Question

1. What is the solution to the inequality?

Answer 1

we are given

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

Since, we have to solve this inequality

so, we will isolate x on anyone side

step-1:

Multiply both sides by 3/2

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

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

$-(2x-\frac{1}{2}) \leq\frac{3}{10}x-\frac{3}{2}$

step-2:

Distribute negative sign

$-2x+\frac{1}{2} \leq\frac{3}{10}x-\frac{3}{2}$

step-3:

Subtract both sides by 1/2

$-2x+\frac{1}{2}-\frac{1}{2} \leq\frac{3}{10}x-\frac{3}{2}-\frac{1}{2}$

$-2x \leq\frac{3}{10}x-2$

step-4:

Subtract both sides by (3/10)x

$-2x-\frac{3}{10}x \leq\frac{3}{10}x-2-\frac{3}{10}x$

$-2x-\frac{3}{10}x \leq-2$

$-\frac{23}{10}x \leq-2$

step-5:

Multiply by sides by -10/23

$-\frac{10}{23}\times -\frac{23}{10}x \geq-\frac{10}{23}\times-2$

now, we can simplify it

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

now, we can write in interval notation as

$[\frac{20}{23},\:\infty \:)$................Answer