Question

If $10<|x+10|\le 25$, then $x$ lies in

• A. $[-45,-20)\cup(0,15]$
• B. $[-35,-25)\cup(0,15]$
• C. $[-35,-20)\cup(0,15]$
• D. $[-35,-20)\cup(0,25]$
• E. $[-35,-10)\cup(0,15]$

Hint:

Math Tip: When dealing with a bounded absolute value like $a<|x|<b$, think of it geometrically as the distance from zero. The value must be between $a$ and $b$ units away from zero on the positive side, OR between $a$ and $b$ units away on the negative side.

Answer 1

The Correct Option is C

Concept:
Algebra - Solving Absolute Value Inequalities.
If $a<|X| \le b$, then the expression inside the absolute value can be either positive or negative, leading to two distinct cases: $a<X \le b$ OR $-b \le X<-a$.
Step 1: Set up the two cases based on absolute value properties.
Given the inequality: $10<|x+10| \le 25$.
Let $X = x+10$. We split this into two separate compound inequalities: $$ \text{Case 1 (Positive): } 10<x+10 \le 25 $$ $$ \text{Case 2 (Negative): } -25 \le x+10<-10 $$
Step 2: Solve Case 1.
$$ 10<x + 10 \le 25 $$ Subtract 10 from all parts of the inequality: $$ 10 - 10<x \le 25 - 10 $$ $$ 0<x \le 15 $$ In interval notation, this gives: $(0, 15]$.
Step 3: Solve Case 2.
$$ -25 \le x + 10<-10 $$ Subtract 10 from all parts of the inequality: $$ -25 - 10 \le x<-10 - 10 $$ $$ -35 \le x<-20 $$ In interval notation, this gives: $[-35, -20)$.
Step 4: Combine the results.
The final solution is the union of the sets found in Case 1 and Case 2: $$ x \in [-35,-20) \cup (0,15] $$