the up side last step and lower first step. how its jump direct.
Answers!
Hello Shashank
We have
2a (p - q) = -d [(p-q)(p+q) - (p-q)]
Let's write it as
2a (p - q) = -d [(p-q)(p+q) - (p-q) x 1]
Taking (p-q) common from left side
2a (p - q) = -d (p - q) [(p+q) - 1]
We will add this step in the answer as well.
Comment
Answer the question...
Hello Shashank
We have
2a (p - q) = -d [(p-q)(p+q) - (p-q)]
Let's write it as
2a (p - q) = -d [(p-q)(p+q) - (p-q) x 1]
Taking (p-q) common from left side
2a (p - q) = -d (p - q) [(p+q) - 1]
We will add this step in the answer as well.
Comment