The solution in interval notation is [-3, -2]
STEP - BY - STEP SOLUTION
What to find?
The solution of the given inequality in interval notation.
Given:
4(x+3) ≥ 0 or 4(x + 4) ≤ 8
Solve the first part of the inequality.
That is;
4(x+3)≥ 0
Open the parenthesis.
4x + 12 ≥ 0
Subtract 12 from both-side of the inequality.
4x + 12 - 12 ≥ 0- 12
4x ≥ -12
Divide both-side of the inequality by 4
[tex]\frac{\cancel{4}x}{\cancel{4}}\ge-\frac{12}{4}[/tex]
x ≥ -3
Proceed to solve the second part of the inequality.
4(x+ 4) ≤ 8
Open the parenthesis.
4x + 16 ≤ 8
Subtract 16 from both-side of the inequality.
4x ≤ 8-16
4x≤ -8
Divide both-side of the inequality by 4
[tex]\frac{\cancel{4}x}{\cancel{4}}\leq-\frac{8}{4}[/tex]
x ≤ -2
Combine the solutions.
-3 ≤ x ≤ -2
This can be written in interval notation as [-3, -2].
