$$\\\frac{(x+1)}{(2-x) }< \frac{x}{(3+x)}\\\\
$I am going to multiply both sides by positive numbers and see if that will help$\\\\
(2-x)^2(3+x)^2*\frac{(x+1)}{(2-x) }<(2-x)^2(3+x)^2 \frac{x}{(3+x)}\\\\
(2-x)(3+x)^2(x+1)<(2-x)^2(3+x) x\\\\
(2-x)(3+x)^2(x+1)-(2-x)^2(3+x) x<0\\\\
(2-x)(3+x)\;[(3+x)(x+1)-(2-x) x]<0\\\\
-(x-2)(x+3)\;[(x+3)(x+1)-x(-x+2)]<0\\\\
-(x-2)(x+3)\;[x^2+x+3x+3+x^2-2x]<0\\\\
-(x-2)(x+3)\;[2x^2+2x+3]<0\\\\$$
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. :)
One thing about probabiliity is that it can be really difficult to interprete what you are being asked for.
The other thing about prob is that although you may be able to understand the correct answer when it is presented to you, you often still cannot understand why your own original answer is incorrect!
It is an interesting and a frustrating feild of methematics. ![]()
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I'm listening to this
https://www.youtube.com/watch?v=Uka8ykFDw2U
Something different for you Chris ![]()