Let \(L(x)<R(x)\) be an inequality. While solving it, which of the following transformations is always equivalent without imposing any additional condition on \(x\)?

Correct answer: \(L(x)+3x-2<R(x)+3x-2\)

Adding the same expression \(3x-2\) to both sides preserves the inequality, so A is always equivalent. In B and C, the factor or divisor may change sign; moreover, C is undefined at \(x=-2\). Exam tip: do not multiply or divide by an expression whose sign is unknown.