To solve the inequality [tex]\( |2x - 2| \leq 6 \)[/tex], we need to consider the nature of absolute value expressions. The absolute value of a number [tex]\( a \)[/tex], denoted [tex]\( |a| \)[/tex], is defined as the distance of [tex]\( a \)[/tex] from 0 on the number line. For any expression [tex]\( |E| \leq b \)[/tex], where [tex]\( E \)[/tex] is an algebraic expression and [tex]\( b \)[/tex] is a positive constant, the inequality can be rewritten without the absolute value as:
[tex]\[ -b \leq E \leq b \][/tex]
Given the inequality [tex]\( |2x - 2| \leq 6 \)[/tex], we can rewrite it as:
[tex]\[ -6 \leq 2x - 2 \leq 6 \][/tex]
This means we need to solve the compound inequality:
[tex]\[ -6 \leq 2x - 2 \leq 6 \][/tex]
Let's break this into two separate inequalities and solve each step-by-step:
### 1. Solving [tex]\(-6 \leq 2x - 2\)[/tex]:
Add 2 to both sides to isolate the term involving [tex]\( x \)[/tex]:
[tex]\[ -6 + 2 \leq 2x - 2 + 2 \][/tex]
[tex]\[ -4 \leq 2x \][/tex]
Now, divide both sides by 2:
[tex]\[ \frac{-4}{2} \leq x \][/tex]
[tex]\[ -2 \leq x \][/tex]
### 2. Solving [tex]\(2x - 2 \leq 6\)[/tex]:
Add 2 to both sides to isolate the term involving [tex]\( x \)[/tex]:
[tex]\[ 2x - 2 + 2 \leq 6 + 2 \][/tex]
[tex]\[ 2x \leq 8 \][/tex]
Now, divide both sides by 2:
[tex]\[ \frac{2x}{2} \leq \frac{8}{2} \][/tex]
[tex]\[ x \leq 4 \][/tex]
### Combining the Results:
We have two inequalities: [tex]\( -2 \leq x \)[/tex] and [tex]\( x \leq 4 \)[/tex]. When we combine these, we obtain:
[tex]\[ -2 \leq x \leq 4 \][/tex]
This means that [tex]\( x \)[/tex] can take any value between -2 and 4, inclusive.
### Final Answer:
The solution to the inequality [tex]\( |2x - 2| \leq 6 \)[/tex] is:
[tex]\[ -2 \leq x \leq 4 \][/tex]
This tells us the range of [tex]\( x \)[/tex] values that satisfy the given inequality.
