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What is the solution set of the inequality l2x + 3l < 1, then the solution set contains…

A. Only negative real numbers
B. Only positive real numbers
C. Both positive and negative real numbers
D. No real numbers

1 Answer

4 votes

Answer: Only negative real numbers (choice A)

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Use the rule that if |x| < k then -k < x < k where k is some positive real number

In this case we have k = 1, so,

|2x+3| < 1

turns into

-1 < 2x+3 < 1

From here we subtract 3 from all three sides

-1-3 < 2x+3-3 < 1-3

-4 < 2x < -2

Then we divide all three sides by 2 to fully isolate x

-4/2 < 2x/2 < -2/2

-2 < x < -1

Therefore, x is some number between -2 and -1. The value of x cannot equal -2. The value of x cannot equal -1 either. To graph this on a number line, we plot open circles at -2 and -1, then we shade the region between those open circles. The open circles tell the reader "do not include this value as part of the solution". Check out the attached image below for a visual.

No matter what x value you pick as a possible solution (eg: x = -1.7) the value will be negative. This is because -1 is to the left of 0 on the number line, so further left is even more negative territory.

What is the solution set of the inequality l2x + 3l < 1, then the solution set-example-1
User Morgan Creighton
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