Question

Solve the inequality |3x+3| + 3 > 15Write the answer in interval notation

Answer 1

Solution:

Given the inequality:

$|3x+3|+3>15$

To solve the inequality,

step 1: Add -3 to both sides of the inequality.

Thus,

$\begin{gathered} |3x+3|+3-3>-3+15 \\ \Rightarrow|3x+3|>12 \end{gathered}$

Step 2: Apply the absolute rule.

According to the absolute rule:

$\mathrm{If}\:|u|\:>\:a,\:a>0\:\mathrm{then}\:u\:\:a$

Thus, from step 1, we have

$\begin{gathered} 3x+312} \\ \end{gathered}$$\begin{gathered} when \\ 3x+3$$\begin{gathered} when \\ 3x+3>12 \\ add\text{ -3 to both sides of the inequality} \\ 3x+3-3>12-3 \\ \Rightarrow3x>9 \\ divide\text{ both sides by the coefficient of x, which is 3} \\ \frac{3x}{3}>\frac{9}{3} \\ \Rightarrow x>3 \end{gathered}$

This implies that

$x3$

Hence, in interval notation, we have:

$\left(-\infty\:,\:-5\right)\cup\left(3,\:\infty\:\right)$