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How do you solve #abs(x+a)<b#?

Answer 1

Eva Adamson

#a-b < x < b-a#

#abs(x+a) < b -> {(x+a < b->x < b-a), (-x+a < b->x > a-b) :}#

#a-b < x < b-a#

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Answer 2

Nathan Axel

To solve the inequality ( |x + a| < b ), where ( a ) and ( b ) are constants:

Subtract ( a ) from both sides: ( |x + a| - a < b - a ).
Split the inequality into two cases: a) If ( x + a \geq 0 ), then the inequality becomes ( x + a < b - a ). b) If ( x + a < 0 ), then the inequality becomes ( -(x + a) < b - a ).
Solve each case separately for ( x ): a) For ( x + a \geq 0 ): ( x < b - 2a ). b) For ( x + a < 0 ): ( -x - a < b - a ).
For case b), multiply both sides by -1 to remove the negative sign: ( x + a > -(b - a) ).
Simplify: ( x + a > -b + a ).
Subtract ( a ) from both sides: ( x > -b ).
Combine the solutions from both cases: a) ( x < b - 2a ) b) ( x > -b )

These are the solutions for the inequality ( |x + a| < b ).

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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 #abs(x+a)&lt;b#?

How do you solve #abs(x+a)<b#?