# Solve the inequality for real $x, \large\frac{x}{3} > \frac{x}{2} $$+1 \begin{array}{1 1} (A)\;(-3,\infty)\\(B)\;(-\infty,-6)\\(C)\;(-4,\infty)\\(D)\; (-9,\infty)\end{array} ## 1 Answer Toolbox: • Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality. • Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality. • If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e '>' sign changes to '<' and '<' changes '>' . The given inequality is \large\frac{x}{3} > \frac{x}{2}$$+1$
Subtracting $\large\frac{x}{2}$ on both sides,
$=> \large\frac{x}{3} -\frac{x}{2} > \frac{ x}{2} $$+1 -\large\frac{x}{2} => \large\frac{x}{3}-\frac{x}{2}$$ > 1$
$=> \large\frac{2x-3x }{6} $$> 1 => \large\frac{-x}{6}$$ > 1$
Multiplying both sides of the inequality by a negative number $-6$.