Question

Subject: Mathematics

Answer 1

$\\ \sf\longmapsto \dfrac{1}{x-1}>4$

• Multiply x-1 on both sides

$\\ \sf\longmapsto \dfrac{1}{x-1}\times (x-1)> 4(x-1)$

$\\ \sf\longmapsto 1>4(x-1)$

$\\ \sf\longmapsto 1> 4x-4$

• Add 4 on both sides

$\\ \sf\longmapsto 1+4> 4x$

$\\ \sf\longmapsto 5> 4x$

• Divide both sides by 4

$\\ \sf\longmapsto \dfrac{5}{4}>x$

$\\ \sf\longmapsto x< \dfrac{5}{4}$

Now

ATQ to the equation $\bf \dfrac{1}{x-1}$ x should be greater than 1 .Because if it becomes 1 then the denominator will be 0 which is impossible.

Hence x>1

Thus

$\\ \sf{:}\!\implies 1$

Now

$\\ \therefore\boxed{\bf x\epsilon \left(1,\dfrac{5}{4}\right)}$

Answer 2

Answer:

$\displaystyle \frac{5}{4} > x > 1$

$\displaystyle x \in \left(1, \frac{5}{4} \right)$

Step-by-step explanation:

we want to figure out the following inequality:

$\displaystyle \frac{1}{x - 1} > 4$

first note, we have the expression x-1 on the denominator, which means x of course CANNOT be 1 in other words x must be greater than 1,we will see the use of it later,

for now multiply both sides by x-1 which yields:

$\displaystyle 1> 4(x - 1)$

distribute:

$\displaystyle 1> 4x - 4$

add 4 to both sides:

$\displaystyle 5> 4x$

divide both sides by 4:

$\displaystyle \frac{5}{4} > x$

as I mentioned before x must be greater than 1 hence,

$\displaystyle \frac{5}{4} > x > 1$

$\displaystyle \xrightarrow{ \rm interval \: notation}x \in \left(1, \frac{5}{4} \right)$

and we're done!