Respuesta :
we are given
[tex]-\frac{2}{3} (2x-\frac{1}{2}) \leq\frac{1}{5}x-1[/tex]
Since, we have to solve this inequality
so, we will isolate x on anyone side
step-1:
Multiply both sides by 3/2
[tex]\frac{3}{2}\times-\frac{2}{3}(2x-\frac{1}{2})\leq\frac{3}{2}\times(\frac{1}{5}x-1)[/tex]
[tex] -(2x-\frac{1}{2})\leq\frac{3}{2}\times\frac{1}{5}x-\frac{3}{2}\times 1[/tex]
[tex] -(2x-\frac{1}{2}) \leq\frac{3}{10}x-\frac{3}{2}[/tex]
step-2:
Distribute negative sign
[tex] -2x+\frac{1}{2} \leq\frac{3}{10}x-\frac{3}{2}[/tex]
step-3:
Subtract both sides by 1/2
[tex] -2x+\frac{1}{2}-\frac{1}{2} \leq\frac{3}{10}x-\frac{3}{2}-\frac{1}{2}[/tex]
[tex] -2x \leq\frac{3}{10}x-2[/tex]
step-4:
Subtract both sides by (3/10)x
[tex]-2x-\frac{3}{10}x \leq\frac{3}{10}x-2-\frac{3}{10}x[/tex]
[tex]-2x-\frac{3}{10}x \leq-2[/tex]
[tex]-\frac{23}{10}x \leq-2[/tex]
step-5:
Multiply by sides by -10/23
[tex]-\frac{10}{23}\times -\frac{23}{10}x \geq-\frac{10}{23}\times-2[/tex]
now, we can simplify it
[tex]x \geq\frac{20}{23}[/tex]
now, we can write in interval notation as
[tex][\frac{20}{23},\:\infty \:)[/tex]................Answer
