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

`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.

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]}

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