Alright, lets get started.
[tex]-\frac{2}{3} (2x-\frac{1}{2} )\leq \frac{1}{5}x -1[/tex]
Distributing [tex]-\frac{2}{3}[/tex] into parenthesis
[tex]-\frac{4x}{3}+ \frac{2}{6} \leq \frac{x}{5} -1[/tex]
[tex]-\frac{4x}{3}+ \frac{1}{3} \leq \frac{x}{5} -1[/tex]
Subtracting [tex]\frac{1}{3}[/tex] from both sides
[tex]-\frac{4x}{3}+ \frac{1}{3}- \frac{1}{3}\leq\frac{x}{5}-1- \frac{1}{3}[/tex]
[tex]-\frac{4x}{3} \leq \frac{x}{5}- \frac{4}{3}[/tex]
Adding [tex]\frac{4}{3}[/tex] in both sides
[tex]-\frac{4x}{3}+ \frac{4}{3} \leq \frac{x}{5} -\frac{4}{3}+ \frac{4}{3}[/tex]
[tex]-\frac{4x}{3} +\frac{4}{3} \leq \frac{x}{5}[/tex]
Adding [tex]\frac{4x}{3}[/tex] in both sides
[tex]-\frac{4x}{3} +\frac{4}{3} +\frac{4x}{3} \leq \frac{x}{5} +\frac{4x}{3}[/tex]
[tex]\frac{4}{3} \leq \frac{x}{5} +\frac{4x}{3}[/tex]
Making common denominator, adding fractions
[tex]\frac{4}{3} \leq \frac{23x}{15}[/tex]
It means
[tex]23x\geq \frac{4*15}{3}[/tex]
[tex]23x\geq 20[/tex]
Dividing 23 in both sides
[tex]x\geq \frac{20}{23}[/tex]
In interval notation
[[tex]\frac{20}{23}[/tex],∞)
This is the answer
Hope it will help :)
