A simple Inequality equation that I cannot solve

$$\\\frac{(x+1)}{(2-x) }< \frac{x}{(3+x)}\\\\
$I am going to multiply both sides by positive numbers and see if that will help$\\\\
(2-x)^2(3+x)^2*\frac{(x+1)}{(2-x) }<(2-x)^2(3+x)^2 \frac{x}{(3+x)}\\\\
(2-x)(3+x)^2(x+1)<(2-x)^2(3+x) x\\\\
(2-x)(3+x)^2(x+1)-(2-x)^2(3+x) x<0\\\\
(2-x)(3+x)\;[(3+x)(x+1)-(2-x) x]<0\\\\
-(x-2)(x+3)\;[(x+3)(x+1)-x(-x+2)]<0\\\\
-(x-2)(x+3)\;[x^2+x+3x+3+x^2-2x]<0\\\\
-(x-2)(x+3)\;[2x^2+2x+3]<0\\\\$$

consider   

  $$\\2x^2+2x+3=0\\\\
\triangle=4-24<0\\\\
$since the discriminant is 0 there are no roots for this.$\\\\
$the axis of symmetry for $y=2x^2+2x+3\;\;is\;\; \frac{-b}{2a}=\frac{-2}{4}=\frac{-1}{2}$$

the roots are    x=2 AND x=-3       

the polynomial if set to y and graphed will finish in the bottom right corner because of the - out to front.

I can see that the polynomial will be less than 0 when  x>2 and when x<-3

I am sorry i probably have not explained this very well, it was a difficult question for this type.

I did not use the graph but I will draw it now to show you.

https://www.desmos.com/calculator/5xfsalfbbg

Feel free to ask questions

CPhill's way was probably better for this one.  :)

(x+1)/(2-x) < x/(3+x)   

This might be solved easiest by some analysis

Let's divide the inequality into three iintervals

(-∞, -3), (-3, 2) and (2, ∞)

Notice that, on the first interval, the function on the left is always less than the function on the right

We can't say what happens at x = -3 because the function on the right isn't defined there

On the second interval, the function on the left is always greater than the function on the right

And at x = 2 the function on the left is undefined

On the third interval, the function on the left is again always less than the function on the right

So, the solution to this problem is  that the intervals (-∞, -3) and (2, ∞)  make the inequality true and the interval (-3, 2) makes it false

Here's a graph that confirms our suspicions.... https://www.desmos.com/calculator/ymvoluxlsa