Sure! Let's solve each of the given inequalities step-by-step.
### Inequality 1: [tex]\( 3(x+3) \geq 9 + 3x \)[/tex]
First, distribute the [tex]\(3\)[/tex] on the left-hand side:
[tex]\[ 3(x+3) = 3x + 9 \][/tex]
So the inequality now looks like:
[tex]\[ 3x + 9 \geq 9 + 3x \][/tex]
Next, subtract [tex]\(3x\)[/tex] from both sides of the inequality:
[tex]\[ 3x + 9 - 3x \geq 9 + 3x - 3x \][/tex]
[tex]\[ 9 \geq 9 \][/tex]
This simplifies to a true statement:
[tex]\[ 9 \geq 9 \][/tex]
Since this statement is always true regardless of the value of [tex]\(x\)[/tex], the inequality [tex]\( 3(x+3) \geq 9 + 3x \)[/tex] holds for all values of [tex]\(x\)[/tex]. Therefore, the inequality has no specific solution constraint and is true for all real numbers.
### Inequality 2: [tex]\( |2v - 1| \leq 7 \)[/tex]
For an absolute value inequality [tex]\( |A| \leq B \)[/tex], we solve it by considering the two cases:
[tex]\[ -B \leq A \leq B \][/tex]
Here [tex]\(A = 2v - 1\)[/tex] and [tex]\(B = 7\)[/tex], so we write:
[tex]\[ -7 \leq 2v - 1 \leq 7 \][/tex]
Now, solve this compound inequality step-by-step.
First, add 1 to all parts of the inequality:
[tex]\[ -7 + 1 \leq 2v - 1 + 1 \leq 7 + 1 \][/tex]
[tex]\[ -6 \leq 2v \leq 8 \][/tex]
Next, divide all parts of the inequality by 2:
[tex]\[ \frac{-6}{2} \leq \frac{2v}{2} \leq \frac{8}{2} \][/tex]
[tex]\[ -3 \leq v \leq 4 \][/tex]
So the solution to the inequality [tex]\( |2v - 1| \leq 7 \)[/tex] is:
[tex]\[ v \in [-3, 4] \][/tex]
### Summary
- Inequality [tex]\(3(x+3) \geq 9+3x\)[/tex] has the solution: All real numbers.
- Inequality [tex]\(|2v - 1| \leq 7\)[/tex] has the solution: [tex]\(-3 \leq v \leq 4\)[/tex].
These are the solutions to the given inequalities.
