Can u show me how to simply this step by step please? This is very important. HELP!!!!!

Question

asked Sep 4, 2021 115k views

1 Answer

Webberpuma · Answer 1 · 2021-09-10T14:30:07+0000

Given:

$$(\left((a^(2)-b^(2))/(a b)\right))/(\left((1)/(b)-(1)/(a)\right))$

To find:

The simplified expression.

Solution:

$$(\left((a^(2)-b^(2))/(a b)\right))/(\left((1)/(b)-(1)/(a)\right))$

Apply the fraction rule:
$((y)/(z))/(x)=(y)/(z \cdot x)$

$$(\left((a^(2)-b^(2))/(a b)\right))/(\left((1)/(b)-(1)/(a)\right))=(a^(2)-b^(2))/(a b\left((1)/(b)-(1)/(a)\right))$ ------------ (1)

Let us simplify
(1)/(b)-(1)/(a) .

LCM of a and b = ab

Adjacent fraction based on the LCM

(1)/(b)-(1)/(a)=(a)/(a b)-(b)/(a b)

=(a-b)/(a b)

Substitute this in equation (1).

$=(a^(2)-b^(2))/((a-b)/(a b) a b)

Common factor ab get canceled.

$=(a^(2)-b^(2))/(a-b)

Apply the algebraic formula:
x^(2)-y^(2)=(x+y)(x-y)

$=((a+b)(a-b))/(a-b)

Cancel the common factor a - b, we get

=a+b

The simplified expression is a + b.