions.
Option D: \( \frac{4x - 20}{4} \) (Seems to be incomplete; assuming it's equal to something)
If we assume it's set equal to 0 for analysis: \[ \frac{4x - 20}{4} = 0 \] Multiply both sides by 4: \[ 4x - 20 = 0 \] Add 20 to both sides: \[ 4x = 20 \] Divide by 4: \[ x = 5 \] This equation has exactly one solution.
From our analysis:
- Option A has one solution: \( x = 18 \)
- Option D (if set to 0) has one solution: \( x = 5 \)
- Options B and C have no solutions.
Thus, both Option A and Option D have exactly one solution, but if you only need one correct response, you could select Option A.