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What is the solution to l x + 2 | <1?

OA. 1
OB. -3
OC. x>3 or x< 1

OD. x>-1 or x < -3

User Arindam
by
8.4k points

1 Answer

5 votes

Answer:

Answer:

x > - 3 or x < -1

I cannot see this answer choice; perhaps you typed it wrong and it is option C. Please check again

Explanation:

The absolute value rule says for a given variableu,
if |u| < a where a is a constant,

then -a < u < a

Here we can represent x + 2 as u giving

|u| < 1

=> - 1 < u < 1

Substituting for u = x + 2 gives

-1 < x + 2 < 1

Subtract 2 from all sides
-1 - 2 < x + 2 - 2 < 1 - 2

-3 < x < -1

This can be represented as
x > -3 or x < -1

I do not see this choice anywhere but that is the correct answer. Check your answer choices

User Arvind S Salunke
by
8.9k points