Question

What is the solution to the inequality?

Answer 1

Alright, lets get started.

$-\frac{2}{3} (2x-\frac{1}{2} )\leq \frac{1}{5}x -1$

Distributing $-\frac{2}{3}$ into parenthesis

$-\frac{4x}{3}+ \frac{2}{6} \leq \frac{x}{5} -1$

$-\frac{4x}{3}+ \frac{1}{3} \leq \frac{x}{5} -1$

Subtracting $\frac{1}{3}$ from both sides

$-\frac{4x}{3}+ \frac{1}{3}- \frac{1}{3}\leq\frac{x}{5}-1- \frac{1}{3}$

$-\frac{4x}{3} \leq \frac{x}{5}- \frac{4}{3}$

Adding $\frac{4}{3}$ in both sides

$-\frac{4x}{3}+ \frac{4}{3} \leq \frac{x}{5} -\frac{4}{3}+ \frac{4}{3}$

$-\frac{4x}{3} +\frac{4}{3} \leq \frac{x}{5}$

Adding $\frac{4x}{3}$ in both sides

$-\frac{4x}{3} +\frac{4}{3} +\frac{4x}{3} \leq \frac{x}{5} +\frac{4x}{3}$

$\frac{4}{3} \leq \frac{x}{5} +\frac{4x}{3}$

Making common denominator, adding fractions

$\frac{4}{3} \leq \frac{23x}{15}$

It means

$23x\geq \frac{4*15}{3}$

$23x\geq 20$

Dividing 23 in both sides

$x\geq \frac{20}{23}$

In interval notation

[$\frac{20}{23}$,∞)

This is the answer

Hope it will help :)