Question

Solve the compound inequality 6b < 36 or 2b + 12 > 6.

Answer 1

Answer:

All real numbers $(-\infty, \infty)$

Step-by-step explanation:

First we solve the following inequality

$6b < 36$

Divide by 6 both sides of the inequality

$b$

The set of solutions is:

$(-\infty, 6)$

Now we solve the following inequality

$2b + 12 > 6$

Subtract 12 on both sides of the inequality

$2b + 12-12 > 6-12$

$2b> -6$

Divide by 2 on both sides of the inequality

$\frac{2}{2}b> -\frac{6}{2}$

$b> -3$

The set of solutions is:

$(-3, \infty)$

Finally, the set of solutions for composite inequality is:

$(-\infty, 6)$  ∪ $(-3, \infty)$

This is: All real numbers $(-\infty, \infty)$

Answer 2

Answer:

Answer is all real numbers.

<~~~~~~~~~~~~~~~~~~~~~~~~~~~~~>

---------(-3)---------------------(6)-------------

Step-by-step explanation:

6b<36

Divide both sides by 6:

b<6

or

2b+12>6

Subtract 12 on both sides:

2b>-6

Divide both sides by 2:

b>-3

So we want to graph b<6 or b>-3:

               o~~~~~~~~~~~~~~~~~~~~~~~~~~ b>-3

~~~~~~~~~~~~~~~~~~~~~~~~o                        b<6

_______(-3)____________(6)___________

So again "or" is a key word! Or means wherever you see shading for either inequality then that is a solution to the compound inequality.  You see shading everywhere so the answer is all real numbers.

<~~~~~~~~~~~~~~~~~~~~~~~~~~~~~>

---------(-3)---------------------(6)-------------