# How do you solve 2/(p-1)>=3/4 using a sign chart?

Jan 6, 2018

$p \in \left(1 , \frac{11}{3}\right]$

#### Explanation:

$\frac{2}{p - 1} \ge \frac{3}{4}$

$\frac{2}{p - 1} - \frac{3}{4} \ge 0$

$\frac{2 \cdot 4 - 3 \left(p - 1\right)}{4 \left(p - 1\right)} \ge 0$

$\frac{11 - 3 p}{4 \left(p - 1\right)} \ge 0$

zero points:
${p}_{1} = \frac{11}{3}$

${p}_{2} = 1$

[[,|,-oo;1,1;11/3,11/3;oo],[11-3p,|,+,+,-],[p-1,|, -,+,+]]

$p \in \left(- \infty , 1\right) \Leftrightarrow$ function is negative

$\textcolor{red}{p \in \left(1 , \frac{11}{3}\right) \Leftrightarrow \text{function is positive}}$

$p \in \left(\frac{11}{3} , \infty\right) \Leftrightarrow$ function is negative

But the function may equal to zero which is only when $p = \frac{11}{3}$. Therefore we include 11/3 to the asnwer.

Jan 6, 2018

The solution is $p \in \left(1 , \frac{11}{3}\right]$

#### Explanation:

Simplify the inequality

$\frac{2}{p - 1} \ge \frac{3}{4}$

$\frac{2}{p - 1} - \frac{3}{4} \ge 0$

Putting on the same denominator

$\frac{\left(2 \cdot 4 - 3 \cdot \left(p - 1\right)\right)}{4 \left(p - 1\right)} \ge 0$

$\frac{8 - 3 p + 3}{4 \left(p - 1\right)} \ge 0$

$\frac{11 - 3 p}{4 \left(p - 1\right)} \ge 0$

Let $f \left(p\right) = \frac{11 - 3 p}{4 \left(p - 1\right)}$

Construct the sign chart

$\textcolor{w h i t e}{a a a a}$$p$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$1$$\textcolor{w h i t e}{a a a a a a a}$$\frac{11}{3}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$p - 1$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$11 - 3 p$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$color(white)(aaaaa)+$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(p\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$-$

Therefore,

$f \left(p\right) \ge 0$ when $p \in \left(1 , \frac{11}{3}\right]$

graph{(11-3x)/(4(x-1)) [-12.66, 12.65, -6.33, 6.33]}

Jan 6, 2018

I never really did them this way. But, here is an excellent tutorial and example: http://www.fmaths.com/signcharts/lesson.php

#### Explanation:

Check the link to see the steps required for this particular problem.

Jan 6, 2018

Ulternate approach

#### Explanation: 