Answer:
[tex]\large\boxed{x\leq1\to x\in(-\infty,\ 1]}[/tex]
Step-by-step explanation:
[tex]2-\dfrac{1}{5}(x-1)\geq\dfrac{2}{3}(2x+1)\qquad\text{multiply both sides by}\ LCM(5,\ 3)=15\\\\(15)(2)-15\!\!\!\!\!\diagup^3\cdot\dfrac{1}{5\!\!\!\!\!\diagup_1}(x-1)\geq15\!\!\!\!\!\diagup^5\cdot\dfrac{2}{3\!\!\!\!\!\diagup_1}(2x+1)\\\\30-3(x-1)\geq10(2x+1)\qquad\text{use the distributive property}\\\\30+(-3)(x)+(-3)(-1)\geq(10)(2x)+(10)(1)\\\\30-3x+3\geq20x+10\\\\33-3x\geq20x+10\qquad\text{subtract 33 from both sides}\\\\-3x\geq20x-23\qquad\text{subtract}\ 20x\ \text{from both sides}[/tex]
[tex]-23x\geq-23\qquad\text{change the signs}\\\\23x\leq23\qquad\text{divide both sides by 23}\\\\x\leq1[/tex]
