# Thread: Nonlinear Inequalities

1. ## Nonlinear Inequalities

Express the solution using interval notation and graph.

#1)
$\displaystyle 4x/2x+3>2$

#2)
$\displaystyle x+2/x+3<x-1/x-2$

Thanks

2. Originally Posted by Ballplaya4237
Express the solution using interval notation and graph.

#1)
$\displaystyle 4x/2x+3>2$

#2)
$\displaystyle x+2/x+3<x-1/x-2$

Thanks
I'll show the first, the second is too ambiguous and needs to be re-formatted:

$\displaystyle \frac{4x}{2x + 3} > 2$.

There's a number of aproaches - the simplest would be to solve $\displaystyle \frac{4x}{2x + 3} = 2$, look at the graph to see the relevant intervals and then write down the answer. But that might be beyond your present scope.

A slower way:

$\displaystyle \frac{4x}{2x + 3} - 2 > 0$

$\displaystyle \Rightarrow \frac{4x}{2x + 3} - \frac{2(2x + 3)}{2x + 3} > 0$

$\displaystyle \Rightarrow \frac{4x - 2(2x + 3)}{2x + 3} > 0$

$\displaystyle \Rightarrow \frac{-6}{2x + 3} > 0$

$\displaystyle \Rightarrow \frac{6}{2x + 3} < 0$

You therefore require 2x + 3 < 0 (think about it) => ......

3. Alright, thanks. Interval notation would be (-Infinity, -3/2), correct?

How would I go about doing the second problem?

4. Originally Posted by Ballplaya4237
Alright, thanks. Interval notation would be (-Infinity, -3/2), correct?

How would I go about doing the second problem?
First you would comply with the implied request that was made:

Originally Posted by mr fantastic
[snip]
the second is too ambiguous and needs to be re-formatted
[snip]

5. $\displaystyle \frac{x+2}{x+3}<\frac{x-1}{x-2}$

Better?

6. Originally Posted by Ballplaya4237
$\displaystyle \frac{x+2}{x+3}<\frac{x-1}{x-2}$

Better?
Much!

$\displaystyle \frac{x+2}{x+3} < \frac{x-1}{x-2}$

$\displaystyle \Rightarrow \frac{x+2}{x+3} - \frac{x-1}{x-2} < 0$

Get common denominator:

$\displaystyle \Rightarrow \frac{(x+2)(x - 2)}{(x+3)(x - 2)} - \frac{(x-1)(x + 3)}{(x-2)(x + 3)} < 0$

$\displaystyle \Rightarrow \frac{(x+2)(x - 2) - (x-1)(x + 3)}{(x+3)(x - 2)} < 0$

Simplify:

$\displaystyle \Rightarrow \frac{-1 - 2x}{(x+3)(x - 2)} < 0$

$\displaystyle \Rightarrow \frac{2x + 1}{(x+3)(x - 2)} > 0$

So you want either:

2x + 1 > 0 AND (x + 3)(x - 2) > 0

OR

2x + 1 < 0 AND (x + 3)(x - 2) < 0

The first is only satisfied when x > 2.

The second is only satisfied when x < -3.

The details are left for you .....

7. You lost me when you got the common denominator. Not sure where you got it from nor why one side is being multiplied by $\displaystyle (x-2)$ and the other side by$\displaystyle (x+2)$.

8. Originally Posted by Ballplaya4237
You lost me when you got the common denominator. Not sure where you got it from nor why one side is being multiplied by $\displaystyle (x-2)$ and the other side by$\displaystyle (x+2)$.
My mistake - the common denominator is (x + 3)(x - 2).

Standby while I fix it up ......

OK, all fixed. Re-read. Are you still lost? Do you understand what's happening with the common denominator business ......

It's the same idea as getting $\displaystyle \frac{1}{3} + \frac{1}{7}$ over a common denominator .....

9. I understand the GCF just fine but I'm completly failing to see how you simplified the numerator $\displaystyle (x+2)(x-2)-(x-1)(x+3)$

10. Originally Posted by Ballplaya4237
I understand the GCF just fine but I'm completly failing to see how you simplified the numerator $\displaystyle (x+2)(x-2)-(x-1)(x+3)$
$\displaystyle (x+2)(x-2)-(x-1)(x+3) = x^2 -2x + 2x - 4 - (x^2 + 3x - x - 3)$

$\displaystyle = x^2 - 4 - (x^2 + 2x - 3) = x^2 - 4 - x^2 - 2x + 3 = -1 - 2x$.

Which I hope is what I had in the numerator before I multiplied both sides of the inequality by -1 (hence reversing the inequality sign and turning -1 - 2x into -(-1 - 2x) = 1 + 2x).

11. Awesome, thanks for the help.