## Precalculus (6th Edition)

The solution is $\Big(-\infty,\dfrac{48}{7}\Big]$
$\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{1}{2}(x+3)\le\dfrac{1}{10}$ Multiply the whole inequality by $30$: $30\Big[\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{1}{2}(x+3)\le\dfrac{1}{10}\Big]$ $10x+(2)(6)x-15(x+3)\le3$ $10x+12x-15x-45\le3$ Take $45$ to the right side and simplify the inequality: $10x+12x-15x\le3+45$ $7x\le48$ Take $7$ to divide the right side: $x\le\dfrac{48}{7}$ Expressing the solution in interval notation: $\Big(-\infty,\dfrac{48}{7}\Big]$