# What is the ans please?

Mar 5, 2018

The correct answer is $\text{option A}$

#### Explanation:

The inequality is

$\frac{1}{4 x - 3} \le x$

Therefore,

$x - \frac{1}{4 x - 3} \ge 0$

Putting on the same denominator

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

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

The correct answer is $\text{option A}$

Mar 5, 2018

$0 \le \frac{4 {x}^{2} - 3 x - 1}{4 x - 3}$ (part a)

#### Explanation:

given

$\textcolor{g r e e n}{\implies \frac{1}{4 x - 3} \le x}$

Subtract $\textcolor{b l u e}{\frac{1}{4 x - 3}}$ from both sides.

$\frac{1}{4 x - 3} - \textcolor{red}{\frac{1}{4 x - 3}} \le x - \textcolor{red}{\frac{1}{4 x - 3}}$

$0 \le \frac{x \left(4 x - 3\right) - 1}{4 x - 3}$

$\textcolor{f i r e b r i c k}{0 \le \frac{4 {x}^{2} - 3 x - 1}{4 x - 3}}$ or part a

Mar 6, 2018

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

#### Explanation:

First I will show you $\textcolor{red}{\frac{1}{4 x - 3} \le x}$

Please observe that it is saying that y is any value that you like but domain of x is $- \frac{1}{4} \le x < \frac{3}{4}$ and $x \ge 1$.

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 $\textcolor{b l u e}{\frac{4 {x}^{2} - 3 x - 1}{4 x - 3} \le 0}$:

Please observe that it is saying that y is any value that you like but domain of x is $- \frac{1}{4} \le x$ and $\frac{3}{4} < x \le 1$; this clearly disagrees with the graph of the original equation.

Here is Selection C $\textcolor{g r e e n}{4 {x}^{2} - 3 x - 1 \ge 0}$

Please observe that it is saying that y is any value that you like but domain of x is $- \frac{1}{4} \le x$ and $x \ge 1$; this clearly disagrees with the graph of the original equation.

Here is Selection D $\textcolor{p u r p \le}{4 {x}^{2} - 3 x - 1 \le 0}$

Please observe that it is saying that y is any value that you like but domain of x is $- \frac{1}{4} \le x \le 1$; this clearly disagrees with the graph of the original equation.

I cannot use the inequality function for selection E $\textcolor{\mathmr{and} a n \ge}{4 {x}^{2} + 3 x + 1 \le 0}$, because $\textcolor{\mathmr{and} a n \ge}{y = 4 {x}^{2} + 3 x + 1}$ 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 $\frac{4 {x}^{2} - 3 x - 1}{4 x - 3} \ge 0$

Please observe that the graph of the original inequality is identical to this graph with the same domain, $- \frac{1}{4} \le x < \frac{3}{4}$ and $x \ge 1$.