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