Solving Rational Inequality

[sunwers]

I need help with a question that I have missed. Here is my question

Solve the rational inequality. Write the answer in interval notation.

(3)/(x+1)>(-x)/(x+1)

Here is my answer in steps of how I solved it

(3)/(x+1)+(x)/(x+1)>0

(3(x+1)/(x+1)+(x(x+1)/(x+1)>0

(3x+3+x^2+x)/(x+1)>0

(x^2+4x+3)/(x+1)>0

(x+3)(x+1)/(x+1)>0

x+3=0
x=-3

x+1=0
x=-1

(-infinity,-5)
Sorry not sure what the infinity symbol is here on the forum
((-4+3)(-4+1))/(-4+1)>0
-1>0 False

(-3,-1)
((-2+3)(-2+1))/(-2+1)>0
1>0 True

(-1,infinity)
((0+3)(0+1))/(0+1)>0
3>0 True

So, I put my answer as (-3,-1)U(-1,infinity)

The correct answer was (-infinity,-3)U(-1,infinity)

The only thing I see that I done differently in the beginning was I moved the equation differently to equal it to zero. Here is how they moved the equation

0>(-x)/(x+1)-(3)/(x+1)

What did I do wrong? Do you move them in a certain direction or can you get the same answer no matter which end you have the zero on... if so how did I miss it?

Thanks,
Sunwers

 

[sunwers]

Is this a typo? I'm stopping here because this step is clearly wrong. (x+3) has magically turned into (x+1). Check this. If it's a typo, edit the post and I'll continue to go through it.

I'm sorry yes it is a typo. Both denominators are (x+1)

 

[sunwers]

both denominators are (x+1)? You're going about it the hard way then.

so you have

\(\displaystyle \dfrac{3}{x+1}>\dfrac{-x}{x+1}\)

\(\displaystyle \dfrac{3+x}{x+1}>0\)

this will be true when both numerator and denominator are both either positive or negative.

Figure out what those intervals are.

LOL, oh I see I did go the hard way. Since the denominators are the same I wouldn't have to multiply both by the denominator. So my intervals would be (-infinity,-3)(-3,-1)(-1,infinity)

so (-infinity,-3)U(-1,infinity) would be the answer because these two are true

(3-4)/(-4+1)>0
1/3>0 True

(3+0)/(0+1)>0
3>0 True

(3-2)/(-2+1)>0
-1>0 False

Does it matter which side you put the zero on when solving these problems?

Thanks for all your help.