Answer:
[tex] \displaystyle \frac{5}{4} > x > 1[/tex]
[tex]\displaystyle x \in \left(1, \frac{5}{4} \right)[/tex]
Step-by-step explanation:
we would like to solve the following rational inequality
[tex] \displaystyle \frac{1}{x - 1} > 4[/tex]
Note that we really CANNOT multiply both sides by x-1 as it can either be negative or positive however there're two methods of addressing this problem. Methods are as follows
Method-1:
In this method we would guess the answer by examining several values of x. Before we do so, we need to rearrange the inequality.
firstly, cancel 4 from both sides:
[tex] \displaystyle \frac{1}{x - 1} - 4> 0[/tex]
simplify:
[tex] \displaystyle \frac{1 - 4(x - 1)}{x - 1} > 0 \\ \\ \frac{1 - 4x + 4}{x - 1} > 0 \\ \\ \frac{ - 4x + 5}{x - 1} > 0 [/tex]
now we can examine different values of x to test where -4x+5/x-1 is greater than 0.
At x = -1, -4x+5/x-1 is less than 0
At x = 0, -4x+5/x-1 is less than 0
At x = 1 , -4x+5/x-1 is undefined
At x = 5/4 ,-4x+5/x-1 is equal to 0
At x = 2 , -4x+5/x-1 is less than 0
It tells us the image that
The inequality is true on the interval (1,5/4)
Method-2:
In this method, we would consider nothing but algebra to solve the inequality. Likewise method-1, we need to rearrange the inequality. As I've already shown how to rearrange the inequality, I am skipping the steps for now. so rearranging the inequality yields
[tex]\dfrac{ - 4x + 5}{x - 1} > 0 [/tex]
owing to algebra, we know that [tex]\frac{ - 4x + 5}{x - 1} [/tex] would be greater than 0 in case
- Both the numerator and denominator is greater than 0
- Both the numerator and denominator is less than 0
thus it can be separated in two conditions
[tex] \begin{cases} - 4x + 5 > 0\\\text{and} \\ x - 1 > 0\end{cases} \text{ \: \: \: or \: \: } \begin{cases} - 4x + 5 < 0\\\text{and} \\ \text{and}\\ x - 1 < 0\end{cases} [/tex]
solve the inequalities:
[tex] \begin{cases} x < \dfrac{5}{4} \\ \text{and} \\ x > 1\end{cases} \text{ \: \: \: or \: \: } \begin{cases} x > \dfrac{5}{4} \\ \text{and}\\ x < 1 \end{cases} [/tex]
solve the "and" inequality or find the interception:
[tex] x \in (1, \frac{5}{4} ) \text{ \: \: \: or \: \: }x \in \varnothing [/tex]
solve the "or" inequality or work out the union:
[tex] x \in (1, \frac{5}{4} ) [/tex]
and we're done!
