solve the following inequality algebraically. |x + 2| ≥ 13

Explanation:

Step1: Split into two cases

Case 1: $x + 2\geq0$ (i.e., $x\geq - 2$), then $|x + 2|=x + 2$. The inequality becomes $x + 2\geq13$.

Step2: Solve Case 1 inequality

Subtract 2 from both sides: $x+2-2\geq13 - 2$ $x\geq11$

Step3: Consider Case 2

Case 2: $x + 2<0$ (i.e., $x<-2$), then $|x + 2|=-(x + 2)$. The inequality becomes $-(x + 2)\geq13$.

Step4: Solve Case 2 inequality

Multiply both sides by - 1 and reverse the inequality sign: $x+2\leq - 13$ Subtract 2 from both sides: $x+2-2\leq - 13-2$ $x\leq - 15$

Answer:

$x\leq - 15$ or $x\geq11$