BECE Mathematics 2011 Paper

Get rid of the fraction by multiplying both sides of the equation by the L.C.M of the denominators. The L.C.M of 3 and 3 is 3 .The denominators cancel out the ones multiplying.$x - \; \frac{1}{3} \geq \; \frac{2}{3} - x = 3x - 1 \geq 2 - 3x$Group the $x$ at the left and the numbers at the right. When one is negative, it becomes positive at the other side and when positive, negative.$x - \; \frac{1}{3} \geq \; \frac{2}{3} - x = 3x + 3x \geq 2 + 1$$x - \; \frac{1}{3} \geq \; \frac{2}{3} - x = 6x \geq 3$Divide both sides by 6$x - \; \frac{1}{3} \geq \; \frac{2}{3} - x = x \geq \frac{3}{6}$Note: Since we are not dividing by a negative number, the inequality sign remains the same.$x - \; \frac{1}{3} \geq \; \frac{2}{3} - x = x \geq \frac{1}{2}$