If (\frac{x-2}{3}\ge \frac{2x+1}{5}) and (x+4\le 10), what is the combined solution?

Correct answer: (x\le -13)

The direct answer is option B: \(x\le -13\). First solve each inequality separately. From \(\frac{x-2}{3}\ge\frac{2x+1}{5}\), multiply by the positive number 15, so the sign does not change: \(5(x-2)\ge3(2x+1)\). Thus \(5x-10\ge6x+3\), so \(-13\ge x\), or \(x\le-13\). The second inequality gives \(x+4\le10\), hence \(x\le6\). “And” means both conditions must hold, so we take their intersection. Every number at most \(-13\) is also at most 6; therefore the combined solution is \(x\le-13\). Option A, \(x\le6\), gives only the second condition and includes values that fail the first. Option B is correct because it satisfies both. Option C, \(-13<x\le6\), wrongly excludes \(-13\) and includes numbers greater than \(-13\). Option D, \(x\ge-13\), reverses the direction and includes unsuitable large values. Remember: for “and”, use the common overlap; multiplying an inequality by a positive number keeps its direction.