How do you solve #|10x - 28| + 2> 24#?

Answer 1
To solve the inequality |10x - 28| + 2 > 24, follow these steps: 1. Subtract 2 from both sides to isolate the absolute value term: |10x - 28| > 22. 2. Break the absolute value inequality into two separate inequalities: 10x - 28 > 22 and -(10x - 28) > 22. 3. Solve each inequality separately. 4. For 10x - 28 > 22: - Add 28 to both sides: 10x > 50. - Divide both sides by 10: x > 5. 5. For -(10x - 28) > 22: - Distribute the negative sign: -10x + 28 > 22. - Subtract 28 from both sides: -10x > -6. - Divide both sides by -10, remembering to reverse the inequality sign because of dividing by a negative number: x < 0.6. 6. Combine the solutions from steps 4 and 5: x < 0.6 or x > 5.
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Answer 2

#x>5, x < 3/5#

Let's first isolate the absolute value term:

#abs(10x-28)+2>24#
#abs(10x-28)>22#
With absolute values, we need to evaluate the positive aspect and negative aspect (keep in mind that with, say #abs(x)=3#, #x# can be 3 and also can be #-3#, and so when we evaluate the absolute value, we say #x=pm3#. We do the same process for all absolute value questions).
#pm(10x-28)>22#

Positive aspect

#10x-28>22#
#10x>50#
#x>5#

Negative aspect

#-(10x-28)>22#
#-10x+28>22#
#-10x> -6#
we divide by #-10# and so we change the direction of the inequality:
#x < 6/10=3/5#

Putting it together:

#x>5, x < 3/5#
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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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