Solve each of the following in equations and represent the solution set on the number line.
$\frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}$ where $x \in \mathbf{R}$.
Given:
$\frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}$, where $x \in R$
Multiplying by 60 on both the sides in the above equation.
$(60) \frac{(2 x-1)}{3} \geq(60) \frac{(3 x-2)}{4}-(60) \frac{(2-x)}{5}$
$20(2 x-1) \geq 15(3 x-2)-12(2-x)$
$40 x-20 \geq 45 x-30-24+12 x$
$40 x-20 \geq 57 x-54$
Now, Adding 20 on both the sides in the above equation
$40 x-20+20 \geq 57 x-54+20$
$40 x \geq 57 x-34$
Now, subtracting 57x from both the sides in the above equation
$40 x-57 x \geq 57 x-34-57 x$
$-17 x \geq-34$
Multiplying by -1 on both sides in the above equation
$(-17 x)(-1) \geq(-34)(-1)$
$17 x \leq 34$
Now, divide by 17 on both sides in the above equation
$\frac{17 x}{17} \leq \frac{34}{17}$
Therefore,
$x \leq 2$