Question

What is the ans please?

Answer 1

The correct answer is `"option A"`

Explanation:

The inequality is

`1/(4x-3)<=x`

Therefore,

`x-1/(4x-3)>=0`

Putting on the same denominator

`(x(4x-3)-1)/(4x-3)>=0`

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

The correct answer is `"option A"`

Answer 2

`0<=(4x^2-3x-1)/(4x-3)` (part a)

Explanation:

given

`color(green)(=> 1/(4x-3)<=x)`

Subtract `color(blue)(1/(4x-3))` from both sides.

`1/(4x-3)-color (red)(1/(4x-3))<=x-color (red)(1/(4x-3))`

`0 <=(x (4x-3)-1)/(4x-3)`

`color (firebrick)(0 <=(4x^2-3x-1)/(4x-3))` or part a

Answer 3

I will use this online graphing calculator to show that the correct answer is A

Explanation:

First I will show you `color(red)(1/(4x-3)<=x)`

Please observe that it is saying that y is any value that you like but domain of x is `-1/4<=x<3/4` and `x>=1`.

I will save the graph of A for after I have shown that selections B through E are NOT the solutions:

Here is Selection B `color(blue)((4x^2-3x-1)/(4x-3)<=0)`:

Please observe that it is saying that y is any value that you like but domain of x is `-1/4<=x` and `3/4<x<=1`; this clearly disagrees with the graph of the original equation.

Here is Selection C `color(green)(4x^2-3x-1>=0)`

Please observe that it is saying that y is any value that you like but domain of x is `-1/4<=x` and `x>=1`; this clearly disagrees with the graph of the original equation.

Here is Selection D `color(purple)(4x^2-3x-1<=0)`

Please observe that it is saying that y is any value that you like but domain of x is `-1/4<=x<=1`; this clearly disagrees with the graph of the original equation.

I cannot use the inequality function for selection E `color(orange)(4x^2+3x+1<=0)`, because `color(orange)(y = 4x^2+3x+1)` is never less than 0 and, therefore, the domain of the inequality is the null set but I will show you the parabola:

Finally, the correct answer Selection A `(4x^2-3x-1)/(4x-3)>=0`

Please observe that the graph of the original inequality is identical to this graph with the same domain, `-1/4<=x<3/4` and `x>=1`.