What interval includes all possible values of x, where -3(6- 2x) ≥ 4x + 12?
(-∞,-3], [-3, ∞), (-∞, 15], [15, ∞)

Solution:

The given inequality is -3(6 - 2x) ≥ 4x + 12

Using the multiplicative distributive property

-18 + 6x ≥ 4x + 12

On rearranging

6x - 4x ≥ 12 + 18

2x ≥ 30

Dividing both sides by 2

x ≥ 15

We can see that as long as x is equal or more than 15 it will satisfy the equation.

The interval is [15, ∞)

Therefore, the interval which includes all possible values of x is [15, ∞).

What interval includes all possible values of x, where -3(6 - 2x) ≥ 4x + 12?

Summary:

The interval which includes all possible values of x, where -3(6 - 2x) ≥ 4x + 12 is [15, ∞).