I have the following inequalities:
-2 < h*2/x < 0
-2.78 <h*2/x < 0
where x is 1 < x < 2 ?
Are the answers:
x > h > 0 and 1.39x > h > 0
or do i have problems with the way I have placed the signs?
and How does the interval affect the outcome?
I have the following inequalities:
-2 < h*2/x < 0
-2.78 <h*2/x < 0
where x is 1 < x < 2 ?
Are the answers:
x > h > 0 and 1.39x > h > 0
or do i have problems with the way I have placed the signs?
and How does the interval affect the outcome?
There are only few rules which you have to observe when dealing with unequalities:
a) adding or subtracting a term doesn't change the relation-sign
b) dividing or multiplying by a positive term doesn't change the relation-sign
c) calculationg the reziprocal of fractions could change the relation-sign depending of the sign of the fractions in question. Examples:
$\displaystyle \dfrac13 > \dfrac14~\implies~3<4$
$\displaystyle -\dfrac13 < -\dfrac14~\implies~-3>-4$
BUT:
$\displaystyle -\dfrac13 < \dfrac14~\implies~-3<4$
To your questions:
$\displaystyle -2<h\cdot \frac2x <0~\wedge~x>0$ that means x is positive
$\displaystyle -2x<2h<0~\wedge~x>0$ divide through by -2 (the divisor is negative). The relation-signs have to be changed. Therefore your answer is correct.
$\displaystyle -2<h\cdot \frac2x <0~\wedge~x>0~\implies~-2x<2h<0$ . Dividing thriough by +2 will yield:
$\displaystyle -1.39x<h<0$
You probably have noticed that I've made a silly mistake. Here comes the corrected calculation:
To your questions:
$\displaystyle -2<h\cdot \frac2x <0~\wedge~x>0$ that means x is positive and h must be negative
$\displaystyle -2x<2h<0~\wedge~x>0$ divide through by -2 (the divisor is negative). The relation-signs have to be changed. Therefore the answer is:
$\displaystyle x>-h>0$