Question

Solve the compound inequality 6b < 24 or 4b + 12 > 4.

b < 4 or b > −2

b < 4 or b > 4

b < 4 or b > 2

b > 4 or b < −2

Answer 1

Solve each inequality and keep the word "or" in between them.
\(6b < 24\) or \(4b + 12 > 4\)

\(\dfrac{6b}{6} \lt \dfrac{24}{6} \) or \(4b + 12 -12 \gt 4 - 12\)

\(b \lt 4\) or \(4b \gt -8 \)

\(b \lt 4\) or \(\dfrac{4b}{4} \gt \dfrac{-8}{4} \)

\(b \lt 4\) or \(b \gt -2\)

Now look at the two inequalities. If b < 4 or b > -2, that means b can be any number. The solutions is all real numbers.

Answer 2

why dont you flip the sign

Answer 3

The sign in an inequality only flips direction when you multiply both sides or divide both sides by a negative number. In this case, we never did that.

Answer 4

For example:

\(-5x \le 30 \)

Here you need to divide both sides by -5, obviously a negative number, so you get this:

\(\dfrac{-5x}{-5} \ge \dfrac{30}{-5} \)

\(\dfrac{\cancel{-5}x}{\cancel{-5}1} \ge -6 \)

\(x \ge -6 \)

Answer 5

the answer is A (b<4 or b>-2)

Answer 6

Yes, it's A.