Question 776554
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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 *[tex \LARGE \frac{1}{4}|x\ -\ 3|] can assume is zero, namely when *[tex \LARGE x = 3].  So when *[tex \LARGE x\ =\ 3], the entire LHS of the inequality evaluates to 2 which is always larger than 1.  Hence, there are no real values of *[tex \LARGE x] that satisfy the given inequality.


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{4}|x\ -\ 3|\ +\ 2\ <\ 1]


Multiply by 4


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |x\ -\ 3|\ +\ 8\ <\ 4]


Add -8 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ |x\ -\ 3|\ <\ -4]


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


Then either


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 3\ <\ -4\ \Right\ x\ <\ -1]


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 3\ >\ 4\ \Right\ x\ >\ 7]


You were correct in your assessment that the *[tex \LARGE <] 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
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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