1. ## Rational inequalities

$\frac{1}{2x-3} > 5$

Can someone show me how the heck i work these out, i don't recall going over them in class and i have a test tonight which i need to know how to do this. I know what the answer is but i am banging my head as to how the heck it is that. I know how to do inequalities but not rational ones.

2. Originally Posted by el123
$\frac{1}{2x-3} > 5$

Can someone show me how the heck i work these out, i don't recall going over them in class and i have a test tonight which i need to know how to do this. I know what the answer is but i am banging my head as to how the heck it is that. I know how to do inequalities but not rational ones.
In order to be greater than 5, $\frac{1}{2x-3}$ must be certainly be greater than 0 so 2x-3 is positive. That means that we can just multiply both sides of the inequality by it to get 1> 5(2x- 3)= 10x- 15. That should be easy to solve.

That was easy because it was clear that the denominator must e positive to satisfy the inequality but usually we cannot be sure of that. More generally, the simplest way to solve complicated inequalities is to first solve the associated equality- here, $\frac{1}{2x-3}= 5$. Solving that is also easy: multiplying both sides by 2x- 3, 1= 5(2x- 3). 10x- 15= 1 gives 10x= 16 so x= 16/10= 8/5.

The point of that is that such an inequality can "swap" directions only where the two sides are equal or where there is a "discontinuity". Rational functions are discontinuous only where the denominator is 0:here where 2x-3= 0 or x= 3/2. The two points at which the inquality can reverse are 3/2 and 8/5. They divide the real number line into three intervals. Checking a single point in each interval will tell you which is ">" and which is "<".

3. the denominator should be $|2x-3|$

Sorry , i tried to edit but net cut out.

so does the aboe still apply if the denominator is an absolute value?

4. Originally Posted by el123
so does the aboe still apply if the denominator is an absolute value?
Hello you have two formula
The firste is :

$a > 0,b > 0:a > b \Leftrightarrow \frac{1}{a} < \frac{1}{b}
$

The second is :

$c > 0:\left| x \right| < c \Leftrightarrow - c < x < c$
In this exercice :

$\frac{1}{{\left| {2x - 3} \right|}} > 5 \Leftrightarrow \left| {2x - 3} \right| < \frac{1}{5}
$

$\left| {2x - 3} \right| < \frac{1}{5} \Leftrightarrow - \frac{1}{5} < 2x - 3 < \frac{1}{5}$

$- \frac{1}{5} < 2x - 3 < \frac{1}{5} \Leftrightarrow - \frac{1}{5} + 3 < 2x - 3 + 3 < \frac{1}{5} + 3$
$
- \frac{1}{5} + 3 < 2x - 3 + 3 < \frac{1}{5} + 3 \Leftrightarrow \frac{{14}}{5} < 2x < \frac{{16}}{5}

$

$\frac{{14}}{5} < 2x < \frac{{16}}{5} \Leftrightarrow \frac{7}{5} < x < \frac{8}{5}$
Conclusion the set of solutions is :

$\frac{7}{5} < x < \frac{8}{5}$

5. Hello, el123!

$\frac{1}{2x-3} \:>\: 5$

Resist the temptation to eliminate the denominator ... It contains an $x.$

Instead, bring all terms to one side and simplify:

. . $\frac{1}{2x-3} - 5 \:>\:0 \quad\Rightarrow\quad \frac{16-10x}{2x-3} \:>\:0$

We have a positive fraction.

A fraction ia positive if:

. . (1) the numerator and denominator are positive.
. . (2) the numerator and denominator are negative.

Examine the two cases.

(1) Both positive: . $\begin{array}{ccccccc}16-10x \:>\:0 & \Rightarrow & x \:<\:\frac{8}{5} \\ \\[-4mm] 2x-3 \:>\:0 & \Rightarrow & x \:>\:\frac{3}{2} \end{array}$

. . .We have: . $(x < 1.6) \wedge (x > 1.5) \quad\Rightarrow\quad{\color{blue} \boxed{1.5 \:<\:x \:<\:1.6}}$

(2) Both negative: . $\begin{array}{ccccccc}16-10x \:<\:0 & \Rightarrow & x \:>\:\frac{8}{5} \\ \\[-4mm] 2x-3 \:<\:0 & \Rightarrow & x \:<\:\frac{3}{2} \end{array}$

. . .We have: . $(x > 1.6) \wedge (x < 1.5)\quad\hdots \text{Impossible}$