Question

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

Answer 1

Given:

The equation is

$\dfrac{???}{x-2}/\dfrac{x^2-1}{x^2-4x+4}=\dfrac{5(x-2)}{x-1}$

To find:

The missing value.

Solution:

Let the missing value be k.

$\dfrac{k}{x-2}/\dfrac{x^2-1}{x^2-4x+4}=\dfrac{5(x-2)}{x-1}$

$\dfrac{k}{x-2}\times \dfrac{x^2-2(x)(2)+2^2}{x^2-1^2}=\dfrac{5(x-2)}{x-1}$

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

$\dfrac{k}{x-2}\times \dfrac{(x-2)^2}{(x-1)(x+1)}=\dfrac{5(x-2)}{x-1}$

$\dfrac{k(x-2)}{(x-1)(x+1)}=\dfrac{5(x-2)}{x-1}$

Cancel out the common factors from both sides.

$\dfrac{k}{x+1}=5$

Multiply both sides by (x+1).

$k=5(x+1)$

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