Respuesta :
Solution:
Given the inequality:
[tex]|3x+3|+3>15[/tex]To solve the inequality,
step 1: Add -3 to both sides of the inequality.
Thus,
[tex]\begin{gathered} |3x+3|+3-3>-3+15 \\ \Rightarrow|3x+3|>12 \end{gathered}[/tex]Step 2: Apply the absolute rule.
According to the absolute rule:
[tex]\mathrm{If}\:|u|\:>\:a,\:a>0\:\mathrm{then}\:u\:<\:-a\:\quad \mathrm{or}\quad \:u\:>\:a[/tex]Thus, from step 1, we have
[tex]\begin{gathered} 3x+3<-12\text{ or 3x+3>12} \\ \end{gathered}[/tex][tex]\begin{gathered} when \\ 3x+3<-12 \\ add\text{ -3 to both sides of the inequality} \\ 3x-3+3<-3-12 \\ \Rightarrow3x<-15 \\ divide\text{ both sides by the coefficient of x, which is 3} \\ \frac{3x}{3}<-\frac{15}{3} \\ \Rightarrow x<-5 \end{gathered}[/tex][tex]\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}[/tex]This implies that
[tex]x<-5\quad \mathrm{or}\quad \:x>3[/tex]Hence, in interval notation, we have:
[tex]\left(-\infty\:,\:-5\right)\cup\left(3,\:\infty\:\right)[/tex]