How do you solve and write the following in interval notation: #-2 ≤ x + 4 # OR #-1 + 3x -8#?

Question

How do you solve and write the following in interval notation: #-2 ≤ x + 4 # OR #-1 + 3x > -8#?

Ethan Adkins · Accepted Answer

#[-6,\infty)# You can rewrite #-2 \le x+4# as #-2-4 \le x+4-4 \implies -6 \le x \implies x \ge -6# Similarly, rewrite #-1+3x>-8# as #-1+3x+1>-8+1 \implies 3x>-7 \implies x> -7/3# Since #-6< -7/3#, any number which is greater than #-7/3# will automatically be greater than #-6#. And since the OR condition requires at least one of the conditions to be true, it is sufficient to ask #x \ge -6# To write it in interval notation, we must find the boundaries of the desired region. We want #x# to be greater than #-6#, but we have no upper bound, which means it's #\infty#. So, the interval is #[-6,\infty)#

Julia Adams · Answer

undefined To solve and write the compound inequality -2 ≤ x + 4 or -1 + 3x > -8 in interval notation:

-2 ≤ x + 4:
Subtract 4 from both sides to isolate x.
-2 - 4 ≤ x
-6 ≤ x

-1 + 3x > -8:
Add 1 to both sides to isolate the term with x.
3x > -8 + 1
3x > -7
Divide both sides by 3 to solve for x.
x > -7/3

Combine the solutions for both inequalities:
x ∈ [-6, ∞) ∪ (-7/3, ∞)

How do you solve and write the following in interval notation: #-2 ≤ x + 4 # OR #-1 + 3x &gt; -8#?

How do you solve and write the following in interval notation: #-2 ≤ x + 4 # OR #-1 + 3x > -8#?