Answer
The solution is $\Big(-\infty,\dfrac{48}{7}\Big]$
Work Step by Step
$\dfrac{1}{3}x+\dfrac{2}{5}x-\dfrac{1}{2}(x+3)\le\dfrac{1}{10}$
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:
$\Big(-\infty,\dfrac{48}{7}\Big]$
