Question

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

Answer 1

Answer:

all values of b

Step-by-step explanation:

6b < 36 or 2b + 12 > 6.

First solve the one on the left

6b < 36

Divide by 6

6b/6 < 36/6

b  <6

Then solve the one on the right

2b + 12 > 6

Subtract 12 from each side

2b+12-12 >6-12

2b >-6

Divide by 2

2b/2 >-6/2

b >-3

b<6 or b >-3

Rewriting

b>-3  or b<6

b > -3 is an open circle at -3 with a line going to the right

b < 6 is an open circle at 6 with a line going to the left

The or means we add the lines together

We have a line going from negative infinity to infinity

all values of b

Answer 2

Answer:

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

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 $(-\infty, \infty)$