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

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

All real numbers  b ∈

Step-by-step explanation:

First we solve the following inequality

Divide by 6 both sides of the inequality

The set of solutions is:

Now we solve the following inequality

Subtract 12 on both sides of the inequality

Divide by 2 on both sides of the inequality

The set of solutions is:

Finally, the set of solutions for composite inequality is:

 ∪

This is: All real numbers

a4 - 3a2 - 1

step-by-step explanation:

step   1   :

equation at the end of step   1   :

  ((a4) -   3a2) -   1

step   2   :

trying to factor by splitting the middle term

2.1     factoring   a4-3a2-1  

the first term is,   a4   its coefficient is   1 .

the middle term is,   -3a2   its coefficient is   -3 .

the last term, "the constant", is   -1