Solve the following inequality for \( x \). \[ a x+14<45 \] A \( x<31 a \) B \( x<\frac{31}{a} \) C \( x<\frac{a}{31} \) D \( x>\frac{31}{a} \)

To solve the inequality \( a x + 14 < 45 \), we first isolate \( ax \) by subtracting 14 from both sides: \[ ax < 45 - 14 \] \[ ax < 31 \] Next, we divide both sides by \( a \). Remember, the direction of the inequality will depend on the sign of \( a \): 1. If \( a > 0 \): \[ x < \frac{31}{a} \] 2. If \( a < 0 \): \[ x > \frac{31}{a} \] The correct answer, assuming \( a > 0 \), is \( x < \frac{31}{a} \), which corresponds to option B. If \( a \) is negative, we have \( x > \frac{31}{a} \), which corresponds to option D. In general, without further restrictions on \( a \), the solution could lead to different interpretations, but for the typical context of this inequality, we see that option B is the standard solution when assuming \( a > 0 \).