solving an inequality

[pmuhs3562]

need help seeing the steps in solving the inequality (x+1)/x>0
I know the solution set is x < -1, x >0
but I'm having trouble seeing how to arrive at this answer using the addition, multiplication, reciprocal and multiplicative inverse properties for inequalities

 

[pka]

need help seeing the steps in solving the inequality (x+1)/x>0
I know the solution set is x < -1, x >0

The critical numbers (number where the sign changes) are [imath]-1~\&~0[/imath].
Thus we need to consider [imath](-\infty,-1], [-1,0],~\&~[0,\infty)[/imath].
On which of those sets is the inequality true? Be sure to check the end points.

 

[Dr.Peterson]

need help seeing the steps in solving the inequality (x+1)/x>0
I know the solution set is x < -1, x >0
but I'm having trouble seeing how to arrive at this answer using the addition, multiplication, reciprocal and multiplicative inverse properties for inequalities

We don't typically use those properties to solve such an inequality. We do something along the lines of @pka's suggestion. (There are a couple different ways to do that.)

If you were told that you must use them, please show what those properties are, as taught to you, and make an attempt at using them (or else show a worked example you were given in which they were used).

 

[Steven G]

You have a fraction that is greater than 0.
So either both the numerator and denominator are positive or they are both negative.

You therefore need to solve both (x+1 >0 AND x>0) OR (x+1<0 AND x<0)

 

[pmuhs3562]

Thank you pka, dr peterson and steven g. I was not told I must use my aforementioned " - - - properities for inequalities" to solve for "x". I just assumed I could use them to solve for "x". I now see I can use Steven G's logic to find numbers where the sign (+to- or -to+) changes and then use pka's method to check which ranges the inequality is true. thumbs up :-0