BECE Mathematics 2011 Paper

Concept:A linear inequality in one variable is solved by isolating $x$ while keeping the inequality sign unchanged when multiplying or dividing by a positive number.Explanation:Start with the inequality: $x - \frac{1}{3} \geq \frac{2}{3} - x$.Add $x$ to both sides to bring all $x$-terms to the left: $x + x - \frac{1}{3} \geq \frac{2}{3}$.This simplifies to $2x - \frac{1}{3} \geq \frac{2}{3}$.Add $\frac{1}{3}$ to both sides to move the constant term to the right: $2x \geq \frac{2}{3} + \frac{1}{3}$.So, $2x \geq 1$.Divide both sides by $2$: $x \geq \frac{1}{2}$.Since we divided by a positive number, the inequality sign remains the same.Answer:$x \geq \frac{1}{2}$, which is option C.