#2**+10 **

$$\\\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. :)

Melody
Mar 16, 2015

#1**+10 **

(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

CPhill
Mar 16, 2015

#2**+10 **

Best Answer

$$\\\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. :)

Melody
Mar 16, 2015