1. ## Inequality problem

Hi I really have no idea how to deal with this - it should reduce to some kind of quadratic. Any help would be much appreciated.

(2+3x)/(4-x) < 2x

2. ## Re: Inequality problem

I wrote out a detailed explanation to a similar problem here: Trying to go back to school to learn calculus (x-2) / (x+4) <7

$\frac{2+3x}{4-x} < 2x$, so

$\frac{2+3x}{4-x} -2x< 0$, so

$\frac{2+3x}{4-x} + \frac{(-2x)(4-x)}{4-x}< 0$, so

$\frac{2+3x}{4-x} + \frac{2x^2 - 8x}{4-x}< 0$, so

$\frac{(2+3x) + (2x^2 - 8x)}{4-x}< 0$, so

$\frac{2x^2 - 5x + 2}{4-x}< 0$, so

$\frac{(2x -1)(x-2)}{4-x}< 0$.

Many ways to proceed now, by the easiest is using the Intermediate Value Theorem. Basically, on intervals where it's continuous (so doesn't include 4), and not 0 (so doesn't include 2 or 1/2), it must remain entirely positive or entirely negative throughout the entire interval. Test a point in there, and you'll know the inequality for the entire interval.

Intervals are: $(-\infty, 1/2), (1/2, 2), (2, 4), (4, \infty)$.

At $x=-1, \frac{(2(-1) -1)((-1)-2)}{4-(-1)} = \frac{( - )( - )}{+} = +$. No good. (Were checking $(-\infty, 1/2)$.)

At $x=1, \frac{(2(1) -1)((1)-2)}{4-(1)} = \frac{( + )( - )}{+} = -$. So interval $(1/2, 2)$ is part of the solution.

At $x=3, \frac{(2(3) -1)((3)-2)}{4-(3)} = \frac{( + )( + )}{+} = +$. No good. (Were checking $(2, 4)$.)

At $x=5, \frac{(2(5) -1)((5)-2)}{4-(5)} = \frac{( + )( + )}{-} = -$. So interval $(4, \infty)$ is part of the solution.

Since this was a strick inequality, we don't have to check the endpoints.

Thus the solution is $x \in (1/2, 2) \cup (4, \infty)$.