Trouble with Inequalities problem

161
0
Hi, can someone tell me when should i change my sign when i have such a problem?

-(75/x) > 15

i find that if -75/x > 15
i could do :
-75 > 15x
-5 > x
x < -5.

or 75/-x > 15
i could do :
75 < -15x
-5 > x

which the answer in the book says -5 < x < 0

Thanks.
 

Mentallic

Homework Helper
3,797
94
Re: Inequalities

The problem is that x can be any real value, which means it can be negative as well. But when you multiply by a negative number, the inequality needs to be changed around. There are a few ways of solving this dilemma:

1) Consider both cases, when x>0 and x<0 ([tex]x \neq 0[/tex])

Solve both these cases normally like you would with any positive or negative number in place for the x. Then the solutions for both cases need to intersect (both must be satisfied).

2) Manipulate the fraction in such a way that you won't have the problem of multiplying through by a negative. Hint: a number squared is always positive.


p.s. graphmatica also makes the same mistake.
 

HallsofIvy

Science Advisor
41,626
821
Re: Inequalities

Hi, can someone tell me when should i change my sign when i have such a problem?

-(75/x) > 15

i find that if -75/x > 15
i could do :
-75 > 15x
-5 > x
x < -5.

or 75/-x > 15
i could do :
75 < -15x
-5 > x

which the answer in the book says -5 < x < 0

Thanks.
Neither -75> 15x nor 75< -15x is correct. You change the direction of the inequality when multiplying by a negative number, but it does NOT follow that "-x" is a negative number or that "x" is a positive number; that depends upon x.

If x is positive, then you can multiply on both sides by x to get -75> 15x so -5> x. But if x is positive, it can't be less than -5. If x is negative, then multiplying on both sides gives -75< 15x so -5< x. Since we have assumed that x is negative, that says -5< x< 0.
 

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