Question

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

Answer 1

`p in (1,11/3]`

Explanation:

`2/(p-1)>=3/4`

`2/(p-1)-3/4>=0`

`(2*4-3(p-1))/(4(p-1))>=0`

`(11-3p)/(4(p-1))>=0`

zero points:
`p_1=11/3`

`p_2=1`

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

`p in (-oo,1)hArr` function is negative

`color(red)(p in (1,11/3)hArr "function is positive")`

`p in (11/3,oo)hArr` function is negative

But the function may equal to zero which is only when `p=11/3`. Therefore we include 11/3 to the asnwer.

Answer 2

The solution is `p in (1,11/3]`

Explanation:

Simplify the inequality

`2/(p-1)>=3/4`

`2/(p-1)-3/4>=0`

Putting on the same denominator

`((2*4-3*(p-1)))/(4(p-1))>=0`

`(8-3p+3)/(4(p-1))>=0`

`(11-3p)/(4(p-1))>=0`

Let `f(p)=(11-3p)/(4(p-1))`

Construct the sign chart

`color(white)(aaaa)` `p` `color(white)(aaaa)` `-oo` `color(white)(aaaaaa)` `1` `color(white)(aaaaaaa)` `11/3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `p-1` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `11-3p` `color(white)(aaa)` `+` `color(white)(aaaa)` ###color(white)(aaaaa)#`+` `color(white)(aa)` `0` `color(white)(aa)` `-`

`color(white)(aaaa)` `f(p)` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+` `color(white)(aa)` `0` `color(white)(aa)` `-`

Therefore,

`f(p) >=0` when `p in (1,11/3]`

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

Answer 3

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.

Answer 4

Ulternate approach

Explanation: