Question

What are the set of all $x$ satisfying the inequality $\dfrac{{4x - 1}}{{3x + 1}} \geqslant 1$ is
A.$\left( { - \infty ,\dfrac{{ - 1}}{3}} \right) \cup \left( {\dfrac{1}{4},\infty } \right)$
B.$\left( { - \infty ,\dfrac{{ - 2}}{3}} \right) \cup \left( {\dfrac{5}{4},\infty } \right)$
C.$\left( { - \infty ,\dfrac{{ - 1}}{3}} \right) \cup \left( {2,\infty } \right)$
D.$\left( { - \infty ,\dfrac{{ - 2}}{3}} \right) \cup \left( {4,\infty } \right)$

Answer 1

Hint: In this problem, quadratic inequality is simply a type of equation that does not have an equal sign and includes the second highest degree. Any number may be added or subtracted from the two sides of inequality in a variation. The resolution of quadratic inequalities is the same as quadratic equation solving.

Formula used:
Inequalities condition,
 First condition$\left\{
  (x - 1) > 0 \Rightarrow x > 1 \\
  (x - 2) > 0 \Rightarrow x > 2 \\
  \right\}x \geqslant 2$
Second condition, $\left\{
  x - 1 < 0 \Rightarrow x < 1 \\
  x - 2 < 0 \Rightarrow x > 2 \\
  \right\}x \leqslant 1$
Therefore, $x \in \left( { - \infty ,1} \right) \cup \left( {2,\infty } \right)$
Where,
Inequalities of the equation mentioned,
$ > $ greater than
$ < $ less than
$ \geqslant $greater than or equal to
$ \leqslant $less than or equal to
The standard formula of quadratic equation, $a{x^2} + bx + c = 0$
Where,
$a,b,c$ are the values, $a$ can’t be $0$
$x$ is a variable,

Complete step-by-step answer:
Given by,
 All $x$ is satisfying inequalities the $\dfrac{{4x - 1}}{{3x + 1}} \geqslant 1$
The inequalities of above equation are
$\Rightarrow$ $\dfrac{{4x - 1}}{{3x + 1}} - 1 \geqslant 0$
On simplifying the above equation,
$\Rightarrow$ $\dfrac{{4x - 1 - 3x - 1}}{{3x + 1}} \geqslant 0$
Performing the both arithmetic and subtraction,
We get,
$\Rightarrow$ $\dfrac{{x - 2}}{{3x + 1}} \geqslant 0$
Applying inequalities,
The equation can be separating,
Then,
$\Rightarrow$ \[x - 2 \geqslant 0,3x + 1 \geqslant 0\]
Changing sign of the above equation,
We get,
$\Rightarrow$ $x \geqslant 2,x > \dfrac{{ - 1}}{3}$
Is negative we need to reverse the inequality,
Or we changing the less than symbol,
$\Rightarrow$ $x \leqslant 2,x < \dfrac{{ - 1}}{3}$
Substituting the inequalities condition,
Belongs to$x \in \left( { - \infty ,\dfrac{{ - 1}}{3}} \right),$$\left( {2,\infty } \right)$
The set of all $x$ is satisfying inequality $\dfrac{{4x - 1}}{{3x + 1}} \geqslant 1$ is $x \in \left( { - \infty ,\dfrac{{ - 1}}{3}} \right),$ $\left( {2,\infty } \right)$
Hence, the option C is correct answer $x \in \left( { - \infty ,\dfrac{{ - 1}}{3}} \right),$ $\left( {2,\infty } \right)$.

Note: By adding, subtracting, multiplying, or separating both sides until you are left with the attribute on its own, several basic inequalities can be solved. But the direction of inequalities will change these things. Multiplying by a negative number or separating both sides. Do not multiply or divide by a variable unless you know it is always positive or always negative.