Question

Find x $x \in R:\dfrac{{14x}}{{x + 1}} - \dfrac{{9x - 30}}{{x - 4}} < 0$ is equal to
A. $\left( { - 1,4} \right)$
B. $\left( {1,4} \right) \cup \left( {5,7} \right)$
C. $\left( {1,7} \right)$
D. $\left( { - 1,1} \right) \cup \left( {4,6} \right)$

Answer 1

Hint: This question is related to linear inequalities. A relation which holds between two values when they are different is known as inequality. Any two real numbers or algebraic expressions when related by ‘$ < $’, ‘$ > $’, ‘$ \leqslant $’ or ‘$ \geqslant $’ form a linear inequality. The notation referred to in the given question is ‘$ < $’ which means that the algebraic expression $\dfrac{{14x}}{{x + 1}} - \dfrac{{9x - 30}}{{x - 4}}$ is less than $0$. There are various types of inequalities: numerical inequalities, variable inequalities, double inequalities, strict inequalities, slack inequalities, linear inequalities in one variable, linear inequalities in two variables and quadratic inequalities. The given question is a type of linear inequalities in one variable.

Complete step by step solution:
Given inequality is $x \in R:\dfrac{{14x}}{{x + 1}} - \dfrac{{9x - 30}}{{x - 4}} < 0$.
Let us try to simplify the given inequality;
$
   \Rightarrow \dfrac{{14x\left( {x - 4} \right) - \left( {9x - 30} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 4} \right)}} < 0 \\
   \Rightarrow \dfrac{{14{x^2} - 56x - \left[ {9{x^2} + 9x - 30x - 30} \right]}}{{\left( {x + 1} \right)\left( {x - 4} \right)}} < 0 \\
   \Rightarrow \dfrac{{14{x^2} - 56x - 9{x^2} - 9x + 30x + 30}}{{\left( {x + 1} \right)\left( {x - 4} \right)}} < 0 \\
   \Rightarrow \dfrac{{5{x^2} - 35x + 30}}{{\left( {x + 1} \right)\left( {x - 4} \right)}} < 0 \\
   \Rightarrow \dfrac{{{x^2} - 7x + 6}}{{\left( {x + 1} \right)\left( {x - 4} \right)}} < 0 \\
 $
Now, we got a quadratic equation and we will solve it using the method of splitting the middle term.
$
   \Rightarrow \dfrac{{{x^2} - 6x - x + 6}}{{\left( {x + 1} \right)\left( {x - 4} \right)}} < 0 \\
   \Rightarrow \dfrac{{x\left( {x - 6} \right) - 1\left( {x - 6} \right)}}{{\left( {x + 1} \right)\left( {x - 4} \right)}} < 0 \\
   \Rightarrow \dfrac{{\left( {x - 6} \right)\left( {x - 1} \right)}}{{\left( {x + 1} \right)\left( {x - 4} \right)}} < 0 \\
 $
As$x > 1\& x > 6$, the denominator can’t be zero which means that $x$ cannot be $ - 1\;or\;4$.
Therefore, we can say that $x \in \left( { - 1,4} \right)$.

So, the correct answer is Option A.

Note: While solving the question, the students should ensure that the inequality symbol should not change by mistake. There are certain rules of inequality which need to be followed while solving inequality problems. Some of them are:
Only equal numbers should be added or subtracted from both the sides of inequality.
Both the sides of inequality can only be multiplied and divided with the same positive number.
When both sides of inequality are multiplied or divided with the same negative number, then the sign of inequality gets reversed.