need help with another non linear inequality (with fractions)

[abel muroi]

I was given another problem...

1 + 2/(x+1) <= 2/x

and here is what i tried out so far...

1 + 2/(x+1) <= 2/x
(x+1)/(x+1) + 2/(x+1) - 2/x <= 0

my question is... how can i find a common denominator for the -2/x? should i just multiply the numerator and denominator with (x + 1)?

 

[HallsofIvy]

I was given another problem...

1 + 2/(x+1) <= 2/x

and here is what i tried out so far...

1 + 2/(x+1) <= 2/x
(x+1)/(x+1) + 2/(x+1) - 2/x <= 0

my question is... how can i find a common denominator for the -2/x? should i just multiply the numerator and denominator with (x + 1)?

Yes, the denominators are x and x+ 1 so the "common" denominator would be x(x+ 1). To get common denominators, you would need to multiply both numerator and denominator of "-2/x" by x+ 1 and multiply numerator and denominator of the other two fractions by x.

 

[Steven G]

I was given another problem...

1 + 2/(x+1) <= 2/x

and here is what i tried out so far...

1 + 2/(x+1) <= 2/x
(x+1)/(x+1) + 2/(x+1) - 2/x <= 0

my question is... how can i find a common denominator for the -2/x? should i just multiply the numerator and denominator with (x + 1)?

You asked how can i find a common denominator for the -2/x. This question makes no sense as you find common denominator between two or more terms, Not one term.
I would proceed as follows. The first term has a denominator of (x+1), so (x+1) will be a factor of the common denominator. The next term has a denominator of (x+1). Since this factor is already part of my common denominator so far, which is (x+1), we kind of ignore this factor. The 3rd term has a factor of x which is NOT part of our common denominator so far. So we also use the factor x in our common denominator. So our common denominator is x(x+1).