Question

What is the missing expression that would make the left-hand-side of the equation equal to the right-hand-side?


`(???)/(x-2)/(x^(2)-1 )/(x^(2) -4x+4) =(5(x-2))/(x-1)`

Answer 1

Given:

The equation is

`(???)/(x-2)/(x^2-1)/(x^2-4x+4)=(5(x-2))/(x-1)`

To find:

The missing value.

Solution:

Let the missing value be k.

`(k)/(x-2)/(x^2-1)/(x^2-4x+4)=(5(x-2))/(x-1)`

`(k)/(x-2)* (x^2-2(x)(2)+2^2)/(x^2-1^2)=(5(x-2))/(x-1)`

Using the formulae
`(a-b)^2=a^2-2ab+b^2` and
`(a-b)(a+b)=a^2-b^2`.

`(k)/(x-2)* ((x-2)^2)/((x-1)(x+1))=(5(x-2))/(x-1)`

`(k(x-2))/((x-1)(x+1))=(5(x-2))/(x-1)`

Cancel out the common factors from both sides.

`(k)/(x+1)=5`

Multiply both sides by (x+1).

`k=5(x+1)`

Therefore, the missing value is 5(x+1).