# How do you solve and graph -2a+3>=6a-1>3a-10?

Feb 24, 2018

Answer is $- 3 < a \le \frac{1}{2}$

#### Explanation:

$- 2 a + 3 \ge 6 a - 1 > 3 a - 10$

means $- 2 a + 3 \ge 6 a - 1$

i.e. $3 + 1 \ge 6 a + 2 a$

or $8 a \le 4$ i.e. $a \le \frac{1}{2}$

Also $6 a - 1 > 3 a - 10$

or $6 a - 3 a > - 10 + 1$

or $3 a > - 9$ i.e. $a > - 3$

Hence answer is $- 3 < a \le \frac{1}{2}$

Feb 24, 2018

$- 3 < a \le \frac{1}{2}$

#### Explanation:

$\textcolor{b l u e}{\text{Consider: } - 2 a + 3 \ge 6 a - 1}$

Add $2 a$ to both sides

$3 \ge 8 a - 1$

$4 \ge 8 a$

Divide both sides by 8

$\frac{1}{2} \ge a$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Consider: } 6 a - 1 > 3 a - 10}$

Subtract $3 a$ from both sides

$3 a - 1 > - 10$

$3 a > - 9$

Divide both sides by 3

$a > - 3$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{putting it all together}}$

In the same order as in the question.

$\frac{1}{2} \ge a > - 3$

Lets 'rotate' it horizontally ${180}^{o}$

color(red)("THE SOLUTION "-> bar(ul(|color(white)("d")-3 < a <= 1/2color(white)("d") |)))

You use a 'hollow' circle at the end for 'greater than' or 'less than'.

You use a 'filled in' circle at the end for 'less than or equal to' or 'greater than or equal to'.

$\approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx \approx$

$\textcolor{b l u e}{\text{IF YOU WERE TO PLOT THESE AND ASSIGN THE VALUES TO y}}$

The feasible region for $y$ is the shaded area

To construct this I used:
$\underbrace{- 2 a + 3 \ge 6 a - 1 > 3 a - 10}$

$- 2 x + 3 \ge y > 3 x - 10$

AND

$\underbrace{- 2 a + 3 \ge 6 a - 1 > 3 a - 10}$
$\textcolor{w h i t e}{\text{dddddd}} y \ge 6 x - 1$

$y \le - 2 x + 3$
$y \ge \textcolor{w h i t e}{-} 6 x - 1$
$y > \textcolor{w h i t e}{-} 3 x - 10$

$x \ge - 3 \leftarrow \text{ Needed to control the lower bound of } a$