Question

Solve the given inequality for real x: ${\displaystyle \frac{1}{2}(\frac{3x}{5}+4)\ge \frac{1}{3}(x-6)}$

Answer 1

The given inequality ${\displaystyle \frac{1}{2}(\frac{3x}{5}+4)\ge \frac{1}{3}(x-6)}$

= ${\displaystyle 3\left(\frac{3x+20}{5}\right)\ge 2(x-6)}$

= ${\displaystyle \frac{9x}{5}+12\ge 2x-12}$

= ${\displaystyle 12+12\ge 2x-\frac{9x}{5}}$

= ${\displaystyle 24\ge \frac{10x-9x}{5}}$

= ${\displaystyle 24\ge \frac{x}{5}}$

= 120 ≥ x

Thus, all real numbers x, which are less than or equal to 120, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–∞, 120].