How do you solve # 2x + 1  <3x + 2#?
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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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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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