Question 1092649
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Problem 1) Solve |x| > x-1


If |x| > x-1, then we have two cases:
Case A: x > x-1
OR
Case B: -x > x-1


The difference between case A and case B is that negative out front the first 'x'.


Let's solve for x in each case


Case A:
x > x-1
x-x > x-1-x
0x > 0x - 1
0 > -1
This is always true so we effectively have infinitely many solutions for this inequality. In other words, pick any number you want for x and plug it into the inequality. You will get a true result.


Case B:
-x > x-1
-x+x > x-1+x
0x > 2x-1
0 > 2x-1
2x-1 < 0
2x-1+1 < 0+1
2x+0 < 1
2x < 1
2x/2 < 1/2
x < 1/2
x < 0.5
So any value smaller than 1/2 or 0.5 will work as a solution for this inequality


So to recap, the two solutions are
Case A ---> all real numbers
Case B ---> any real number smaller than 0.5


Since we have the keyword "or" as part of the absolute value definition, this means we combine the two results to get the final answer to be "the set of all real numbers".


We can plug any number in for x, into |x| > x-1, and we'll get a true result. It doesn't matter what number you pick


To graph this, simply draw a number line and shade everything on the number line. This indicates you're including all real numbers.


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Problem 2) Solve 7-|2x| > 5


7-|2x| > 5
7-|2x|+|2x| > 5+|2x|
7 > 5+|2x|
7-5 > 5+|2x|-5
2 > |2x|
|2x| < 2
-2 < 2x < 2
-2/2 < 2x/2 < 2/2
-1 < x < 1


To graph this, draw a number line; then plot -1 and 1. Make sure there are open circles at -1 and 1. Finally shade between the open circles (don't shade over the open circles). 
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