SOLUTION: I'm confused by the outcome I get from the problem 1/4|x-3|+2 <1 since < in the original equation is supposed to represent "and", yet my answer when graphed shows an "or" trend on

Question 776554: I'm confused by the outcome I get from the problem 1/4|x-3|+2 <1 since < in the original equation is supposed to represent "and", yet my answer when graphed shows an "or" trend on my number line. When I solved it, I ended out with x <7 and x>-1. If you see anything I've done wrong, please correct me, and I thank you in advance for your input.
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I'm confused by the outcome I get from the problem 1/4|x-3|+2 <1 since < in the original equation is supposed to represent "and", yet my answer when graphed shows an "or" trend on my number line. When I solved it, I ended out with x <7 and x>-1. If you see anything I've done wrong, please correct me, and I thank you in advance for your input.
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Absolute value problems usually give 2 answers, thus the "or."
----
1/4|x-3|+2 <1
Multiply by 4
|x-3| + 8 < 4
Subtract 8
|x-3| < -4
--> no solution
Absolute value cannot be negative.

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Your problem is that you are ignoring the elephant in the living room. No matter what happens inside of the absolute value bars, the absolute value quantity will be greater than or equal to zero. So the smallest value that can assume is zero, namely when . So when , the entire LHS of the inequality evaluates to 2 which is always larger than 1. Hence, there are no real values of that satisfy the given inequality.

So let's solve the thing and see if we get a result compatible with our intuition.

Multiply by 4

Add -8 to both sides:

(((actually, we could have quit right here. Right?)))

Then either

Or

You were correct in your assessment that the in the original inequality indicated that you needed to find the intersection of two solution sets ("and" as you put it). So, any value in the solution set of the original inequality must be simultaneously less than minus 1 and greater than 7. Since there are no real numbers that fit these criteria, the solution set to the original inequality is indeed the null set.

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it

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