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How do you solve #| 2x + 1 | <3x + 2#?

Answer 1

Eva Abramson

#x > -3/5#

#| 2x + 1 | <3x + 2->{(2x+1 < 3x + 2->x > -1), (2x+1 < -3x-2->x > -3/5):}#

The more restrictive is #x > -3/5# so is the solution

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Answer 2

Avery Alexander

To solve the inequality |2x + 1| < 3x + 2:

Split the inequality into two cases: Case 1: ( 2x + 1 < 3x + 2 ) Case 2: ( -(2x + 1) < 3x + 2 )
Solve each case separately: For Case 1: ( 2x + 1 < 3x + 2 ) ( -x < 1 ) ( x > -1 )

For Case 2: ( -(2x + 1) < 3x + 2 ) ( -2x - 1 < 3x + 2 ) ( -5x < 3 ) ( x > -\frac{3}{5} )
Combine the solutions from both cases: ( x > -\frac{3}{5} ) and ( x > -1 )
Since both conditions are satisfied when ( x > -1 ), the solution to the inequality is ( x > -1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #| 2x + 1 | &lt;3x + 2#?

How do you solve #| 2x + 1 | <3x + 2#?