- #1
NumbersBelow
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Hi, I'm stuck on the following problem:
|(4/x)| > 5
I split it up into two cases, case 1 is x > 0, case 2 is x < 0
*This is where things go wrong, I tested on a number line and I know that the correct solution should be -4/5 < x < 4/5.. but doesn't the sign have to get switched back again because I'm dividing by a negative?
I tried doing it another way:
*Now for some reason I'm getting x > 4/5 which is false.
Any help would be very appreciated!
|(4/x)| > 5
I split it up into two cases, case 1 is x > 0, case 2 is x < 0
Code:
Case 1: Case 2:
4/x > 5 4/x < -5
4 > 5x 4 < -5x
5x < 4 -5x > 4
x < 4/5 *x < -4/5
*This is where things go wrong, I tested on a number line and I know that the correct solution should be -4/5 < x < 4/5.. but doesn't the sign have to get switched back again because I'm dividing by a negative?
I tried doing it another way:
Code:
-5 < 4/x < 5
-5x < 4 < 5x
-5x < 4 4 < 5x
x > -4/5 5x > 4
*x > 4/5
*Now for some reason I'm getting x > 4/5 which is false.
Any help would be very appreciated!