How do you solve the inequality #abs(x + 2) <=11#?

Answer 1

Determine your boundary conditions to solve for the minimum and maximum values of x, then write out the final inequality #-13lexle9#

We need to determine the boundaries where this inequality is true. Because we're dealing with an absolute value, the two scenarios look like this:

#x+2le11#
#x+2ge-11#

Notice I flipped the sign for the opposite boundary. This is because we're dealing with the negative solution, and when you flip a sign on an inequality, the direction of the comparator flips as well.

Now, let's solve for x in both expressions:

#x+2le11#
#xcancel(+2-2)le11-2#
#color(red)(xle9)#
#x+2ge-11#
#xcancel(+2-2)ge-11-2#
#color(blue)(xge-13)#

Now that we know our bounds, we can write it as a single inequality:

#color(green)(-13lexle9)#
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Answer 2

To solve the inequality ( |x + 2| \leq 11 ), you'll consider two cases: when ( x + 2 ) is positive and when it's negative.

Case 1: ( x + 2 \geq 0 ) In this case, the absolute value function is just ( x + 2 ), so the inequality becomes: [ x + 2 \leq 11 ]

Case 2: ( x + 2 < 0 ) In this case, the absolute value function is ( -(x + 2) ), so the inequality becomes: [ -(x + 2) \leq 11 ]

Now, solve each inequality separately.

Case 1: [ x + 2 \leq 11 ] [ x \leq 11 - 2 ] [ x \leq 9 ]

Case 2: [ -(x + 2) \leq 11 ] [ -x - 2 \leq 11 ] [ -x \leq 11 + 2 ] [ -x \leq 13 ] [ x \geq -13 ]

So, combining both cases, the solution to the inequality ( |x + 2| \leq 11 ) is: [ -13 \leq x \leq 9 ]

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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.

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.

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.

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.

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