# Rational inequalities

• Aug 19th 2009, 05:21 PM
el123
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.
• Aug 19th 2009, 05:40 PM
HallsofIvy
Quote:

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 "<".
• Aug 19th 2009, 05:42 PM
el123
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?
• Aug 20th 2009, 04:05 AM
dhiab
Quote:

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 :

$\frac{7}{5} < x < \frac{8}{5}$
• Aug 20th 2009, 07:50 AM
Soroban
Hello, el123!

Quote:

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

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